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 \input{preamble} % OK, start here. % \begin{document} \title{Varieties} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we start studying varieties and more generally schemes over a field. A fundamental reference is \cite{EGA}. \section{Notation} \label{section-notation} \noindent Throughout this chapter we use the letter $k$ to denote the ground field. \section{Varieties} \label{section-varieties} \noindent In the Stacks project we will use the following as our definition of a variety. \begin{definition} \label{definition-variety} Let $k$ be a field. A {\it variety} is a scheme $X$ over $k$ such that $X$ is integral and the structure morphism $X \to \Spec(k)$ is separated and of finite type. \end{definition} \noindent This definition has the following drawback. Suppose that $k \subset k'$ is an extension of fields. Suppose that $X$ is a variety over $k$. Then the base change $X_{k'} = X \times_{\Spec(k)} \Spec(k')$ is not necessarily a variety over $k'$. This phenomenon (in greater generality) will be discussed in detail in the following sections. The product of two varieties need not be a variety (this is really the same phenomenon). Here is an example. \begin{example} \label{example-product-not-a-variety} Let $k = \mathbf{Q}$. Let $X = \Spec(\mathbf{Q}(i))$ and $Y = \Spec(\mathbf{Q}(i))$. Then the product $X \times_{\Spec(k)} Y$ of the varieties $X$ and $Y$ is not a variety, since it is reducible. (It is isomorphic to the disjoint union of two copies of $X$.) \end{example} \noindent If the ground field is algebraically closed however, then the product of varieties is a variety. This follows from the results in the algebra chapter, but there we treat much more general situations. There is also a simple direct proof of it which we present here. \begin{lemma} \label{lemma-product-varieties} Let $k$ be an algebraically closed field. Let $X$, $Y$ be varieties over $k$. Then $X \times_{\Spec(k)} Y$ is a variety over $k$. \end{lemma} \begin{proof} The morphism $X \times_{\Spec(k)} Y \to \Spec(k)$ is of finite type and separated because it is the composition of the morphisms $X \times_{\Spec(k)} Y \to Y \to \Spec(k)$ which are separated and of finite type, see Morphisms, Lemmas \ref{morphisms-lemma-base-change-finite-type} and \ref{morphisms-lemma-composition-finite-type} and Schemes, Lemma \ref{schemes-lemma-separated-permanence}. To finish the proof it suffices to show that $X \times_{\Spec(k)} Y$ is integral. Let $X = \bigcup_{i = 1, \ldots, n} U_i$, $Y = \bigcup_{j = 1, \ldots, m} V_j$ be finite affine open coverings. If we can show that each $U_i \times_{\Spec(k)} V_j$ is integral, then we are done by Properties, Lemmas \ref{properties-lemma-characterize-reduced}, \ref{properties-lemma-characterize-irreducible}, and \ref{properties-lemma-characterize-integral}. This reduces us to the affine case. \medskip\noindent The affine case translates into the following algebra statement: Suppose that $A$, $B$ are integral domains and finitely generated $k$-algebras. Then $A \otimes_k B$ is an integral domain. To get a contradiction suppose that $$(\sum\nolimits_{i = 1, \ldots, n} a_i \otimes b_i) (\sum\nolimits_{j = 1, \ldots, m} c_j \otimes d_j) = 0$$ in $A \otimes_k B$ with both factors nonzero in $A \otimes_k B$. We may assume that $b_1, \ldots, b_n$ are $k$-linearly independent in $B$, and that $d_1, \ldots, d_m$ are $k$-linearly independent in $B$. Of course we may also assume that $a_1$ and $c_1$ are nonzero in $A$. Hence $D(a_1c_1) \subset \Spec(A)$ is nonempty. By the Hilbert Nullstellensatz (Algebra, Theorem \ref{algebra-theorem-nullstellensatz}) we can find a maximal ideal $\mathfrak m \subset A$ contained in $D(a_1c_1)$ and $A/\mathfrak m = k$ as $k$ is algebraically closed. Denote $\overline{a}_i, \overline{c}_j$ the residue classes of $a_i, c_j$ in $A/\mathfrak m = k$. Then equation above becomes $$(\sum\nolimits_{i = 1, \ldots, n} \overline{a}_i b_i) (\sum\nolimits_{j = 1, \ldots, m} \overline{c}_j d_j) = 0$$ which is a contradiction with $\mathfrak m \in D(a_1c_1)$, the linear independence of $b_1, \ldots, b_n$ and $d_1, \ldots, d_m$, and the fact that $B$ is a domain. \end{proof} \section{Varieties and rational maps} \label{section-varieties-rational-maps} \noindent Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$. We will use the phrase {\it rational map of varieties from $X$ to $Y$} to mean a $\Spec(k)$-rational map from the scheme $X$ to the scheme $Y$ as defined in Morphisms, Definition \ref{morphisms-definition-rational-map}. As is customary, the phrase rational map of varieties'' does not refer to the (common) base field of the varieties, even though for general schemes we make the distinction between rational maps and rational maps over a given base. \medskip\noindent The title of this section refers to the following fundamental theorem. \begin{theorem} \label{theorem-varieties-rational-maps} Let $k$ be a field. The category of varieties and dominant rational maps is equivalent to the category of finitely generated field extensions $K/k$. \end{theorem} \begin{proof} Let $X$ and $Y$ be varieties with generic points $x \in X$ and $y \in Y$. Recall that dominant rational maps from $X$ to $Y$ are exactly those rational maps which map $x$ to $y$ (Morphisms, Definition \ref{morphisms-definition-dominant-rational} and discussion following). Thus given a dominant rational map $X \supset U \to Y$ we obtain a map of function fields $$k(Y) = \kappa(y) = \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} = \kappa(x) = k(X)$$ Conversely, such a $k$-algebra map (which is automatically local as the source and target are fields) determines (uniquely) a dominant rational map by Morphisms, Lemma \ref{morphisms-lemma-rational-map-finite-presentation}. In this way we obtain a fully faithful functor. To finish the proof it suffices to show that every finitely generated field extension $K/k$ is in the essential image. Since $K/k$ is finitely generated, there exists a finite type $k$-algebra $A \subset K$ such that $K$ is the fraction field of $A$. Then $X = \Spec(A)$ is a variety whose function field is $K$. \end{proof} \noindent Let $k$ be a field. Let $X$ and $Y$ be varieties over $k$. We will use the phrase {\it $X$ and $Y$ are birational varieties} to mean $X$ and $Y$ are $\Spec(k)$-birational as defined in Morphisms, Definition \ref{morphisms-definition-birational}. As is customary, the phrase birational varieties'' does not refer to the (common) base field of the varieties, even though for general irreducible schemes we make the distinction between being birational and being birational over a given base. \begin{lemma} \label{lemma-birational-varieties} Let $X$ and $Y$ be varieties over a field $k$. The following are equivalent \begin{enumerate} \item $X$ and $Y$ are birational varieties, \item the function fields $k(X)$ and $k(Y)$ are isomorphic, \item there exist nonempty opens of $X$ and $Y$ which are isomorphic as varieties, \item there exists an open $U \subset X$ and a birational morphism $U \to Y$ of varieties. \end{enumerate} \end{lemma} \begin{proof} This is a special case of Morphisms, Lemma \ref{morphisms-lemma-criterion-birational-finite-presentation}. \end{proof} \section{Change of fields and local rings} \label{section-local-rings} \noindent Some preliminary results on what happens to local rings under an extension of ground fields. \begin{lemma} \label{lemma-change-fields-flat} Let $K/k$ be an extension of fields. Let $X$ be scheme over $k$ and set $Y = X_K$. If $y \in Y$ with image $x \in X$, then \begin{enumerate} \item $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$ is a faithfully flat local ring homomorphism, \item with $\mathfrak p_0 = \Ker(\kappa(x) \otimes_k K \to \kappa(y))$ we have $\kappa(y) = \kappa(\mathfrak p_0)$, \item $\mathcal{O}_{Y, y} = (\mathcal{O}_{X, x} \otimes_k K)_\mathfrak p$ where $\mathfrak p \subset \mathcal{O}_{X, x} \otimes_k K$ is the inverse image of $\mathfrak p_0$. \item we have $\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y} = (\kappa(x) \otimes_k K)_{\mathfrak p_0}$ \end{enumerate} \end{lemma} \begin{proof} We may assume $X = \Spec(A)$ is affine. Then $Y = \Spec(A \otimes_k K)$. Since $K$ is flat over $k$, we see that $A \to A \otimes_k K$ is flat. Hence $Y \to X$ is flat and we get the first statement if we also use Algebra, Lemma \ref{algebra-lemma-local-flat-ff}. The second statement follows from Schemes, Lemma \ref{schemes-lemma-points-fibre-product}. Now $y$ corresponds to a prime ideal $\mathfrak q \subset A \otimes_k K$ and $x$ to $\mathfrak r = A \cap \mathfrak q$. Then $\mathfrak p_0$ is the kernel of the induced map $\kappa(\mathfrak r) \otimes_k K \to \kappa(\mathfrak q)$. The map on local rings is $$A_\mathfrak r \longrightarrow (A \otimes_k K)_\mathfrak q$$ We can factor this map through $A_\mathfrak r \otimes_k K = (A \otimes_k K)_{\mathfrak r}$ to get $$A_\mathfrak r \longrightarrow A_\mathfrak r \otimes_k K \longrightarrow (A \otimes_k K)_\mathfrak q$$ and then the second arrow is a localization at some prime. This prime ideal is the inverse image of $\mathfrak p_0$ (details omitted) and this proves (3). To see (4) use (3) and that localization and $- \otimes_k K$ are exact functors. \end{proof} \begin{lemma} \label{lemma-change-fields-algebraic-dim} Notation as in Lemma \ref{lemma-change-fields-flat}. Assume $X$ is locally of finite type over $k$. Then $$\dim(\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y}) = \text{trdeg}_k(\kappa(x)) - \text{trdeg}_K(\kappa(y)) = \dim(\mathcal{O}_{Y, y}) - \dim(\mathcal{O}_{X, x})$$ \end{lemma} \begin{proof} This is a restatement of Algebra, Lemma \ref{algebra-lemma-inequalities-under-field-extension}. \end{proof} \begin{lemma} \label{lemma-change-fields-algebraic-unramified} Notation as in Lemma \ref{lemma-change-fields-flat}. Assume $X$ is locally of finite type over $k$, that $\dim(\mathcal{O}_{X, x}) = \dim(\mathcal{O}_{Y, y})$ and that $\kappa(x) \otimes_k K$ is reduced (for example if $\kappa(x)/k$ is separable or $K/k$ is separable). Then $\mathfrak m_x \mathcal{O}_{Y, y} = \mathfrak m_y$. \end{lemma} \begin{proof} (The parenthetical statement follows from Algebra, Lemma \ref{algebra-lemma-separable-extension-preserves-reducedness}.) Combining Lemmas \ref{lemma-change-fields-flat} and \ref{lemma-change-fields-algebraic-dim} we see that $\mathcal{O}_{Y, y}/\mathfrak m_x \mathcal{O}_{Y, y}$ has dimension $0$ and is reduced. Hence it is a field. \end{proof} \section{Geometrically reduced schemes} \label{section-geometrically-reduced} \noindent If $X$ is a reduced scheme over a field, then it can happen that $X$ becomes nonreduced after extending the ground field. This does not happen for geometrically reduced schemes. \begin{definition} \label{definition-geometrically-reduced} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$ be a point. \begin{enumerate} \item Let $x \in X$ be a point. We say $X$ is {\it geometrically reduced at $x$} if for any field extension $k \subset k'$ and any point $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is reduced. \item We say $X$ is {\it geometrically reduced} over $k$ if $X$ is geometrically reduced at every point of $X$. \end{enumerate} \end{definition} \noindent This may seem a little mysterious at first, but it is really the same thing as the notion discussed in the algebra chapter. Here are some basic results explaining the connection. \begin{lemma} \label{lemma-geometrically-reduced-at-point} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent \begin{enumerate} \item $X$ is geometrically reduced at $x$, and \item the ring $\mathcal{O}_{X, x}$ is geometrically reduced over $k$ (see Algebra, Definition \ref{algebra-definition-geometrically-reduced}). \end{enumerate} \end{lemma} \begin{proof} Assume (1). This in particular implies that $\mathcal{O}_{X, x}$ is reduced. Let $k \subset k'$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes_k k'$. By Algebra, Lemma \ref{algebra-lemma-p-ring-map} its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$. By assumption this is a reduced ring. Hence we deduce (2) by Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}. \medskip\noindent Assume (2). Let $k \subset k'$ be a field extension. Since $\Spec(k') \to \Spec(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$. Hence it is reduced by assumption and (1) is proved. \end{proof} \noindent The notion isn't interesting in characteristic zero. \begin{lemma} \label{lemma-perfect-reduced} Let $X$ be a scheme over a perfect field $k$ (e.g.\ $k$ has characteristic zero). Let $x \in X$. If $\mathcal{O}_{X, x}$ is reduced, then $X$ is geometrically reduced at $x$. If $X$ is reduced, then $X$ is geometrically reduced over $k$. \end{lemma} \begin{proof} The first statement follows from Lemma \ref{lemma-geometrically-reduced-at-point} and Algebra, Lemma \ref{algebra-lemma-separable-extension-preserves-reducedness} and the definition of a perfect field (Algebra, Definition \ref{algebra-definition-perfect}). The second statement follows from the first. \end{proof} \begin{lemma} \label{lemma-geometrically-reduced} Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$. The following are equivalent \begin{enumerate} \item $X$ is geometrically reduced, \item $X_{k'}$ is reduced for every field extension $k \subset k'$, \item $X_{k'}$ is reduced for every finite purely inseparable field extension $k \subset k'$, \item $X_{k^{1/p}}$ is reduced, \item $X_{k^{perf}}$ is reduced, \item $X_{\bar k}$ is reduced, \item for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is geometrically reduced (see Algebra, Definition \ref{algebra-definition-geometrically-reduced}). \end{enumerate} \end{lemma} \begin{proof} Assume (1). Then for every field extension $k \subset k'$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is reduced. In other words $X_{k'}$ is reduced. Hence (2). \medskip\noindent Assume (2). Let $U \subset X$ be an affine open. Then for every field extension $k \subset k'$ the scheme $X_{k'}$ is reduced, hence $U_{k'} = \Spec(\mathcal{O}(U)\otimes_k k')$ is reduced, hence $\mathcal{O}(U)\otimes_k k'$ is reduced (see Properties, Section \ref{properties-section-integral}). In other words $\mathcal{O}(U)$ is geometrically reduced, so (7) holds. \medskip\noindent Assume (7). For any field extension $k \subset k'$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_X(U) \otimes_k k'$ where $U$ is affine open in $X$ (see Schemes, Section \ref{schemes-section-fibre-products}). Hence $X_{k'}$ is reduced. So (1) holds. \medskip\noindent This proves that (1), (2), and (7) are equivalent. These are equivalent to (3), (4), (5), and (6) because we can apply Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension} to $\mathcal{O}_X(U)$ for $U \subset X$ affine open. \end{proof} \begin{lemma} \label{lemma-check-only-finite-inseparable-extensions} Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent \begin{enumerate} \item $X$ is geometrically reduced at $x$, \item $\mathcal{O}_{X_{k'}, x'}$ is reduced for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ the unique point lying over $x$, \item $\mathcal{O}_{X_{k^{1/p}}, x'}$ is reduced for $x' \in X_{k'}$ the unique point lying over $x$, and \item $\mathcal{O}_{X_{k^{perf}}, x'}$ is reduced for $x' \in X_{k^{perf}}$ the unique point lying over $x$. \end{enumerate} \end{lemma} \begin{proof} Note that if $k \subset k'$ is purely inseparable, then $X_{k'} \to X$ induces a homeomorphism on underlying topological spaces, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. Whence the uniqueness of $x'$ lying over $x$ mentioned in the statement. Moreover, in this case $\mathcal{O}_{X_{k'}, x'} = \mathcal{O}_{X, x} \otimes_k k'$. Hence the lemma follows from Lemma \ref{lemma-geometrically-reduced-at-point} above and Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}. \end{proof} \begin{lemma} \label{lemma-geometrically-reduced-upstairs} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. The following are equivalent \begin{enumerate} \item $X$ is geometrically reduced at $x$, \item $X_{k'}$ is geometrically reduced at $x'$. \end{enumerate} In particular, $X$ is geometrically reduced over $k$ if and only if $X_{k'}$ is geometrically reduced over $k'$. \end{lemma} \begin{proof} It is clear that (1) implies (2). Assume (2). Let $k \subset k''$ be a finite purely inseparable field extension and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is unique). We can find a common field extension $k \subset k'''$ (i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$) and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$. Consider the map of local rings $$\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.$$ This is a flat local ring homomorphism and hence faithfully flat. By (2) we see that the local ring on the right is reduced. Thus by Algebra, Lemma \ref{algebra-lemma-descent-reduced} we conclude that $\mathcal{O}_{X_{k''}, x''}$ is reduced. Thus by Lemma \ref{lemma-check-only-finite-inseparable-extensions} we conclude that $X$ is geometrically reduced at $x$. \end{proof} \begin{lemma} \label{lemma-geometrically-reduced-any-base-change} Let $k$ be a field. Let $X$, $Y$ be schemes over $k$. \begin{enumerate} \item If $X$ is geometrically reduced at $x$, and $Y$ reduced, then $X \times_k Y$ is reduced at every point lying over $x$. \item If $X$ geometrically reduced over $k$ and $Y$ reduced. Then $X \times_k Y$ is reduced. \end{enumerate} \end{lemma} \begin{proof} Combine, Lemmas \ref{lemma-geometrically-reduced-at-point} and \ref{lemma-geometrically-reduced} and Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-any-reduced-base-change}. \end{proof} \begin{lemma} \label{lemma-generic-points-geometrically-reduced} Let $k$ be a field. Let $X$ be a scheme over $k$. \begin{enumerate} \item If $x' \leadsto x$ is a specialization and $X$ is geometrically reduced at $x$, then $X$ is geometrically reduced at $x'$. \item If $x \in X$ such that (a) $\mathcal{O}_{X, x}$ is reduced, and (b) for each specialization $x' \leadsto x$ where $x'$ is a generic point of an irreducible component of $X$ the scheme $X$ is geometrically reduced at $x'$, then $X$ is geometrically reduced at $x$. \item If $X$ is reduced and geometrically reduced at all generic points of irreducible components of $X$, then $X$ is geometrically reduced. \end{enumerate} \end{lemma} \begin{proof} Part (1) follows from Lemma \ref{lemma-geometrically-reduced-at-point} and the fact that if $A$ is a geometrically reduced $k$-algebra, then $S^{-1}A$ is a geometrically reduced $k$-algebra for any multiplicative subset $S$ of $A$, see Algebra, Lemma \ref{algebra-lemma-geometrically-reduced-permanence}. \medskip\noindent Let $A = \mathcal{O}_{X, x}$. The assumptions (a) and (b) of (2) imply that $A$ is reduced, and that $A_{\mathfrak q}$ is geometrically reduced over $k$ for every minimal prime $\mathfrak q$ of $A$. Hence $A$ is geometrically reduced over $k$, see Algebra, Lemma \ref{algebra-lemma-generic-points-geometrically-reduced}. Thus $X$ is geometrically reduced at $x$, see Lemma \ref{lemma-geometrically-reduced-at-point}. \medskip\noindent Part (3) follows trivially from part (2). \end{proof} \begin{lemma} \label{lemma-Noetherian-geometrically-reduced-at-point} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. Assume $X$ locally Noetherian and geometrically reduced at $x$. Then there exists an open neighbourhood $U \subset X$ of $x$ which is geometrically reduced over $k$. \end{lemma} \begin{proof} Assume $X$ locally Noetherian and geometrically reduced at $x$. By Properties, Lemma \ref{properties-lemma-ring-affine-open-injective-local-ring} we can find an affine open neighbourhood $U \subset X$ of $x$ such that $R = \mathcal{O}_X(U) \to \mathcal{O}_{X, x}$ is injective. By Lemma \ref{lemma-geometrically-reduced-at-point} the assumption means that $\mathcal{O}_{X, x}$ is geometrically reduced over $k$. By Algebra, Lemma \ref{algebra-lemma-subalgebra-separable} this implies that $R$ is geometrically reduced over $k$, which in turn implies that $U$ is geometrically reduced. \end{proof} \begin{example} \label{example-not-geometrically-reduced} Let $k = \mathbf{F}_p(s, t)$, i.e., a purely transcendental extension of the prime field. Consider the variety $X = \Spec(k[x, y]/(1 + sx^p + ty^p))$. Let $k \subset k'$ be any extension such that both $s$ and $t$ have a $p$th root in $k'$. Then the base change $X_{k'}$ is not reduced. Namely, the ring $k'[x, y]/(1 + s x^p + ty^p)$ contains the element $1 + s^{1/p}x + t^{1/p}y$ whose $p$th power is zero but which is not zero (since the ideal $(1 + sx^p + ty^p)$ certainly does not contain any nonzero element of degree $< p$). \end{example} \begin{lemma} \label{lemma-finite-extension-geometrically-reduced} Let $k$ be a field. Let $X \to \Spec(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite purely inseparable extension $k \subset k'$ such that $(X_{k'})_{red}$ is geometrically reduced over $k'$. \end{lemma} \begin{proof} To prove this lemma we may replace $X$ by its reduction $X_{red}$. Hence we may assume that $X$ is reduced and locally of finite type over $k$. Let $x_1, \ldots, x_n \in X$ be the generic points of the irreducible components of $X$. Note that for every purely inseparable algebraic extension $k \subset k'$ the morphism $(X_{k'})_{red} \to X$ is a homeomorphism, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. Hence the points $x'_1, \ldots, x'_n$ lying over $x_1, \ldots, x_n$ are the generic points of the irreducible components of $(X_{k'})_{red}$. As $X$ is reduced the local rings $K_i = \mathcal{O}_{X, x_i}$ are fields, see Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}. As $X$ is locally of finite type over $k$ the field extensions $k \subset K_i$ are finitely generated field extensions. Finally, the local rings $\mathcal{O}_{(X_{k'})_{red}, x'_i}$ are the fields $(K_i \otimes_k k')_{red}$. By Algebra, Lemma \ref{algebra-lemma-make-separable} we can find a finite purely inseparable extension $k \subset k'$ such that $(K_i \otimes_k k')_{red}$ are separable field extensions of $k'$. In particular each $(K_i \otimes_k k')_{red}$ is geometrically reduced over $k'$ by Algebra, Lemma \ref{algebra-lemma-characterize-separable-field-extensions}. At this point Lemma \ref{lemma-generic-points-geometrically-reduced} part (3) implies that $(X_{k'})_{red}$ is geometrically reduced. \end{proof} \section{Geometrically connected schemes} \label{section-geometrically-connected} \noindent If $X$ is a connected scheme over a field, then it can happen that $X$ becomes disconnected after extending the ground field. This does not happen for geometrically connected schemes. \begin{definition} \label{definition-geometrically-connected} Let $X$ be a scheme over the field $k$. We say $X$ is {\it geometrically connected} over $k$ if the scheme $X_{k'}$ is connected for every field extension $k'$ of $k$. \end{definition} \noindent By convention a connected topological space is nonempty; hence a fortiori geometrically connected schemes are nonempty. Here is an example of a variety which is not geometrically connected. \begin{example} \label{example-not-geometrically-irreducible} Let $k = \mathbf{Q}$. The scheme $X = \Spec(\mathbf{Q}(i))$ is a variety over $\Spec(\mathbf{Q})$. But the base change $X_{\mathbf{C}}$ is the spectrum of $\mathbf{C} \otimes_{\mathbf{Q}} \mathbf{Q}(i) \cong \mathbf{C} \times \mathbf{C}$ which is the disjoint union of two copies of $\Spec(\mathbf{C})$. So in fact, this is an example of a non-geometrically connected variety. \end{example} \begin{lemma} \label{lemma-geometrically-connected-check-after-extension} Let $X$ be a scheme over the field $k$. Let $k \subset k'$ be a field extension. Then $X$ is geometrically connected over $k$ if and only if $X_{k'}$ is geometrically connected over $k'$. \end{lemma} \begin{proof} If $X$ is geometrically connected over $k$, then it is clear that $X_{k'}$ is geometrically connected over $k'$. For the converse, note that for any field extension $k \subset k''$ there exists a common field extension $k' \subset k'''$ and $k'' \subset k'''$. As the morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of a surjective morphism between spectra of fields) we see that the connectedness of $X_{k'''}$ implies the connectedness of $X_{k''}$. Thus if $X_{k'}$ is geometrically connected over $k'$ then $X$ is geometrically connected over $k$. \end{proof} \begin{lemma} \label{lemma-bijection-connected-components} Let $k$ be a field. Let $X$, $Y$ be schemes over $k$. Assume $X$ is geometrically connected over $k$. Then the projection morphism $$p : X \times_k Y \longrightarrow Y$$ induces a bijection between connected components. \end{lemma} \begin{proof} The scheme theoretic fibres of $p$ are connected, since they are base changes of the geometrically connected scheme $X$ by field extensions. Moreover the scheme theoretic fibres are homeomorphic to the set theoretic fibres, see Schemes, Lemma \ref{schemes-lemma-fibre-topological}. By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open} the map $p$ is open. Thus we may apply Topology, Lemma \ref{topology-lemma-connected-fibres-connected-components} to conclude. \end{proof} \begin{lemma} \label{lemma-affine-geometrically-connected} Let $k$ be a field. Let $A$ be a $k$-algebra. Then $X = \Spec(A)$ is geometrically connected over $k$ if and only if $A$ is geometrically connected over $k$ (see Algebra, Definition \ref{algebra-definition-geometrically-connected}). \end{lemma} \begin{proof} Immediate from the definitions. \end{proof} \begin{lemma} \label{lemma-separably-closed-field-connected-components} Let $k \subset k'$ be an extension of fields. Let $X$ be a scheme over $k$. Assume $k$ separably algebraically closed. Then the morphism $X_{k'} \to X$ induces a bijection of connected components. In particular, $X$ is geometrically connected over $k$ if and only if $X$ is connected. \end{lemma} \begin{proof} Since $k$ is separably algebraically closed we see that $k'$ is geometrically connected over $k$, see Algebra, Lemma \ref{algebra-lemma-separably-closed-connected-implies-geometric}. Hence $Z = \Spec(k')$ is geometrically connected over $k$ by Lemma \ref{lemma-affine-geometrically-connected} above. Since $X_{k'} = Z \times_k X$ the result is a special case of Lemma \ref{lemma-bijection-connected-components}. \end{proof} \begin{lemma} \label{lemma-characterize-geometrically-connected} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $\overline{k}$ be a separable algebraic closure of $k$. Then $X$ is geometrically connected if and only if the base change $X_{\overline{k}}$ is connected. \end{lemma} \begin{proof} Assume $X_{\overline{k}}$ is connected. Let $k \subset k'$ be a field extension. There exists a field extension $\overline{k} \subset \overline{k}'$ such that $k'$ embeds into $\overline{k}'$ as an extension of $k$. By Lemma \ref{lemma-separably-closed-field-connected-components} we see that $X_{\overline{k}'}$ is connected. Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude that $X_{k'}$ is connected as desired. \end{proof} \begin{lemma} \label{lemma-descend-open} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $A$ be a $k$-algebra. Let $V \subset X_A$ be a quasi-compact open. Then there exists a finitely generated $k$-subalgebra $A' \subset A$ and a quasi-compact open $V' \subset X_{A'}$ such that $V = V'_A$. \end{lemma} \begin{proof} We remark that if $X$ is also quasi-separated this follows from Limits, Lemma \ref{limits-lemma-descend-opens}. Let $U_1, \ldots, U_n$ be finitely many affine opens of $X$ such that $V \subset \bigcup U_{i, A}$. Say $U_i = \Spec(R_i)$. Since $V$ is quasi-compact we can find finitely many $f_{ij} \in R_i \otimes_k A$, $j = 1, \ldots, n_i$ such that $V = \bigcup_i \bigcup_{j = 1, \ldots, n_i} D(f_{ij})$ where $D(f_{ij}) \subset U_{i, A}$ is the corresponding standard open. (We do not claim that $V \cap U_{i, A}$ is the union of the $D(f_{ij})$, $j = 1, \ldots, n_i$.) It is clear that we can find a finitely generated $k$-subalgebra $A' \subset A$ such that $f_{ij}$ is the image of some $f_{ij}' \in R_i \otimes_k A'$. Set $V' = \bigcup D(f_{ij}')$ which is a quasi-compact open of $X_{A'}$. Denote $\pi : X_A \to X_{A'}$ the canonical morphism. We have $\pi(V) \subset V'$ as $\pi(D(f_{ij})) \subset D(f_{ij}')$. If $x \in X_A$ with $\pi(x) \in V'$, then $\pi(x) \in D(f_{ij}')$ for some $i, j$ and we see that $x \in D(f_{ij})$ as $f_{ij}'$ maps to $f_{ij}$. Thus we see that $V = \pi^{-1}(V')$ as desired. \end{proof} \noindent Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. For example $\overline{k}$ could be the separable algebraic closure of $k$. For any $\sigma \in \text{Gal}(\overline{k}/k)$ we get a corresponding automorphism $\Spec(\sigma) : \Spec(\overline{k}) \longrightarrow \Spec(\overline{k})$. Note that $\Spec(\sigma) \circ \Spec(\tau) = \Spec(\tau \circ \sigma)$. Hence we get an action $$\text{Gal}(\overline{k}/k)^{opp} \times \Spec(\overline{k}) \longrightarrow \Spec(\overline{k})$$ of the opposite group on the scheme $\Spec(\overline{k})$. Let $X$ be a scheme over $k$. Since $X_{\overline{k}} = \Spec(\overline{k}) \times_{\Spec(k)} X$ by definition we see that the action above induces a canonical action \begin{equation} \label{equation-galois-action-base-change-kbar} \text{Gal}(\overline{k}/k)^{opp} \times X_{\overline{k}} \longrightarrow X_{\overline{k}}. \end{equation} \begin{lemma} \label{lemma-Galois-action-quasi-compact-open} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $\overline{k}$ be a (possibly infinite) Galois extension of $k$. Let $V \subset X_{\overline{k}}$ be a quasi-compact open. Then \begin{enumerate} \item there exists a finite subextension $k \subset k' \subset \overline{k}$ and a quasi-compact open $V' \subset X_{k'}$ such that $V = (V')_{\overline{k}}$, \item there exists an open subgroup $H \subset \text{Gal}(\overline{k}/k)$ such that $\sigma(V) = V$ for all $\sigma \in H$. \end{enumerate} \end{lemma} \begin{proof} By Lemma \ref{lemma-descend-open} there exists a finite subextension $k \subset k' \subset \overline{k}$ and an open $V' \subset X_{k'}$ which pulls back to $V$. This proves (1). Since $\text{Gal}(\overline{k}/k')$ is open in $\text{Gal}(\overline{k}/k)$ part (2) is clear as well. \end{proof} \begin{lemma} \label{lemma-closed-fixed-by-Galois} Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. Let $X$ be a scheme over $k$. Let $\overline{T} \subset X_{\overline{k}}$ have the following properties \begin{enumerate} \item $\overline{T}$ is a closed subset of $X_{\overline{k}}$, \item for every $\sigma \in \text{Gal}(\overline{k}/k)$ we have $\sigma(\overline{T}) = \overline{T}$. \end{enumerate} Then there exists a closed subset $T \subset X$ whose inverse image in $X_{\overline{k}}$ is $\overline{T}$. \end{lemma} \begin{proof} This lemma immediately reduces to the case where $X = \Spec(A)$ is affine. In this case, let $\overline{I} \subset A \otimes_k \overline{k}$ be the radical ideal corresponding to $\overline{T}$. Assumption (2) implies that $\sigma(\overline{I}) = \overline{I}$ for all $\sigma \in \text{Gal}(\overline{k}/k)$. Pick $x \in \overline{I}$. There exists a finite Galois extension $k \subset k'$ contained in $\overline{k}$ such that $x \in A \otimes_k k'$. Set $G = \text{Gal}(k'/k)$. Set $$P(T) = \prod\nolimits_{\sigma \in G} (T - \sigma(x)) \in (A \otimes_k k')[T]$$ It is clear that $P(T)$ is monic and is actually an element of $(A \otimes_k k')^G[T] = A[T]$ (by basic Galois theory). Moreover, if we write $P(T) = T^d + a_1T^{d - 1} + \ldots + a_0$ the we see that $a_i \in I := A \cap \overline{I}$. By Algebra, Lemma \ref{algebra-lemma-polynomials-divide} we see that $x$ is contained in the radical of $I(A \otimes_k \overline{k})$. Hence $\overline{I}$ is the radical of $I(A \otimes_k \overline{k})$ and setting $T = V(I)$ is a solution. \end{proof} \begin{lemma} \label{lemma-characterize-geometrically-disconnected} Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent \begin{enumerate} \item $X$ is geometrically connected, \item for every finite separable field extension $k \subset k'$ the scheme $X_{k'}$ is connected. \end{enumerate} \end{lemma} \begin{proof} It follows immediately from the definition that (1) implies (2). Assume that $X$ is not geometrically connected. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. By Lemma \ref{lemma-characterize-geometrically-connected} it follows that $X_{\overline{k}}$ is disconnected. Say $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open, closed, and nonempty. \medskip\noindent Suppose that $W \subset X$ is any quasi-compact open. Then $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are open and closed in $W_{\overline{k}}$. In particular $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are quasi-compact, and by Lemma \ref{lemma-Galois-action-quasi-compact-open} both $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are defined over a finite subextension and invariant under an open subgroup of $\text{Gal}(\overline{k}/k)$. We will use this without further mention in the following. \medskip\noindent Pick $W_0 \subset X$ quasi-compact open such that both $W_{0, \overline{k}} \cap \overline{U}$ and $W_{0, \overline{k}} \cap \overline{V}$ are nonempty. Choose a finite subextension $k \subset k' \subset \overline{k}$ and a decomposition $W_{0, k'} = U_0' \amalg V_0'$ into open and closed subsets such that $W_{0, \overline{k}} \cap \overline{U} = (U'_0)_{\overline{k}}$ and $W_{0, \overline{k}} \cap \overline{V} = (V'_0)_{\overline{k}}$. Let $H = \text{Gal}(\overline{k}/k') \subset \text{Gal}(\overline{k}/k)$. In particular $\sigma(W_{0, \overline{k}} \cap \overline{U}) = W_{0, \overline{k}} \cap \overline{U}$ and similarly for $\overline{V}$. \medskip\noindent Having chosen $W_0$, $k'$ as above, for every quasi-compact open $W \subset X$ we set $$U_W = \bigcap\nolimits_{\sigma \in H} \sigma(W_{\overline{k}} \cap \overline{U}), \quad V_W = \bigcup\nolimits_{\sigma \in H} \sigma(W_{\overline{k}} \cap \overline{V}).$$ Now, since $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are fixed by an open subgroup of $\text{Gal}(\overline{k}/k)$ we see that the union and intersection above are finite. Hence $U_W$ and $V_W$ are both open and closed. Also, by construction $W_{\bar k} = U_W \amalg V_W$. \medskip\noindent We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W_{\overline{k}} \cap U_{W'} = U_W$ and $W_{\overline{k}} \cap V_{W'} = V_W$. Verification omitted. Hence we see that upon defining $U = \bigcup_{W \subset X} U_W$ and $V = \bigcup_{W \subset X} V_W$ we obtain $X_{\overline{k}} = U \amalg V$ is a disjoint union of open and closed subsets. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $W_0 \cap \overline{U}$ by construction. Finally, $U, V \subset X_{\bar k}$ are closed and $H$-invariant by construction. Hence by Lemma \ref{lemma-closed-fixed-by-Galois} we have $U = (U')_{\bar k}$, and $V = (V')_{\bar k}$ for some closed $U', V' \subset X_{k'}$. Clearly $X_{k'} = U' \amalg V'$ and we see that $X_{k'}$ is disconnected as desired. \end{proof} \begin{lemma} \label{lemma-tricky} Let $k$ be a field. Let $k \subset \overline{k}$ be a (possibly infinite) Galois extension. Let $f : T \to X$ be a morphism of schemes over $k$. Assume $T_{\overline{k}}$ connected and $X_{\overline{k}}$ disconnected. Then $X$ is disconnected. \end{lemma} \begin{proof} Write $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open and closed. Denote $\overline{f} : T_{\overline{k}} \to X_{\overline{k}}$ the base change of $f$. Since $T_{\overline{k}}$ is connected we see that $T_{\overline{k}}$ is contained in either $\overline{f}^{-1}(\overline{U})$ or $\overline{f}^{-1}(\overline{V})$. Say $T_{\overline{k}} \subset \overline{f}^{-1}(\overline{U})$. \medskip\noindent Fix a quasi-compact open $W \subset X$. There exists a finite Galois subextension $k \subset k' \subset \overline{k}$ such that $\overline{U} \cap W_{\overline{k}}$ and $\overline{V} \cap W_{\overline{k}}$ come from quasi-compact opens $U', V' \subset W_{k'}$. Then also $W_{k'} = U' \amalg V'$. Consider $$U'' = \bigcap\nolimits_{\sigma \in \text{Gal}(k'/k)} \sigma(U'), \quad V'' = \bigcup\nolimits_{\sigma \in \text{Gal}(k'/k)} \sigma(V').$$ These are Galois invariant, open and closed, and $W_{k'} = U'' \amalg V''$. By Lemma \ref{lemma-closed-fixed-by-Galois} we get open and closed subsets $U_W, V_W \subset W$ such that $U'' = (U_W)_{k'}$, $V'' = (V_W)_{k'}$ and $W = U_W \amalg V_W$. \medskip\noindent We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W \cap U_{W'} = U_W$ and $W \cap V_{W'} = V_W$. Verification omitted. Hence we see that upon defining $U = \bigcup_{W \subset X} U_W$ and $V = \bigcup_{W \subset X} V_W$ we obtain $X = U \amalg V$. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $f(T)$ by construction. \end{proof} \begin{lemma} \label{lemma-geometrically-connected-criterion} Let $k$ be a field. Let $T \to X$ be a morphism of schemes over $k$. Assume $T$ is geometrically connected and $X$ connected. Then $X$ is geometrically connected. \end{lemma} \begin{proof} This is a reformulation of Lemma \ref{lemma-tricky}. \end{proof} \begin{lemma} \label{lemma-geometrically-connected-if-connected-and-point} Let $k$ be a field. Let $X$ be a scheme over $k$. Assume $X$ is connected and has a point $x$ such that $k$ is algebraically closed in $\kappa(x)$. Then $X$ is geometrically connected. In particular, if $X$ has a $k$-rational point and $X$ is connected, then $X$ is geometrically connected. \end{lemma} \begin{proof} Set $T = \Spec(\kappa(x))$. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. The assumption on $k \subset \kappa(x)$ implies that $T_{\overline{k}}$ is irreducible, see Algebra, Lemma \ref{algebra-lemma-field-extension-geometrically-irreducible}. Hence by Lemma \ref{lemma-geometrically-connected-criterion} we see that $X_{\overline{k}}$ is connected. By Lemma \ref{lemma-characterize-geometrically-connected} we conclude that $X$ is geometrically connected. \end{proof} \begin{lemma} \label{lemma-inverse-image-connected-component} Let $k \subset K$ be an extension of fields. Let $X$ be a scheme over $k$. For every connected component $T$ of $X$ the inverse image $T_K \subset X_K$ is a union of connected components of $X_K$. \end{lemma} \begin{proof} This is a purely topological statement. Denote $p : X_K \to X$ the projection morphism. Let $T \subset X$ be a connected component of $X$. Let $t \in T_K = p^{-1}(T)$. Let $C \subset X_K$ be a connected component containing $t$. Then $p(C)$ is a connected subset of $X$ which meets $T$, hence $p(C) \subset T$. Hence $C \subset T_K$. \end{proof} \noindent The following lemma will be superseded by the stronger Lemma \ref{lemma-image-connected-component} below. \begin{lemma} \label{lemma-image-connected-component-finite-extension} Let $k \subset K$ be a finite extension of fields and let $X$ be a scheme over $k$. Denote by $p : X_K \to X$ the projection morphism. For every connected component $T$ of $X_K$ the image $p(T)$ is a connected component of $X$. \end{lemma} \begin{proof} The image $p(T)$ is contained in some connected component $X'$ of $X$. Consider $X'$ as a closed subscheme of $X$ in any way. Then $T$ is also a connected component of $X'_K = p^{-1}(X')$ and we may therefore assume that $X$ is connected. The morphism $p$ is open (Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}), closed (Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}) and the fibers of $p$ are finite sets (Morphisms, Lemma \ref{morphisms-lemma-finite-quasi-finite}). Thus we may apply Topology, Lemma \ref{topology-lemma-finite-fibre-connected-components} to conclude. \end{proof} \begin{lemma}[Gabber] \label{lemma-image-connected-component} \begin{reference} Email from Ofer Gabber dated June 4, 2016 \end{reference} Let $k \subset K$ be an extension of fields. Let $X$ be a scheme over $k$. Denote $p : X_K \to X$ the projection morphism. Let $\overline{T} \subset X_K$ be a connected component. Then $p(\overline{T})$ is a connected component of $X$. \end{lemma} \begin{proof} When $k \subset K$ is finite this is Lemma \ref{lemma-image-connected-component-finite-extension}. In general the proof is more difficult. \medskip\noindent Let $T \subset X$ be the connected component of $X$ containing the image of $\overline{T}$. We may replace $X$ by $T$ (with the induced reduced subscheme structure). Thus we may assume $X$ is connected. Let $A = H^0(X, \mathcal{O}_X)$. Let $L \subset A$ be the maximal weakly \'etale $k$-subalgebra, see More on Algebra, Lemma \ref{more-algebra-lemma-max-weakly-etale-subalgebra}. Since $A$ does not have any nontrivial idempotents we see that $L$ is a field and a separable algebraic extension of $k$ by More on Algebra, Lemma \ref{more-algebra-lemma-class-weakly-etale-over-field}. Observe that $L$ is also the maximal weakly \'etale $L$-subalgebra of $A$ (because any weakly \'etale $L$-algebra is weakly \'etale over $k$ by More on Algebra, Lemma \ref{more-algebra-lemma-composition-weakly-etale}). By Schemes, Lemma \ref{schemes-lemma-morphism-into-affine} we obtain a factorization $X \to \Spec(L) \to \Spec(k)$ of the structure morphism. \medskip\noindent Let $L'/L$ be a finite separable extension. By Cohomology of Schemes, Lemma \ref{coherent-lemma-finite-locally-free-base-change-cohomology} we have $$A \otimes_L L' = H^0(X \times_{\Spec(L)} \Spec(L'), \mathcal{O}_{X \times_{\Spec(L)} \Spec(L')})$$ The maximal weakly \'etale $L'$-subalgebra of $A \otimes_L L'$ is $L \otimes_L L' = L'$ by More on Algebra, Lemma \ref{more-algebra-lemma-change-fields-max-weakly-etale-subalgebra}. In particular $A \otimes_L L'$ does not have nontrivial idempotents (such an idempotent would generate a weakly \'etale subalgebra) and we conclude that $X \times_{\Spec(L)} \Spec(L')$ is connected. By Lemma \ref{lemma-characterize-geometrically-disconnected} we conclude that $X$ is geometrically connected over $L$. \medskip\noindent Let's give $\overline{T}$ the reduced induced scheme structure and consider the composition $$\overline{T} \xrightarrow{i} X_K = X \times_{\Spec(k)} \Spec(K) \xrightarrow{\pi} \Spec(L \otimes_k K)$$ The image is contained in a connected component of $\Spec(L \otimes_k K)$. Since $K \to L \otimes_k K$ is integral we see that the connected components of $\Spec(L \otimes_k K)$ are points and all points are closed, see Algebra, Lemma \ref{algebra-lemma-integral-over-field}. Thus we get a quotient field $L \otimes_k K \to E$ such that $\overline{T}$ maps into $\Spec(E) \subset \Spec(L \otimes_k K)$. Hence $i(\overline{T}) \subset \pi^{-1}(\Spec(E))$. But $$\pi^{-1}(\Spec(E)) = (X \times_{\Spec(k)} \Spec(K)) \times_{\Spec(L \otimes_k K)} \Spec(E) = X \times_{\Spec(L)} \Spec(E)$$ which is connected because $X$ is geometrically connected over $L$. Then we get the equality $\overline{T} = X \times_{\Spec(L)} \Spec(E)$ (set theoretically) and we conclude that $\overline{T} \to X$ is surjective as desired. \end{proof} \noindent Let $X$ be a scheme. We denote $\pi_0(X)$ the set of connected components of $X$. \begin{lemma} \label{lemma-galois-action-connected-components} Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. There is an action $$\text{Gal}(\overline{k}/k)^{opp} \times \pi_0(X_{\overline{k}}) \longrightarrow \pi_0(X_{\overline{k}})$$ with the following properties: \begin{enumerate} \item An element $\overline{T} \in \pi_0(X_{\overline{k}})$ is fixed by the action if and only if there exists a connected component $T \subset X$, which is geometrically connected over $k$, such that $T_{\overline{k}} = \overline{T}$. \item For any field extension $k \subset k'$ with separable algebraic closure $\overline{k}'$ the diagram $$\xymatrix{ \text{Gal}(\overline{k}'/k') \times \pi_0(X_{\overline{k}'}) \ar[r] \ar[d] & \pi_0(X_{\overline{k}'}) \ar[d] \\ \text{Gal}(\overline{k}/k) \times \pi_0(X_{\overline{k}}) \ar[r] & \pi_0(X_{\overline{k}}) }$$ is commutative (where the right vertical arrow is a bijection according to Lemma \ref{lemma-separably-closed-field-connected-components}). \end{enumerate} \end{lemma} \begin{proof} The action (\ref{equation-galois-action-base-change-kbar}) of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its connected components. Connected components are always closed (Topology, Lemma \ref{topology-lemma-connected-components}). Hence if $\overline{T}$ is as in (1), then by Lemma \ref{lemma-closed-fixed-by-Galois} there exists a closed subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$. Note that $T$ is geometrically connected over $k$, see Lemma \ref{lemma-characterize-geometrically-connected}. To see that $T$ is a connected component of $X$, suppose that $T \subset T'$, $T \not = T'$ where $T'$ is a connected component of $X$. In this case $T'_{k'}$ strictly contains $\overline{T}$ and hence is disconnected. By Lemma \ref{lemma-tricky} this means that $T'$ is disconnected! Contradiction. \medskip\noindent We omit the proof of the functoriality in (2). \end{proof} \begin{lemma} \label{lemma-galois-action-connected-components-continuous} Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. Assume \begin{enumerate} \item $X$ is quasi-compact, and \item the connected components of $X_{\overline{k}}$ are open. \end{enumerate} Then \begin{enumerate} \item[(a)] $\pi_0(X_{\overline{k}})$ is finite, and \item[(b)] the action of $\text{Gal}(\overline{k}/k)$ on $\pi_0(X_{\overline{k}})$ is continuous. \end{enumerate} Moreover, assumptions (1) and (2) are satisfied when $X$ is of finite type over $k$. \end{lemma} \begin{proof} Since the connected components are open, cover $X_{\overline{k}}$ (Topology, Lemma \ref{topology-lemma-connected-components}) and $X_{\overline{k}}$ is quasi-compact, we conclude that there are only finitely many of them. Thus (a) holds. By Lemma \ref{lemma-descend-open} these connected components are each defined over a finite subextension of $k \subset \overline{k}$ and we get (b). If $X$ is of finite type over $k$, then $X_{\overline{k}}$ is of finite type over $\overline{k}$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-finite-type}). Hence $X_{\overline{k}}$ is a Noetherian scheme (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). Thus $X_{\overline{k}}$ has finitely many irreducible components (Properties, Lemma \ref{properties-lemma-Noetherian-irreducible-components}) and a fortiori finitely many connected components (which are therefore open). \end{proof} \section{Geometrically irreducible schemes} \label{section-geometrically-irreducible} \noindent If $X$ is an irreducible scheme over a field, then it can happen that $X$ becomes reducible after extending the ground field. This does not happen for geometrically irreducible schemes. \begin{definition} \label{definition-geometrically-irreducible} Let $X$ be a scheme over the field $k$. We say $X$ is {\it geometrically irreducible} over $k$ if the scheme $X_{k'}$ is irreducible\footnote{An irreducible space is nonempty.} for any field extension $k'$ of $k$. \end{definition} \begin{lemma} \label{lemma-geometrically-irreducible-check-after-extension} Let $X$ be a scheme over the field $k$. Let $k \subset k'$ be a field extension. Then $X$ is geometrically irreducible over $k$ if and only if $X_{k'}$ is geometrically irreducible over $k'$. \end{lemma} \begin{proof} If $X$ is geometrically irreducible over $k$, then it is clear that $X_{k'}$ is geometrically irreducible over $k'$. For the converse, note that for any field extension $k \subset k''$ there exists a common field extension $k' \subset k'''$ and $k'' \subset k'''$. As the morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of a surjective morphism between spectra of fields) we see that the irreducibility of $X_{k'''}$ implies the irreducibility of $X_{k''}$. Thus if $X_{k'}$ is geometrically irreducible over $k'$ then $X$ is geometrically irreducible over $k$. \end{proof} \begin{lemma} \label{lemma-separably-closed-irreducible} Let $X$ be a scheme over a separably closed field $k$. If $X$ is irreducible, then $X_K$ is irreducible for any field extension $k \subset K$. I.e., $X$ is geometrically irreducible over $k$. \end{lemma} \begin{proof} Use Properties, Lemma \ref{properties-lemma-characterize-irreducible} and Algebra, Lemma \ref{algebra-lemma-separably-closed-irreducible}. \end{proof} \begin{lemma} \label{lemma-bijection-irreducible-components} Let $k$ be a field. Let $X$, $Y$ be schemes over $k$. Assume $X$ is geometrically irreducible over $k$. Then the projection morphism $$p : X \times_k Y \longrightarrow Y$$ induces a bijection between irreducible components. \end{lemma} \begin{proof} First, note that the scheme theoretic fibres of $p$ are irreducible, since they are base changes of the geometrically irreducible scheme $X$ by field extensions. Moreover the scheme theoretic fibres are homeomorphic to the set theoretic fibres, see Schemes, Lemma \ref{schemes-lemma-fibre-topological}. By Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open} the map $p$ is open. Thus we may apply Topology, Lemma \ref{topology-lemma-irreducible-fibres-irreducible-components} to conclude. \end{proof} \begin{lemma} \label{lemma-geometrically-irreducible-local} \begin{slogan} Geometric irreductibility is Zariski local modulo connectedness. \end{slogan} Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent \begin{enumerate} \item $X$ is geometrically irreducible over $k$, \item for every nonempty affine open $U$ the $k$-algebra $\mathcal{O}_X(U)$ is geometrically irreducible over $k$ (see Algebra, Definition \ref{algebra-definition-geometrically-irreducible}), \item $X$ is irreducible and there exists an affine open covering $X = \bigcup U_i$ such that each $k$-algebra $\mathcal{O}_X(U_i)$ is geometrically irreducible, and \item there exists an open covering $X = \bigcup_{i \in I} X_i$ with $I \not = \emptyset$ such that $X_i$ is geometrically irreducible for each $i$ and such that $X_i \cap X_j \not = \emptyset$ for all $i, j \in I$. \end{enumerate} Moreover, if $X$ is geometrically irreducible so is every nonempty open subscheme of $X$. \end{lemma} \begin{proof} An affine scheme $\Spec(A)$ over $k$ is geometrically irreducible if and only if $A$ is geometrically irreducible over $k$; this is immediate from the definitions. Recall that if a scheme is irreducible so is every nonempty open subscheme of $X$, any two nonempty open subsets have a nonempty intersection. Also, if every affine open is irreducible then the scheme is irreducible, see Properties, Lemma \ref{properties-lemma-characterize-irreducible}. Hence the final statement of the lemma is clear, as well as the implications (1) $\Rightarrow$ (2), (2) $\Rightarrow$ (3), and (3) $\Rightarrow$ (4). If (4) holds, then for any field extension $k'/k$ the scheme $X_{k'}$ has a covering by irreducible opens which pairwise intersect. Hence $X_{k'}$ is irreducible. Hence (4) implies (1). \end{proof} \begin{lemma} \label{lemma-geometrically-irreducible-function-field} Let $X$ be a geometrically irreducible scheme over the field $k$. Let $\xi \in X$ be its generic point. Then $\kappa(\xi)$ is a geometrically irreducible over $k$. \end{lemma} \begin{proof} Combining Lemma \ref{lemma-geometrically-irreducible-local} and Algebra, Lemma \ref{algebra-lemma-subalgebra-geometrically-irreducible} we see that $\mathcal{O}_{X, \xi}$ is geometrically irreducible over $k$. Since $\mathcal{O}_{X, \xi} \to \kappa(\xi)$ is a surjection with locally nilpotent kernel (see Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}) it follows that $\kappa(\xi)$ is geometrically irreducible, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. \end{proof} \begin{lemma} \label{lemma-separably-closed-field-irreducible-components} Let $k \subset k'$ be an extension of fields. Let $X$ be a scheme over $k$. Set $X' = X_{k'}$. Assume $k$ separably algebraically closed. Then the morphism $X' \to X$ induces a bijection of irreducible components. \end{lemma} \begin{proof} Since $k$ is separably algebraically closed we see that $k'$ is geometrically irreducible over $k$, see Algebra, Lemma \ref{algebra-lemma-separably-closed-irreducible-implies-geometric}. Hence $Z = \Spec(k')$ is geometrically irreducible over $k$. by Lemma \ref{lemma-geometrically-irreducible-local} above. Since $X' = Z \times_k X$ the result is a special case of Lemma \ref{lemma-bijection-irreducible-components}. \end{proof} \begin{lemma} \label{lemma-characterize-geometrically-irreducible} \begin{slogan} Geometric irreducibility can be tested over a separable algebraic closure of the base field. \end{slogan} Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent: \begin{enumerate} \item $X$ is geometrically irreducible over $k$, \item for every finite separable field extension $k \subset k'$ the scheme $X_{k'}$ is irreducible, and \item $X_{\overline{k}}$ is irreducible, where $k \subset \overline{k}$ is a separable algebraic closure of $k$. \end{enumerate} \end{lemma} \begin{proof} Assume $X_{\overline{k}}$ is irreducible, i.e., assume (3). Let $k \subset k'$ be a field extension. There exists a field extension $\overline{k} \subset \overline{k}'$ such that $k'$ embeds into $\overline{k}'$ as an extension of $k$. By Lemma \ref{lemma-separably-closed-field-irreducible-components} we see that $X_{\overline{k}'}$ is irreducible. Since $X_{\overline{k}'} \to X_{k'}$ is surjective we conclude that $X_{k'}$ is irreducible. Hence (1) holds. \medskip\noindent Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Assume not (3), i.e., assume $X_{\overline{k}}$ is reducible. Our goal is to show that also $X_{k'}$ is reducible for some finite subextension $k \subset k' \subset \overline{k}$. Let $X = \bigcup_{i \in I} U_i$ be an affine open covering with $U_i$ not empty. If for some $i$ the scheme $U_i$ is reducible, or if for some pair $i \not = j$ the intersection $U_i \cap U_j$ is empty, then $X$ is reducible (Properties, Lemma \ref{properties-lemma-characterize-irreducible}) and we are done. In particular we may assume that $U_{i, \overline{k}} \cap U_{j, \overline{k}}$ for all $i, j \in I$ is nonempty and we conclude that $U_{i, \overline{k}}$ has to be reducible for some $i$. According to Algebra, Lemma \ref{algebra-lemma-geometrically-irreducible} this means that $U_{i, k'}$ is reducible for some finite separable field extension $k \subset k'$. Hence also $X_{k'}$ is reducible. Thus we see that (2) implies (3). \medskip\noindent The implication (1) $\Rightarrow$ (2) is immediate. This proves the lemma. \end{proof} \begin{lemma} \label{lemma-inverse-image-irreducible} Let $k \subset K$ be an extension of fields. Let $X$ be a scheme over $k$. For every irreducible component $T$ of $X$ the inverse image $T_K \subset X_K$ is a union of irreducible components of $X_K$. \end{lemma} \begin{proof} Let $T \subset X$ be an irreducible component of $X$. The morphism $T_K \to T$ is flat, so generalizations lift along $T_K \to T$. Hence every $\xi \in T_K$ which is a generic point of an irreducible component of $T_K$ maps to the generic point $\eta$ of $T$. If $\xi' \leadsto \xi$ is a specialization in $X_K$ then $\xi'$ maps to $\eta$ since there are no points specializing to $\eta$ in $X$. Hence $\xi' \in T_K$ and we conclude that $\xi = \xi'$. In other words $\xi$ is the generic point of an irreducible component of $X_K$. This means that the irreducible components of $T_K$ are all irreducible components of $X_K$. \end{proof} \noindent For a scheme $X$ we denote $\text{IrredComp}(X)$ the set of irreducible components of $X$. \begin{lemma} \label{lemma-image-irreducible} Let $k \subset K$ be an extension of fields. Let $X$ be a scheme over $k$. For every irreducible component $\overline{T} \subset X_K$ the image of $\overline{T}$ in $X$ is an irreducible component in $X$. This defines a canonical map $$\text{IrredComp}(X_K) \longrightarrow \text{IrredComp}(X)$$ which is surjective. \end{lemma} \begin{proof} Consider the diagram $$\xymatrix{ X_K \ar[d] & X_{\overline{K}} \ar[d] \ar[l] \\ X & X_{\overline{k}} \ar[l] }$$ where $\overline{K}$ is the separable algebraic closure of $K$, and where $\overline{k}$ is the separable algebraic closure of $k$. By Lemma \ref{lemma-separably-closed-field-irreducible-components} the morphism $X_{\overline{K}} \to X_{\overline{k}}$ induces a bijection between irreducible components. Hence it suffices to show the lemma for the morphisms $X_{\overline{k}} \to X$ and $X_{\overline{K}} \to X_K$. In other words we may assume that $K = \overline{k}$. \medskip\noindent The morphism $p : X_{\overline{k}} \to X$ is integral, flat and surjective. Flatness implies that generalizations lift along $p$, see Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}. Hence generic points of irreducible components of $X_{\overline{k}}$ map to generic points of irreducible components of $X$. Integrality implies that $p$ is universally closed, see Morphisms, Lemma \ref{morphisms-lemma-integral-universally-closed}. Hence we conclude that the image $p(\overline{T})$ of an irreducible component is a closed irreducible subset which contains a generic point of an irreducible component of $X$, hence $p(\overline{T})$ is an irreducible component of $X$. This proves the first assertion. If $T \subset X$ is an irreducible component, then $p^{-1}(T) =T_K$ is a nonempty union of irreducible components, see Lemma \ref{lemma-inverse-image-irreducible}. Each of these necessarily maps onto $T$ by the first part. Hence the map is surjective. \end{proof} \begin{lemma} \label{lemma-galois-action-irreducible-components} Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. There is an action $$\text{Gal}(\overline{k}/k)^{opp} \times \text{IrredComp}(X_{\overline{k}}) \longrightarrow \text{IrredComp}(X_{\overline{k}})$$ with the following properties: \begin{enumerate} \item An element $\overline{T} \in \text{IrredComp}(X_{\overline{k}})$ is fixed by the action if and only if there exists an irreducible component $T \subset X$, which is geometrically irreducible over $k$, such that $T_{\overline{k}} = \overline{T}$. \item For any field extension $k \subset k'$ with separable algebraic closure $\overline{k}'$ the diagram $$\xymatrix{ \text{Gal}(\overline{k}'/k') \times \text{IrredComp}(X_{\overline{k}'}) \ar[r] \ar[d] & \text{IrredComp}(X_{\overline{k}'}) \ar[d] \\ \text{Gal}(\overline{k}/k) \times \text{IrredComp}(X_{\overline{k}}) \ar[r] & \text{IrredComp}(X_{\overline{k}}) }$$ is commutative (where the right vertical arrow is a bijection according to Lemma \ref{lemma-separably-closed-field-irreducible-components}). \end{enumerate} \end{lemma} \begin{proof} The action (\ref{equation-galois-action-base-change-kbar}) of $\text{Gal}(\overline{k}/k)$ on $X_{\overline{k}}$ induces an action on its irreducible components. Irreducible components are always closed (Topology, Lemma \ref{topology-lemma-connected-components}). Hence if $\overline{T}$ is as in (1), then by Lemma \ref{lemma-closed-fixed-by-Galois} there exists a closed subset $T \subset X$ such that $\overline{T} = T_{\overline{k}}$. Note that $T$ is geometrically irreducible over $k$, see Lemma \ref{lemma-characterize-geometrically-irreducible}. To see that $T$ is an irreducible component of $X$, suppose that $T \subset T'$, $T \not = T'$ where $T'$ is an irreducible component of $X$. Let $\overline{\eta}$ be the generic point of $\overline{T}$. It maps to the generic point $\eta$ of $T$. Then the generic point $\xi \in T'$ specializes to $\eta$. As $X_{\overline{k}} \to X$ is flat there exists a point $\overline{\xi} \in X_{\overline{k}}$ which maps to $\xi$ and specializes to $\overline{\eta}$. It follows that the closure of the singleton $\{\overline{\xi}\}$ is an irreducible closed subset of $X_{\overline{\xi}}$ which strictly contains $\overline{T}$. This is the desired contradiction. \medskip\noindent We omit the proof of the functoriality in (2). \end{proof} \begin{lemma} \label{lemma-orbit-irreducible-components} Let $k$ be a field, with separable algebraic closure $\overline{k}$. Let $X$ be a scheme over $k$. The fibres of the map $$\text{IrredComp}(X_{\overline{k}}) \longrightarrow \text{IrredComp}(X)$$ of Lemma \ref{lemma-image-irreducible} are exactly the orbits of $\text{Gal}(\overline{k}/k)$ under the action of Lemma \ref{lemma-galois-action-irreducible-components}. \end{lemma} \begin{proof} Let $T \subset X$ be an irreducible component of $X$. Let $\eta \in T$ be its generic point. By Lemmas \ref{lemma-inverse-image-irreducible} and \ref{lemma-image-irreducible} the generic points of irreducible components of $\overline{T}$ which map into $T$ map to $\eta$. By Algebra, Lemma \ref{algebra-lemma-Galois-orbit} the Galois group acts transitively on all of the points of $X_{\overline{k}}$ mapping to $\eta$. Hence the lemma follows. \end{proof} \begin{lemma} \label{lemma-galois-action-irreducible-components-locally-finite-type} Let $k$ be a field. Assume $X \to \Spec(k)$ locally of finite type. In this case \begin{enumerate} \item the action $$\text{Gal}(\overline{k}/k)^{opp} \times \text{IrredComp}(X_{\overline{k}}) \longrightarrow \text{IrredComp}(X_{\overline{k}})$$ is continuous if we give $\text{IrredComp}(X_{\overline{k}})$ the discrete topology, \item every irreducible component of $X_{\overline{k}}$ can be defined over a finite extension of $k$, and \item given any irreducible component $T \subset X$ the scheme $T_{\overline{k}}$ is a finite union of irreducible components of $X_{\overline{k}}$ which are all in the same $\text{Gal}(\overline{k}/k)$-orbit. \end{enumerate} \end{lemma} \begin{proof} Let $\overline{T}$ be an irreducible component of $X_{\overline{k}}$. We may choose an affine open $U \subset X$ such that $\overline{T} \cap U_{\overline{k}}$ is not empty. Write $U = \Spec(A)$, so $A$ is a finite type $k$-algebra, see Morphisms, Lemma \ref{morphisms-lemma-locally-finite-type-characterize}. Hence $A_{\overline{k}}$ is a finite type $\overline{k}$-algebra, and in particular Noetherian. Let $\mathfrak p = (f_1, \ldots, f_n)$ be the prime ideal corresponding to $\overline{T} \cap U_{\overline{k}}$. Since $A_{\overline{k}} = A \otimes_k \overline{k}$ we see that there exists a finite subextension $k \subset k' \subset \overline{k}$ such that each $f_i \in A_{k'}$. It is clear that $\text{Gal}(\overline{k}/k')$ fixes $\overline{T}$, which proves (1). \medskip\noindent Part (2) follows by applying Lemma \ref{lemma-galois-action-irreducible-components} (1) to the situation over $k'$ which implies the irreducible component $\overline{T}$ is of the form $T'_{\overline{k}}$ for some irreducible $T' \subset X_{k'}$. \medskip\noindent To prove (3), let $T \subset X$ be an irreducible component. Choose an irreducible component $\overline{T} \subset X_{\overline{k}}$ which maps to $T$, see Lemma \ref{lemma-image-irreducible}. By the above the orbit of $\overline{T}$ is finite, say it is $\overline{T}_1, \ldots, \overline{T}_n$. Then $\overline{T}_1 \cup \ldots \cup \overline{T}_n$ is a $\text{Gal}(\overline{k}/k)$-invariant closed subset of $X_{\overline{k}}$ hence of the form $W_{\overline{k}}$ for some $W \subset X$ closed by Lemma \ref{lemma-closed-fixed-by-Galois}. Clearly $W = T$ and we win. \end{proof} \begin{lemma} \label{lemma-finite-extension-geometrically-irreducible-components} Let $k$ be a field. Let $X \to \Spec(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite separable extension $k \subset k'$ such that every irreducible component of $X_{k'}$ is geometrically irreducible over $k'$. \end{lemma} \begin{proof} Let $\overline{k}$ be a separable algebraic closure of $k$. The assumption that $X$ has finitely many irreducible components combined with Lemma \ref{lemma-galois-action-irreducible-components-locally-finite-type} (3) shows that $X_{\overline{k}}$ has finitely many irreducible components $\overline{T}_1, \ldots, \overline{T}_n$. By Lemma \ref{lemma-galois-action-irreducible-components-locally-finite-type} (2) there exists a finite extension $k \subset k' \subset \overline{k}$ and irreducible components $T_i \subset X_{k'}$ such that $\overline{T}_i = T_{i, \overline{k}}$ and we win. \end{proof} \begin{lemma} \label{lemma-irreducible-components-geometrically-irreducible} Let $X$ be a scheme over the field $k$. Assume $X$ has finitely many irreducible components which are all geometrically irreducible. Then $X$ has finitely many connected components each of which is geometrically connected. \end{lemma} \begin{proof} This is clear because a connected component is a union of irreducible components. Details omitted. \end{proof} \section{Geometrically integral schemes} \label{section-geometrically-integral} \noindent If $X$ is an irreducible scheme over a field, then it can happen that $X$ becomes reducible after extending the ground field. This does not happen for geometrically irreducible schemes. \begin{definition} \label{definition-geometrically-integral} Let $X$ be a scheme over the field $k$. \begin{enumerate} \item Let $x \in X$. We say $X$ is {\it geometrically pointwise integral at $x$} if for every field extension $k \subset k'$ and every $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is integral. \item We say $X$ is {\it geometrically pointwise integral} if $X$ is geometrically pointwise integral at every point. \item We say $X$ is {\it geometrically integral} over $k$ if the scheme $X_{k'}$ is integral for every field extension $k'$ of $k$. \end{enumerate} \end{definition} \noindent The distinction between notions (2) and (3) is necessary. For example if $k = \mathbf{R}$ and $X = \Spec(\mathbf{C}[x])$, then $X$ is geometrically pointwise integral over $\mathbf{R}$ but of course not geometrically integral. \begin{lemma} \label{lemma-geometrically-integral} Let $k$ be a field. Let $X$ be a scheme over $k$. Then $X$ is geometrically integral over $k$ if and only if $X$ is both geometrically reduced and geometrically irreducible over $k$. \end{lemma} \begin{proof} See Properties, Lemma \ref{properties-lemma-characterize-integral}. \end{proof} \begin{lemma} \label{lemma-proper-geometrically-reduced-global-sections} Let $k$ be a field. Let $X$ be a proper scheme over $k$. \begin{enumerate} \item $A = H^0(X, \mathcal{O}_X)$ is a finite dimensional $k$-algebra, \item $A = \prod_{i = 1, \ldots, n} A_i$ is a product of Artinian local $k$-algebras, one factor for each connected component of $X$, \item if $X$ is reduced, then $A = \prod_{i = 1, \ldots, n} k_i$ is a product of fields, each a finite extension of $k$, \item if $X$ is geometrically reduced, then $k_i$ is finite separable over $k$, \item if $X$ is geometrically irreducible, then $A$ is geometrically irreducible over $k$, \item if $X$ is geometrically integral, then $A = k$. \end{enumerate} \end{lemma} \begin{proof} By Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-over-affine-cohomology-finite} we see that $A = H^0(X, \mathcal{O}_X)$ is a finite dimensional $k$-algebra. This proves (1). \medskip\noindent Then $A$ is a product of local rings by Algebra, Lemma \ref{algebra-lemma-finite-dimensional-algebra} and Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring}. If $X = Y \amalg Z$ with $Y$ and $Z$ open in $X$, then we obtain an idempotent $e \in A$ by taking the section of $\mathcal{O}_X$ which is $1$ on $Y$ and $0$ on $Z$. Conversely, if $e \in A$ is an idempotent, then we get a corresponding decomposition of $X$. Finally, as $X$ has a Noetherian underlying topological space its connected components are open. Hence the connected components of $X$ correspond $1$-to-$1$ with primitive idempotents of $A$. This proves (2). \medskip\noindent If $X$ is reduced, then $A$ is reduced. Hence the local rings $A_i = k_i$ are reduced and therefore fields (for example by Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}). This proves (3). \medskip\noindent If $X$ is geometrically reduced, then same thing is true for $A \otimes_k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ (see Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology} for equality). This implies that $k_i \otimes_k \overline{k}$ is a product of fields and hence $k_i/k$ is separable for example by Algebra, Lemmas \ref{algebra-lemma-characterize-separable-field-extensions} and \ref{algebra-lemma-geometrically-reduced-finite-purely-inseparable-extension}. This proves (4). \medskip\noindent If $X$ is geometrically irreducible, then $A \otimes_k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ is a zero dimensional local ring by part (2) and hence its spectrum has one point, in particular it is irreducible. Thus $A$ is geometrically irreducible. This proves (5). \medskip\noindent If $X$ is geometrically irreducible and geometrically reduced, then $A = k_1$ is a field and the extension $k_1/k$ is finite separable and geometrically irreducible. However, then $k_1 \otimes_k \overline{k}$ is a product of $[k_1 : k]$ copies of $\overline{k}$ and we conclude that $k_1 = k$. \end{proof} \section{Geometrically normal schemes} \label{section-geometrically-normal} \noindent In Properties, Definition \ref{properties-definition-normal} we have defined the notion of a normal scheme. This notion is defined even for non-Noetherian schemes. Hence, contrary to our discussion of geometrically regular'' schemes we consider all field extensions of the ground field. \begin{definition} \label{definition-geometrically-normal} Let $X$ be a scheme over the field $k$. \begin{enumerate} \item Let $x \in X$. We say $X$ is {\it geometrically normal at $x$} if for every field extension $k \subset k'$ and every $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is normal. \item We say $X$ is {\it geometrically normal} over $k$ if $X$ is geometrically normal at every $x \in X$. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-geometrically-normal-at-point} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent \begin{enumerate} \item $X$ is geometrically normal at $x$, \item for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ lying over over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is normal, and \item the ring $\mathcal{O}_{X, x}$ is geometrically normal over $k$ (see Algebra, Definition \ref{algebra-definition-geometrically-normal}). \end{enumerate} \end{lemma} \begin{proof} It is clear that (1) implies (2). Assume (2). Let $k \subset k'$ be a finite purely inseparable field extension (for example $k = k'$). Consider the ring $\mathcal{O}_{X, x} \otimes_k k'$. By Algebra, Lemma \ref{algebra-lemma-p-ring-map} its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$. By assumption this is a normal ring. Hence we deduce (3) by Algebra, Lemma \ref{algebra-lemma-geometrically-normal}. \medskip\noindent Assume (3). Let $k \subset k'$ be a field extension. Since $\Spec(k') \to \Spec(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$. Hence it is normal by assumption and (1) is proved. \end{proof} \begin{lemma} \label{lemma-geometrically-normal} Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent \begin{enumerate} \item $X$ is geometrically normal, \item $X_{k'}$ is a normal scheme for every field extension $k'/k$, \item $X_{k'}$ is a normal scheme for every finitely generated field extension $k'/k$, \item $X_{k'}$ is a normal scheme for every finite purely inseparable field extension $k'/k$, \item for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is geometrically normal (see Algebra, Definition \ref{algebra-definition-geometrically-normal}), and \item $X_{k^{perf}}$ is a normal scheme. \end{enumerate} \end{lemma} \begin{proof} Assume (1). Then for every field extension $k \subset k'$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is normal. By definition this means that $X_{k'}$ is normal. Hence (2). \medskip\noindent It is clear that (2) implies (3) implies (4). \medskip\noindent Assume (4) and let $U \subset X$ be an affine open subscheme. Then $U_{k'}$ is a normal scheme for any finite purely inseparable extension $k \subset k'$ (including $k = k'$). This means that $k' \otimes_k \mathcal{O}(U)$ is a normal ring for all finite purely inseparable extensions $k \subset k'$. Hence $\mathcal{O}(U)$ is a geometrically normal $k$-algebra by definition. Hence (4) implies (5). \medskip\noindent Assume (5). For any field extension $k \subset k'$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_X(U) \otimes_k k'$ where $U$ is affine open in $X$ (see Schemes, Section \ref{schemes-section-fibre-products}). Hence $X_{k'}$ is normal. So (1) holds. \medskip\noindent The equivalence of (5) and (6) follows from the definition of geometrically normal algebras and the equivalence (just proved) of (3) and (4). \end{proof} \begin{lemma} \label{lemma-geometrically-normal-upstairs} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. The following are equivalent \begin{enumerate} \item $X$ is geometrically normal at $x$, \item $X_{k'}$ is geometrically normal at $x'$. \end{enumerate} In particular, $X$ is geometrically normal over $k$ if and only if $X_{k'}$ is geometrically normal over $k'$. \end{lemma} \begin{proof} It is clear that (1) implies (2). Assume (2). Let $k \subset k''$ be a finite purely inseparable field extension and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is unique). We can find a common field extension $k \subset k'''$ (i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$) and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$. Consider the map of local rings $$\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.$$ This is a flat local ring homomorphism and hence faithfully flat. By (2) we see that the local ring on the right is normal. Thus by Algebra, Lemma \ref{algebra-lemma-descent-normal} we conclude that $\mathcal{O}_{X_{k''}, x''}$ is normal. By Lemma \ref{lemma-geometrically-normal-at-point} we see that $X$ is geometrically normal at $x$. \end{proof} \begin{lemma} \label{lemma-fibre-product-normal} Let $k$ be a field. Let $X$ be a geometrically normal scheme over $k$ and let $Y$ be a normal scheme over $k$. Then $X \times_k Y$ is a normal scheme. \end{lemma} \begin{proof} This reduces to Algebra, Lemma \ref{algebra-lemma-geometrically-normal-tensor-normal} by Lemma \ref{lemma-geometrically-normal}. \end{proof} \begin{lemma} \label{lemma-base-change-normal-by-separable} Let $k$ be a field. Let $X$ be a normal scheme over $k$. Let $K/k$ be a separable field extension. Then $X_K$ is a normal scheme. \end{lemma} \begin{proof} Follows from Lemma \ref{lemma-fibre-product-normal} and Algebra, Lemma \ref{algebra-lemma-separable-field-extension-geometrically-normal}. \end{proof} \section{Change of fields and locally Noetherian schemes} \label{section-locally-Noetherian} \noindent Let $X$ a locally Noetherian scheme over a field $k$. It is not always that case that $X_{k'}$ is locally Noetherian too. For example if $X = \Spec(\overline{\mathbf{Q}})$ and $k = \mathbf{Q}$, then $X_{\overline{\mathbf{Q}}}$ is the spectrum of $\overline{\mathbf{Q}} \otimes_{\mathbf{Q}} \overline{\mathbf{Q}}$ which is not Noetherian. (Hint: It has too many idempotents). But if we only base change using finitely generated field extensions then the Noetherian property is preserved. (Or if $X$ is locally of finite type over $k$, since this property is preserved under base change.) \begin{lemma} \label{lemma-locally-Noetherian-base-change} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k \subset k'$ be a finitely generated field extension. Then $X$ is locally Noetherian if and only if $X_{k'}$ is locally Noetherian. \end{lemma} \begin{proof} Using Properties, Lemma \ref{properties-lemma-locally-Noetherian} we reduce to the case where $X$ is affine, say $X = \Spec(A)$. In this case we have to prove that $A$ is Noetherian if and only if $A_{k'}$ is Noetherian. Since $A \to A_{k'} = k' \otimes_k A$ is faithfully flat, we see that if $A_{k'}$ is Noetherian, then so is $A$, by Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}. Conversely, if $A$ is Noetherian then $A_{k'}$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-Noetherian-field-extension}. \end{proof} \section{Geometrically regular schemes} \label{section-geometrically-regular} \noindent A geometrically regular scheme over a field $k$ is a locally Noetherian scheme over $k$ which remains regular upon suitable changes of base field. A finite type scheme over $k$ is geometrically regular if and only if it is smooth over $k$ (see Lemma \ref{lemma-geometrically-regular-smooth}). The notion of geometric regularity is most interesting in situations where smoothness cannot be used such as formal fibres (insert future reference here). \medskip\noindent In the following definition we restrict ourselves to locally Noetherian schemes, since the property of being a regular local ring is only defined for Noetherian local rings. By Lemma \ref{lemma-geometrically-normal} above, if we restrict ourselves to finitely generated field extensions then this property is preserved under change of base field. This comment will be used without further reference in this section. In particular the following definition makes sense. \begin{definition} \label{definition-geometrically-regular} Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. \begin{enumerate} \item Let $x \in X$. We say $X$ is {\it geometrically regular at $x$} over $k$ if for every finitely generated field extension $k \subset k'$ and any $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is regular. \item We say $X$ is {\it geometrically regular over $k$} if $X$ is geometrically regular at all of its points. \end{enumerate} \end{definition} \noindent A similar definition works to define geometrically Cohen-Macaulay, $(R_k)$, and $(S_k)$ schemes over a field. We will add a section for these separately as needed. \begin{lemma} \label{lemma-geometrically-regular-at-point} Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. Let $x \in X$. The following are equivalent \begin{enumerate} \item $X$ is geometrically regular at $x$, \item for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ lying over over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is regular, and \item the ring $\mathcal{O}_{X, x}$ is geometrically regular over $k$ (see Algebra, Definition \ref{algebra-definition-geometrically-regular}). \end{enumerate} \end{lemma} \begin{proof} It is clear that (1) implies (2). Assume (2). This in particular implies that $\mathcal{O}_{X, x}$ is a regular local ring. Let $k \subset k'$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes_k k'$. By Algebra, Lemma \ref{algebra-lemma-p-ring-map} its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes_k k'$. By assumption this is a regular ring. Hence we deduce (3) from the definition of a geometrically regular ring. \medskip\noindent Assume (3). Let $k \subset k'$ be a field extension. Since $\Spec(k') \to \Spec(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma \ref{morphisms-lemma-base-change-surjective}). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes_k k'$. Hence it is regular by assumption and (1) is proved. \end{proof} \begin{lemma} \label{lemma-geometrically-regular} Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. The following are equivalent \begin{enumerate} \item $X$ is geometrically regular, \item $X_{k'}$ is a regular scheme for every finitely generated field extension $k \subset k'$, \item $X_{k'}$ is a regular scheme for every finite purely inseparable field extension $k \subset k'$, \item for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is geometrically regular (see Algebra, Definition \ref{algebra-definition-geometrically-regular}), and \item there exists an affine open covering $X = \bigcup U_i$ such that each $\mathcal{O}_X(U_i)$ is geometrically regular over $k$. \end{enumerate} \end{lemma} \begin{proof} Assume (1). Then for every finitely generated field extension $k \subset k'$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is regular. By Properties, Lemma \ref{properties-lemma-characterize-regular} this means that $X_{k'}$ is regular. Hence (2). \medskip\noindent It is clear that (2) implies (3). \medskip\noindent Assume (3) and let $U \subset X$ be an affine open subscheme. Then $U_{k'}$ is a regular scheme for any finite purely inseparable extension $k \subset k'$ (including $k = k'$). This means that $k' \otimes_k \mathcal{O}(U)$ is a regular ring for all finite purely inseparable extensions $k \subset k'$. Hence $\mathcal{O}(U)$ is a geometrically regular $k$-algebra and we see that (4) holds. \medskip\noindent It is clear that (4) implies (5). Let $X = \bigcup U_i$ be an affine open covering as in (5). For any field extension $k \subset k'$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_X(U_i) \otimes_k k'$ (see Schemes, Section \ref{schemes-section-fibre-products}). Hence $X_{k'}$ is regular. So (1) holds. \end{proof} \begin{lemma} \label{lemma-geometrically-regular-upstairs} Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a finitely generated field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. The following are equivalent \begin{enumerate} \item $X$ is geometrically regular at $x$, \item $X_{k'}$ is geometrically regular at $x'$. \end{enumerate} In particular, $X$ is geometrically regular over $k$ if and only if $X_{k'}$ is geometrically regular over $k'$. \end{lemma} \begin{proof} It is clear that (1) implies (2). Assume (2). Let $k \subset k''$ be a finite purely inseparable field extension and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is unique). We can find a common, finitely generated, field extension $k \subset k'''$ (i.e.\ with both $k' \subset k'''$ and $k'' \subset k'''$) and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$. Consider the map of local rings $$\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.$$ This is a flat local ring homomorphism of Noetherian local rings and hence faithfully flat. By (2) we see that the local ring on the right is regular. Thus by Algebra, Lemma \ref{algebra-lemma-flat-under-regular} we conclude that $\mathcal{O}_{X_{k''}, x''}$ is regular. By Lemma \ref{lemma-geometrically-regular-at-point} we see that $X$ is geometrically regular at $x$. \end{proof} \noindent The following lemma is a geometric variant of Algebra, Lemma \ref{algebra-lemma-geometrically-regular-descent}. \begin{lemma} \label{lemma-flat-under-geometrically-regular} Let $k$ be a field. Let $f : X \to Y$ be a morphism of locally Noetherian schemes over $k$. Let $x \in X$ be a point and set $y = f(x)$. If $X$ is geometrically regular at $x$ and $f$ is flat at $x$ then $Y$ is geometrically regular at $y$. In particular, if $X$ is geometrically regular over $k$ and $f$ is flat and surjective, then $Y$ is geometrically regular over $k$. \end{lemma} \begin{proof} Let $k'$ be finite purely inseparable extension of $k$. Let $f' : X_{k'} \to Y_{k'}$ be the base change of $f$. Let $x' \in X_{k'}$ be the unique point lying over $x$. If we show that $Y_{k'}$ is regular at $y' = f'(x')$, then $Y$ is geometrically regular over $k$ at $y'$, see Lemma \ref{lemma-geometrically-regular}. By Morphisms, Lemma \ref{morphisms-lemma-base-change-module-flat} the morphism $X_{k'} \to Y_{k'}$ is flat at $x'$. Hence the ring map $$\mathcal{O}_{Y_{k'}, y'} \longrightarrow \mathcal{O}_{X_{k'}, x'}$$ is a flat local homomorphism of local Noetherian rings with right hand side regular by assumption. Hence the left hand side is a regular local ring by Algebra, Lemma \ref{algebra-lemma-flat-under-regular}. \end{proof} \begin{lemma} \label{lemma-geometrically-regular-smooth} Let $k$ be a field. Let $X$ be a scheme of finite type over $k$. Let $x \in X$. Then $X$ is geometrically regular at $x$ if and only if $X \to \Spec(k)$ is smooth at $x$ (Morphisms, Definition \ref{morphisms-definition-smooth}). \end{lemma} \begin{proof} The question is local around $x$, hence we may assume that $X = \Spec(A)$ for some finite type $k$-algebra. Let $x$ correspond to the prime $\mathfrak p$. \medskip\noindent If $A$ is smooth over $k$ at $\mathfrak p$, then we may localize $A$ and assume that $A$ is smooth over $k$. In this case $k' \otimes_k A$ is smooth over $k'$ for all extension fields $k'/k$, and each of these Noetherian rings is regular by Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}. \medskip\noindent Assume $X$ is geometrically regular at $x$. Consider the residue field $K := \kappa(x) = \kappa(\mathfrak p)$ of $x$. It is a finitely generated extension of $k$. By Algebra, Lemma \ref{algebra-lemma-make-separable} there exists a finite purely inseparable extension $k \subset k'$ such that the compositum $k'K$ is a separable field extension of $k'$. Let $\mathfrak p' \subset A' = k' \otimes_k A$ be a prime ideal lying over $\mathfrak p$. It is the unique prime lying over $\mathfrak p$, see Algebra, Lemma \ref{algebra-lemma-p-ring-map}. Hence the residue field $K' := \kappa(\mathfrak p')$ is the compositum $k'K$. By assumption the local ring $(A')_{\mathfrak p'}$ is regular. Hence by Algebra, Lemma \ref{algebra-lemma-separable-smooth} we see that $k' \to A'$ is smooth at $\mathfrak p'$. This in turn implies that $k \to A$ is smooth at $\mathfrak p$ by Algebra, Lemma \ref{algebra-lemma-smooth-field-change-local}. The lemma is proved. \end{proof} \begin{example} \label{example-geometrically-reduced-not-normal} Let $k =\mathbf{F}_p(t)$. It is quite easy to give an example of a regular variety $V$ over $k$ which is not geometrically reduced. For example we can take $\Spec(k[x]/(x^p - t))$. In fact, there exists an example of a regular variety $V$ which is geometrically reduced, but not even geometrically normal. Namely, take for $p > 2$ the scheme $V = \Spec(k[x, y]/(y^2 - x^p + t))$. This is a variety as the polynomial $y^2 - x^p + t \in k[x, y]$ is irreducible. The morphism $V \to \Spec(k)$ is smooth at all points except at the point $v_0 \in V$ corresponding to the maximal ideal $(y, x^p - t)$ (because $2y$ is invertible). In particular we see that $V$ is (geometrically) regular at all points, except possibly $v_0$. The local ring $$\mathcal{O}_{V, v_0} = \left(k[x, y]/(y^2 - x^p + t)\right)_{(y, x^p - t)}$$ is a domain of dimension $1$. Its maximal ideal is generated by $1$ element, namely $y$. Hence it is a discrete valuation ring and regular. Let $k' = k[t^{1/p}]$. Denote $t' = t^{1/p} \in k'$, $V' = V_{k'}$, $v'_0 \in V'$ the unique point lying over $v_0$. Over $k'$ we can write $x^p - t = (x - t')^p$, but the polynomial $y^2 - (x - t')^p$ is still irreducible and $V'$ is still a variety. But the element $$\frac{y}{x - t'} \in f.f.(\mathcal{O}_{V', v'_0})$$ is integral over $\mathcal{O}_{V', v'_0}$ (just compute its square) and not contained in it, so $V'$ is not normal at $v'_0$. This concludes the example. \end{example} \section{Change of fields and the Cohen-Macaulay property} \label{section-CM} \noindent The following lemma says that it does not make sense to define geometrically Cohen-Macaulay schemes, since these would be the same as Cohen-Macaulay schemes. \begin{lemma} \label{lemma-CM-base-change} Let $X$ be a locally Noetherian scheme over the field $k$. Let $k \subset k'$ be a finitely generated field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. Then we have $$\mathcal{O}_{X, x}\text{ is Cohen-Macaulay} \Leftrightarrow \mathcal{O}_{X_{k'}, x'}\text{ is Cohen-Macaulay}$$ If $X$ is locally of finite type over $k$, the same holds for any field extension $k \subset k'$. \end{lemma} \begin{proof} The first case of the lemma follows from Algebra, Lemma \ref{algebra-lemma-CM-geometrically-CM}. The second case of the lemma is equivalent to Algebra, Lemma \ref{algebra-lemma-extend-field-CM-locus}. \end{proof} \section{Change of fields and the Jacobson property} \label{section-overfield} \noindent A scheme locally of finite type over a field has plenty of closed points, namely it is Jacobson. Moreover, the residue fields are finite extensions of the ground field. \begin{lemma} \label{lemma-locally-finite-type-Jacobson} Let $X$ be a scheme which is locally of finite type over $k$. Then \begin{enumerate} \item for any closed point $x \in X$ the extension $k \subset \kappa(x)$ is algebraic, and \item $X$ is a Jacobson scheme (Properties, Definition \ref{properties-definition-jacobson}). \end{enumerate} \end{lemma} \begin{proof} A scheme is Jacobson if and only if it has an affine open covering by Jacobson schemes, see Properties, Lemma \ref{properties-lemma-locally-jacobson}. The property on residue fields at closed points is also local on $X$. Hence we may assume that $X$ is affine. In this case the result is a consequence of the Hilbert Nullstellensatz, see Algebra, Theorem \ref{algebra-theorem-nullstellensatz}. It also follows from a combination of Morphisms, Lemmas \ref{morphisms-lemma-jacobson-finite-type-points}, \ref{morphisms-lemma-Jacobson-universally-Jacobson}, and \ref{morphisms-lemma-ubiquity-Jacobson-schemes}. \end{proof} \noindent It turns out that if $X$ is not locally of finite type, then we can achieve the same result after making a suitably large base field extension. \begin{lemma} \label{lemma-make-Jacobson} Let $X$ be a scheme over a field $k$. For any field extension $k \subset K$ whose cardinality is large enough we have \begin{enumerate} \item for any closed point $x \in X_K$ the extension $K \subset \kappa(x)$ is algebraic, and \item $X_K$ is a Jacobson scheme (Properties, Definition \ref{properties-definition-jacobson}). \end{enumerate} \end{lemma} \begin{proof} Choose an affine open covering $X = \bigcup U_i$. By Algebra, Lemma \ref{algebra-lemma-base-change-Jacobson} and Properties, Lemma \ref{properties-lemma-affine-jacobson} there exist cardinals $\kappa_i$ such that $U_{i, K}$ has the desired properties over $K$ if $\#(K) \geq \kappa_i$. Set $\kappa = \max\{\kappa_i\}$. Then if the cardinality of $K$ is larger than $\kappa$ we see that each $U_{i, K}$ satisfies the conclusions of the lemma. Hence $X_K$ is Jacobson by Properties, Lemma \ref{properties-lemma-locally-jacobson}. The statement on residue fields at closed points of $X_K$ follows from the corresponding statements for residue fields of closed points of the $U_{i, K}$. \end{proof} \section{Change of fields and ample invertible sheaves} \label{section-change-fields-ample} \noindent The following result is typical for the results in this section. \begin{lemma} \label{lemma-ample-after-field-extension} Let $k$ be a field. Let $X$ be a scheme over $k$. If there exists an ample invertible sheaf on $X_K$ for some field extension $k \subset K$, then $X$ has an ample invertible sheaf. \end{lemma} \begin{proof} Let $k \subset K$ be a field extension such that $X_K$ has an ample invertible sheaf $\mathcal{L}$. The morphism $X_K \to X$ is surjective. Hence $X$ is quasi-compact as the image of a quasi-compact scheme (Properties, Definition \ref{properties-definition-ample}). Since $X_K$ is quasi-separated (by Properties, Lemma \ref{properties-lemma-affine-s-opens-cover-quasi-separated}) we see that $X$ is quasi-separated: If $U, V \subset X$ are affine open, then $(U \cap V)_K = U_K \cap V_K$ is quasi-compact and $(U \cap V)_K \to U \cap V$ is surjective. Thus Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} applies. \medskip\noindent Write $K = \colim A_i$ as the colimit of the subalgebras of $K$ which are of finite type over $k$. Denote $X_i = X \times_{\Spec(k)} \Spec(A_i)$. Since $X_K = \lim X_i$ we find an $i$ and an invertible sheaf' $\mathcal{L}_i$ on $X_i$ whose pullback to $X_K$ is $\mathcal{L}$ (Limits, Lemma \ref{limits-lemma-descend-invertible-modules}; here and below we use that $X$ is quasi-compact and quasi-separated as just shown). By Limits, Lemma \ref{limits-lemma-limit-ample} we may assume $\mathcal{L}_i$ is ample after possibly increasing $i$. Fix such an $i$ and let $\mathfrak m \subset A_i$ be a maximal ideal. By the Hilbert Nullstellensatz (Algebra, Theorem \ref{algebra-theorem-nullstellensatz}) the residue field $k' = A_i/\mathfrak m$ is a finite extension of $k$. Hence $X_{k'} \subset X_i$ is a closed subscheme hence has an ample invertible sheaf (Properties, Lemma \ref{properties-lemma-ample-on-closed}). Since $X_{k'} \to X$ is finite locally free we conclude that $X$ has an ample invertible sheaf by Divisors, Proposition \ref{divisors-proposition-push-down-ample}. \end{proof} \begin{lemma} \label{lemma-quasi-affine-after-field-extension} Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is quasi-affine for some field extension $k \subset K$, then $X$ is quasi-affine. \end{lemma} \begin{proof} Let $k \subset K$ be a field extension such that $X_K$ is quasi-affine. The morphism $X_K \to X$ is surjective. Hence $X$ is quasi-compact as the image of a quasi-compact scheme (Properties, Definition \ref{properties-definition-quasi-affine}). Since $X_K$ is quasi-separated (as an open subscheme of an affine scheme) we see that $X$ is quasi-separated: If $U, V \subset X$ are affine open, then $(U \cap V)_K = U_K \cap V_K$ is quasi-compact and $(U \cap V)_K \to U \cap V$ is surjective. Thus Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} applies. \medskip\noindent Write $K = \colim A_i$ as the colimit of the subalgebras of $K$ which are of finite type over $k$. Denote $X_i = X \times_{\Spec(k)} \Spec(A_i)$. Since $X_K = \lim X_i$ we find an $i$ such that $X_i$ is quasi-affine (Limits, Lemma \ref{limits-lemma-limit-quasi-affine}; here we use that $X$ is quasi-compact and quasi-separated as just shown). By the Hilbert Nullstellensatz (Algebra, Theorem \ref{algebra-theorem-nullstellensatz}) the residue field $k' = A_i/\mathfrak m$ is a finite extension of $k$. Hence $X_{k'} \subset X_i$ is a closed subscheme hence is quasi-affine (Properties, Lemma \ref{properties-lemma-quasi-affine-locally-closed}). Since $X_{k'} \to X$ is finite locally free we conclude by Divisors, Lemma \ref{divisors-lemma-push-down-quasi-affine}. \end{proof} \begin{lemma} \label{lemma-quasi-projective-after-field-extension} Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is quasi-projective over $K$ for some field extension $k \subset K$, then $X$ is quasi-projective over $k$. \end{lemma} \begin{proof} By definition a morphism of schemes $g : Y \to T$ is quasi-projective if it is locally of finite type, quasi-compact, and there exists a $g$-ample invertible sheaf on $Y$. Let $k \subset K$ be a field extension such that $X_K$ is quasi-projective over $K$. Let $\Spec(A) \subset X$ be an affine open. Then $U_K$ is an affine open subscheme of $X_K$, hence $A_K$ is a $K$-algebra of finite type. Then $A$ is a $k$-algebra of finite type by Algebra, Lemma \ref{algebra-lemma-finite-type-descends}. Hence $X \to \Spec(k)$ is locally of finite type. Since $X_K \to \Spec(K)$ is quasi-compact, we see that $X_K$ is quasi-compact, hence $X$ is quasi-compact, hence $X \to \Spec(k)$ is of finite type. By Morphisms, Lemma \ref{morphisms-lemma-finite-type-over-affine-ample-very-ample} we see that $X_K$ has an ample invertible sheaf. Then $X$ has an ample invertible sheaf by Lemma \ref{lemma-ample-after-field-extension}. Hence $X \to \Spec(k)$ is quasi-projective by Morphisms, Lemma \ref{morphisms-lemma-finite-type-over-affine-ample-very-ample}. \end{proof} \noindent The following lemma is a special case of Descent, Lemma \ref{descent-lemma-descending-property-proper}. \begin{lemma} \label{lemma-proper-after-field-extension} Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is proper over $K$ for some field extension $k \subset K$, then $X$ is proper over $k$. \end{lemma} \begin{proof} Let $k \subset K$ be a field extension such that $X_K$ is proper over $K$. Recall that this implies $X_K$ is separated and quasi-compact (Morphisms, Definition \ref{morphisms-definition-proper}). The morphism $X_K \to X$ is surjective. Hence $X$ is quasi-compact as the image of a quasi-compact scheme (Properties, Definition \ref{properties-definition-ample}). Since $X_K$ is separated we see that $X$ is quasi-separated: If $U, V \subset X$ are affine open, then $(U \cap V)_K = U_K \cap V_K$ is quasi-compact and $(U \cap V)_K \to U \cap V$ is surjective. Thus Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} applies. \medskip\noindent Write $K = \colim A_i$ as the colimit of the subalgebras of $K$ which are of finite type over $k$. Denote $X_i = X \times_{\Spec(k)} \Spec(A_i)$. By Limits, Lemma \ref{limits-lemma-eventually-proper} there exists an $i$ such that $X_i \to \Spec(A_i)$ is proper. Here we use that $X$ is quasi-compact and quasi-separated as just shown. Choose a maximal ideal $\mathfrak m \subset A_i$. By the Hilbert Nullstellensatz (Algebra, Theorem \ref{algebra-theorem-nullstellensatz}) the residue field $k' = A_i/\mathfrak m$ is a finite extension of $k$. The base change $X_{k'} \to \Spec(k')$ is proper (Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}). Since $k \subset k'$ is finite both $X_{k'} \to X$ and the composition $X_{k'} \to \Spec(k)$ are proper as well (Morphisms, Lemmas \ref{morphisms-lemma-finite-proper}, \ref{morphisms-lemma-base-change-proper}, and \ref{morphisms-lemma-composition-proper}). The first implies that $X$ is separated over $k$ as $X_{k'}$ is separated (Morphisms, Lemma \ref{morphisms-lemma-image-universally-closed-separated}). The second implies that $X \to \Spec(k)$ is proper by Morphisms, Lemma \ref{morphisms-lemma-image-proper-is-proper}. \end{proof} \begin{lemma} \label{lemma-projective-after-field-extension} Let $k$ be a field. Let $X$ be a scheme over $k$. If $X_K$ is projective over $K$ for some field extension $k \subset K$, then $X$ is projective over $k$. \end{lemma} \begin{proof} A scheme over $k$ is projective over $k$ if and only if it is quasi-projective and proper over $k$. See Morphisms, Lemma \ref{morphisms-lemma-projective-is-quasi-projective-proper}. Thus the lemma follows from Lemmas \ref{lemma-quasi-projective-after-field-extension} and \ref{lemma-proper-after-field-extension}. \end{proof} \section{Tangent spaces} \label{section-tangent-spaces} \noindent In this section we define the tangent space of a morphism of schemes at a point of the source using points with values in dual numbers. \begin{definition} \label{definition-dual-numbers} For any ring $R$ the {\it dual numbers} over $R$ is the $R$-algebra denoted $R[\epsilon]$. As an $R$-module it is free with basis $1$, $\epsilon$ and the $R$-algebra structure comes from setting $\epsilon^2 = 0$. \end{definition} \noindent Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point with image $s = f(x)$ in $S$. Consider the solid commutative diagram \begin{equation} \label{equation-tangent-space} \vcenter{ \xymatrix{ \Spec(\kappa(x)) \ar[r] \ar[dr] \ar@/^1pc/[rr] & \Spec(\kappa(x)[\epsilon]) \ar@{.>}[r] \ar[d]& X \ar[d] \\ & \Spec(\kappa(s)) \ar[r] & S } } \end{equation} with the curved arrow being the canonical morphism of $\Spec(\kappa(x))$ into $X$. \begin{lemma} \label{lemma-tangent-space} The set of dotted arrows making (\ref{equation-tangent-space}) commute has a canonical $\kappa(x)$-vector space structure. \end{lemma} \begin{proof} Set $\kappa = \kappa(x)$. Observe that we have a pushout in the category of schemes $$\Spec(\kappa[\epsilon]) \amalg_{\Spec(\kappa)} \Spec(\kappa[\epsilon]) = \Spec(\kappa[\epsilon_1, \epsilon_2])$$ where $\kappa[\epsilon_1, \epsilon_2]$ is the $\kappa$-algebra with basis $1, \epsilon_1, \epsilon_2$ and $\epsilon_1^2 = \epsilon_1\epsilon_2 = \epsilon_2^2 = 0$. This follows immediately from the corresponding result for rings and the description of morphisms from spectra of local rings to schemes in Schemes, Lemma \ref{schemes-lemma-morphism-from-spec-local-ring}. Given two arrows $\theta_1, \theta_2 : \Spec(\kappa[\epsilon]) \to X$ we can consider the morphism $$\theta_1 + \theta_2 : \Spec(\kappa[\epsilon]) \to \Spec(\kappa[\epsilon_1, \epsilon_2]) \xrightarrow{\theta_1, \theta_2} X$$ where the first arrow is given by $\epsilon_i \mapsto \epsilon$. On the other hand, given $\lambda \in \kappa$ there is a self map of $\Spec(\kappa[\epsilon])$ corresponding to the $\kappa$-algebra endomorphism of $\kappa[\epsilon]$ which sends $\epsilon$ to $\lambda \epsilon$. Precomposing $\theta : \Spec(\kappa[\epsilon]) \to X$ by this selfmap gives $\lambda \theta$. The reader can verify the axioms of a vector space by verifying the existence of suitable commutative diagrams of schemes. We omit the details. (An alternative proof would be to express everything in terms of local rings and then verify the vector space axioms on the level of ring maps.) \end{proof} \begin{definition} \label{definition-tangent-space} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. The set of dotted arrows making (\ref{equation-tangent-space}) commute with its canonical $\kappa(x)$-vector space structure is called the {\it tangent space of $X$ over $S$ at $x$} and we denote it $T_{X/S, x}$. An element of this space is called a {\it tangent vector} of $X/S$ at $x$. \end{definition} \noindent Since tangent vectors at $x \in X$ live in the scheme theoretic fibre $X_s$ of $f : X \to S$ over $s = f(x)$, we get a canonical identification \begin{equation} \label{equation-tangent-space-fibre} T_{X/S, x} = T_{X_s/s, x} \end{equation} This pleasing definition involving the functor of points has the following algebraic description, which suggests defining the {\it cotangent space of $X$ over $S$ at $x$} as the $\kappa(x)$-vector space $$T^*_{X/S, x} = \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$$ simply because it is canonically $\kappa(x)$-dual to the tangent space of $X$ over $S$ at $x$. \begin{lemma} \label{lemma-tangent-space-cotangent-space} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. There is a canonical isomorphism $$T_{X/S, x} = \Hom_{\mathcal{O}_{X, x}}(\Omega_{X/S, x}, \kappa(x))$$ of vector spaces over $\kappa(x)$. \end{lemma} \begin{proof} Set $\kappa = \kappa(x)$. Given $\theta \in T_{X/S, x}$ we obtain a map $$\theta^*\Omega_{X/S} \to \Omega_{\Spec(\kappa[\epsilon])/\Spec(\kappa(s))} \to \Omega_{\Spec(\kappa[\epsilon])/\Spec(\kappa)}$$ Taking sections we obtain an $\mathcal{O}_{X, x}$-linear map $\xi_\theta: \Omega_{X/S, x} \to \kappa \text{d}\epsilon$, i.e., an element of the right hand side of the formula of the lemma. To show that $\theta \mapsto \xi_\theta$ is an isomorphism we can replace $S$ by $s$ and $X$ by the scheme theoretic fibre $X_s$. Indeed, both sides of the formula only depend on the scheme theoretic fibre; this is clear for $T_{X/S, x}$ and for the RHS see Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}. We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$ as this does not change $T_{X/S, x}$ (Schemes, Lemma \ref{schemes-lemma-morphism-from-spec-local-ring}) nor $\Omega_{X/S, x}$ (Modules, Lemma \ref{modules-lemma-stalk-module-differentials}). \medskip\noindent Let $(A, \mathfrak m, \kappa)$ be a local ring over a field $k$. To finish the proof we have to show that any $A$-linear map $\xi : \Omega_{A/k} \to \kappa$ comes from a unique $k$-algebra map $\varphi : A \to \kappa[\epsilon]$ agreeing with the canonical map $c : A \to \kappa$ modulo $\epsilon$. Write $\varphi(a) = c(a) + D(a) \epsilon$ the reader sees that $a \mapsto D(a)$ is a $k$-derivation. Using the universal property of $\Omega_{A/k}$ we see that each $D$ corresponds to a unique $\xi$ and vice versa. This finishes the proof. \end{proof} \begin{lemma} \label{lemma-tangent-space-rational-point} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point and let $s = f(x) \in S$. Assume that $\kappa(x) = \kappa(s)$. Then there are canonical isomorphisms $$\mathfrak m_x/(\mathfrak m_x^2 + \mathfrak m_s\mathcal{O}_{X, x}) = \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$$ and $$T_{X/S, x} = \Hom_{\kappa(x)}( \mathfrak m_x/(\mathfrak m_x^2 + \mathfrak m_s\mathcal{O}_{X, x}), \kappa(x))$$ This works more generally if $\kappa(x)/\kappa(s)$ is a separable algebraic extension. \end{lemma} \begin{proof} The second isomorphism follows from the first by Lemma \ref{lemma-tangent-space-cotangent-space}. For the first, we can replace $S$ by $s$ and $X$ by $X_s$, see Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}. We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$, see Modules, Lemma \ref{modules-lemma-stalk-module-differentials}. Thus we have to show the following algebra fact: let $(A, \mathfrak m, \kappa)$ be a local ring over a field $k$ such that $\kappa/k$ is separable algebraic. Then the canonical map $$\mathfrak m/\mathfrak m^2 \longrightarrow \Omega_{A/k} \otimes \kappa$$ is an isomorphism. Observe that $\mathfrak m/\mathfrak m^2 = H_1(\NL_{\kappa/A})$. By Algebra, Lemma \ref{algebra-lemma-exact-sequence-NL} it suffices to show that $\Omega_{\kappa/k} = 0$ and $H_1(\NL_{\kappa/k}) = 0$. Since $\kappa$ is the union of its finite separable extensions in $k$ it suffices to prove this when $\kappa$ is a finite separable extension of $k$ (Algebra, Lemma \ref{algebra-lemma-colimits-NL}). In this case the ring map $k \to \kappa$ is \'etale and hence $\NL_{\kappa/k} = 0$ (more or less by definition, see Algebra, Section \ref{algebra-section-etale}). \end{proof} \begin{lemma} \label{lemma-map-tangent-spaces} Let $f : X \to Y$ be a morphism of schemes over a base scheme $S$. Let $x \in X$ be a point. Set $y = f(x)$. If $\kappa(y) = \kappa(x)$, then $f$ induces a natural linear map $$\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}.$$ which is dual to the linear map $\Omega_{Y/S, y} \otimes \kappa(y) \to \Omega_{X/S, \kappa(x)}$ via the identifications of Lemma \ref{lemma-tangent-space-cotangent-space}. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-tangent-space-product} Let $X$, $Y$ be schemes over a base $S$. Let $x \in X$ and $y \in Y$ with the same image point $s \in S$ such that $\kappa(s) = \kappa(x)$ and $\kappa(s) = \kappa(y)$. There is a canonical isomorphism $$T_{X \times_S Y/S, (x, y)} = T_{X/S, x} \oplus T_{Y/S, y}$$ The map from left to right is induced by the maps on tangent spaces coming from the projections $X \times_S Y \to X$ and $X \times_S Y \to Y$. The map from right to left is induced by the maps $1 \times y : X_s \to X_s \times_s Y_s$ and $x \times 1 : Y_s \to X_s \times_s Y_s$ via the identification (\ref{equation-tangent-space-fibre}) of tangent spaces with tangent spaces of fibres. \end{lemma} \begin{proof} The direct sum decomposition follows from Morphisms, Lemma \ref{morphisms-lemma-differential-product} via Lemma \ref{lemma-tangent-space-rational-point}. Compatibility with the maps comes from Lemma \ref{lemma-map-tangent-spaces}. \end{proof} \begin{lemma} \label{lemma-injective-tangent-spaces-unramified} Let $f : X \to Y$ be a morphism of schemes locally of finite type over a base scheme $S$. Let $x \in X$ be a point. Set $y = f(x)$ and assume that $\kappa(y) = \kappa(x)$. Then the following are equivalent \begin{enumerate} \item $\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}$ is injective, and \item $f$ is unramified at $x$. \end{enumerate} \end{lemma} \begin{proof} The morphism $f$ is locally of finite type by Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}. The map $\text{d}f$ is injective, if and only if $\Omega_{Y/S, y} \otimes \kappa(y) \to \Omega_{X/S, x} \otimes \kappa(x)$ is surjective (Lemma \ref{lemma-map-tangent-spaces}). The exact sequence $f^*\Omega_{Y/S} \to \Omega_{X/S} \to \Omega_{X/Y} \to 0$ (Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials}) then shows that this happens if and only if $\Omega_{X/Y, x} \otimes \kappa(x) = 0$. Hence the result follows from Morphisms, Lemma \ref{morphisms-lemma-unramified-at-point}. \end{proof} \section{Generically finite morphisms} \label{section-generically-finite} \noindent In this section we revisit the notion of a generically finite morphism of schemes as studied in Morphisms, Section \ref{morphisms-section-generically-finite}. \begin{lemma} \label{lemma-quasi-finite-in-codim-1} Let $f : X \to Y$ be locally of finite type. Let $y \in Y$ be a point such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$. Assume in addition one of the following conditions is satisfied \begin{enumerate} \item for every generic point $\eta$ of an irreducible component of $X$ the field extension $\kappa(\eta) \supset \kappa(f(\eta))$ is finite (or algebraic), \item for every generic point $\eta$ of an irreducible component of $X$ such that $f(\eta) \leadsto y$ the field extension $\kappa(\eta) \supset \kappa(f(\eta))$ is finite (or algebraic), \item $f$ is quasi-finite at every generic point of an irreducible component of $X$, \item $Y$ is locally Noetherian and $f$ is quasi-finite at a dense set of points of $X$, \item add more here. \end{enumerate} Then $f$ is quasi-finite at every point of $X$ lying over $y$. \end{lemma} \begin{proof} Condition (4) implies $X$ is locally Noetherian (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}). The set of points at which morphism is quasi-finite is open (Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-points-open}). A dense open of a locally Noetherian scheme contains all generic point of irreducible components, hence (4) implies (3). Condition (3) implies condition (1) by Morphisms, Lemma \ref{morphisms-lemma-residue-field-quasi-finite}. Condition (1) implies condition (2). Thus it suffices to prove the lemma in case (2) holds. \medskip\noindent Assume (2) holds. Recall that $\Spec(\mathcal{O}_{Y, y})$ is the set of points of $Y$ specializing to $y$, see Schemes, Lemma \ref{schemes-lemma-specialize-points}. Combined with Morphisms, Lemma \ref{morphisms-lemma-base-change-quasi-finite} this shows we may replace $Y$ by $\Spec(\mathcal{O}_{Y, y})$. Thus we may assume $Y = \Spec(B)$ where $B$ is a Noetherian local ring of dimension $\leq 1$ and $y$ is the closed point. \medskip\noindent Let $X = \bigcup X_i$ be the irreducible components of $X$ viewed as reduced closed subschemes. If we can show each fibre $X_{i, y}$ is a discrete space, then $X_y = \bigcup X_{i, y}$ is discrete as well and we conclude that $X \to Y$ is quasi-finite at all points of $X_y$ by Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}. Thus we may assume $X$ is an integral scheme. \medskip\noindent If $X \to Y$ maps the generic point $\eta$ of $X$ to $y$, then $X$ is the spectrum of a finite extension of $\kappa(y)$ and the result is true. Assume that $X$ maps $\eta$ to a point corresponding to a minimal prime $\mathfrak q$ of $B$ different from $\mathfrak m_B$. We obtain a factorization $X \to \Spec(B/\mathfrak q) \to \Spec(B)$. Let $x \in X$ be a point lying over $y$. By the dimension formula (Morphisms, Lemma \ref{morphisms-lemma-dimension-formula}) we have $$\dim(\mathcal{O}_{X, x}) \leq \dim(B/\mathfrak q) + \text{trdeg}_{\kappa(\mathfrak q)}(R(X)) - \text{trdeg}_{\kappa(y)} \kappa(x)$$ We know that $\dim(B/\mathfrak q) = 1$, that the generic point of $X$ is not equal to $x$ and specializes to $x$ and that $R(X)$ is algebraic over $\kappa(\mathfrak q)$. Thus we get $$1 \leq 1 - \text{trdeg}_{\kappa(y)} \kappa(x)$$ Hence every point $x$ of $X_y$ is closed in $X_y$ by Morphisms, Lemma \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre} and hence $X \to Y$ is quasi-finite at every point $x$ of $X_y$ by Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize} (which also implies that $X_y$ is a discrete topological space). \end{proof} \begin{lemma} \label{lemma-finite-in-codim-1} Let $f : X \to Y$ be a proper morphism. Let $y \in Y$ be a point such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$. Assume in addition one of the following conditions is satisfied \begin{enumerate} \item for every generic point $\eta$ of an irreducible component of $X$ the field extension $\kappa(\eta) \supset \kappa(f(\eta))$ is finite (or algebraic), \item for every generic point $\eta$ of an irreducible component of $X$ such that $f(\eta) \leadsto y$ the field extension $\kappa(\eta) \supset \kappa(f(\eta))$ is finite (or algebraic), \item $f$ is quasi-finite at every generic point of $X$, \item $Y$ is locally Noetherian and $f$ is quasi-finite at a dense set of points of $X$, \item add more here. \end{enumerate} Then there exists an open neighbourhood $V \subset Y$ of $y$ such that $f^{-1}(V) \to V$ is finite. \end{lemma} \begin{proof} By Lemma \ref{lemma-quasi-finite-in-codim-1} the morphism $f$ is quasi-finite at every point of the fibre $X_y$. Hence $X_y$ is a discrete topological space (Morphisms, Lemma \ref{morphisms-lemma-quasi-finite-at-point-characterize}). As $f$ is proper the fibre $X_y$ is quasi-compact, i.e., finite. Thus we can apply Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood} to conclude. \end{proof} \begin{lemma} \label{lemma-modification-normal-iso-over-codimension-1} Let $X$ be a Noetherian scheme. Let $f : Y \to X$ be a birational proper morphism of schemes with $Y$ reduced. Let $U \subset X$ be the maximal open over which $f$ is an isomorphism. Then $U$ contains \begin{enumerate} \item every point of codimension $0$ in $X$, \item every $x \in X$ of codimension $1$ on $X$ such that $\mathcal{O}_{X, x}$ is a discrete valuation ring, \item every $x \in X$ such that the fibre of $Y \to X$ over $x$ is finite and such that $\mathcal{O}_{X, x}$ is normal, and \item every $x \in X$ such that $f$ is quasi-finite at some $y \in Y$ lying over $x$ and $\mathcal{O}_{X, x}$ is normal. \end{enumerate} \end{lemma} \begin{proof} Part (1) follows from Morphisms, Lemma \ref{morphisms-lemma-birational-isomorphism-over-dense-open}. Part (2) follows from part (3) and Lemma \ref{lemma-finite-in-codim-1} (and the fact that finite morphisms have finite fibres). \medskip\noindent Part (3) follows from part (4) and Morphisms, Lemma \ref{morphisms-lemma-finite-fibre} but we will also give a direct proof. Let $x \in X$ be as in (3). By Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood} we may assume $f$ is finite. We may assume $X$ affine. This reduces us to the case of a finite birational morphism of Noetherian affine schemes $Y \to X$ and $x \in X$ such that $\mathcal{O}_{X, x}$ is a normal domain. Since $\mathcal{O}_{X, x}$ is a domain and $X$ is Noetherian, we may replace $X$ by an affine open of $x$ which is integral. Then, since $Y \to X$ is birational and $Y$ is reduced we see that $Y$ is integral. Writing $X = \Spec(A)$ and $Y = \Spec(B)$ we see that $A \subset B$ is a finite inclusion of domains having the same field of fractions. If $\mathfrak p \subset A$ is the prime corresponding to $x$, then $A_\mathfrak p$ being normal implies that $A_\mathfrak p \subset B_\mathfrak p$ is an equality. Since $B$ is a finite $A$-module, we see there exists an $a \in A$, $a \not \in \mathfrak p$ such that $A_a \to B_a$ is an isomorphism. \medskip\noindent Let $x \in X$ and $y \in Y$ be as in (4). After replacing $X$ by an affine open neighbourhood we may assume $X = \Spec(A)$ and $A \subset \mathcal{O}_{X, x}$, see Properties, Lemma \ref{properties-lemma-ring-affine-open-injective-local-ring}. Then $A$ is a domain and hence $X$ is integral. Since $f$ is birational and $Y$ is reduced it follows that $Y$ is integral too. Consider the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$. This is a ring map which is essentially of finite type, the residue field extension is finite, and $\dim(\mathcal{O}_{Y, y}/\mathfrak m_x\mathcal{O}_{Y, y}) = 0$ (to see this trace through the definitions of quasi-finite maps in Morphisms, Definition \ref{morphisms-definition-quasi-finite} and Algebra, Definition \ref{algebra-definition-quasi-finite}). By Algebra, Lemma \ref{algebra-lemma-essentially-finite-type-fibre-dim-zero} $\mathcal{O}_{Y, y}$ is the localization of a finite $\mathcal{O}_{X, x}$-algebra $B$. Of course we may replace $B$ by the image of $B$ in $\mathcal{O}_{Y, y}$ and assume that $B$ is a domain with the same fraction field as $\mathcal{O}_{Y, y}$. Then $\mathcal{O}_{X, x} \subset B$ have the same fraction field as $f$ is birational. Since $\mathcal{O}_{X, x}$ is normal, we conclude that $\mathcal{O}_{X, x} = B$ (because finite implies integral), in particular, we see that $\mathcal{O}_{X, x} = \mathcal{O}_{Y, y}$. By Morphisms, Lemma \ref{morphisms-lemma-morphism-defined-local-ring} after shrinking $X$ we may assume there is a section $X \to Y$ of $f$ mapping $x$ to $y$ and inducing the given isomorphism on local rings. Since $X \to Y$ is closed (by Schemes, Lemma \ref{schemes-lemma-section-immersion}) necessarily maps the generic point of $X$ to the generic point of $Y$ it follows that the image of $X \to Y$ is $Y$. Then $Y = X$ and we've proved what we wanted to show. \end{proof} \section{Variants of Noether normalization} \label{section-noether-normalization} \noindent Noether normalization is the statement that if $k$ is a field and $A$ is a finite type $k$ algebra of dimension $d$, then there exists a finite injective $k$-algebra homomorphism $k[x_1, \ldots, x_d] \to A$. See Algebra, Lemma \ref{algebra-lemma-Noether-normalization}. Geometrically this means there is a finite surjective morphism $\Spec(A) \to \mathbf{A}^d_k$ over $\Spec(k)$. \begin{lemma} \label{lemma-noether-normalization} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ with image $s \in S$. Let $V \subset S$ be an affine open neighbourhood of $s$. If $f$ is locally of finite type and $\dim_x(X_s) = d$, then there exists an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a factorization $$U \xrightarrow{\pi} \mathbf{A}^d_V \to V$$ of $f|_U : U \to V$ such that $\pi$ is quasi-finite. \end{lemma} \begin{proof} This follows from Algebra, Lemma \ref{algebra-lemma-quasi-finite-over-polynomial-algebra}. \end{proof} \begin{lemma} \label{lemma-noether-normalization-affine} Let $f : X \to S$ be a finite type morphism of affine schemes. Let $s \in S$. If $\dim(X_s) = d$, then there exists a factorization $$X \xrightarrow{\pi} \mathbf{A}^d_S \to S$$ of $f$ such that the morphism $\pi_s : X_s \to \mathbf{A}^d_{\kappa(s)}$ of fibres over $s$ is finite. \end{lemma} \begin{proof} Write $S = \Spec(A)$ and $X = \Spec(B)$ and let $A \to B$ be the ring map corresponding to $f$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. We can choose a surjection $A[x_1, \ldots, x_r] \to B$. By Algebra, Lemma \ref{algebra-lemma-Noether-normalization} there exist elements $y_1, \ldots, y_d \in A$ in the $\mathbf{Z}$-subalgebra of $A$ generated by $x_1, \ldots, x_r$ such that the $A$-algebra homomorphism $A[t_1, \ldots, t_d] \to B$ sending $t_i$ to $y_i$ induces a finite $\kappa(\mathfrak p)$-algebra homomorphism $\kappa(\mathfrak p)[t_1, \ldots, t_d] \to B \otimes_A \kappa(\mathfrak p)$. This proves the lemma. \end{proof} \begin{lemma} \label{lemma-geometric-structure-unramified} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. Let $V = \Spec(A)$ be an affine open neighbourhood of $f(x)$ in $S$. If $f$ is unramified at $x$, then there exist exists an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ such that we have a commutative diagram $$\xymatrix{ X \ar[d] & U \ar[l] \ar[rd] \ar[r]^-j & \Spec(A[t]_{g'}/(g)) \ar[d] \ar[r] & \Spec(A[t]) = \mathbf{A}^1_V \ar[ld] \\ Y & & V \ar[ll] }$$ where $j$ is an immersion, $g \in A[t]$ is a monic polynomial, and $g'$ is the derivative of $g$ with respect to $t$. If $f$ is \'etale at $x$, then we may choose the diagram such that $j$ is an open immersion. \end{lemma} \begin{proof} The unramified case is a translation of Algebra, Proposition \ref{algebra-proposition-unramified-locally-standard}. In the \'etale case this is a translation of Algebra, Proposition \ref{algebra-proposition-etale-locally-standard} or equivalently it follows from Morphisms, Lemma \ref{morphisms-lemma-etale-locally-standard-etale} although the statements differ slightly. \end{proof} \begin{lemma} \label{lemma-unramfied-over-affine} Let $f : X \to S$ be a finite type morphism of affine schemes. Let $x \in X$ with image $s \in S$. Let $$r = \dim_{\kappa(x)} \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x) = \dim_{\kappa(x)} \Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) = \dim_{\kappa(x)} T_{X/S, x}$$ Then there exists a factorization $$X \xrightarrow{\pi} \mathbf{A}^r_S \to S$$ of $f$ such that $\pi$ is unramified at $x$. \end{lemma} \begin{proof} By Morphisms, Lemma \ref{morphisms-lemma-finite-type-differentials} the first dimension is finite. The first equality follows as the restriction of $\Omega_{X/S}$ to the fibre is the module of differentials from Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}. The last equality follows from Lemma \ref{lemma-tangent-space-cotangent-space}. Thus we see that the statement makes sense. \medskip\noindent To prove the lemma write $S = \Spec(A)$ and $X = \Spec(B)$ and let $A \to B$ be the ring map corresponding to $f$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$. Choose a surjection of $A$-algebras $A[x_1, \ldots, x_t] \to B$. Since $\Omega_{B/A}$ is generated by $\text{d}x_1, \ldots, \text{d}x_t$ we see that their images in $\Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$ generate this as a $\kappa(x)$-vector space. After renumbering we may assume that $\text{d}x_1, \ldots, \text{d}x_r$ map to a basis of $\Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$. We claim that $P = A[x_1, \ldots, x_r] \to B$ is unramified at $\mathfrak q$. To see this it suffices to show that $\Omega_{B/P, \mathfrak q} = 0$ (Algebra, Lemma \ref{algebra-lemma-unramified}). Note that $\Omega_{B/P}$ is the quotient of $\Omega_{B/A}$ by the submodule generated by $\text{d}x_1, \ldots, \text{d}x_r$. Hence $\Omega_{B/P, \mathfrak q} \otimes_{B_\mathfrak q} \kappa(\mathfrak q) = 0$ by our choice of $x_1, \ldots, x_r$. By Nakayama's lemma, more precisely Algebra, Lemma \ref{algebra-lemma-NAK} part (2) which applies as $\Omega_{B/P}$ is finite (see reference above), we conclude that $\Omega_{B/P, \mathfrak q} = 0$. \end{proof} \begin{lemma} \label{lemma-immersion-into-affine} Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ with image $s \in S$. Let $V \subset S$ be an affine open neighbourhood of $s$. If $f$ is locally of finite type and $$r = \dim_{\kappa(x)} \Omega_{X/S, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x) = \dim_{\kappa(x)} \Omega_{X_s/s, x} \otimes_{\mathcal{O}_{X_s, x}} \kappa(x) = \dim_{\kappa(x)} T_{X/S, x}$$ then there exist \begin{enumerate} \item an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a factorization $$U \xrightarrow{j} \mathbf{A}^{r + 1}_V \to V$$ of $f|_U$ such that $j$ is an immersion, or \item an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ and a factorization $$U \xrightarrow{j} D \to V$$ of $f|_U$ such that $j$ is a closed immersion and $D \to V$ is smooth of relative dimension $r$. \end{enumerate} \end{lemma} \begin{proof} Pick any affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$. Apply Lemma \ref{lemma-unramfied-over-affine} to $U \to V$ to get $U \to \mathbf{A}^r_V \to V$ as in the statement of that lemma. By Lemma \ref{lemma-geometric-structure-unramified} we get a factorization $$U \xrightarrow{j} D \xrightarrow{j'} \mathbf{A}^{r + 1}_V \xrightarrow{p} \mathbf{A}^r_V \to V$$ where $j$ and $j'$ are immersions, $p$ is the projection, and $p \circ j'$ is standard \'etale. Thus we see in particular that (1) and (2) hold. \end{proof} \section{Dimension of fibres} \label{section-dimension-fibres} \noindent We have already seen that dimension of fibres of finite type morphisms typically jump up. In this section we discuss the phenomenon that in codimension $1$ this does not happen. More generally, we discuss how much the dimension of a fibre can jump. Here is a list of related results: \begin{enumerate} \item For a finite type morphism $X \to S$ the set of $x \in X$ with $\dim_x(X_{f(x)}) \leq d$ is open, see Algebra, Lemma \ref{algebra-lemma-dimension-fibres-bounded-open-upstairs} and Morphisms, Lemma \ref{morphisms-lemma-openness-bounded-dimension-fibres}. \item We have the dimension formula, see Algebra, Lemma \ref{algebra-lemma-dimension-formula} and Morphisms, Lemma \ref{morphisms-lemma-dimension-formula}. \item Constant fibre dimension for an integral finite type scheme dominating a valuation ring, see Algebra, Lemma \ref{algebra-lemma-finite-type-domain-over-valuation-ring-dim-fibres}. \item If $X \to S$ is of finite type and is quasi-finite at every generic point of $X$, then $X \to S$ is quasi-finite in codimension $1$, see Algebra, Lemma \ref{algebra-lemma-finite-in-codim-1} and Lemma \ref{lemma-quasi-finite-in-codim-1}. \end{enumerate} The last result mentioned above generalizes as follows. \begin{lemma} \label{lemma-dimension-fibre-in-codim-1} Let $f : X \to Y$ be locally of finite type. Let $x \in X$ be a point with image $y \in Y$ such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$. Let $d \geq 0$ be an integer such that for every generic point $\eta$ of an irreducible component of $X$ which contains $x$, we have $\dim_\eta(X_{f(\eta)}) = d$. Then $\dim_x(X_y) = d$. \end{lemma} \begin{proof} Recall that $\Spec(\mathcal{O}_{Y, y})$ is the set of points of $Y$ specializing to $y$, see Schemes, Lemma \ref{schemes-lemma-specialize-points}. Thus we may replace $Y$ by $\Spec(\mathcal{O}_{Y, y})$