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 \input{preamble} % OK, start here. % \begin{document} \title{Algebraic Curves} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we develop some of the theory of algebraic curves. A reference covering algebraic curves over the complex numbers is the book \cite{ACGH}. \medskip\noindent What we already know. Besides general algebraic geometry, we have already proved some specific results on algebraic curves. Here is a list. \begin{enumerate} \item We have discussed affine opens of and ample invertible sheaves on $1$ dimensional Noetherian schemes in Varieties, Section \ref{varieties-section-dimension-one}. \item We have seen a curve is either affine or projective in Varieties, Section \ref{varieties-section-curves}. \item We have discussed degrees of locally free modules on proper curves in Varieties, Section \ref{varieties-section-divisors-curves}. \item We have discussed the Picard scheme of a nonsingular projective curve over an algebraically closed field in Picard Schemes of Curves, Section \ref{pic-section-introduction}. \end{enumerate} \section{Curves and function fields} \label{section-curves-function-fields} \noindent In this section we elaborate on the results of Varieties, Section \ref{varieties-section-varieties-rational-maps} in the case of curves. \begin{lemma} \label{lemma-extend-over-dvr} Let $k$ be a field. Let $X$ be a curve and $Y$ a proper variety. Let $U \subset X$ be a nonempty open and let $f : U \to Y$ be a morphism. If $x \in X$ is a closed point such that $\mathcal{O}_{X, x}$ is a discrete valuation ring, then there exists an open $U \subset U' \subset X$ containing $x$ and a morphism of varieties $f' : U' \to Y$ extending $f$. \end{lemma} \begin{proof} This is a special case of Morphisms, Lemma \ref{morphisms-lemma-extend-across}. \end{proof} \begin{lemma} \label{lemma-extend-over-normal-curve} Let $k$ be a field. Let $X$ be a normal curve and $Y$ a proper variety. The set of rational maps from $X$ to $Y$ is the same as the set of morphisms $X \to Y$. \end{lemma} \begin{proof} This is clear from Lemma \ref{lemma-extend-over-dvr} as every local ring is a discrete valuation ring (for example by Varieties, Lemma \ref{varieties-lemma-regular-point-on-curve}). \end{proof} \begin{lemma} \label{lemma-flat} Let $k$ be a field. Let $f : X \to Y$ be a nonconstant morphism of curves over $k$. If $Y$ is normal, then $f$ is flat. \end{lemma} \begin{proof} Pick $x \in X$ mapping to $y \in Y$. Then $\mathcal{O}_{Y, y}$ is either a field or a discrete valuation ring (Varieties, Lemma \ref{varieties-lemma-regular-point-on-curve}). Since $f$ is nonconstant it is dominant (as it must map the generic point of $X$ to the generic point of $Y$). This implies that $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective (Morphisms, Lemma \ref{morphisms-lemma-dominant-between-integral}). Hence $\mathcal{O}_{X, x}$ is torsion free as a $\mathcal{O}_{Y, y}$-module and therefore $\mathcal{O}_{X, x}$ is flat as a $\mathcal{O}_{Y, y}$-module by More on Algebra, Lemma \ref{more-algebra-lemma-valuation-ring-torsion-free-flat}. \end{proof} \begin{lemma} \label{lemma-finite} Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes over $k$. Assume \begin{enumerate} \item $Y$ is separated over $k$, \item $X$ is proper of dimension $\leq 1$ over $k$, \item $f(Z)$ has at least two points for every irreducible component $Z \subset X$ of dimension $1$. \end{enumerate} Then $f$ is finite. \end{lemma} \begin{proof} The morphism $f$ is proper by Morphisms, Lemma \ref{morphisms-lemma-image-proper-scheme-closed}. Thus $f(X)$ is closed and images of closed points are closed. Let $y \in Y$ be the image of a closed point in $X$. Then $f^{-1}(\{y\})$ is a closed subset of $X$ not containing any of the generic points of irreducible components of dimension $1$ by condition (3). It follows that $f^{-1}(\{y\})$ is finite. Hence $f$ is finite over an open neighbourhood of $y$ by More on Morphisms, Lemma \ref{more-morphisms-lemma-proper-finite-fibre-finite-in-neighbourhood} (if $Y$ is Noetherian, then you can use the easier Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-finite-fibre-finite-in-neighbourhood}). Since we've seen above that there are enough of these points $y$, the proof is complete. \end{proof} \begin{lemma} \label{lemma-extend-to-completion} Let $k$ be a field. Let $X \to Y$ be a morphism of varieties with $Y$ proper and $X$ a curve. There exists a factorization $X \to \overline{X} \to Y$ where $X \to \overline{X}$ is an open immersion and $\overline{X}$ is a projective curve. \end{lemma} \begin{proof} This is clear from Lemma \ref{lemma-extend-over-dvr} and Varieties, Lemma \ref{varieties-lemma-reduced-dim-1-projective-completion}. \end{proof} \noindent Here is the main theorem of this section. We will say a morphism $f : X \to Y$ of varieties is {\it constant} if the image $f(X)$ consists of a single point $y$ of $Y$. If this happens then $y$ is a closed point of $Y$ (since the image of a closed point of $X$ will be a closed point of $Y$). \begin{theorem} \label{theorem-curves-rational-maps} Let $k$ be a field. The following categories are canonically equivalent \begin{enumerate} \item The category of finitely generated field extensions $K/k$ of transcendence degree $1$. \item The category of curves and dominant rational maps. \item The category of normal projective curves and nonconstant morphisms. \item The category of nonsingular projective curves and nonconstant morphisms. \item The category of regular projective curves and nonconstant morphisms. \item The category of normal proper curves and nonconstant morphisms. \end{enumerate} \end{theorem} \begin{proof} The equivalence between categories (1) and (2) is the restriction of the equivalence of Varieties, Theorem \ref{varieties-theorem-varieties-rational-maps}. Namely, a variety is a curve if and only if its function field has transcendence degree $1$, see for example Varieties, Lemma \ref{varieties-lemma-dimension-locally-algebraic}. \medskip\noindent The categories in (3), (4), (5), and (6) are the same. First of all, the terms regular'' and nonsingular'' are synonyms, see Properties, Definition \ref{properties-definition-regular}. Being normal and regular are the same thing for Noetherian $1$-dimensional schemes (Properties, Lemmas \ref{properties-lemma-regular-normal} and \ref{properties-lemma-normal-dimension-1-regular}). See Varieties, Lemma \ref{varieties-lemma-regular-point-on-curve} for the case of curves. Thus (3) is the same as (5). Finally, (6) is the same as (3) by Varieties, Lemma \ref{varieties-lemma-dim-1-proper-projective}. \medskip\noindent If $f : X \to Y$ is a nonconstant morphism of nonsingular projective curves, then $f$ sends the generic point $\eta$ of $X$ to the generic point $\xi$ of $Y$. Hence we obtain a morphism $k(Y) = \mathcal{O}_{Y, \xi} \to \mathcal{O}_{X, \eta} = k(X)$ in the category (1). Conversely, suppose that we have a map $k(Y) \to k(X)$ in the category (1). Then we obtain a morphism $U \to Y$ for some nonempty open $U \subset X$. By Lemma \ref{lemma-extend-over-dvr} this extends to all of $X$ and we obtain a morphism in the category (5). Thus we see that there is a fully faithful functor (5)$\to$(1). \medskip\noindent To finish the proof we have to show that every $K/k$ in (1) is the function field of a normal projective curve. We already know that $K = k(X)$ for some curve $X$. After replacing $X$ by its normalization (which is a variety birational to $X$) we may assume $X$ is normal (Varieties, Lemma \ref{varieties-lemma-normalization-locally-algebraic}). Then we choose $X \to \overline{X}$ with $\overline{X} \setminus X = \{x_1, \ldots, x_n\}$ as in Varieties, Lemma \ref{varieties-lemma-reduced-dim-1-projective-completion}. Since $X$ is normal and since each of the local rings $\mathcal{O}_{\overline{X}, x_i}$ is normal we conclude that $\overline{X}$ is a normal projective curve as desired. (Remark: We can also first compactify using Varieties, Lemma \ref{varieties-lemma-dim-1-projective-completion} and then normalize using Varieties, Lemma \ref{varieties-lemma-normalization-locally-algebraic}. Doing it this way we avoid using the somewhat tricky Morphisms, Lemma \ref{morphisms-lemma-relative-normalization-normal-codim-1}.) \end{proof} \begin{definition} \label{definition-normal-projective-model} Let $k$ be a field. Let $X$ be a curve. A {\it nonsingular projective model of $X$} is a pair $(Y, \varphi)$ where $Y$ is a nonsingular projective curve and $\varphi : k(X) \to k(Y)$ is an isomorphism of function fields. \end{definition} \noindent A nonsingular projective model is determined up to unique isomorphism by Theorem \ref{theorem-curves-rational-maps}. Thus we often say the nonsingular projective model''. We usually drop $\varphi$ from the notation. Warning: it needn't be the case that $Y$ is smooth over $k$ but Lemma \ref{lemma-nonsingular-model-smooth} shows this can only happen in positive characteristic. \begin{lemma} \label{lemma-nonsingular-model-smooth} Let $k$ be a field. Let $X$ be a curve and let $Y$ be the nonsingular projective model of $X$. If $k$ is perfect, then $Y$ is a smooth projective curve. \end{lemma} \begin{proof} See Varieties, Lemma \ref{varieties-lemma-regular-point-on-curve} for example. \end{proof} \begin{lemma} \label{lemma-smooth-models} Let $k$ be a field. Let $X$ be a geometrically irreducible curve over $k$. For a field extension $K/k$ denote $Y_K$ a nonsingular projective model of $(X_K)_{red}$. \begin{enumerate} \item If $X$ is proper, then $Y_K$ is the normalization of $X_K$. \item There exists $K/k$ finite purely inseparable such that $Y_K$ is smooth. \item Whenever $Y_K$ is smooth\footnote{Or even geometrically reduced.} we have $H^0(Y_K, \mathcal{O}_{Y_K}) = K$. \item Given a commutative diagram $$\xymatrix{ \Omega & K' \ar[l] \\ K \ar[u] & k \ar[l] \ar[u] }$$ of fields such that $Y_K$ and $Y_{K'}$ are smooth, then $Y_\Omega = (Y_K)_\Omega = (Y_{K'})_\Omega$. \end{enumerate} \end{lemma} \begin{proof} Let $X'$ be a nonsingular projective model of $X$. Then $X'$ and $X$ have isomorphic nonempty open subschemes. In particular $X'$ is geometrically irreducible as $X$ is (some details omitted). Thus we may assume that $X$ is projective. \medskip\noindent Assume $X$ is proper. Then $X_K$ is proper and hence the normalization $(X_K)^\nu$ is proper as a scheme finite over a proper scheme (Varieties, Lemma \ref{varieties-lemma-normalization-locally-algebraic} and Morphisms, Lemmas \ref{morphisms-lemma-finite-proper} and \ref{morphisms-lemma-composition-proper}). On the other hand, $X_K$ is irreducible as $X$ is geometrically irreducible. Hence $X_K^\nu$ is proper, normal, irreducible, and birational to $(X_K)_{red}$. This proves (1) because a proper curve is projective (Varieties, Lemma \ref{varieties-lemma-dim-1-proper-projective}). \medskip\noindent Proof of (2). As $X$ is proper and we have (1), we can apply Varieties, Lemma \ref{varieties-lemma-finite-extension-geometrically-normal} to find $K/k$ finite purely inseparable such that $Y_K$ is geometrically normal. Then $Y_K$ is geometrically regular as normal and regular are the same for curves (Properties, Lemma \ref{properties-lemma-normal-dimension-1-regular}). Then $Y$ is a smooth variety by Varieties, Lemma \ref{varieties-lemma-geometrically-regular-smooth}. \medskip\noindent If $Y_K$ is geometrically reduced, then $Y_K$ is geometrically integral (Varieties, Lemma \ref{varieties-lemma-geometrically-integral}) and we see that $H^0(Y_K, \mathcal{O}_{Y_K}) = K$ by Varieties, Lemma \ref{varieties-lemma-regular-functions-proper-variety}. This proves (3) because a smooth variety is geometrically reduced (even geometrically regular, see Varieties, Lemma \ref{varieties-lemma-geometrically-regular-smooth}). \medskip\noindent If $Y_K$ is smooth, then for every extension $\Omega/K$ the base change $(Y_K)_\Omega$ is smooth over $\Omega$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-smooth}). Hence it is clear that $Y_\Omega = (Y_K)_\Omega$. This proves (4). \end{proof} \section{Linear series} \label{section-linear-series} \noindent We deviate from the classical story (see Remark \ref{remark-classical-linear-series}) by defining linear series in the following manner. \begin{definition} \label{definition-linear-series} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $d \geq 0$ and $r \geq 0$. A {\it linear series of degree $d$ and dimension $r$} is a pair $(\mathcal{L}, V)$ where $\mathcal{L}$ in an invertible $\mathcal{O}_X$-module of degree $d$ (Varieties, Definition \ref{varieties-definition-degree-invertible-sheaf}) and $V \subset H^0(X, \mathcal{L})$ is a $k$-subvector space of dimension $r + 1$. We will abbreviate this by saying $(\mathcal{L}, V)$ is a {\it $\mathfrak g^r_d$} on $X$. \end{definition} \noindent We will mostly use this when $X$ is a nonsingular proper curve. In fact, the definition above is just one way to generalize the classical definition of a $\mathfrak g^r_d$. For example, if $X$ is a proper curve, then one can generalize linear series by allowing $\mathcal{L}$ to be a torsion free coherent $\mathcal{O}_X$-module of rank $1$. On a nonsingular curve every torsion free coherent module is locally free, so this agrees with our notion for nonsingular proper curves. \medskip\noindent The following lemma explains the geometric meaning of linear series for proper nonsingular curves. \begin{lemma} \label{lemma-linear-series} Let $k$ be a field. Let $X$ be a nonsingular proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^r_d$ on $X$. Then there exists a morphism $$\varphi : X \longrightarrow \mathbf{P}^r_k = \text{Proj}(k[T_0, \ldots, T_r])$$ of varieties over $k$ and a map $\alpha : \varphi^*\mathcal{O}_{\mathbf{P}^r_k}(1) \to \mathcal{L}$ such that $\varphi^*T_0, \ldots, \varphi^*T_r$ are sent to a basis of $V$ by $\alpha$. \end{lemma} \begin{proof} Let $s_0, \ldots, s_r \in V$ be a $k$-basis. Since $X$ is nonsingular the image $\mathcal{L}' \subset \mathcal{L}$ of the map $s_0, \ldots, s_r : \mathcal{O}_X^{\oplus r} \to \mathcal{L}$ is an invertible $\mathcal{O}_X$-module for example by Divisors, Lemma \ref{divisors-lemma-torsion-free-over-regular-dim-1}. Then we use Constructions, Lemma \ref{constructions-lemma-projective-space} to get a morphism $$\varphi = \varphi_{(\mathcal{L}', (s_0, \ldots, s_r))} : X \longrightarrow \mathbf{P}^r_k$$ as in the statement of the lemma. \end{proof} \begin{lemma} \label{lemma-linear-series-trivial-existence} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. If $X$ has a $\mathfrak g^r_d$, then $X$ has a $\mathfrak g^s_d$ for all $0 \leq s \leq r$. \end{lemma} \begin{proof} This is true because a vector space $V$ of dimension $r + 1$ over $k$ has a linear subspace of dimension $s + 1$ for all $0 \leq s \leq r$. \end{proof} \begin{lemma} \label{lemma-g1d} Let $k$ be a field. Let $X$ be a nonsingular proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^1_d$ on $X$. Then the morphism $\varphi : X \to \mathbf{P}^1_k$ of Lemma \ref{lemma-linear-series} has degree $\leq d$. \end{lemma} \begin{proof} By Lemma \ref{lemma-linear-series} we see that $\mathcal{L}' = \varphi^*\mathcal{O}_{\mathbf{P}^1_k}(1)$ has a nonzero map $\mathcal{L}' \to \mathcal{L}$. Hence by Varieties, Lemma \ref{varieties-lemma-check-invertible-sheaf-trivial} we see that $\deg(\mathcal{L}') \leq d$. On the other hand, we have $$\deg(\mathcal{L}') = \deg(X/\mathbf{P}^1_k) \deg(\mathcal{O}_{\mathbf{P}^1_k}(1))$$ by Varieties, Lemma \ref{varieties-lemma-degree-pullback-map-proper-curves}. This finishes the proof as the degree of $\mathcal{O}_{\mathbf{P}^1_k}(1)$ is $1$. \end{proof} \begin{lemma} \label{lemma-grd-inequalities} Let $k$ be a field. Let $X$ be a proper curve over $k$ with $H^0(X, \mathcal{O}_X) = k$. If $X$ has a $\mathfrak g^r_d$, then $r \leq d$. If equality holds, then $H^1(X, \mathcal{O}_X) = 0$, i.e., the genus of $X$ (Definition \ref{definition-genus}) is $0$. \end{lemma} \begin{proof} Let $(\mathcal{L}, V)$ be a $\mathfrak g^r_d$. Since this will only increase $r$, we may assume $V = H^0(X, \mathcal{L})$. Choose a nonzero element $s \in V$. Then the zero scheme of $s$ is an effective Cartier divisor $D \subset X$, we have $\mathcal{L} = \mathcal{O}_X(D)$, and we have a short exact sequence $$0 \to \mathcal{O}_X \to \mathcal{L} \to \mathcal{L}|_D \to 0$$ see Divisors, Lemma \ref{divisors-lemma-characterize-OD} and Remark \ref{divisors-remark-ses-regular-section}. By Varieties, Lemma \ref{varieties-lemma-degree-effective-Cartier-divisor} we have $\deg(D) = \deg(\mathcal{L}) = d$. Since $D$ is an Artinian scheme we have $\mathcal{L}|_D \cong \mathcal{O}_D$\footnote{In our case this follows from Divisors, Lemma \ref{divisors-lemma-finite-trivialize-invertible-upstairs} as $D \to \Spec(k)$ is finite.}. Thus $$\dim_k H^0(D, \mathcal{L}|_D) = \dim_k H^0(D, \mathcal{O}_D) = \deg(D) = d$$ On the other hand, by assumption $\dim_k H^0(X, \mathcal{O}_X) = 1$ and $\dim H^0(X, \mathcal{L}) = r + 1$. We conclude that $r + 1 \leq 1 + d$, i.e., $r \leq d$ as in the lemma. \medskip\noindent Assume equality holds. Then $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_D)$ is surjective. If we knew that $H^1(X, \mathcal{L})$ was zero, then we would conclude that $H^1(X, \mathcal{O}_X)$ is zero by the long exact cohomology sequence and the proof would be complete. Our strategy will be to replace $\mathcal{L}$ by a large power which has vanishing. As $\mathcal{L}|_D$ is the trivial invertible module (see above), we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_D$. Consider the multiplication map $$\mu : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}) \longrightarrow H^0(X, \mathcal{L}^{\otimes 2})$$ and consider the short exact sequence $$0 \to \mathcal{L} \xrightarrow{s} \mathcal{L}^{\otimes 2} \to \mathcal{L}^{\otimes 2}|_D \to 0$$ Since $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_D)$ is surjective and since $t$ maps to a trivialization of $\mathcal{L}|_D$ we see that $\mu(H^0(X, \mathcal{L}) \otimes t)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes 2})$ surjecting onto the global sections of $\mathcal{L}^{\otimes 2}|_D$. Thus we see that $$\dim H^0(X, \mathcal{L}^{\otimes 2}) = r + 1 + d = 2r + 1 = \deg(\mathcal{L}^{\otimes 2}) + 1$$ Ok, so $\mathcal{L}^{\otimes 2}$ has the same property as $\mathcal{L}$, i.e., that the dimension of the space of global sections is equal to the degree plus one. Since $\mathcal{L}$ is ample (Varieties, Lemma \ref{varieties-lemma-ample-curve}) there exists some $n_0$ such that $\mathcal{L}^{\otimes n}$ has vanishing $H^1$ for all $n \geq n_0$ (Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-proper-ample}). Thus applying the argument above to $\mathcal{L}^{\otimes n}$ with $n = 2^m$ for some sufficiently large $m$ we conclude the lemma is true. \end{proof} \begin{remark}[Classical definition] \label{remark-classical-linear-series} Let $X$ be a smooth projective curve over an algebraically closed field $k$. We say two effective Cartier divisors $D, D' \subset X$ are {\it linearly equivalent} if and only if $\mathcal{O}_X(D) \cong \mathcal{O}_X(D')$ as $\mathcal{O}_X$-modules. Since $\text{Pic}(X) = \text{Cl}(X)$ (Divisors, Lemma \ref{divisors-lemma-local-rings-UFD-c1-bijective}) we see that $D$ and $D'$ are linearly equivalent if and only if the Weil divisors associated to $D$ and $D'$ define the same element of $\text{Cl}(X)$. Given an effective Cartier divisor $D \subset X$ of degree $d$ the {\it complete linear system} or {\it complete linear series} $|D|$ of $D$ is the set of effective Cartier divisors $E \subset X$ which are linearly equivalent to $D$. Another way to say it is that $|D|$ is the set of closed points of the fibre of the morphism $$\gamma_d : \underline{\text{Hilb}}^d_{X/k} \longrightarrow \underline{\text{Pic}}^d_{X/k}$$ (Picard Schemes of Curves, Lemma \ref{pic-lemma-picard-pieces}) over the closed point corresponding to $\mathcal{O}_X(D)$. This gives $|D|$ a natural scheme structure and it turns out that $|D| \cong \mathbf{P}^m_k$ with $m + 1 = h^0(\mathcal{O}_X(D))$. In fact, more canonically we have $$|D| = \mathbf{P}(H^0(X, \mathcal{O}_X(D))^\wedge)$$ where $(-)^\wedge$ indicates $k$-linear dual and $\mathbf{P}$ is as in Intersection Theory, Section \ref{intersection-section-projection}. In this language a {\it linear system} or a {\it linear series} on $X$ is a closed subvariety $L \subset |D|$ which can be cut out by linear equations. If $L$ has dimension $r$, then $L = \mathbf{P}(V^\wedge)$ where $V \subset H^0(X, \mathcal{O}_X(D))$ is a linear subspace of dimension $r + 1$. Thus the classical linear series $L \subset |D|$ corresponds to the linear series $(\mathcal{O}_X(D), V)$ as defined above. \end{remark} \section{Riemann-Roch and duality} \label{section-Riemann-Roch} \noindent Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. In Varieties, Section \ref{varieties-section-divisors-curves} we have defined the degree of a locally free $\mathcal{O}_X$-module $\mathcal{E}$ of constant rank by the formula \begin{equation} \label{equation-degree} \deg(\mathcal{E}) = \chi(X, \mathcal{E}) - \text{rank}(\mathcal{E})\chi(X, \mathcal{O}_X) \end{equation} see Varieties, Definition \ref{varieties-definition-degree-invertible-sheaf}. In the chapter on Chow Homology we defined the first chern class of $\mathcal{E}$ as an operation on cycles (Chow Homology, Section \ref{chow-section-intersecting-chern-classes}) and we proved that \begin{equation} \label{equation-degree-c1} \deg(\mathcal{E}) = \deg(c_1(\mathcal{E}) \cap [X]_1) \end{equation} see Chow Homology, Lemma \ref{chow-lemma-degree-vector-bundle}. Combining (\ref{equation-degree}) and (\ref{equation-degree-c1}) we obtain our first version of the Riemann-Roch formula \begin{equation} \label{equation-rr} \chi(X, \mathcal{E}) = \deg(c_1(\mathcal{E}) \cap [X]_1) + \text{rank}(\mathcal{E})\chi(X, \mathcal{O}_X) \end{equation} If $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then we can also consider the numerical intersection $(\mathcal{L} \cdot X)$ as defined in Varieties, Definition \ref{varieties-definition-intersection-number}. However, this does not give anything new as \begin{equation} \label{equation-numerical-degree} (\mathcal{L} \cdot X) = \deg(\mathcal{L}) \end{equation} by Varieties, Lemma \ref{varieties-lemma-intersection-numbers-and-degrees-on-curves}. If $\mathcal{L}$ is ample, then this integer is positive and is called the degree \begin{equation} \label{equation-degree-X} \deg_\mathcal{L}(X) = (\mathcal{L} \cdot X) = \deg(\mathcal{L}) \end{equation} of $X$ with respect to $\mathcal{L}$, see Varieties, Definition \ref{varieties-definition-degree}. \medskip\noindent To obtain a true Riemann-Roch theorem we would like to write $\chi(X, \mathcal{O}_X)$ as the degree of a canonical zero cycle on $X$. We refer to \cite{F} for a fully general version of this. We will use duality to get a formula in the case where $X$ is Gorenstein; however, in some sense this is a cheat (for example because this method cannot work in higher dimension). \begin{lemma} \label{lemma-duality-dim-1} Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. There exists a dualizing complex $\omega_X^\bullet$ with the following properties \begin{enumerate} \item $H^i(\omega_X^\bullet)$ is nonzero only for $i = -1, 0$, \item $\omega_X = H^{-1}(\omega_X^\bullet)$ is a coherent Cohen-Macaulay module whose support is the irreducible components of dimension $1$, \item for $x \in X$ closed, the module $H^0(\omega_{X, x}^\bullet)$ is nonzero if and only if either \begin{enumerate} \item $\dim(\mathcal{O}_{X, x}) = 0$ or \item $\dim(\mathcal{O}_{X, x}) = 1$ and $\mathcal{O}_{X, x}$ is not Cohen-Macaulay, \end{enumerate} \item there are functorial isomorphisms $\text{Ext}^i_X(K, \omega_X^\bullet) = \Hom_k(H^{-i}(X, K), k)$ compatible with shifts for $K \in D_\QCoh(X)$, \item there are functorial isomorphisms $\Hom(\mathcal{F}, \omega_X) = \Hom_k(H^1(X, \mathcal{F}), k)$ for $\mathcal{F}$ quasi-coherent on $X$, \item if $X \to \Spec(k)$ is smooth of relative dimension $1$, then $\omega_X \cong \Omega_{X/k}$. \end{enumerate} \end{lemma} \begin{proof} Denote $f : X \to \Spec(k)$ the structure morphism. We start with the relative dualizing complex $$\omega_X^\bullet = \omega_{X/k}^\bullet = a(\mathcal{O}_{\Spec(k)}) = f^!\mathcal{O}_{\Spec(k)}$$ as described in Dualizing Complexes, Remark \ref{dualizing-remark-relative-dualizing-complex}. Then property (4) holds by construction. Observe that $\omega_X^\bullet$ is also the dualizing complex normalized relative to $\omega_{\Spec(k)}^\bullet = \mathcal{O}_{\Spec(k)}$, i.e., it is the dualizing complex $\omega_X^\bullet$ as in Dualizing Complexes, Example \ref{dualizing-example-proper-over-local} with $A = k$ and $\omega_A = k[0]$. Parts (1) and (2) follow from Dualizing Complexes, Lemma \ref{dualizing-lemma-vanishing-good-dualizing}. For a closed point $x \in X$ we see that $\omega_{X, x}^\bullet$ is a normalized dualizing complex over $\mathcal{O}_{X, x}$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-good-dualizing-normalized}. Assertion (3) then follows from Dualizing Complexes, Lemma \ref{dualizing-lemma-apply-CM}. Assertion (5) follows from Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing-module-proper-over-A} for coherent $\mathcal{F}$ and in general by unwinding (4) for $K = \mathcal{F}[0]$ and $i = -1$. Assertion (6) follows from Dualizing Complexes, Lemma \ref{dualizing-lemma-smooth-proper}. \end{proof} \begin{lemma} \label{lemma-duality-dim-1-CM} Let $X$ be a proper scheme over a field $k$ which is Cohen-Macaulay and equidimensional of dimension $1$. There exists a dualizing module $\omega_X$ with the following properties \begin{enumerate} \item $\omega_X$ is a coherent Cohen-Macaulay module whose support is $X$, \item there are functorial isomorphisms $\text{Ext}^i_X(K, \omega_X[1]) = \Hom_k(H^{-i}(X, K), k)$ compatible with shifts for $K \in D_\QCoh(X)$, \item there are functorial isomorphisms $\text{Ext}^{1 + i}(\mathcal{F}, \omega_X) = \Hom_k(H^{-i}(X, \mathcal{F}), k)$ for $\mathcal{F}$ quasi-coherent on $X$. \end{enumerate} \end{lemma} \begin{proof} Let us take $\omega_X$ normalized as in as in Dualizing Complexes, Example \ref{dualizing-example-equidimensional-over-field}. Then the statements follow from Lemma \ref{lemma-duality-dim-1} and the fact that $\omega_X^\bullet = \omega_X[1]$ as $X$ is Cohen-Macualay (Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing-module-CM-scheme}). \end{proof} \begin{remark} \label{remark-rework-duality-locally-free} Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. Let $\omega_X^\bullet$ be as in Lemma \ref{lemma-duality-dim-1}. If $\mathcal{E}$ is a finite locally free $\mathcal{O}_X$-module with dual $\mathcal{E}^\wedge$ then we have canonical isomorphisms $$\Hom_k(H^{-i}(X, \mathcal{E}), k) = H^i(X, \mathcal{E}^\wedge \otimes_{\mathcal{O}_X}^\mathbf{L} \omega_X^\bullet)$$ This follows from the lemma and Cohomology, Lemma \ref{cohomology-lemma-dual-perfect-complex}. If $X$ is Cohen-Macaulay and equidimensional of dimension $1$, then we have canonical isomorphisms $$\Hom_k(H^{-i}(X, \mathcal{E}), k) = H^{1 - i}(X, \mathcal{E}^\wedge \otimes_{\mathcal{O}_X} \omega_X)$$ where $\omega_X$ is as in Lemma \ref{lemma-duality-dim-1-CM}. In particular if $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then we have $$\dim_k H^0(X, \mathcal{L}) = \dim_k H^1(X, \mathcal{L}^{\otimes -1} \otimes_{\mathcal{O}_X} \omega_X)$$ and $$\dim_k H^1(X, \mathcal{L}) = \dim_k H^0(X, \mathcal{L}^{\otimes -1} \otimes_{\mathcal{O}_X} \omega_X)$$ \end{remark} \noindent We can use Lemmas \ref{lemma-duality-dim-1} and \ref{lemma-duality-dim-1-CM} to get a relation between the euler characteristic of $\mathcal{O}_X$ and the euler characteristic of the dualizing complex or the dualizing module. \begin{lemma} \label{lemma-euler} Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. With $\omega_X^\bullet$ as in Lemma \ref{lemma-duality-dim-1} we have $$\chi(X, \mathcal{O}_X) = \chi(X, \omega_X^\bullet)$$ If $X$ is Cohen-Macaulay and equidimensional of dimension $1$, then $$\chi(X, \mathcal{O}_X) = - \chi(X, \omega_X)$$ with $\omega_X$ as in Lemma \ref{lemma-duality-dim-1-CM}. \end{lemma} \begin{proof} We define the right hand side of the first formula as follows: $$\chi(X, \omega_X^\bullet) = \sum\nolimits_{i \in \mathbf{Z}} (-1)^i\dim_k H^i(X, \omega_X^\bullet)$$ This is well defined because $\omega_X^\bullet$ is in $D^b_{\textit{Coh}}(\mathcal{O}_X)$, but also because $$H^i(X, \omega_X^\bullet) = \text{Ext}^i(\mathcal{O}_X, \omega_X^\bullet) = H^{-i}(X, \mathcal{O}_X)$$ which is always finite dimensional and nonzero only if $i = 0, -1$. This of course also proves the first formula. The second is a consequence of the first because $\omega_X^\bullet = \omega_X[1]$ in the CM case. \end{proof} \noindent We will use Lemma \ref{lemma-euler} to get the desired formula for $\chi(X, \mathcal{O}_X)$ in the case that $\omega_X$ is invertible, i.e., that $X$ is Gorenstein. The statement is that $-1/2$ of the first chern class of $\omega_X$ capped with the cycle $[X]_1$ associated to $X$ is a natural zero cycle on $X$ with half-integer coefficients whose degree is $\chi(X, \mathcal{O}_X)$. The occurence of fractions in the statement of Riemann-Roch cannot be avoided. \begin{lemma}[Riemann-Roch] \label{lemma-rr} Let $X$ be a proper scheme over a field $k$ which is Gorenstein and equidimensional of dimension $1$. Let $\omega_X$ be as in Lemma \ref{lemma-duality-dim-1-CM}. Then \begin{enumerate} \item $\omega_X$ is an invertible $\mathcal{O}_X$-module, \item $\deg(\omega_X) = -2\chi(X, \mathcal{O}_X)$, \item for a locally free $\mathcal{O}_X$-module $\mathcal{E}$ of constant rank we have $$\chi(X, \mathcal{E}) = \deg(\mathcal{E}) - \textstyle{\frac{1}{2}} \text{rank}(\mathcal{E}) \deg(\omega_X)$$ and $\dim_k(H^i(X, \mathcal{E})) = \dim_k(H^{1 - i}(X, \mathcal{E}^\wedge \otimes_{\mathcal{O}_X} \omega_X))$ for all $i \in \mathbf{Z}$. \end{enumerate} \end{lemma} \begin{proof} It follows more or less from the definition of the Gorenstein property that the dualizing sheaf is invertible, see Dualizing Complexes, Section \ref{dualizing-section-gorenstein}. By (\ref{equation-rr}) applied to $\omega_X$ we have $$\chi(X, \omega_X) = \deg(c_1(\omega_X) \cap [X]_1) + \chi(X, \mathcal{O}_X)$$ Combined with Lemma \ref{lemma-euler} this gives $$2\chi(X, \mathcal{O}_X) = - \deg(c_1(\omega_X) \cap [X]_1) = - \deg(\omega_X)$$ the second equality by (\ref{equation-degree-c1}). Putting this back into (\ref{equation-rr}) for $\mathcal{E}$ gives the displayed formula of the lemma. The symmetry in dimensions is a consequence of duality for $X$, see Remark \ref{remark-rework-duality-locally-free}. \end{proof} \section{Some vanishing results} \label{section-vanishing} \begin{lemma} \label{lemma-automatic} Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_X) = k$. Then $X$ is Cohen-Macaulay and equidimensional of dimension $1$. \end{lemma} \begin{proof} Since $\Gamma(X, \mathcal{O}_X) = k$ has no nontrivial idempotents, we see that $X$ is connected. This already shows that $X$ is equidimensional of dimension $1$ (any irreducible component of dimension $0$ would be a connected component). Let $\mathcal{I} \subset \mathcal{O}_X$ be the maximal coherent submodule supported in closed points. Then $\mathcal{I}$ exists (Divisors, Lemma \ref{divisors-lemma-remove-embedded-points}) and is globally generated (Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}). Since $1 \in \Gamma(X, \mathcal{O}_X)$ is not a section of $\mathcal{I}$ we conclude that $\mathcal{I} = 0$. Thus $X$ does not have embedded points (Divisors, Lemma \ref{divisors-lemma-remove-embedded-points}). Thus $X$ has $(S_1)$ by Divisors, Lemma \ref{divisors-lemma-S1-no-embedded}. Hence $X$ is Cohen-Macaulay. \end{proof} \noindent In this section we work in the following situation. \begin{situation} \label{situation-Cohen-Macaulay-curve} Here $k$ is a field, $X$ is a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_X) = k$. \end{situation} \noindent By Lemma \ref{lemma-automatic} the scheme $X$ is Cohen-Macaulay and equidimensional of dimension $1$. We denote $\omega_X$ the dualizing module of $X$ as in Dualizing Complexes, Example \ref{dualizing-example-equidimensional-over-field}. Then Lemmas \ref{lemma-duality-dim-1} and \ref{lemma-duality-dim-1-CM} show that $\omega_X$ has nonvanishing $H^1$ and in fact $\dim_k H^1(X, \omega_X) = 1$. It turns out that anything slightly more positive'' than $\omega_X$ has vanishing $H^1$. \begin{lemma} \label{lemma-vanishing} In Situation \ref{situation-Cohen-Macaulay-curve}. Given an exact sequence $$\omega_X \to \mathcal{F} \to \mathcal{Q} \to 0$$ of coherent $\mathcal{O}_X$-modules with $H^1(X, \mathcal{Q}) = 0$ (for example if $\dim(\text{Supp}(\mathcal{Q})) = 0$), then either $H^1(X, \mathcal{F}) = 0$ or $\mathcal{F} = \omega_X \oplus \mathcal{Q}$. \end{lemma} \begin{proof} (The parenthetical statement follows from Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-support-dimension-0}.) Since $H^0(X, \mathcal{O}_X) = k$ is dual to $H^1(X, \omega_X)$ (see Section \ref{section-Riemann-Roch}) we see that $\dim H^1(X, \omega_X) = 1$. The sheaf $\omega_X$ represents the functor $\mathcal{F} \mapsto \Hom_k(H^1(X, \mathcal{F}), k)$ on the category of coherent $\mathcal{O}_X$-modules (Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing-module-proper-over-A}). Consider an exact sequence as in the statement of the lemma and assume that $H^1(X, \mathcal{F}) \not = 0$. Since $H^1(X, \mathcal{Q}) = 0$ we see that $H^1(X, \omega_X) \to H^1(X, \mathcal{F})$ is an isomorphism. By the universal property of $\omega_X$ stated above, we conclude there is a map $\mathcal{F} \to \omega_X$ whose action on $H^1$ is the inverse of this isomorphism. The composition $\omega_X \to \mathcal{F} \to \omega_X$ is the identity (by the universal property) and the lemma is proved. \end{proof} \begin{lemma} \label{lemma-vanishing-twist} In Situation \ref{situation-Cohen-Macaulay-curve}. Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module which is globally generated and not isomorphic to $\mathcal{O}_X$. Then $H^1(X, \omega_X \otimes \mathcal{L}) = 0$. \end{lemma} \begin{proof} By duality as discussed in Section \ref{section-Riemann-Roch} we have to show that $H^0(X, \mathcal{L}^{\otimes - 1}) = 0$. If not, then we can choose a global section $t$ of $\mathcal{L}^{\otimes - 1}$ and a global section $s$ of $\mathcal{L}$ such that $st \not = 0$. However, then $st$ is a constant multiple of $1$, by our assumption that $H^0(X, \mathcal{O}_X) = k$. It follows that $\mathcal{L} \cong \mathcal{O}_X$, which is a contradiction. \end{proof} \begin{lemma} \label{lemma-globally-generated} In Situation \ref{situation-Cohen-Macaulay-curve}. Given an exact sequence $$\omega_X \to \mathcal{F} \to \mathcal{Q} \to 0$$ of coherent $\mathcal{O}_X$-modules with $\dim(\text{Supp}(\mathcal{Q})) = 0$ and $\dim_k H^0(X, \mathcal{Q}) \geq 2$ and such that there is no nonzero submodule $\mathcal{Q}' \subset \mathcal{F}$ such that $\mathcal{Q}' \to \mathcal{Q}$ is injective. Then the submodule of $\mathcal{F}$ generated by global sections surjects onto $\mathcal{Q}$. \end{lemma} \begin{proof} Let $\mathcal{F}' \subset \mathcal{F}$ be the submodule generated by global sections and the image of $\omega_X \to \mathcal{F}$. Since $\dim_k H^0(X, \mathcal{Q}) \geq 2$ and $\dim_k H^1(X, \omega_X) = \dim_k H^0(X, \mathcal{O}_X) = 1$, we see that $\mathcal{F}' \to \mathcal{Q}$ is not zero and $\omega_X \to \mathcal{F}'$ is not an isomorphism. Hence $H^1(X, \mathcal{F}') = 0$ by Lemma \ref{lemma-vanishing} and our assumption on $\mathcal{F}$. Consider the short exact sequence $$0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{Q}/\Im(\mathcal{F}' \to \mathcal{Q}) \to 0$$ If the quotient on the right is nonzero, then we obtain a contradiction because then $H^0(X, \mathcal{F})$ is bigger than $H^0(X, \mathcal{F}')$. \end{proof} \noindent Here is an example global generation statement. \begin{lemma} \label{lemma-globally-generated-curve} In Situation \ref{situation-Cohen-Macaulay-curve} assume that $X$ is integral. Let $0 \to \omega_X \to \mathcal{F} \to \mathcal{Q} \to 0$ be a short exact sequence of coherent $\mathcal{O}_X$-modules with $\mathcal{F}$ torsion free, $\dim(\text{Supp}(\mathcal{Q})) = 0$, and $\dim_k H^0(X, \mathcal{Q}) \geq 2$. Then $\mathcal{F}$ is globally generated. \end{lemma} \begin{proof} Consider the submodule $\mathcal{F}'$ generated by the global sections. By Lemma \ref{lemma-globally-generated} we see that $\mathcal{F}' \to \mathcal{Q}$ is surjective, in particular $\mathcal{F}' \not = 0$. Since $X$ is a curve, we see that $\mathcal{F}' \subset \mathcal{F}$ is an inclusion of rank $1$ sheaves, hence $\mathcal{Q}' = \mathcal{F}/\mathcal{F}'$ is supported in finitely many points. To get a contradiction, assume that $\mathcal{Q}'$ is nonzero. Then we see that $H^1(X, \mathcal{F}') \not = 0$. Then we get a nonzero map $\mathcal{F}' \to \omega_X$ by the universal property (Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing-module-proper-over-A}). The image of the composition $\mathcal{F}' \to \omega_X \to \mathcal{F}$ is generated by global sections, hence is inside of $\mathcal{F}'$. Thus we get a nonzero self map $\mathcal{F}' \to \mathcal{F}'$. Since $\mathcal{F}'$ is torsion free of rank $1$ on a proper curve this has to be an automorphism (details omitted). But then this implies that $\mathcal{F}'$ is contained in $\omega_X \subset \mathcal{F}$ contradicting the surjectivity of $\mathcal{F}' \to \mathcal{Q}$. \end{proof} \begin{lemma} \label{lemma-tensor-omega-with-globally-generated-invertible} In Situation \ref{situation-Cohen-Macaulay-curve}. Let $\mathcal{L}$ be a very ample invertible $\mathcal{O}_X$-module with $\deg(\mathcal{L}) \geq 2$. Then $\omega_X \otimes_{\mathcal{O}_X} \mathcal{L}$ is globally generated. \end{lemma} \begin{proof} Assume $k$ is algebraically closed. Let $x \in X$ be a closed point. Let $C_i \subset X$ be the irreducible components and for each $i$ let $x_i \in C_i$ be the generic point. By Varieties, Lemma \ref{varieties-lemma-very-ample-vanish-at-point} we can choose a section $s \in H^0(X, \mathcal{L})$ such that $s$ vanishes at $x$ but not at $x_i$ for all $i$. The corresponding module map $s : \mathcal{O}_X \to \mathcal{L}$ is injective with cokernel $\mathcal{Q}$ supported in finitely many points and with $H^0(X, \mathcal{Q}) \geq 2$. Consider the corresponding exact sequence $$0 \to \omega_X \to \omega_X \otimes \mathcal{L} \to \omega_X \otimes \mathcal{Q} \to 0$$ By Lemma \ref{lemma-globally-generated} we see that the module generated by global sections surjects onto $\omega_X \otimes \mathcal{Q}$. Since $x$ was arbitrary this proves the lemma. Some details omitted. \medskip\noindent We will reduce the case where $k$ is not algebraically closed, to the algebraically closed field case. We suggest the reader skip the rest of the proof. Choose an algebraic closure $\overline{k}$ of $k$ and consider the base change $X_{\overline{k}}$. Let us check that $X_{\overline{k}} \to \Spec(\overline{k})$ is an example of Situation \ref{situation-Cohen-Macaulay-curve}. By flat base change (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}) we see that $H^0(X_{\overline{k}}, \mathcal{O}) = \overline{k}$. By Varieties, Lemma \ref{varieties-lemma-CM-base-change} we see that $X_{\overline{k}}$ is Cohen-Macaulay. The scheme $X_{\overline{k}}$ is proper over $\overline{k}$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}) and equidimensional of dimension $1$ (Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}). The pullback of $\omega_X$ to $X_{\overline{k}}$ is the dualizing module of $X_{\overline{k}}$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-more-base-change}. The pullback of $\mathcal{L}$ to $X_{\overline{k}}$ is very ample (Morphisms, Lemma \ref{morphisms-lemma-very-ample-base-change}). The degree of the pullback of $\mathcal{L}$ to $X_{\overline{k}}$ is equal to the degree of $\mathcal{L}$ on $X$ (Varieties, Lemma \ref{varieties-lemma-degree-base-change}). Finally, we see that $\omega_X \otimes \mathcal{L}$ is globally generated if and only if its base change is so (Varieties, Lemma \ref{varieties-lemma-globally-generated-base-change}). In this way we see that the result follows from the result in the case of an algebraically closed ground field. \end{proof} \section{The genus of a curve} \label{section-genus} \noindent If $X$ is a smooth projective curve over an algebraically closed field, then we've previously defined the genus of $X$ as the dimension of $H^1(X, \mathcal{O}_X)$, see Picard Schemes of Curves, Definition \ref{pic-definition-genus}. Let us generalize this as follows. \begin{definition} \label{definition-genus} Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_X) = k$. Then the {\it genus} of $X$ is $g = \dim_k H^1(X, \mathcal{O}_X)$. \end{definition} \noindent This is sometimes called the {\it arithmetic genus} of $X$. In the literature the arithmetic genus of a proper curve $X$ over $k$ is sometimes defined as $$p_a(X) = 1 - \chi(X, \mathcal{O}_X) = 1 - \dim_k H^0(X, \mathcal{O}_X) + \dim_k H^1(X, \mathcal{O}_X)$$ This agrees with our definition when it applies because we assume $H^0(X, \mathcal{O}_X) = k$. But note that \begin{enumerate} \item $p_a(X)$ can be negative, and \item $p_a(X)$ depends on the base field $k$ and should be written $p_a(X/k)$. \end{enumerate} For example if $k = \mathbf{Q}$ and $X = \mathbf{P}^1_{\mathbf{Q}(i)}$ then $p_a(X/\mathbf{Q}) = -1$ and $p_a(X/\mathbf{Q}(i)) = 0$. \medskip\noindent The assumption that $H^0(X, \mathcal{O}_X) = k$ in our definition has two consequences. On the one hand, it means there is no confusion about the base field. On the other hand, it implies the scheme $X$ is Cohen-Macaulay and equidimensional of dimension $1$ (Lemma \ref{lemma-automatic}). Letting $\omega_X$ be the dualizing module as in Dualizing Complexes, Example \ref{dualizing-example-equidimensional-over-field} we see that \begin{equation} \label{equation-genus} g = \dim_k H^1(X, \mathcal{O}_X) = \dim_k H^0(X, \omega_X) \end{equation} by duality (see Remark \ref{remark-rework-duality-locally-free}). \medskip\noindent If $X$ is proper over $k$ of dimension $\leq 1$ and $H^0(X, \mathcal{O}_X)$ is not equal to the ground field $k$, instead of using the arithmetic genus $p_a(X)$ given by the displayed formula above we shall use the invariant $\chi(X, \mathcal{O}_X)$. In fact, it is advocated in \cite[page 276]{FAC} and \cite[Introduction]{Hirzebruch} that we should call $\chi(X, \mathcal{O}_X)$ the arithmetic genus. \begin{lemma} \label{lemma-genus-base-change} Let $k'/k$ be a field extension. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_X) = k$. Then $X_{k'}$ is a proper scheme over $k'$ having dimension $1$ and $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$. Moreover the genus of $X_{k'}$ is equal to the genus of $X$. \end{lemma} \begin{proof} The dimension of $X_{k'}$ is $1$ for example by Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}. The morphism $X_{k'} \to \Spec(k')$ is proper by Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}. The equality $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$ follows from Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}. The equality of the genus follows from the same lemma. \end{proof} \begin{lemma} \label{lemma-genus-gorenstein} Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_X) = k$. If $X$ is Gorenstein, then $$\deg(\omega_X) = 2g - 2$$ where $g$ is the genus of $X$ and $\omega_X$ is as in Lemma \ref{lemma-duality-dim-1-CM}. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-rr}. \end{proof} \begin{lemma} \label{lemma-genus-smooth} Let $X$ be a smooth proper curve over a field $k$ with $H^0(X, \mathcal{O}_X) = k$. Then $$\dim_k H^0(X, \Omega_{X/k}) = g \quad\text{and}\quad \deg(\Omega_{X/k}) = 2g - 2$$ where $g$ is the genus of $X$. \end{lemma} \begin{proof} By Lemma \ref{lemma-duality-dim-1} we have $\Omega_{X/k} = \omega_X$. Hence the formulas hold by (\ref{equation-genus}) and Lemma \ref{lemma-genus-gorenstein}. \end{proof} \section{Plane curves} \label{section-plane-curves} \noindent Let $k$ be a field. A {\it plane curve} will be a curve $X$ which is isomorphic to a closed subscheme of $\mathbf{P}^2_k$. Often the embedding $X \to \mathbf{P}^2_k$ will be considered given. By Divisors, Example \ref{divisors-example-closed-subscheme-of-proj} a curve is determined by the corresponding homogeneous ideal $$I(X) = \Ker\left( k[T_0, T_2, T_2] \longrightarrow \bigoplus \Gamma(X, \mathcal{O}_X(n)) \right)$$ Recall that in this situation we have $$X = \text{Proj}(k[T_0, T_2, T_2]/I)$$ as closed subschemes of $\mathbf{P}^2_k$. For more general information on these constructions we refer the reader to Divisors, Example \ref{divisors-example-closed-subscheme-of-proj} and the references therein. It turns out that $I(X) = (F)$ for some homogeneous polynomial $F \in k[T_0, T_1, T_2]$, see Lemma \ref{lemma-equation-plane-curve}. Since $X$ is irreducible, it follows that $F$ is irreducible, see Lemma \ref{lemma-plane-curve}. Moreover, looking at the short exact sequence $$0 \to \mathcal{O}_{\mathbf{P}^2_k}(-d) \xrightarrow{F} \mathcal{O}_{\mathbf{P}^2_k} \to \mathcal{O}_X \to 0$$ where $d = \deg(F)$ we find that $H^0(X, \mathcal{O}_X) = k$ and that $X$ has genus $(d - 1)(d - 2)/2$, see proof of Lemma \ref{lemma-genus-plane-curve}. \medskip\noindent To find smooth plane curves it is easiest to write explicit equations. Let $p$ denote the characteristic of $k$. If $p$ does not divide $d$, then we can take $$F = T_0^d + T_1^d + T_2^d$$ The corresponding curve $X = V_+(F)$ is called the {\it Fermat curve} of degree $d$. It is smooth because on each standard affine piece $D_+(T_i)$ we obtain a curve isomorphic to the affine curve $$\Spec(k[x, y]/(x^d + y^d + 1))$$ The ring map $k \to k[x, y]/(x^d + y^d + 1)$ is smooth by Algebra, Lemma \ref{algebra-lemma-relative-global-complete-intersection-smooth} as $d x^{d - 1}$ and $d y^{d - 1}$ generate the unit ideal in $k[x, y]/(x^d + y^d + 1)$. If $p | d$ but $p \not = 3$ then you can use the equation $$F = T_0^{d - 1}T_1 + T_1^{d - 1}T_2 + T_2^{d - 1}T_0$$ Namely, on the affine pieces you get $x + x^{d - 1}y + y^{d - 1}$ with derivatives $1 - x^{d - 2}y$ and $x^{d - 1} - y^{d - 2}$ whose common zero set (of all three) is empty\footnote{Namely, as $x^{d - 1} = y^{d - 2}$, then $0 = x + x^{d - 1}y + y^{d - 1} = x + 2 x^{d - 1} y$. Since $x \not = 0$ because $1 = x^{d - 2}y$ we get $0 = 1 + 2x^{d - 2}y = 3$ which is absurd unless $3 = 0$.}. We leave it to the reader to make examples in characteristic $3$. \medskip\noindent More generally for any field $k$ and any $n$ and $d$ there exists a smooth hypersurface of degree $d$ in $\mathbf{P}^n_k$, see for example \cite{Poonen}. \medskip\noindent Of course, in this way we only find smooth curves whose genus is a triangular number. To get smooth curves of an arbitrary genus one can look for smooth curves lying on $\mathbf{P}^1 \times \mathbf{P}^1$ (insert future reference here). \begin{lemma} \label{lemma-equation-plane-curve} Let $Z \subset \mathbf{P}^2_k$ be a closed subscheme which is equidimensional of dimension $1$ and has no embedded points (equivalently $Z$ is Cohen-Macaulay). Then the ideal $I(Z) \subset k[T_0, T_1, T_2]$ corresponding to $Z$ is principal. \end{lemma} \begin{proof} This is a special case of Divisors, Lemma \ref{divisors-lemma-equation-codim-1-in-projective-space} (see also Varieties, Lemma \ref{varieties-lemma-equation-codim-1-in-projective-space}). The parenthetical statement follows from the fact that a $1$ dimensional Noetherian scheme is Cohen-Macaulay if and only if it has no embedded points, see Divisors, Lemma \ref{divisors-lemma-noetherian-dim-1-CM-no-embedded-points}. \end{proof} \begin{lemma} \label{lemma-plane-curve} Let $Z \subset \mathbf{P}^2_k$ be as in Lemma \ref{lemma-equation-plane-curve} and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. Then $Z$ is a curve if and only if $F$ is irreducible. \end{lemma} \begin{proof} If $F$ is reducible, say $F = F' F''$ then let $Z'$ be the closed subscheme of $\mathbf{P}^2_k$ defined by $F'$. It is clear that $Z' \subset Z$ and that $Z' \not = Z$. Since $Z'$ has dimension $1$ as well, we conclude that either $Z$ is not reduced, or that $Z$ is not irreducible. Conversely, write $Z = \sum a_i D_i$ where $D_i$ are the irreducible components of $Z$, see Divisors, Lemmas \ref{divisors-lemma-codim-1-part} and \ref{divisors-lemma-codimension-1-is-effective-Cartier}. Let $F_i \in k[T_0, T_1, T_2]$ be the homogeneous polynomial generating the ideal of $D_i$. Then it is clear that $F$ and $\prod F_i^{a_i}$ cut out the same closed subscheme of $\mathbf{P}^2_k$. Hence $F = \lambda \prod F_i^{a_i}$ for some $\lambda \in k^*$ because both generate the ideal of $Z$. Thus we see that if $F$ is irreducible, then $Z$ is a prime divisor, i.e., a curve. \end{proof} \begin{lemma} \label{lemma-genus-plane-curve} Let $Z \subset \mathbf{P}^2_k$ be as in Lemma \ref{lemma-equation-plane-curve} and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. Then $H^0(Z, \mathcal{O}_Z) = k$ and the genus of $Z$ is $(d - 1)(d - 2)/2$ where $d = \deg(F)$. \end{lemma} \begin{proof} Let $S = k[T_0, T_1, T_2]$. There is an exact sequence of graded modules $$0 \to S(-d) \xrightarrow{F} S \to S/(F) \to 0$$ Denote $i : Z \to \mathbf{P}^2_k$ the given closed immersion. Applying the exact functor $\widetilde{\ }$ (Constructions, Lemma \ref{constructions-lemma-proj-sheaves}) we obtain $$0 \to \mathcal{O}_{\mathbf{P}^2_k}(-d) \to \mathcal{O}_{\mathbf{P}^2_k} \to i_*\mathcal{O}_Z \to 0$$ because $F$ generates the ideal of $Z$. Note that the cohomology groups of $\mathcal{O}_{\mathbf{P}^2_k}(-d)$ and $\mathcal{O}_{\mathbf{P}^2_k}$ are given in Cohomology of Schemes, Lemma \ref{coherent-lemma-cohomology-projective-space-over-ring}. On the other hand, we have $H^q(Z, \mathcal{O}_Z) = H^q(\mathbf{P}^2_k, i_*\mathcal{O}_Z)$ by Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}. Applying the long exact cohomology sequence we first obtain that $$k = H^0(\mathbf{P}^2_k, \mathcal{O}_{\mathbf{P}^2_k}) \longrightarrow H^0(Z, \mathcal{O}_Z)$$ is an isomorphism and next that the boundary map $$H^1(Z, \mathcal{O}_Z) \longrightarrow H^2(\mathbf{P}^2_k, \mathcal{O}_{\mathbf{P}^2_k}(-d)) \cong k[T_0, T_1, T_2]_{d - 3}$$ is an isomorphism. Since it is easy to see that the dimension of this is $(d - 1)(d - 2)/2$ the proof is finished. \end{proof} \begin{lemma} \label{lemma-smooth-plane-curve-point-over-separable} Let $Z \subset \mathbf{P}^2_k$ be as in Lemma \ref{lemma-equation-plane-curve} and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. If $Z \to \Spec(k)$ is smooth in at least one point and $k$ is infinite, then there exists a closed point $z \in Z$ contained in the smooth locus such that $\kappa(z)/k$ is finite separable of degree at most $d$. \end{lemma} \begin{proof} Suppose that $z' \in Z$ is a point where $Z \to \Spec(k)$ is smooth. After renumbering the coordinates if necessary we may assume $z'$ is contained in $D_+(T_0)$. Set $f = F(1, x, y) \in k[x, y]$. Then $Z \cap D_+(X_0)$ is isomorphic to the spectrum of $k[x, y]/(f)$. Let $f_x, f_y$ be the partial derivatives of $f$ with respect to $x, y$. Since $z'$ is a smooth point of $Z/k$ we see that either $f_x$ or $f_y$ is nonzero in $z'$ (see discussion in Algebra, Section \ref{algebra-section-smooth}). After renumbering the coordinates we may assume $f_y$ is not zero at $z'$. Hence there is a nonempty open subscheme $V \subset Z \cap D_{+}(X_0)$ such that the projection $$p : V \longrightarrow \Spec(k[x])$$ is \'etale. Because the degree of $f$ as a polynomial in $y$ is at most $d$, we see that the degrees of the fibres of the projection $p$ are at most $d$ (see discussion in Morphisms, Section \ref{morphisms-section-universally-bounded}). Moreover, as $p$ is \'etale the image of $p$ is an open $U \subset \Spec(k[x])$. Finally, since $k$ is infinite, the set of $k$-rational points $U(k)$ of $U$ is infinite, in particular not empty. Pick any $t \in U(k)$ and let $z \in V$ be a point mapping to $t$. Then $z$ works. \end{proof} \section{Curves of genus zero} \label{section-genus-zero} \noindent Later we will need to know what a proper genus zero curve looks like. It turns out that a Gorenstein proper genus zero curve is a plane curve of degree $2$, i.e., a conic. \begin{lemma} \label{lemma-genus-zero-pic} Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_X) = k$. If $X$ has genus $0$, then every invertible $\mathcal{O}_X$-module $\mathcal{L}$ of degree $0$ is trivial. \end{lemma} \begin{proof} Namely, we have $\dim_k H^0(X, \mathcal{L}) \geq 0 + 1 - 0 = 1$ by Riemann-Roch (Lemma \ref{lemma-rr}), hence $\mathcal{L}$ has a nonzero section, hence $\mathcal{L} \cong \mathcal{O}_X$ by Varieties, Lemma \ref{varieties-lemma-check-invertible-sheaf-trivial}. \end{proof} \begin{lemma} \label{lemma-genus-zero-positive-degree} Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_X) = k$. Assume $X$ has genus $0$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module of degree $d > 0$. Then we have \begin{enumerate} \item $\dim_k H^0(X, \mathcal{L}) = d + 1$ and $\dim_k H^1(X, \mathcal{L}) = 0$, \item $\mathcal{L}$ is very ample and defines a closed immersion into $\mathbf{P}^d_k$. \end{enumerate} \end{lemma} \begin{proof} By definition of degree and genus we have $$\dim_k H^0(X, \mathcal{L}) - \dim_k H^1(X, \mathcal{L}) = d + 1$$ Let $s$ be a nonzero section of $\mathcal{L}$. Then the zero scheme of $s$ is an effective Cartier divisor $D \subset X$, we have $\mathcal{L} = \mathcal{O}_X(D)$ and we have a short exact sequence $$0 \to \mathcal{O}_X \to \mathcal{L} \to \mathcal{L}|_D \to 0$$ see Divisors, Lemma \ref{divisors-lemma-characterize-OD} and Remark \ref{divisors-remark-ses-regular-section}. Since $H^1(X, \mathcal{O}_X) = 0$ by assumption, we see that $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_D)$ is surjective. As $\mathcal{L}|_D$ is generated by global sections (because $\dim(D) = 0$, see Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}) we conclude that the invertible module $\mathcal{L}$ is generated by global sections. In fact, since $D$ is an Artinian scheme we have $\mathcal{L}|_D \cong \mathcal{O}_D$\footnote{In our case this follows from Divisors, Lemma \ref{divisors-lemma-finite-trivialize-invertible-upstairs} as $D \to \Spec(k)$ is finite.} and hence we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_D$. The short exact sequence also shows that $H^1(X, \mathcal{L}) = 0$. \medskip\noindent For $n \geq 1$ consider the multiplication map $$\mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes n}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1})$$ We claim this is surjective. To see this we consider the short exact sequence $$0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to \mathcal{L}^{\otimes n + 1}|_D \to 0$$ The sections of $\mathcal{L}^{\otimes n + 1}$ coming from the left in this sequence are in the image of $\mu_n$. On the other hand, since $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_D)$ is surjective and since $t^n$ maps to a trivialization of $\mathcal{L}^{\otimes n}|_D$ we see that $\mu_n(H^0(X, \mathcal{L}) \otimes t^n)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes n + 1})$ surjecting onto the global sections of $\mathcal{L}^{\otimes n + 1}|_D$. This proves the claim. \medskip\noindent Observe that $\mathcal{L}$ is ample by Varieties, Lemma \ref{varieties-lemma-ample-curve}. Hence Morphisms, Lemma \ref{morphisms-lemma-proper-ample-is-proj} gives an isomorphism $$X \longrightarrow \text{Proj}\left( \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\right)$$ Since the maps $\mu_n$ are surjective for all $n \geq 1$ we see that the graded algebra on the right hand side is a quotient of the symmetric algebra on $H^0(X, \mathcal{L})$. Choosing a $k$-basis $s_0, \ldots, s_d$ of $H^0(X, \mathcal{L})$ we see that it is a quotient of a polynomial algebra in $d + 1$ variables. Since quotients of graded rings correspond to closed immersions of $\text{Proj}$ (Constructions, Lemma \ref{constructions-lemma-surjective-graded-rings-generated-degree-1-map-proj}) we find a closed immersion $X \to \mathbf{P}^d_k$. We omit the verification that this morphism is the morphism of Constructions, Lemma \ref{constructions-lemma-projective-space} associated to the sections $s_0, \ldots, s_d$ of $\mathcal{L}$. \end{proof} \begin{lemma} \label{lemma-genus-zero} Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_X) = k$. If $X$ is Gorenstein and has genus $0$, then $X$ is isomorphic to a plane curve of degree $2$. \end{lemma} \begin{proof} Consider the invertible sheaf $\mathcal{L} = \omega_X^{\otimes -1}$ where $\omega_X$ is as in Lemma \ref{lemma-duality-dim-1-CM}. Then $\deg(\omega_X) = -2$ by Lemma \ref{lemma-genus-gorenstein} and hence $\deg(\mathcal{L}) = 2$. By Lemma \ref{lemma-genus-zero-positive-degree} we conclude that choosing a basis $s_0, s_1, s_2$ of the $k$-vector space of global sections of $\mathcal{L}$ we obtain a closed immersion $$\varphi_{(\mathcal{L}, (s_0, s_1, s_2))} : X \longrightarrow \mathbf{P}^2_k$$ Thus $X$ is a plane curve of some degree $d$. Let $F \in k[T_0, T_1, T_2]_d$ be its equation (Lemma \ref{lemma-equation-plane-curve}). Because the genus of $X$ is $0$ we see that $d$ is $1$ or $2$ (Lemma \ref{lemma-genus-plane-curve}). Observe that $F$ restricts to the zero section on $\varphi(X)$ and hence $F(s_0, s_1, s_2)$ is the zero section of $\mathcal{L}^{\otimes 2}$. Because $s_0, s_1, s_2$ are linearly independent we see that $F$ cannot be linear, i.e., $d = \deg(F) \geq 2$. Thus $d = 2$ and the proof is complete. \end{proof} \begin{proposition}[Characterization of the projective line] \label{proposition-projective-line} Let $k$ be a field. Let $X$ be a proper curve over $k$. The following are equivalent \begin{enumerate} \item $X \cong \mathbf{P}^1_k$, \item $X$ is smooth and geometrically irreducible over $k$, $X$ has genus $0$, and $X$ has an invertible module of odd degree, \item $X$ is geometrically integral over $k$, $X$ has genus $0$, $X$ is Gorenstein, and $X$ has an invertible sheaf of odd degree, \item $H^0(X, \mathcal{O}_X) = k$, $X$ has genus $0$, $X$ is Gorenstein, and $X$ has an invertible sheaf of odd degree, \item $X$ is geometrically integral over $k$, $X$ has genus $0$, and $X$ has an invertible $\mathcal{O}_X$-module of degree $1$, \item $H^0(X, \mathcal{O}_X) = k$, $X$ has genus $0$, and $X$ has an invertible $\mathcal{O}_X$-module of degree $1$, \item $H^1(X, \mathcal{O}_X) = 0$ and $X$ has an invertible $\mathcal{O}_X$-module of degree $1$, \item $H^1(X, \mathcal{O}_X) = 0$ and $X$ has a $k$-rational points $x_1, \ldots, x_n$ such that $\mathcal{O}_{X, x_i}$ is normal and $\gcd([\kappa(x_i) : k]) = 1$, and \item add more here. \end{enumerate} \end{proposition} \begin{proof} We will prove that each condition (2) -- (8) implies (1) and we omit the verification that (1) implies (2) -- (8). \medskip\noindent Assume (2). A smooth scheme over $k$ is geometrically reduced (Varieties, Lemma \ref{varieties-lemma-smooth-geometrically-normal}) and regular (Varieties, Lemma \ref{varieties-lemma-smooth-regular}). Hence $X$ is Gorenstein (Dualizing Complexes, Lemma \ref{dualizing-lemma-regular-gorenstein}). Thus we reduce to (3). \medskip\noindent Assume (3). Since $X$ is geometrically integral over $k$ we have $H^0(X, \mathcal{O}_X) = k$ by Varieties, Lemma \ref{varieties-lemma-regular-functions-proper-variety}. and we reduce to (4). \medskip\noindent Assume (4). Since $X$ is Gorenstein the dualizing module $\omega_X$ as in Lemma \ref{lemma-duality-dim-1-CM} has degree $\deg(\omega_X) = -2$ by Lemma \ref{lemma-genus-gorenstein}. Combined with the assumed existence of an odd degree invertible module, we conclude there exists an invertible module of degree $1$. In this way we reduce to (6). \medskip\noindent Assume (5). Since $X$ is geometrically integral over $k$ we have $H^0(X, \mathcal{O}_X) = k$ by Varieties, Lemma \ref{varieties-lemma-regular-functions-proper-variety}. and we reduce to (6). \medskip\noindent Assume (6). Then $X \cong \mathbf{P}^1_k$ by Lemma \ref{lemma-genus-zero-positive-degree}. \medskip\noindent Assume (7). Observe that $\kappa = H^0(X, \mathcal{O}_X)$ is a field finite over $k$ by Varieties, Lemma \ref{varieties-lemma-regular-functions-proper-variety}. If $d = [\kappa : k] > 1$, then every invertible sheaf has degree divisible by $d$ and there cannot be an invertible sheaf of degree $1$. Hence $d = 1$ and we reduce to case (6). \medskip\noindent Assume (8). Observe that $\kappa = H^0(X, \mathcal{O}_X)$ is a field finite over $k$ by Varieties, Lemma \ref{varieties-lemma-regular-functions-proper-variety}. Since $\kappa \subset \kappa(x_i)$ we see that $k = \kappa$ by the assumption on the gcd of the degrees. The same condition allows us to find integers $a_i$ such that $1 = \sum a_i[\kappa(x_i) : k]$. Because $x_i$ defines an effective Cartier divisor on $X$ by Varieties, Lemma \ref{varieties-lemma-regular-point-on-curve} we can consider the invertible module $\mathcal{O}_X(\sum a_i x_i)$. By our choice of $a_i$ the degree of $\mathcal{L}$ is $1$. Thus $X \cong \mathbf{P}^1_k$ by Lemma \ref{lemma-genus-zero-positive-degree}. \end{proof} \section{Geometric genus} \label{section-geometric-genus} \noindent If $X$ is a proper and {\bf smooth} curve over $k$ with $H^0(X, \mathcal{O}_X) = k$, then $$p_g(X) = \dim_k H^0(X, \Omega_{X/k})$$ is called the {\it geometric genus} of $X$. By Lemma \ref{lemma-genus-smooth} the geometric genus of $X$ agrees with the (arithmetic) genus. However, in higher dimensions there is a difference between the geometric genus and the arithmetic genus, see Remark \ref{remark-genus-higher-dimension}. \medskip\noindent For singular curves, we will define the geometric genus as follows. \begin{definition} \label{definition-geometric-genus} Let $k$ be a field. Let $X$ be a geometrically irreducible curve over $k$. The {\it geometric genus} of $X$ is the genus of a smooth projective model of $X$ possibly defined over an extension field of $k$ as in Lemma \ref{lemma-smooth-models}. \end{definition} \noindent If $k$ is perfect, then the nonsingular projective model $Y$ of $X$ is smooth (Lemma \ref{lemma-nonsingular-model-smooth}) and the geometric genus of $X$ is just the genus of $Y$. But if $k$ is not perfect, this may not be true. In this case we choose an extension $K/k$ such that the nonsingular projective model $Y_K$ of $(X_K)_{red}$ is a smooth projective curve and we define the geometric genus of $X$ to be the genus of $Y_K$. This is well defined by Lemmas \ref{lemma-smooth-models} and \ref{lemma-genus-base-change}. \begin{remark} \label{remark-genus-higher-dimension} Suppose that $X$ is a $d$-dimensional proper smooth variety over an algebraically closed field $k$. Then the {\it arithmetic genus} is often defined as $p_a(X) = (-1)^d(\chi(X, \mathcal{O}_X) - 1)$ and the {\it geometric genus} as $p_g(X) = \dim_k H^0(X, \Omega^d_{X/k})$. In this situation the arithmetic genus and the geometric genus no longer agree even though it is still true that $\omega_X \cong \Omega_{X/k}^d$. For example, if $d = 2$, then we have \begin{align*} p_a(X) - p_g(X) & = h^0(X, \mathcal{O}_X) - h^1(X, \mathcal{O}_X) + h^2(X, \mathcal{O}_X) - 1 - h^0(X, \Omega^2_{X/k}) \\ & = - h^1(X, \mathcal{O}_X) + h^2(X, \mathcal{O}_X) - h^0(X, \omega_X) \\ & = - h^1(X, \mathcal{O}_X) \end{align*} where $h^i(X, \mathcal{F}) = \dim_k H^i(X, \mathcal{F})$ and where the last equality follows from duality. Hence for a surface the difference $p_g(X) - p_a(X)$ is always nonnegative; it is sometimes called the irregularity of the surface. If $X = C_1 \times C_2$ is a product of smooth projective curves of genus $g_1$ and $g_2$, then the irregularity is $g_1 + g_2$. \end{remark} \section{Riemann-Hurwitz} \label{section-riemann-hurewitz} \noindent Let $k$ be a field. Let $f : X \to Y$ be a morphism of smooth curves over $k$. Then we obtain a canonical exact sequence $$f^*\Omega_{Y/k} \xrightarrow{\text{d}f} \Omega_{X/k} \longrightarrow \Omega_{X/Y} \longrightarrow 0$$ by Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials}. Since $X$ and $Y$ are smooth, the sheaves $\Omega_{X/k}$ and $\Omega_{Y/k}$ are invertible modules, see Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}. Assume the first map is nonzero, i.e., assume $f$ is generically \'etale, see Lemma \ref{lemma-generically-etale}. Let $R \subset X$ be the closed subscheme cut out by the different $\mathfrak{D}_f$ of $f$. By Dualizing Complexes, Lemma \ref{dualizing-lemma-discriminant-quasi-finite-morphism-smooth} this is the same as the vanishing locus of $\text{d}f$, it is an effective Cartier divisor, and we get $$f^*\Omega_{Y/k} \otimes_{\mathcal{O}_X} \mathcal{O}_X(R) = \Omega_{X/k}$$ In particular, if $X$, $Y$ are projective with $k = H^0(Y, \mathcal{O}_Y) = H^0(X, \mathcal{O}_X)$ and $X$, $Y$ have genus $g_X$, $g_Y$, then we get the Riemann-Hurwitz formula \begin{align*} 2g_X - 2 & = \deg(\Omega_{X/k}) \\ & = \deg(f^*\Omega_{Y/k} \otimes_{\mathcal{O}_X} \mathcal{O}_X(R)) \\ & = \deg(f) \deg(\Omega_{Y/k}) + \deg(R) \\ & = \deg(f) (2g_Y - 2) + \deg(R) \end{align*} The first and last equality by Lemma \ref{lemma-genus-smooth}. The second equality by the isomorphism of invertible sheaves given above. The third equality by additivity of degrees (Varieties, Lemma \ref{varieties-lemma-degree-tensor-product}), the formula for the degree of a pullback (Varieties, Lemma \ref{varieties-lemma-degree-pullback-map-proper-curves}), and finally the formula for the degree of $\mathcal{O}_X(R)$ (Varieties, Lemma \ref{varieties-lemma-degree-effective-Cartier-divisor}). \medskip\noindent To use the Riemann-Hurwitz formula we need to compute $\deg(R) = \dim_k \Gamma(R, \mathcal{O}_R)$. By the structure of zero dimensional schemes over $k$ (see for example Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0}), we see that $R$ is a finite disjoint union of spectra of Artinian local rings $R = \coprod_{x \in R} \Spec(\mathcal{O}_{R, x})$ with each $\mathcal{O}_{R, x}$ of finite dimension over $k$. Thus $$\deg(R) = \sum\nolimits_{x \in R} \dim_k \mathcal{O}_{R, x} = \sum\nolimits_{x \in R} d_x [\kappa(x) : k]$$ with $$d_x = \text{length}_{\mathcal{O}_{R, x}} \mathcal{O}_{R, x} = \text{length}_{\mathcal{O}_{X, x}} \mathcal{O}_{R, x}$$ the multiplicity of $x$ in $R$ (see Algebra, Lemma \ref{algebra-lemma-pushdown-module}). Let $x \in X$ be a closed point with image $y \in Y$. Looking at stalks we obtain an exact sequence $$\Omega_{Y/k, y} \to \Omega_{X/k, x} \to \Omega_{X/Y, x} \to 0$$ Choosing local generators $\eta_x$ and $\eta_y$ of the (free rank $1$) modules $\Omega_{X/k, x}$ and $\Omega_{Y/k, y}$ we see that $\eta_y \mapsto h \eta_x$ for some nonzero $h \in \mathcal{O}_{X, x}$. By definition $R$ is cut out by $h$. By the exact sequence we see that $\Omega_{X/Y, x} \cong \mathcal{O}_{X, x}/h\mathcal{O}_{X, x}$ as $\mathcal{O}_{X, x}$-modules. Thus we find the following equalities \begin{align*} d_x & = \text{length}_{\mathcal{O}_{X, x}}(\mathcal{O}_{X, x}/h\mathcal{O}_{X, x}) \\ & = \text{length}_{\mathcal{O}_{X, x}}(\Omega_{X/Y, x}) \\ & = \text{ord}_{\mathcal{O}_{X, x}}(h) \\ & = \text{ord}_{\mathcal{O}_{X, x}}(\eta_y/\eta_x") \end{align*} The first equality by our definition of $d_x$. The second by the above. The third equality is the definition of $\text{ord}$, see Algebra, Definition \ref{algebra-definition-ord}. The fourth equality is a mnemonic. Since $\mathcal{O}_{X, x}$ is a discrete valuation ring, the integer $\text{ord}_{\mathcal{O}_{X, x}}(h)$ just the valuation of $h$. \medskip\noindent Here is a case where one can calculate'' the multiplicity $d_x$ in terms of other invariants. Namely, if $\kappa(x)$ is separable over $k$, then we may choose $\eta_x = \text{d}s$ and $\eta_y = \text{d}t$ where $s$ and $t$ are uniformizers in $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ (Lemma \ref{lemma-uniformizer-works}). Then $t \mapsto u s^{e_x}$ for some unit $u \in \mathcal{O}_{X, x}$ where $e_x$ is the ramification index of the extension $\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}$. Hence we get $$\eta_y = \text{d}t = \text{d}(u s^{e_x}) = e s^{e_x - 1} u \text{d}s + s^{e_x} \text{d}u$$ Writing $\text{d}u = w \text{d}s$ for some $w \in \mathcal{O}_{X, x}$ we see that $$\eta_y/\eta_x" = e s^{e_x - 1} u + s^{e_x} w = (e_x u + s w)s^{e_x - 1}$$ We conclude that the order of vanishing of this is $e_x - 1$ unless the characteristic of $\kappa(x)$ is $p > 0$ and $p$ divides $e_x$ in which case the order of vanishing is $> e_x - 1$. \medskip\noindent Combining all of the above we find that if $k$ has characteristic zero, then $$2g_x - 2 = (2g_Y - 2)\deg(f) + \sum\nolimits_{x \in X} (e_x - 1)[\kappa(x) : k]$$ where $e_x$ is the ramification index of $\mathcal{O}_{X, x}$ over $\mathcal{O}_{Y, f(x)}$. This precise formula will hold if and only if all the ramification is tame, i.e., when the residue field extensions $\kappa(x)/\kappa(y)$ are separable and $e_x$ is prime to the characteristic of $k$, although the arguments above are insufficient to prove this. We refer the reader to Lemma \ref{lemma-rhe} and its proof. \begin{lemma} \label{lemma-generically-etale} \begin{slogan} A morphism of smooth curves is separable iff it is etale almost everywhere \end{slogan} Let $k$ be a field. Let $f : X \to Y$ be a morphism of smooth curves over $k$. The following are equivalent \begin{enumerate} \item $\text{d}f : f^*\Omega_{Y/k} \to \Omega_{X/k}$ is nonzero, \item $\Omega_{X/Y}$ is supported on a proper closed subset of $X$, \item there exists a nonempty open $U \subset X$ such that $f|_U : U \to Y$ is unramified, \item there exists a nonempty open $U \subset X$ such that $f|_U : U \to Y$ is \'etale, \item the extension $k(Y) \subset k(X)$ of function fields is finite separable. \end{enumerate} \end{lemma} \begin{proof} Since $X$ and $Y$ are smooth, the sheaves $\Omega_{X/k}$ and $\Omega_{Y/k}$ are invertible modules, see Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}. Using the exact sequence $$f^*\Omega_{Y/k} \longrightarrow \Omega_{X/k} \longrightarrow \Omega_{X/Y} \longrightarrow 0$$ of Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials} we see that (1) and (2) are equivalent and equivalent to the condition that $f^*\Omega_{Y/k} \to \Omega_{X/k}$ is nonzero in the generic point. The equivalence of (2) and (3) follows from Morphisms, Lemma \ref{morphisms-lemma-unramified-omega-zero}. The equivalence between (3) and (4) follows from Morphisms, Lemma \ref{morphisms-lemma-flat-unramified-etale} and the fact that flatness is automatic (Lemma \ref{lemma-flat}). To see the equivalence of (5) and (4) use Algebra, Lemma \ref{algebra-lemma-smooth-at-generic-point}. Some details omitted. \end{proof} \begin{lemma} \label{lemma-rh} Let $f : X \to Y$ be a morphism of smooth proper curves over a field $k$ which satisfies the equivalent conditions of Lemma \ref{lemma-generically-etale}. If $k = H^0(Y, \mathcal{O}_Y) = H^0(X, \mathcal{O}_X)$ and $X$ and $Y$ have genus $g_X$ and $g_y$, then $$2g_X - 2 = (2g_Y - 2) \deg(f) + \deg(R)$$ where $R \subset X$ is the effective Cartier divisor cut out by the different of $f$. \end{lemma} \begin{proof} See discussion above; we used Dualizing Complexes, Lemma \ref{dualizing-lemma-discriminant-quasi-finite-morphism-smooth}, Lemma \ref{lemma-genus-smooth}, and Varieties, Lemmas \ref{varieties-lemma-degree-tensor-product} and \ref{varieties-lemma-degree-pullback-map-proper-curves}. \end{proof} \begin{lemma} \label{lemma-uniformizer-works} Let $X \to \Spec(k)$ be smooth of relative dimension $1$ at a closed point $x \in X$. If $\kappa(x)$ is separable over $k$, then for any uniformizer $s$ in the discrete valuation ring $\mathcal{O}_{X, x}$ the element $\text{d}s$ freely generates $\Omega_{X/k, x}$ over $\mathcal{O}_{X, x}$. \end{lemma} \begin{proof} The ring $\mathcal{O}_{X, x}$ is a discrete valuation ring by Algebra, Lemma \ref{algebra-lemma-characterize-smooth-over-field}. Since $x$ is closed $\kappa(x)$ is finite over $k$. Hence if $\kappa(x)/k$ is separable, then any uniformizer $s$ maps to a nonzero element of $\Omega_{X/k, x} \otimes_{\mathcal{O}_{X, x}} \kappa(x)$ by Algebra, Lemma \ref{algebra-lemma-computation-differential}. Since $\Omega_{X/k, x}$ is free of rank $1$ over $\mathcal{O}_{X, x}$ the result follows. \end{proof} \begin{lemma} \label{lemma-rhe} Notation and assumptions as in Lemma \ref{lemma-rh}. For a closed point $x \in X$ let $d_x$ be the multiplicity of $x$ in $R$. Then $$2g_X - 2 = (2g_Y - 2) \deg(f) + \sum\nolimits d_x [\kappa(x) : k]$$ Moreover, we have the following results \begin{enumerate} \item $d_x = \text{length}_{\mathcal{O}_{X, x}}(\Omega_{X/Y, x})$, \item $d_x \geq e_x - 1$ where $e_x$ is the ramification index of $\mathcal{O}_{X, x}$ over $\mathcal{O}_{Y, y}$, \item $d_x = e_x - 1$ if and only if $\mathcal{O}_{X, x}$ is tamely ramified over $\mathcal{O}_{Y, y}$. \end{enumerate} \end{lemma} \begin{proof} By Lemma \ref{lemma-rh} and the discussion above (which used Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0} and Algebra, Lemma \ref{algebra-lemma-pushdown-module}) it suffices to prove the results on the multiplicity $d_x$ of $x$ in $R$. Part (1) was proved in the discussion above. In the discussion above we proved (2) and (3) only in the case where $\kappa(x)$ is separable over $k$. In the rest of the proof we give a uniform treatment of (2) and (3) using material on differents of quasi-finite Gorenstein morphisms. \medskip\noindent First, observe that $f$ is a quasi-finite Gorenstein morphism. This is true for example because $f$ is a flat quasi-finite morphism and $X$ is Gorenstein (see Dualizing Complexes, Lemma \ref{dualizing-lemma-flat-morphism-from-gorenstein-scheme}) or because it was shown in the proof of Dualizing Complexes, Lemma \ref{dualizing-lemma-discriminant-quasi-finite-morphism-smooth} (which we used above). Thus $\omega_{X/Y}$ is invertible by Dualizing Complexes, Lemma \ref{dualizing-lemma-gorenstein-quasi-finite} and the same remains true after replacing $X$ by opens and after performing a base change by some $Y' \to Y$. We will use this below without further mention. \medskip\noindent Choose affine opens $U \subset X$ and $V \subset Y$ such that $x \in U$, $y \in V$, $f(U) \subset V$, and $x$ is the only point of $U$ lying over $y$. Write $U = \Spec(A)$ and $V = \Spec(B)$. Then $R \cap U$ is the different of $f|_U : U \to V$. By Dualizing Complexes, Lemma \ref{dualizing-lemma-base-change-different} formation of the different commutes with arbitrary base change in our case. By our choice of $U$ and $V$ we have $$A \otimes_B \kappa(y) = \mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \kappa(y) = \mathcal{O}_{X, x}/(s^{e_x})$$ where $e_x$ is the ramification index as in the statement of the lemma. Let $C = \mathcal{O}_{X, x}/(s^{e_x})$ viewed as a finite algebra over $\kappa(y)$. Let $\mathfrak{D}_{C/\kappa(y)}$ be the different of $C$ over $\kappa(y)$ in the sense of Dualizing Complexes, Definition \ref{dualizing-definition-different}. It suffices to show: $\mathfrak{D}_{C/\kappa(y)}$ is nonzero if and only if the the extension $\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}$ is tamely ramified and in the tamely ramified case $\mathfrak{D}_{C/\kappa(y)}$ is equal to the ideal generated by $s^{e_x - 1}$ in $C$. Recall that tame ramification means exactly that $\kappa(x)/\kappa(y)$ is separable and that the characteristic of $\kappa(y)$ does not divide $e_x$. On the other hand, the different of $C/\kappa(y)$ is nonzero if and only if $\tau_{C/\kappa(y)} \in \omega_{C/\kappa(y)}$ is nonzero. Namely, since $\omega_{C/\kappa(y)}$ is an invertible $C$-module (as the base change of $\omega_{A/B}$) it is free of rank $1$, say with generator $\lambda$. Write $\tau_{C/\kappa(y)} = h\lambda$ for some $h \in C$. Then $\mathfrak{D}_{C/\kappa(y)} = (h) \subset C$ whence the claim. By Dualizing Complexes, Lemma \ref{dualizing-lemma-tau-nonzero} we have $\tau_{C/\kappa(y)} \not = 0$ if and only if $\kappa(x)/\kappa(y)$ is separable and $e_x$ is prime to the characteristic. Finally, even if $\tau_{C/\kappa(y)}$ is nonzero, then it is still the case that $s \tau_{C/\kappa(y)} = 0$ because $s\tau_{C/\kappa(y)} : C \to \kappa(y)$ sends $c$ to the trace of the nilpotent operator $sc$ which is zero. Hence $sh = 0$, hence $h \in (s^{e_x - 1})$ which proves that $\mathfrak{D}_{C/\kappa(y)} \subset (s^{e_x - 1})$ always. Since $(s^{e_x - 1}) \subset C$ is the smallest nonzero ideal, we have proved the final assertion. \end{proof} \section{Inseparable maps} \label{section-inseparable} \noindent Some remarks on the behaviour of the genus under inseparable maps. \begin{lemma} \label{lemma-dominated-by-smooth} Let $k$ be a field. Let $f : X \to Y$ be a surjective morphism of curves over $k$. If $X$ is smooth over $k$ and $Y$ is normal, then $Y$ is smooth over $k$. \end{lemma} \begin{proof} Let $y \in Y$. Pick $x \in X$ mapping to $y$. By Varieties, Lemma \ref{varieties-lemma-flat-under-smooth} it suffices to show that $f$ is flat at $x$. This follows from Lemma \ref{lemma-flat}. \end{proof} \begin{lemma} \label{lemma-purely-inseparable} Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper nonsingular curves over $k$. If the extension $k(Y) \subset k(X)$ of function fields is purely inseparable, then there exists a factorization $$X = X_0 \to X_1 \to \ldots \to X_n = Y$$ such that each $X_i$ is a proper nonsingular curve and $X_i \to X_{i + 1}$ is a degree $p$ morphism with $k(X_{i + 1}) \subset k(X_i)$ inseparable. \end{lemma} \begin{proof} This follows from Theorem \ref{theorem-curves-rational-maps} and the fact that a finite purely inseparable extension of fields can always be gotten as a sequence of (inseparable) extensions of degree $p$, see Fields, Lemma \ref{fields-lemma-finite-purely-inseparable}. \end{proof} \begin{lemma} \label{lemma-inseparable-deg-p-smooth} Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper nonsingular curves over $k$. If $X$ is smooth and $k(Y) \subset k(X)$ is inseparable of degree $p$, then there is a unique isomorphism $Y = X^{(p)}$ such that $f$ is $F_{X/k}$. \end{lemma} \begin{proof} The relative frobenius morphism $F_{X/k} : X \to X^{(p)}$ is constructed in Varieties, Section \ref{varieties-section-frobenius}. Observe that $X^{(p)}$ is a smooth proper curve over $k$ as a base change of $X$. The morphism $F_{X/k}$ has degree $p$ by Varieties, Lemma \ref{varieties-lemma-inseparable-deg-p-smooth}. Thus $k(X^{(p)})$ and $k(Y)$ are both subfields of $k(X)$ with $[k(X) : k(Y)] = [k(X) : k(X^{(p)})] = p$. To prove the lemma it suffices to show that $k(Y) = k(X^{(p)})$ inside $k(X)$. See Theorem \ref{theorem-curves-rational-maps}. \medskip\noindent Write $K = k(X)$. Consider the map $\text{d} : K \to \Omega_{K/k}$. It follows from Lemma \ref{lemma-generically-etale} that both $k(Y)$ is contained in the kernel of $\text{d}$. By Varieties, Lemma \ref{varieties-lemma-relative-frobenius-omega} we see that $k(X^{(p)})$ is in the kernel of $\text{d}$. Since $X$ is a smooth curve we know that $\Omega_{K/k}$ is a vector space of dimension $1$ over $K$. Then More on Algebra, Lemma \ref{more-algebra-lemma-p-basis}. implies that $\Ker(\text{d}) = kK^p$ and that $[K : kK^p] = p$. Thus $k(Y) = kK^p = k(X^{(p)})$ for reasons of degree. \end{proof} \begin{lemma} \label{lemma-purely-inseparable-smooth} Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper nonsingular curves over $k$. If $X$ is smooth and $k(Y) \subset k(X)$ is purely inseparable, then there is a unique $n \geq 0$ and a unique isomorphism $Y = X^{(p^n)}$ such that $f$ is the $n$-fold relative Frobenius of $X/k$. \end{lemma} \begin{proof} The $n$-fold relative Frobenius of $X/k$ is defined in Varieties, Remark \ref{varieties-remark-n-fold-relative-frobenius}. The lemma follows by combining Lemmas \ref{lemma-inseparable-deg-p-smooth} and \ref{lemma-purely-inseparable}. \end{proof} \begin{lemma} \label{lemma-purely-inseparable-smooth-genus} Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper nonsingular curves over $k$. Assume \begin{enumerate} \item $X$ is smooth, \item $H^0(X, \mathcal{O}_X) = k$, \item $k(X)/k(Y)$ is purely inseparable. \end{enumerate} Then $Y$ is smooth, $H^0(Y, \mathcal{O}_Y) = k$, and the genus of $Y$ is equal to the genus of $X$. \end{lemma} \begin{proof} By Lemma \ref{lemma-purely-inseparable-smooth} we see that $Y = X^{(p^n)}$ is the base change of $X$ by $F_{\Spec(k)}^n$. Thus $Y$ is smooth and the result on the cohomology and genus follows from Lemma \ref{lemma-genus-base-change}. \end{proof} \begin{example} \label{example-inseparable} This example will show that the genus can change under a purely inseparable morphism of nonsingular projective curves. Let $k$ be a field of characteristic $3$. Assume there exists an element $a \in k$ which is not a $3$rd power. For example $k = \mathbf{F}_3(a)$ would work. Let $X$ be the plane curve with homogeneous equation $$F = T_1^2T_0 - T_2^3 + aT_0^3$$ as in Section \ref{section-plane-curves}. On the affine piece $D_+(T_0)$ using coordinates $x = T_1/T_0$ and $y = T_2/T_0$ we obtain $x^2 - y^3 + a = 0$ which defines a nonsingular affine curve. Moreover, the point at infinity $(0 : 1: 0)$ is a smooth point. Hence $X$ is a nonsingular projective curve of genus $1$ (Lemma \ref{lemma-genus-plane-curve}). On the other hand, consider the morphism $f : X \to \mathbf{P}^1_k$ which on $D_+(T_0)$ sends $(x, y)$ to $y \in \mathbf{A}^1_k \subset \mathbf{P}^1_k$. Then $f$ is a morphism of proper nonsingular curves over $k$ inducing an inseparable function field extension of degree $p$ but the genus of $X$ is $1$ and the genus of $\mathbf{P}^1_k$ is $0$. \end{example} \begin{proposition} \label{proposition-unwind-morphism-smooth} Let $k$ be a field of characteristic $p > 0$. Let $f : X \to Y$ be a nonconstant morphism of proper smooth curves over $k$. Then we can factor $f$ as $$X \longrightarrow X^{(p^n)} \longrightarrow Y$$ where $X^{(p^n)} \to Y$ is a nonconstant morphism of proper smooth curves inducing a separable field extension $k(X^{(p^n)})/k(Y)$, we have $$X^{(p^n)} = X \times_{\Spec(k), F_{\Spec(k)}^n} \Spec(k),$$ and $X \to X^{(p^n)}$ is the $n$-fold relative frobenius of $X$. \end{proposition} \begin{proof} By Fields, Lemma \ref{fields-lemma-separable-first} there is a subextension $k(X)/E/k(Y)$ such that $k(X)/E$ is purely inseparable and $E/k(Y)$ is separable. By Theorem \ref{theorem-curves-rational-maps} this corresponds to a factorization $X \to Z \to Y$ of $f$ with $Z$ a nonsingular proper curve. Apply Lemma \ref{lemma-purely-inseparable-smooth} to the morphism $X \to Z$ to conclude. \end{proof} \begin{lemma} \label{lemma-inseparable-linear-system} Let $k$ be a field of characteristic $p > 0$. Let $X$ be a smooth proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^r_d$ with $r \geq 1$. Then one of the following two is true \begin{enumerate} \item there exists a $\mathfrak g^1_d$ whose corresponding morphism $X \to \mathbf{P}^1_k$ (Lemma \ref{lemma-linear-series}) is generically \'etale (i.e., is as in Lemma \ref{lemma-generically-etale}), or \item there exists a $\mathfrak g^r_{d'}$ on $X^{(p)}$ where $d' \leq d/p$. \end{enumerate} \end{lemma} \begin{proof} Pick two $k$-linearly independent elements $s, t \in V$. Then $f = s/t$ is the rational function defining the morphism $X \to \mathbf{P}^1_k$ corresponding to the linear series $(\mathcal{L}, ks + kt)$. If this morphism is not generically \'etale, then $f \in k(X^{(p)})$ by Proposition \ref{proposition-unwind-morphism-smooth}. Now choose a basis $s_0, \ldots, s_r$ of $V$ and let $\mathcal{L}' \subset \mathcal{L}$ be the invertible sheaf generated by $s_0, \ldots, s_r$. Set $f_i = s_i/s_0$ in $k(X)$. If for each pair $(s_0, s_i)$ we have $f_i \in k(X^{(p)})$, then the morphism $$\varphi = \varphi_{(\mathcal{L}', (s_0, \ldots, s_r)} : X \longrightarrow \mathbf{P}^1_k = \text{Proj}(k[T_0, \ldots, T_r])$$ factors through $X^{(p)}$ as this is true over the affine open $D_+(T_0)$ and we can extend the morphism over the affine part to the whole of the smooth curve $X^{(p)}$ by Lemma \ref{lemma-extend-over-normal-curve}. Introducing notation, say we have the factorization $$X \xrightarrow{F_{X/k}} X^{(p)} \xrightarrow{\psi} \mathbf{P}^r_k$$ of $\varphi$. Then $\mathcal{N} = \psi^*\mathcal{O}_{\mathbf{P}^1_k}(1)$ is an invertible $\mathcal{O}_{X^{(p)}}$-module with $\mathcal{L}' = F_{X/k}^*\mathcal{N}$ and with $\psi^*T_0, \ldots, \psi^*T_r$ $k$-linearly independent (as they pullback to $s_0, \ldots, s_r$ on $X$). Finally, we have $$d = \deg(\mathcal{L}) \geq \deg(\mathcal{L}') = \deg(F_{X/k}) \deg(\mathcal{N}) = p \deg(\mathcal{N})$$ as desired. Here we used Varieties, Lemmas \ref{varieties-lemma-check-invertible-sheaf-trivial}, \ref{varieties-lemma-degree-pullback-map-proper-curves}, and \ref{varieties-lemma-inseparable-deg-p-smooth}. \end{proof} \begin{lemma} \label{lemma-point-over-separable-extension} Let $k$ be a field. Let $X$ be a smooth proper curve over $k$ with $H^0(X, \mathcal{O}_X) = k$ and genus $g \geq 2$. Then there exists a closed point $x \in X$ with $\kappa(x)/k$ separable of degree $\leq 2g - 2$. \end{lemma} \begin{proof} Set $\omega = \Omega_{X/k}$. By Lemma \ref{lemma-genus-smooth} this has degree $2g - 2$ and has $g$ global sections. Thus we have a $\mathfrak g^{g - 1}_{2g - 2}$. By the trivial Lemma \ref{lemma-linear-series-trivial-existence} there exists a $\mathfrak g^1_{2g - 2}$ and by Lemma \ref{lemma-g1d} we obtain a morphism $$\varphi : X \longrightarrow \mathbf{P}^1_k$$ of some degree $d \leq 2g - 2$. Since $\varphi$ is flat (Lemma \ref{lemma-flat}) and finite (Lemma \ref{lemma-finite}) it is finite locally free of degree $d$ (Morphisms, Lemma \ref{morphisms-lemma-finite-flat}). Pick any rational point $t \in \mathbf{P}^1_k$ and any point $x \in X$ with $\varphi(x) = t$. Then $$d \geq [\kappa(x) : \kappa(t)] = [\kappa(x) : k]$$ for example by Morphisms, Lemmas \ref{morphisms-lemma-finite-locally-free-universally-bounded} and \ref{morphisms-lemma-characterize-universally-bounded}. Thus if $k$ is perfect (for example has characteristic zero or is finite) then the lemma is proved. Thus we reduce to the case discussed in the next paragraph. \medskip\noindent Assume that $k$ is an infinite field of characteristic $p > 0$. As above we will use that $X$ has a $\mathfrak g^{g - 1}_{2g - 2}$. The smooth proper curve $X^{(p)}$ has the same genus as $X$. Hence its genus is $> 0$. We conclude that $X^{(p)}$ does not have a $\mathfrak g^{g - 1}_d$ for any $d \leq g - 1$ by Lemma \ref{lemma-grd-inequalities}. Applying Lemma \ref{lemma-inseparable-linear-system} to our $\mathfrak g^{g - 1}_{2g - 2}$ (and noting that $2g - 2/p \leq g - 1$) we conclude that possibility (2) does not occur. Hence we obtain a morphism $$\varphi : X \longrightarrow \mathbf{P}^1_k$$ which is generically \'etale (in the sense of the lemma) and has degree $\leq 2g - 2$. Let $U \subset X$ be the nonempty open subscheme where $\varphi$ is \'etale. Then $\varphi(U) \subset \mathbf{P}^1_k$ is a nonempty Zariski open and we can pick a $k$-rational point $t \in \varphi(U)$ as $k$ is infinite. Let $u \in U$ be a point with $\varphi(u) = t$. Then $\kappa(u)/\kappa(t)$ is separable (Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}), $\kappa(t) = k$, and $[\kappa(u) : k] \leq 2g - 2$ as before. \end{proof} \noindent The following lemma does not really belong in this section but we don't know a good place for it elsewhere. \begin{lemma} \label{lemma-ramification-to-algebraic-closure} Let $X$ be a smooth curve over a field $k$. Let $\overline{x} \in X_{\overline{k}}$ be a closed point with image $x \in X$. The ramification index of $\mathcal{O}_{X, x} \subset \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is the inseparable degree of $\kappa(x)/k$. \end{lemma} \begin{proof} After shrinking $X$ we may assume there is an \'etale morphism $\pi : X \to \mathbf{A}^1_k$, see Morphisms, Lemma \ref{morphisms-lemma-smooth-etale-over-affine-space}. Then we can consider the diagram of local rings $$\xymatrix{ \mathcal{O}_{X_{\overline{k}}, \overline{x}} & \mathcal{O}_{\mathbf{A}^1_{\overline{k}}, \pi(\overline{x})} \ar[l] \\ \mathcal{O}_{X, x} \ar[u] & \mathcal{O}_{\mathbf{A}^1_k, \pi(x)} \ar[l] \ar[u] }$$ The horizontal arrows have ramification index $1$ as they correspond to \'etale morphisms. Moreover, the extension $\kappa(x)/\kappa(\pi(x))$ is separable hence $\kappa(x)$ and $\kappa(\pi(x))$ have the same inseparable degree over $k$. By multiplicativity of ramification indices it suffices to prove the result when $x$ is a point of the affine line. \medskip\noindent Assume $X = \mathbf{A}^1_k$. In this case, the local ring of $X$ at $x$ looks like $$\mathcal{O}_{X, x} = k[t]_{(P)}$$ where $P$ is an irreducible monic polynomial over $k$. Then $P(t) = Q(t^q)$ for some separable polynomial $Q \in k[t]$, see Fields, Lemma \ref{fields-lemma-irreducible-polynomials}. Observe that $\kappa(x) = k[t]/(P)$ has inseparable degree $q$ over $k$. On the other hand, over $\overline{k}$ we can factor $Q(t) = \prod (t - \alpha_i)$ with $\alpha_i$ pairwise distinct. Write $\alpha_i = \beta_i^q$ for some unique $\beta_i \in \overline{k}$. Then our point $\overline{x}$ corresponds to one of the $\beta_i$ and we conclude because the ramification index of $$k[t]_{(P)} \longrightarrow \overline{k}[t]_{(t - \beta_i)}$$ is indeed equal to $q$ as the uniformizer $P$ maps to $(t - \beta_i)^q$ times a unit. \end{proof} \section{Glueing and squishing} \label{section-glueing-squishing} \noindent Below we will indicate $k[\epsilon]$ the algebra of dual numbers over $k$ as defined in Varieties, Definition \ref{varieties-definition-dual-numbers}. \begin{lemma} \label{lemma-no-in-between-over-k} Let $k$ be an algebraically closed field. Let $k \subset A$ be a ring extension such that $A$ has exactly two $k$-sub algebras, then either $A = k \times k$ or $A = k[\epsilon]$. \end{lemma} \begin{proof} The assumption means $k \not = A$ and any subring $k \subset C \subset A$ is equal to either $k$ or $A$. Let $t \in A$, $t \not \in k$. Then $A$ is generated by $t$ over $k$. Hence $A = k[x]/I$ for some ideal $I$. If $I = (0)$, then we have the subalgebra $k[x^2]$ which is not allowed. Otherwise $I$ is generated by a monic polynomial $P$. Write $P = \prod_{i = 1}^d (t - a_i)$. If $d > 2$, then the subalgebra generated by $(t - a_1)(t - a_2)$ gives a contradiction. Thus $d = 2$. If $a_1 \not = a_2$, then $A = k \times k$, if $a_1 = a_2$, then $A = k[\epsilon]$. \end{proof} \begin{example}[Glueing points] \label{example-glue-points} Let $k$ be an algebraically closed field. Let $f : X' \to X$ be a morphism of algebraic $k$-schemes. We say $X$ is obtained by glueing $a$ and $b$ in $X'$ if the following are true: \begin{enumerate} \item $a, b \in X'(k)$ are distinct points which map to the same point $x \in X(k)$, \item $f$ is finite and $f^{-1}(X \setminus \{x\}) \to X \setminus \{x\}$ is an isomorphism, \item there is a short exact sequence $$0 \to \mathcal{O}_X \to f_*\mathcal{O}_{X'} \xrightarrow{a - b} x_*k \to 0$$ where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}$ to the difference $h(a) - h(b) \in k$. \end{enumerate} If this is the case, then there also is a short exact sequence $$0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{ab^{-1}} x_*k^* \to 0$$ where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}^*$ to the multiplicative difference $h(a)h(b)^{-1} \in k^*$. \end{example} \begin{example}[Squishing a tangent vector] \label{example-squish-tangent-vector} Let $k$ be an algebraically closed field. Let $f : X' \to X$ be a morphism of algebraic $k$-schemes. We say $X$ is obtained by squishing the tangent vector $\vartheta$ in $X'$ if the following are true: \begin{enumerate} \item $\vartheta : \Spec(k[\epsilon]) \to X'$ is a closed immersion over $k$ such that $f \circ \vartheta$ factors through a point $x \in X(k)$, \item $f$ is finite and $f^{-1}(X \setminus \{x\}) \to X \setminus \{x\}$ is an isomorphism, \item there is a short exact sequence $$0 \to \mathcal{O}_X \to f_*\mathcal{O}_{X'} \xrightarrow{\vartheta} x_*k \to 0$$ where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}$ to the coefficient of $\epsilon$ in $\vartheta^\sharp(h) \in k[\epsilon]$. \end{enumerate} If this is the case, then there also is a short exact sequence $$0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{\vartheta} x_*k \to 0$$ where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}^*$ to $\text{d}\log(\vartheta^\sharp(h))$ where $\text{d}\log : k[\epsilon]^* \to k$ is the homomorphism of abelian groups sending $a + b\epsilon$ to $b/a \in k$. \end{example} \begin{lemma} \label{lemma-factor-almost-isomorphism} Let $k$ be an algebraically closed field. Let $f : X' \to X$ be a finite morphism algebraic $k$-schemes such that $\mathcal{O}_X \subset f_*\mathcal{O}_{X'}$ and such that $f$ is an isomorphism away from a finite set of points. Then there is a factorization $$X' = X_n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = X$$ such that each $X_i \to X_{i - 1}$ is either the glueing of two points or the squishing of a tangent vector (see Examples \ref{example-glue-points} and \ref{example-squish-tangent-vector}). \end{lemma} \begin{proof} Let $U \subset X$ be the maximal open set over which $f$ is an isomorphism. Then $X \setminus U = \{x_1, \ldots, x_n\}$ with $x_i \in X(k)$. We will consider factorizations $X' \to Y \to X$ of $f$ such that both morphisms are finite and $$\mathcal{O}_X \subset g_*\mathcal{O}_Y \subset f_*\mathcal{O}_{X'}$$ where $g : Y \to X$ is the given morphism. By assumption $\mathcal{O}_{X, x} \to (f_*\mathcal{O}_{X'})_x$ is an isomorphism onless $x = x_i$ for some $i$. Hence the cokernel $$f_*\mathcal{O}_{X'}/\mathcal{O}_X = \bigoplus \mathcal{Q}_i$$ is a direct sum of skyscraper sheaves $\mathcal{Q}_i$ supported at $x_1, \ldots, x_n$. Because the displayed quotient is a coherent $\mathcal{O}_X$-module, we conclude that $\mathcal{Q}_i$ has finite length over $\mathcal{O}_{X, x_i}$. Hence we can argue by induction on the sum of these lengths, i.e., the length of the whole cokernel. \medskip\noindent If $n > 1$, then we can define an $\mathcal{O}_X$-subalgebra $\mathcal{A} \subset f_*\mathcal{O}_{X'}$ by taking the inverse image of $\mathcal{Q}_1$. This will give a nontrivial factorization and we win by induction. \medskip\noindent Assume $n = 1$. We abbreviate $x = x_1$. Consider the finite $k$-algebra extension $$A = \mathcal{O}_{X, x} \subset (f_*\mathcal{O}_{X'})_x = B$$ Note that $\mathcal{Q} = \mathcal{Q}_1$ is the skyscraper sheaf with value $B/A$. We have a $k$-subalgebra $A \subset A + \mathfrak m_A B \subset B$. If both inclusions are strict, then we obtain a nontrivial factorization and we win by induction as above. If $A + \mathfrak m_A B = B$, then $A = B$ by Nakayama, then $f$ is an isomorphism and there is nothing to prove. We conclude that we may assume $B = A + \mathfrak m_A B$. Set $C = B/\mathfrak m_A B$. If $C$ has more than $2$ $k$-subalgebras, then we obtain a subalgebra between $A$ and $B$ by taking the inverse image in $B$. Thus we may assume $C$ has exactly $2$ $k$-subalgebras. Thus $C = k \times k$ or $C = k[\epsilon]$ by Lemma \ref{lemma-no-in-between-over-k}. In this case $f$ is correspondingly the glueing two points or the squishing of a tangent vector. \end{proof} \begin{lemma} \label{lemma-glue-points} Let $k$ be an algebraically closed field. If $f : X \to X'$ is the glueing of two points $a, b$ as in Example \ref{example-glue-points}, then there is an exact sequence $$k^* \to \text{Pic}(X) \to \text{Pic}(X') \to 0$$ The first map is zero if $a$ and $b$ are on different connected components of $X'$ and injective if $X'$ is proper and $a$ and $b$ are on the same connected component of $X'$. \end{lemma} \begin{proof} The map $\text{Pic}(X) \to \text{Pic}(X')$ is surjective by Varieties, Lemma \ref{varieties-lemma-surjective-pic-birational-finite}. Using the short exact sequence $$0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{ab^{-1}} x_*k^* \to 0$$ we obtain $$H^0(X', \mathcal{O}_{X'}^*) \xrightarrow{ab^{-1}} k^* \to H^1(X, \mathcal{O}_X^*) \to H^1(X, f_*\mathcal{O}_{X'}^*)$$ We have $H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)$ (for example by the Leray spectral sequence, see Cohomology, Lemma \ref{cohomology-lemma-Leray}). Hence the kernel of $\text{Pic}(X) \to \text{Pic}(X')$ is the cokernel of $ab^{-1} : H^0(X', \mathcal{O}_{X'}^*) \to k^*$. If $a$ and $b$ are on different connected components of $X'$, then $ab^{-1}$ is surjective (also for example if $X'$ is affine). Because $k$ is algebraically closed any regular function on a reduced connected proper scheme over $k$ comes from an element of $k$, see Varieties, Lemma \ref{varieties-lemma-proper-geometrically-reduced-global-sections}. Thus $ab^{-1}$ is zero if $X'$ is proper and $a$ and $b$ are on the same connected component. \end{proof} \begin{lemma} \label{lemma-squish-tangent-vector} Let $k$ be an algebraically closed field. If $f : X \to X'$ is the squishing of a tangent vector $\vartheta$ as in Example \ref{example-squish-tangent-vector}, then there is an exact sequence $$(k, +) \to \text{Pic}(X) \to \text{Pic}(X') \to 0$$ and the first map is injective if $X'$ is proper and reduced. \end{lemma} \begin{proof} The map $\text{Pic}(X) \to \text{Pic}(X')$ is surjective by Varieties, Lemma \ref{varieties-lemma-surjective-pic-birational-finite}. Using the short exact sequence $$0 \to \mathcal{O}_X^* \to f_*\mathcal{O}_{X'}^* \xrightarrow{\vartheta} x_*k \to 0$$ of Example \ref{example-squish-tangent-vector} we obtain $$H^0(X', \mathcal{O}_{X'}^*) \xrightarrow{\vartheta} k \to H^1(X, \mathcal{O}_X^*) \to H^1(X, f_*\mathcal{O}_{X'}^*)$$ We have $H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)$ (for example by the Leray spectral sequence, see Cohomology, Lemma \ref{cohomology-lemma-Leray}). Hence the kernel of $\text{Pic}(X) \to \text{Pic}(X')$ is the cokernel of the map $\vartheta : H^0(X', \mathcal{O}_{X'}^*) \to k$. Because $k$ is algebraically closed any regular function on a reduced connected proper scheme over $k$ comes from an element of $k$, see Varieties, Lemma \ref{varieties-lemma-proper-geometrically-reduced-global-sections}. Thus the final statement of the lemma. \end{proof} \section{Multicross and nodal singularities} \label{section-multicross} \noindent In this section we discuss the simplest possible curve singularities. \medskip\noindent Let $k$ be a field. Consider the complete local $k$-algebra \begin{equation} \label{equation-multicross} A = \{(f_1, \ldots, f_n) \in k[[t]] \times \ldots \times k[[t]] \mid f_1(0) = \ldots = f_n(0)\} \end{equation} In the language introduced in Varieties, Definition \ref{varieties-definition-wedge} we see that $A$ is a wedge of $n$ copies of the power series ring in $1$ variable over $k$. Observe that $k[[t]] \times \ldots \times k[[t]]$ is the integral closure of $A$ in its total ring of fractions. Hence the $\delta$-invariant of $A$ is $n - 1$. There is an isomorphism $$k[[x_1, \ldots, x_n]]/(\{x_ix_j\}_{i \not = j}) \longrightarrow A$$ obtained by sending $x_i$ to $(0, \ldots, 0, t, 0, \ldots, 0)$ in $A$. It follows that $\dim(A) = 1$ and $\dim_k \mathfrak m/\mathfrak m^2 = n$. In particular, $A$ is regular if and only if $n = 1$. \begin{lemma} \label{lemma-multicross-algebra} Let $k$ be a separably closed field. Let $A$ be a $1$-dimensional reduced Nagata local $k$-algebra with residue field $k$. Then $$\delta\text{-invariant }A \geq \text{number of branches of }A - 1$$ If equality holds, then $A^\wedge$ is as in (\ref{equation-multicross}). \end{lemma} \begin{proof} Since the residue field of $A$ is separably closed, the number of branches of $A$ is equal to the number of geometric branches of $A$, see More on Algebra, Definition \ref{more-algebra-definition-number-of-branches}. The inequality holds by Varieties, Lemma \ref{varieties-lemma-delta-number-branches-inequality}. Assume equality holds. We may replace $A$ by the completion of $A$; this does not change the number of branches or the $\delta$-invariant, see More on Algebra, Lemma \ref{more-algebra-lemma-one-dimensional-number-of-branches} and Varieties, Lemma \ref{varieties-lemma-delta-same-after-completion}. Then $A$ is strictly henselian, see Algebra, Lemma \ref{algebra-lemma-complete-henselian}. By Varieties, Lemma \ref{varieties-lemma-delta-number-branches-inequality-sh} we see that $A$ is a wedge of complete discrete valuation rings. Each of these is isomorphic to $k[[t]]$ by Algebra, Lemma \ref{algebra-lemma-regular-complete-containing-coefficient-field}. Hence $A$ is as in (\ref{equation-multicross}). \end{proof} \begin{definition} \label{definition-multicross} Let $k$ be an algebraically closed field. Let $X$ be an algebraic $1$-dimensional $k$-scheme. Let $x \in X$ be a closed point. We say $x$ defines a {\it multicross singularity} if the completion $\mathcal{O}_{X, x}^\wedge$ is isomorphic to (\ref{equation-multicross}) for some $n \geq 2$. We say $x$ is a {\it node}, or an {\it ordinary double point}, or {\it defines a nodal singularity} if $n = 2$. \end{definition} \noindent These singularities are in some sense the simplest kind of singularities one can have on a curve over an algebraically closed field. \begin{lemma} \label{lemma-multicross} Let $k$ be an algebraically closed field. Let $X$ be a reduced algebraic $1$-dimensional $k$-scheme. Let $x \in X$. The following are equivalent \begin{enumerate} \item $x$ defines a multicross singularity, \item the $\delta$-invariant of $X$ at $x$ is the number of branches of $X$ at $x$ minus $1$, \item there is a sequence of morphisms $U_n \to U_{n - 1} \to \ldots \to U_0 = U \subset X$ where $U$ is an open neighbourhood of $x$, where $U_n$ is nonsingular, and where each $U_i \to U_{i - 1}$ is the glueing of two points as in Example \ref{example-glue-points}. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (2) is Lemma \ref{lemma-multicross-algebra}. \medskip\noindent Assume (3). We will argue by descending induction on $i$ that all singularities of $U_i$ are multicross. This is true for $U_n$ as $U_n$ has no singular points. If $U_i$ is gotten from $U_{i + 1}$ by glueing $a, b \in U_{i + 1}$ to a point $c \in U_i$, then we see that $$\mathcal{O}_{U_i, c}^\wedge \subset \mathcal{O}_{U_{i + 1}, a}^\wedge \times \mathcal{O}_{U_{i + 1}, b}^\wedge$$ is the set of elements having the same residue classes in $k$. Thus the number of branches at $c$ is the sum of the number of branches at $a$ and $b$, and the $\delta$-invariant at $c$ is the sum of the $\delta$-invariants at $a$ and $b$ plus $1$ (because the displayed inclusion has codimension $1$). This proves that (2) holds as desired. \medskip\noindent Assume the equivalent conditions (1) and (2). We may choose an open $U \subset X$ such that $x$ is the only singular point of $U$. Then we apply Lemma \ref{lemma-factor-almost-isomorphism} to the normalization morphism $$U^\nu = U_n \to U_{n - 1} \to \ldots \to U_1 \to U_0 = U$$ All we have to do is show that in none of the steps we are squishing a tangent vector. Suppose $U_{i + 1} \to U_i$ is the smallest $i$ such that this is the squishing of a tangent vector $\theta$ at $u' \in U_{i + 1}$ lying over $u \in U_i$. Arguing as above, we see that $u_i$ is a multicross singularity (because the maps $U_i \to \ldots \to U_0$ are glueing of pairs of points). But now the number of branches at $u'$ and $u$ is the same and the $\delta$-invariant of $U_i$ at $u$ is $1$ bigger than the $\delta$-invariant of $U_{i + 1}$ at $u'$. By Lemma \ref{lemma-multicross-algebra} this implies that $u$ cannot be a multicross singularity which is a contradiction. \end{proof} \begin{lemma} \label{lemma-multicross-gorenstein-is-nodal} Let $k$ be an algebraically closed field. Let $X$ be a reduced algebraic $1$-dimensional $k$-scheme. Let $x \in X$ be a multicross singularity (Definition \ref{definition-multicross}). If $X$ is Gorenstein, then $x$ is a node. \end{lemma} \begin{proof} The map $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^\wedge$ is flat and unramified in the sense that $\kappa(x) = \mathcal{O}_{X, x}^\wedge/\mathfrak m_x \mathcal{O}_{X, x}^\wedge$. (See More on Algebra, Section \ref{more-algebra-section-permanence-completion}.) Thus $X$ is Gorenstein implies $\mathcal{O}_{X, x}$ is Gorenstein, implies $\mathcal{O}_{X, x}^\wedge$ is Gorenstein by Dualizing Complexes, Lemma \ref{dualizing-lemma-flat-under-gorenstein}. Thus it suffices to show that the ring $A$ in (\ref{equation-multicross}) with $n \geq 2$ is Gorenstein if and only if $n = 2$. \medskip\noindent If $n = 2$, then $A = k[[x, y]]/(xy)$ is a complete intersection and hence Gorenstein. For example this follows from Dualizing Complexes, Lemma \ref{dualizing-lemma-gorenstein-lci} applied to $k[[x, y]] \to A$ and the fact that the regular local ring $k[[x, y]]$ is Gorenstein by Dualizing Complexes, Lemma \ref{dualizing-lemma-regular-gorenstein}. \medskip\noindent Assume $n > 2$. If $A$ where Gorenstein, then $A$ would be a dualizing complex over $A$ (Dualizing Complexes, Definition \ref{dualizing-definition-gorenstein}). Then $R\Hom(k, A)$ would be equal to $k[n]$ for some $n \in \mathbf{Z}$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-find-function}. It would follow that $\text{Ext}^1_A(k, A) \cong k$ or $\text{Ext}^1_A(k, A) = 0$ (depending on the value of $n$; in fact $n$ has to be $-1$ but it doesn't matter to us here). Using the exact sequence $$0 \to \mathfrak m_A \to A \to k \to 0$$ we find that $$\text{Ext}^1_A(k, A) = \Hom_A(\mathfrak m_A, A)/A$$ where $A \to \Hom_A(\mathfrak m_A, A)$ is given by $a \mapsto (a' \mapsto aa')$. Let $e_i \in \Hom_A(\mathfrak m_A, A)$ be the element that sends $(f_1, \ldots, f_n) \in \mathfrak m_A$ to $(0, \ldots, 0, f_i, 0, \ldots, 0)$. The reader verifies easily that $e_1, \ldots, e_{n - 1}$ are $k$-linearly independent in $\Hom_A(\mathfrak m_A, A)/A$. Thus $\dim_k \text{Ext}^1_A(k, A) \geq n - 1 \geq 2$ which finishes the proof. (Observe that $e_1 + \ldots + e_n$ is the image of $1$ under the map $A \to \Hom_A(\mathfrak m_A, A)$.) \end{proof} \section{Torsion in the Picard group} \label{section-torsion-in-pic} \noindent In this section we bound the torsion in the Picard group of a $1$-dimensional proper scheme over a field. We will use this in our study of semistable reduction for curves. \medskip\noindent There does not seem to be an elementary way to obtain the result of Lemma \ref{lemma-torsion-picard-smooth-projective}. Analyzing the proof there are two key ingredients: (1) there is an abelian variety classifying degree zero invertible sheaves on a smooth projective curve and (2) the structure of torsion points on an abelian variety can be determined. \begin{lemma} \label{lemma-torsion-picard-smooth-projective} Let $k$ be an algebraically closed field. Let $X$ be a smooth projective curve of genus $g$ over $k$. \begin{enumerate} \item If $n \geq 1$ is invertible in $k$, then $\text{Pic}(X)[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}$. \item If the characteristic of $k$ is $p > 0$, then there exists an integer $0 \leq f \leq g$ such that $\text{Pic}(X)[p^m] \cong (\mathbf{Z}/p^m\mathbf{Z})^{\oplus f}$ for all $m \geq 1$. \end{enumerate} \end{lemma} \begin{proof} Let $\text{Pic}^0(X) \subset \text{Pic}(X)$ denote the subgroup of invertible sheaves of degree $0$. In other words, there is a short exact sequence $$0 \to \text{Pic}^0(X) \to \text{Pic}(X) \xrightarrow{\deg} \mathbf{Z} \to 0.$$ The group $\text{Pic}^0(X)$ is the $k$-points of the group scheme $\underline{\text{Pic}}^0_{X/k}$, see Picard Schemes of Curves, Lemma \ref{pic-lemma-picard-pieces}. The same lemma tells us that $\underline{\text{Pic}}^0_{X/k}$ is a $g$-dimensional abelian variety over $k$ as defined in Groupoids, Definition \ref{groupoids-definition-abelian-variety}. Thus we conclude by the results of Groupoids, Proposition \ref{groupoids-proposition-review-abelian-varieties}. \end{proof} \begin{lemma} \label{lemma-torsion-picard-becomes-visible} Let $k$ be a field. Let $n$ be prime to the characteristic of $k$. Let $X$ be a smooth proper curve over $k$ with $H^0(X, \mathcal{O}_X) = k$ and of genus $g$. \begin{enumerate} \item If $g = 1$ then there exists a finite separable extension $k'/k$ such that $X_{k'}$ has a $k'$-rational point and $\text{Pic}(X_{k'})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2}$. \item If $g \geq 2$ then there exists a finite separable extension $k'/k$ with $[k' : k] \leq (2g - 2)(n^{2g})!$ such that $X_{k'}$ has a $k'$-rational point and $\text{Pic}(X_{k'})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}$. \end{enumerate} \end{lemma} \begin{proof} Assume $g \geq 2$. First we may choose a finite separable extension of degree at most $2g - 2$ such that $X$ acquires a rational point. Thus we may assume $X$ has a $k$-rational point $x \in X(k)$ but now we have to prove the lemma with $[k' : k] \leq \leq (n^{2g})!$. Let $k \subset k^{sep} \subset \overline{k}$ be a separable algebraic closure inside an algebraic closure. By Lemma \ref{lemma-torsion-picard-smooth-projective} we have $$\text{Pic}(X_{\overline{k}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}$$ By Picard Schemes of Curves, Lemma \ref{pic-lemma-torsion-descends} we conclude that $$\text{Pic}(X_{k^{sep}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}$$ By Picard Schemes of Curves, Lemma \ref{pic-lemma-torsion-descends} there is a continuous action $$\text{Gal}(k^{sep}/k) \longrightarrow \text{Aut}(\text{Pic}(X_{k^{sep}})[n]$$ and the lemma is true for the fixed field $k'$ of the kernel of this map. The kernel is open because the action is continuous which implies that $k'/k$ is finite. By Galois theory $\text{Gal}(k'/k)$ is the image of the displayed arrow. Since the permutation group of a set of cardinality $n^{2g}$ has cardinality $(n^{2g})!$ we conclude by Galois theory that $[k' : k] \leq (n^{2g})!$. (Of course this proves the lemma with the bound $|\text{GL}_{2g}(\mathbf{Z}/n\mathbf{Z})|$, but all we want here is that there is some bound.) \medskip\noindent If the genus is $1$, then there is no upper bound on the degree of a finite separable field extension over which $X$ acquires a rational point (details omitted). Still, there is such an extension for example by Varieties, Lemma \ref{varieties-lemma-smooth-separable-closed-points-dense}. The rest of the proof is the same as in the case of $g \geq 2$. \end{proof} \begin{proposition} \label{proposition-torsion-picard-reduced-proper} Let $k$ be an algebraically closed field. Let $X$ be a proper scheme over $k$ which is reduced, connected, and has dimension $1$. Let $g$ be the genus of $X$ and let $g_{geom}$ be the sum of the geometric genera of the irreducible components of $X$. For any prime $\ell$ different from the characteristic of $k$ we have $$\dim_{\mathbf{F}_\ell} \text{Pic}(X)[\ell] \leq g + g_{geom}$$ and equality holds if and only if all the singularities of $X$ are multicross. \end{proposition} \begin{proof} Let $\nu : X^\nu \to X$ be the normalization (Varieties, Lemma \ref{varieties-lemma-prepare-delta-invariant}). Choose a factorization $$X^\nu = X_n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = X$$ as in Lemma \ref{lemma-factor-almost-isomorphism}. Let us denote $h^0_i = \dim_k H^0(X_i, \mathcal{O}_{X_i})$ and $h^1_i = \dim_k H^1(X_i, \mathcal{O}_{X_i})$. By Lemmas \ref{lemma-glue-points} and \ref{lemma-squish-tangent-vector} for each $n > i \geq 0$ we have one of the following there possibilities \begin{enumerate} \item $X_i$ is obtained by glueing $a, b \in X_{i + 1}$ which are on different connected components: in this case $\text{Pic}(X_i) = \text{Pic}(X_{i + 1})$, $h^0_{i + 1} = h^0_i + 1$, $h^1_{i + 1} = h^1_i$, \item $X_i$ is obtained by glueing $a, b \in X_{i + 1}$ which are on the same connected component: in this case there is a short exact sequence $$0 \to k^* \to \text{Pic}(X_i) \to \text{Pic}(X_{i + 1}) \to 0,$$ and $h^0_{i + 1} = h^0_i$, $h^1_{i + 1} = h^1_i - 1$, \item $X_i$ is obtained by squishing a tangent vector in $X_{i + 1}$: in this case there is a short exact sequence $$0 \to (k, +) \to \text{Pic}(X_i) \to \text{Pic}(X_{i + 1}) \to 0,$$ and $h^0_{i + 1} = h^0_i$, $h^1_{i + 1} = h^1_i - 1$. \end{enumerate} To prove the statements on dimensions of cohomology groups of the structure sheaf, use the exact sequences in Examples \ref{example-glue-points} and \ref{example-squish-tangent-vector}. Since $k$ is algebraically closed of characteristic prime to $\ell$ we see that $(k, +)$ and $k^*$ are $\ell$-divisible and with $\ell$-torsion $(k, +)[\ell] = 0$ and $k^*[\ell] \cong \mathbf{F}_\ell$. Hence $$\dim_{\mathbf{F}_\ell} \text{Pic}(X_{i + 1})[\ell] - \dim_{\mathbf{F}_\ell}\text{Pic}(X_i)[\ell]$$ is zero, except in case (2) where it is equal to $-1$. At the end of this process we get the normalization $X^\nu = X_n$ which is a disjoint union of smooth projective curves over $k$. Hence we have \begin{enumerate} \item $h^1_n = g_{geom}$ and \item $\dim_{\mathbf{F}_\ell} \text{Pic}(X_n)[\ell] = 2g_{geom}$. \end{enumerate} The last equality by Lemma \ref{lemma-torsion-picard-smooth-projective}. Since $g = h^1_0$ we see that the number of steps of type (2) and (3) is at most $h^1_0 - h^1_n = g - g_{geom}$. By our comptation of the differences in ranks we conclude that $$\dim_{\mathbf{F}_\ell} \text{Pic}(X)[\ell] \leq g - g_{geom} + 2g_{geom} = g + g_{geom}$$ and equality holds if and only if no steps of type (3) occur. This indeed means that all singularities of $X$ are multicross by Lemma \ref{lemma-multicross}. Conversely, if all the singularities are multicross, then Lemma \ref{lemma-multicross} guarantees that we can find a sequence $X^\nu = X_n \to \ldots \to X_0 = X$ as above such that no steps of type (3) occur in the sequence and we find equality holds in the lemma (just glue the local sequences for each point to find one that works for all singular points of $x$; some details omitted). \end{proof} \section{Genus versus geometric genus} \label{section-genus-geometric-genus} \noindent Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. We define $g_{geom}(X/k)$ to be the sum of the geometric genera of the irreducible components of $X_{\overline{k}}$ which have dimension $1$. \begin{lemma} \label{lemma-bound-geometric-genus} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Then $$g_{geom}(X/k) = \sum\nolimits_{C \subset X} g_{geom}(C/k)$$ where the sum is over irreducible components $C \subset X$ of dimension $1$. \end{lemma} \begin{proof} This is immediate from the definition and the fact that an irreducible component $\overline{Z}$ of $X_{\overline{k}}$ maps onto an irreducible component $Z$ of $X$ (Varieties, Lemma \ref{varieties-lemma-image-irreducible}) of the same dimension (Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change} applied to the generic point of $\overline{Z}$). \end{proof} \begin{lemma} \label{lemma-geometric-genus-normalization} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Then \begin{enumerate} \item We have $g_{geom}(X/k) = g_{geom}(X_{red}/k)$. \item If $X' \to X$ is a birational proper morphism, then $g_{geom}(X'/k) = g_{geom}(X/k)$. \item If $X^\nu \to X$ is the normalization morphism, then $g_{geom}(X^\nu/k) = g_{geom}(X/k)$. \end{enumerate} \end{lemma} \begin{proof} Part (1) is immediate from Lemma \ref{lemma-bound-geometric-genus}. If $X' \to X$ is proper birational, then it is finite and an isomorphism over a dense open (see Varieties, Lemmas \ref{varieties-lemma-finite-in-codim-1} and \ref{varieties-lemma-modification-normal-iso-over-codimension-1}). Hence $X'_{\overline{k}} \to X_{\overline{k}}$ is an isomorphism over a dense open. Thus the irreducible components of $X'_{\overline{k}}$ and $X_{\overline{k}}$ are in bijective correspondence and the corresponding components have isomorphic function fields. In particular these components have isomorphic nonsingular projective models and hence have the same geometric genera. This proves (2). Part (3) follows from (1) and (2) and the fact that $X^\nu \to X_{red}$ is birational (Morphisms, Lemma \ref{morphisms-lemma-normalization-birational}). \end{proof} \begin{lemma} \label{lemma-genus-goes-down} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $f : Y \to X$ be a finite morphism such that there exists a dense open $U \subset X$ over which $f$ is a closed immersion. Then $$\dim_k H^1(X, \mathcal{O}_X) \geq \dim_k H^1(Y, \mathcal{O}_Y)$$ \end{lemma} \begin{proof} Consider the exact sequence $$0 \to \mathcal{G} \to \mathcal{O}_X \to f_*\mathcal{O}_Y \to \mathcal{F} \to 0$$ of coherent sheaves on $X$. By assumption $\mathcal{F}$ is supported in finitely many closed points and hence has vanishing higher cohomology (Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}). On the other hand, we have $H^2(X, \mathcal{G}) = 0$ by Cohomology, Proposition \ref{cohomology-proposition-vanishing-Noetherian}. It follows formally that the induced map $H^1(X, \mathcal{O}_X) \to H^1(X, f_*\mathcal{O}_Y)$ is surjective. Since $H^1(X, f_*\mathcal{O}_Y) = H^1(Y, \mathcal{O}_Y)$ (Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}) we conclude the lemma holds. \end{proof} \begin{lemma} \label{lemma-genus-normalization} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. If $X' \to X$ is a birational proper morphism, then $$\dim_k H^1(X, \mathcal{O}_X) \geq \dim_k H^1(X', \mathcal{O}_{X'})$$ If $X$ is reduced, $H^0(X, \mathcal{O}_X) \to H^0(X', \mathcal{O}_{X'})$ is surjective, and equality holds, then $X' = X$. \end{lemma} \begin{proof} If $f : X' \to X$ is proper birational, then it is finite and an isomorphism over a dense open (see Varieties, Lemmas \ref{varieties-lemma-finite-in-codim-1} and \ref{varieties-lemma-modification-normal-iso-over-codimension-1}). Thus the inequality by Lemma \ref{lemma-genus-goes-down}. Assume $X$ is reduced. Then $\mathcal{O}_X \to f_*\mathcal{O}_{X'}$ is injective and we obtain a short exact sequence $$0 \to \mathcal{O}_X \to f_*\mathcal{O}_{X'} \to \mathcal{F} \to 0$$ Under the assumptions given in the second statement, we conclude from the long exact cohomology sequence that $H^0(X, \mathcal{F}) = 0$. Then $\mathcal{F} = 0$ because $\mathcal{F}$ is generated by global sections (Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}). and $\mathcal{O}_X = f_*\mathcal{O}_{X'}$. Since $f$ is affine this implies $X = X'$. \end{proof} \begin{lemma} \label{lemma-bound-geometric-genus-curve} Let $k$ be a field. Let $C$ be a proper curve over $k$. Set $\kappa = H^0(C, \mathcal{O}_C)$. Then $$[\kappa : k]_s \dim_\kappa H^1(C, \mathcal{O}_C) \geq g_{geom}(C/k)$$ \end{lemma} \begin{proof} Varieties, Lemma \ref{varieties-lemma-regular-functions-proper-variety} implies $\kappa$ is a field and a finite extension of $k$. By Fields, Lemma \ref{fields-lemma-separable-degree} we have $[\kappa : k]_s = |\Mor_k(\kappa, \overline{k})|$ and hence $\Spec(\kappa \otimes_k \overline{k})$ has $[\kappa : k]_s$ points each with residue field $\overline{k}$. Thus $$C_{\overline{k}} = \bigcup\nolimits_{t \in \Spec(\kappa \otimes_k \overline{k})} C_t$$ (set theoretic union). Here $C_t = C \times_{\Spec(\kappa), t} \Spec(\overline{k})$ where we view $t$ as a $k$-algebra map $t : \kappa \to \overline{k}$. The conclusion is that $g_{geom}(C/k) = \sum_t g_{geom}(C_t/\overline{k})$ and the sum is over an index set of size $[\kappa : k]_s$. We have $$H^0(C_t, \mathcal{O}_{C_t}) = \overline{k} \quad\text{and}\quad \dim_{\overline{k}} H^1(C_t, \mathcal{O}_{C_t}) = \dim_\kappa H^1(C, \mathcal{O}_C)$$ by cohomology and base change (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}). Observe that the normalization $C_t^\nu$ is the disjoint union of the nonsingular projective models of the irreducible components of $C_t$ (Morphisms, Lemma \ref{morphisms-lemma-normalization-in-terms-of-components}). Hence $\dim_{\overline{k}} H^1(C_t^\nu, \mathcal{O}_{C_t^\nu})$ is equal to $g_{geom}(C_t/\overline{k})$. By Lemma \ref{lemma-genus-goes-down} we have $$\dim_{\overline{k}} H^1(C_t, \mathcal{O}_{C_t}) \geq \dim_{\overline{k}} H^1(C_t^\nu, \mathcal{O}_{C_t^\nu})$$ and this finishes the proof. \end{proof} \begin{lemma} \label{lemma-bound-torsion-simple} Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $\ell$ be a prime number invertible in $k$. Then $$\dim_{\mathbf{F}_\ell} \text{Pic}(X)[\ell] \leq \dim_k H^1(X, \mathcal{O}_X) + g_{geom}(X/k)$$ where $g_{geom}(X/k)$ is as defined above. \end{lemma} \begin{proof} The map $\text{Pic}(X) \to \text{Pic}(X_{\overline{k}})$ is injective by Varieties, Lemma \ref{varieties-lemma-change-fields-pic}. By Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology} $\dim_k H^1(X, \mathcal{O}_X)$ equals $\dim_{\overline{k}} H^1(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$. Hence we may assume $k$ is algebraically closed. \medskip\noindent Let $X_{red}$ be the reduction of $X$. Then the surjection $\mathcal{O}_X \to \mathcal{O}_{X_{red}}$ induces a surjection $H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_{X_{red}})$ because cohomology of quasi-coherent sheaves vanishes in degrees $\geq 2$ by Cohomology, Proposition \ref{cohomology-proposition-vanishing-Noetherian}. Since $X_{red} \to X$ induces an isomorphism on irreducible components over $\overline{k}$ and an isomorphism on $\ell$-torsion in Picard groups (Picard Schemes of Curves, Lemma \ref{pic-lemma-torsion-descends}) we may replace $X$ by $X_{red}$. In this way we reduce to Proposition \ref{proposition-torsion-picard-reduced-proper}. \end{proof} \section{Nodal curves} \label{section-nodal} \noindent We have already defined ordinary double points over algebraically closed fields, see Definition \ref{definition-multicross}. Namely, if $x \in X$ is a closed point of a $1$-dimensional algebraic scheme over an algebraically closed field $k$, then $x$ is an ordinary double point if and only if $$\mathcal{O}_{X, x}^\wedge \cong k[[x, y]]/(xy)$$ See discussion following (\ref{equation-multicross}) in Section \ref{section-multicross}. \begin{definition} \label{definition-nodal} Let $k$ be a field. Let $X$ be a $1$-dimensional locally algebraic $k$-scheme. \begin{enumerate} \item We say a closed point $x \in X$ is a {\it node}, or an {\it ordinary double point}, or {\it defines a nodal singularity} if there exists an ordinary double point $\overline{x} \in X_{\overline{k}}$ mapping to $x$. \item We say the {\it singularities of $X$ are at-worst-nodal} if all closed points of $X$ are either in the smooth locus of the structure morphism $X \to \Spec(k)$ or are ordinary double points. \end{enumerate} \end{definition} \noindent Often a $1$-dimensional algebraic scheme $X$ is called a {\it nodal curve} if the singularities of $X$ are at worst nodal. Sometimes a nodal curve is required to be proper. Since a nodal curve so defined need not be irreducible, this conflicts with our earlier definition of a curve as a variety of dimension $1$. \begin{lemma} \label{lemma-reduced-quotient-regular-ring-dim-2} Let $(A, \mathfrak m)$ be a regular local ring of dimension $2$. Let $I \subset \mathfrak m$ be an ideal. \begin{enumerate} \item If $A/I$ is reduced, then $I = (0)$, $I = \mathfrak m$, or $I = (f)$ for some nonzero $f \in \mathfrak m$. \item If $A/I$ has depth $1$, then $I = (f)$ for some nonzero $f \in \mathfrak m$. \end{enumerate} \end{lemma} \begin{proof} Assume $I \not = 0$. Write $I = (f_1, \ldots, f_r)$. As $A$ is a UFD (More on Algebra, Lemma \ref{more-algebra-lemma-regular-local-UFD}) we can write $f_i = fg_i$ where $f$ is the gcd of $f_1, \ldots, f_r$. Thus the gcd of $g_1, \ldots, g_r$ is $1$ which means that there is no height $1$ prime ideal over $g_1, \ldots, g_r$. Since $\dim(A) = 2$ this implies that $V(g_1, \ldots, g_r) = \{\mathfrak m\}$, i.e., $\mathfrak m = \sqrt{(g_1, \ldots, g_r)}$. \medskip\noindent Assume $A/I$ reduced, i.e., $I$ radical. If $f$ is a unit, then since $I$ is radical we see that $I = \mathfrak m$. If $f \in \mathfrak m$, then we see that $f^n$ maps to zero in $A/I$. Hence $f \in I$ by reducedness and we conclude $I = (f)$. \medskip\noindent Assume $A/I$ has depth $1$. Then $\mathfrak m$ is not an associated prime of $A/I$. Since the class of $f$ modulo $I$ is annihilated by $g_1, \ldots, g_r$, this implies that the class of $f$ is zero in $A/I$. Thus $I = (f)$ as desired. \end{proof} \noindent Let $\kappa$ be a field and let $V$ be a vector space over $\kappa$. We will say $q \in \text{Sym}^2_\kappa(V)$ is {\it nondegenerate} if the induced $\kappa$-linear map $V^\vee \to V$ is an isomorphism. If $q = \sum_{i \leq j} a_{ij} x_i x_j$ for some $\kappa$-basis $x_1, \ldots, x_n$ of $V$, then this means that the determinant of the matrix $$\left( \begin{matrix} 2a_{11} & a_{12} & \ldots \\ a_{12} & 2a_{22} & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right)$$ is nonzero. This is equivalent to the condition that the partial derivatives of $q$ with respect to the $x_i$ cut out $0$ scheme theoretically. \begin{lemma} \label{lemma-nodal-algebraic} Let $k$ be a field. Let $(A, \mathfrak m, \kappa)$ be a Noetherian local $k$-algebra. The following are equivalent \begin{enumerate} \item $\kappa/k$ is separable, $A$ is reduced, $\dim_\kappa(\mathfrak m/\mathfrak m^2) = 2$, and there exists a nondegenerate $q \in \text{Sym}^2_\kappa(\mathfrak m/\mathfrak m^2)$ which maps to zero in $\mathfrak m^2/\mathfrak m^3$, \item $\kappa/k$ is separable, $\text{depth}(A) = 1$, $\dim_\kappa(\mathfrak m/\mathfrak m^2) = 2$, and there exists a nondegenerate $q \in \text{Sym}^2_\kappa(\mathfrak m/\mathfrak m^2)$ which maps to zero in $\mathfrak m^2/\mathfrak m^3$, \item $\kappa/k$ is separable, $A^\wedge \cong \kappa[[x, y]]/(ax^2 + bxy + cy^2)$ as a $k$-algebra where $ax^2 + bxy + cy^2$ is a nondegenerate quadratic form over $\kappa$. \end{enumerate} \end{lemma} \begin{proof} Assume (3). Then $A^\wedge$ is reduced because $ax^2 + bxy + cy^2$ is either irreducible or a product of two nonassociated prime elements. Hence $A \subset A^\wedge$ is reduced. It follows that (1) is true. \medskip\noindent Assume (1). Then $A$ cannot be Artinian, since it would not be reduced because $\mathfrak m \not = (0)$. Hence $\dim(A) \geq 1$, hence $\text{depth}(A) \geq 1$ by Algebra, Lemma \ref{algebra-lemma-criterion-reduced}. On the other hand $\dim(A) = 2$ implies $A$ is regular which contradicts the existence of $q$ by Algebra, Lemma \ref{algebra-lemma-regular-graded}. Thus $\dim(A) \leq 1$ and we conclude $\text{depth}(A) = 1$ by Algebra, Lemma \ref{algebra-lemma-bound-depth}. It follows that (2) is true. \medskip\noindent Assume (2). Since the depth of $A$ is the same as the depth of $A^\wedge$ (More on Algebra, Lemma \ref{more-algeb