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 \input{preamble} % OK, start here. % \begin{document} \title{Fields} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter, we shall discuss the theory of fields. Recall that a {\it field} is a ring in which all nonzero elements are invertible. Equivalently, the only two ideals of a field are $(0)$ and $(1)$ since any nonzero element is a unit. Consequently fields will be the simplest cases of much of the theory developed later. \medskip\noindent The theory of field extensions has a different feel from standard commutative algebra since, for instance, any morphism of fields is injective. Nonetheless, it turns out that questions involving rings can often be reduced to questions about fields. For instance, any domain can be embedded in a field (its quotient field), and any {\it local ring} (that is, a ring with a unique maximal ideal; we have not defined this term yet) has associated to it its residue field (that is, its quotient by the maximal ideal). A knowledge of field extensions will thus be useful. \section{Basic definitions} \label{section-definitions} \noindent Because we have placed this chapter before the chapter discussing commutative algebra we need to introduce some of the basic definitions here before we discuss these in greater detail in the algebra chapters. \begin{definition} \label{definition-field} An {\it field} is a nonzero ring where every nonzero element is invertible. Given a field a {\it subfield} is a subring that is itself a field. \end{definition} \noindent For a field $k$, we write $k^*$ for the subset $k \setminus \{0\}$. This generalizes the usual notation $R^*$ that refers to the group of invertible elements in a ring $R$. \begin{definition} \label{definition-domain} A {\it domain} or an {\it integral domain} is a nonzero ring where $0$ is the only zerodivisor. \end{definition} \section{Examples of fields} \label{section-examples} \noindent To get started, let us begin by providing several examples of fields. The reader should recall that if $R$ is a ring and $I \subset R$ an ideal, then $R/I$ is a field precisely when $I$ is a maximal ideal. \begin{example}[Rational numbers] \label{example-rational-numbers} The rational numbers form a field. It is called the {\it field of rational numbers} and denoted $\mathbf{Q}$. \end{example} \begin{example}[Prime fields] \label{example-prime-field} If $p$ is a prime number, then $\mathbf{Z}/(p)$ is a field, denoted $\mathbf{F}_p$. Indeed, $(p)$ is a maximal ideal in $\mathbf{Z}$. Thus, fields may be finite: $\mathbf{F}_p$ contains $p$ elements. \end{example} \begin{example} \label{example-quotient-polymial-ring} In a principal ideal domain, an ideal generated by an irreducible element is maximal. Now, if $k$ is a field, then the polynomial ring $k[x]$ is a principal ideal domain. It follows that if $P \in k[x]$ is an irreducible polynomial (that is, a nonconstant polynomial that does not admit a factorization into terms of smaller degrees), then $k[x]/(P)$ is a field. It contains a copy of $k$ in a natural way. This is a very general way of constructing fields. For instance, the complex numbers $\mathbf{C}$ can be constructed as $\mathbf{R}[x]/(x^2 + 1)$. \end{example} \begin{example}[Quotient fields] \label{example-quotient-field} Recall that, given a domain $A$, there is an imbedding $A \to K(A)$ into a field $K(A)$ constructed from $A$ in exactly the same manner that $\mathbf{Q}$ is constructed from $\mathbf{Z}$. Formally the elements of $K(A)$ are (equivalence classes of) fractions $a/b$, $a, b \in A$, $b \not = 0$. As usual $a/b = a'/b'$ if and only if $ab' = ba'$. This is called the {\it quotient field} or {\it field of fractions} or the {\it fraction field} of $A$. The quotient field has the following universal property: given an injective ring map $\varphi : A \to K$ to a field $K$, there is a unique map $\psi : K(A) \to K$ making $$\xymatrix{ K(A) \ar[r]_\psi & K \\ A \ar[u] \ar[ru]_\varphi }$$ commute. Indeed, it is clear how to define such a map: we set $\psi(a/b) = \varphi(a)\varphi(b)^{-1}$ where injectivity of $\varphi$ assures that $\varphi(b) \not = 0$ if $b \not = 0$. \end{example} \begin{example}[Field of rational functions] \label{example-field-of-rational-functions} If $k$ is a field, then we can consider the field $k(x)$ of rational functions over $k$. This is the quotient field of the polynomial ring $k[x]$. In other words, it is the set of quotients $F/G$ for $F, G \in k[x]$, $G \not = 0$ with the obvious equivalence relation. \end{example} \begin{example} \label{example-field-of-meromorphic-functions} Let $X$ be a Riemann surface. Let $\mathbf{C}(X)$ denote the set of meromorphic functions on $X$. Then $\mathbf{C}(X)$ is a ring under multiplication and addition of functions. It turns out that in fact $\mathbf{C}(X)$ is a field. Namely, if a nonzero function $f(z)$ is meromorphic, so is $1/f(z)$. For example, let $S^2$ be the Riemann sphere; then we know from complex analysis that the ring of meromorphic functions $\mathbf{C}(S^2)$ is the field of rational functions $\mathbf{C}(z)$. \end{example} \section{Vector spaces} \label{section-vector-spaces} \noindent One reason fields are so nice is that the theory of modules over fields (i.e. vector spaces), is very simple. \begin{lemma} \label{lemma-vector-space-is-free} If $k$ is a field, then every $k$-module is free. \end{lemma} \begin{proof} Indeed, by linear algebra we know that a $k$-module (i.e. vector space) $V$ has a {\it basis} $\mathcal{B} \subset V$, which defines an isomorphism from the free vector space on $\mathcal{B}$ to $V$. \end{proof} \begin{lemma} \label{lemma-field-semi-simple} Every exact sequence of modules over a field splits. \end{lemma} \begin{proof} This follows from Lemma \ref{lemma-vector-space-is-free} as every vector space is a projective module. \end{proof} \noindent This is another reason why much of the theory in future chapters will not say very much about fields, since modules behave in such a simple manner. Note that Lemma \ref{lemma-field-semi-simple} is a statement about the {\it category} of $k$-modules (for $k$ a field), because the notion of exactness is inherently arrow-theoretic, i.e., makes use of purely categorical notions, and can in fact be phrased within a so-called {\it abelian category}. \medskip\noindent Henceforth, since the study of modules over a field is linear algebra, and since the ideal theory of fields is not very interesting, we shall study what this chapter is really about: {\it extensions} of fields. \section{The characteristic of a field} \label{section-more-fields} \noindent In the category of rings, there is an {\it initial object} $\mathbf{Z}$: any ring $R$ has a map from $\mathbf{Z}$ into it in precisely one way. For fields, there is no such initial object. Nonetheless, there is a family of objects such that every field can be mapped into in exactly one way by exactly one of them, and in no way by the others. \medskip\noindent Let $F$ be a field. Think of $F$ as a ring to get a ring map $f : \mathbf{Z} \to F$. The image of this ring map is a domain (as a subring of a field) hence the kernel of $f$ is a prime ideal in $\mathbf{Z}$. Hence the kernel of $f$ is either $(0)$ or $(p)$ for some prime number $p$. \medskip\noindent In the first case we see that $f$ is injective, and in this case we think of $\mathbf{Z}$ as a subring of $F$. Moreover, since every nonzero element of $F$ is invertible we see that it makes sense to talk about $p/q \in F$ for $p, q \in \mathbf{Z}$ with $q \not = 0$. Hence in this case we may and we do think of $\mathbf{Q}$ as a subring of $F$. One can easily see that this is the smallest subfield of $F$ in this case. \medskip\noindent In the second case, i.e., when $\Ker(f) = (p)$ we see that $\mathbf{Z}/(p) = \mathbf{F}_p$ is a subring of $F$. Clearly it is the smallest subfield of $F$. \medskip\noindent Arguing in this way we see that every field contains a smallest subfield which is either $\mathbf{Q}$ or finite equal to $\mathbf{F}_p$ for some prime number $p$. \begin{definition} \label{definition-characteristic} The {\it characteristic} of a field $F$ is $0$ if $\mathbf{Z} \subset F$, or is a prime $p$ if $p = 0$ in $F$. The {\it prime subfield of $F$} is the smallest subfield of $F$ which is either $\mathbf{Q} \subset F$ if the characteristic is zero, or $\mathbf{F}_p \subset F$ if the characteristic is $p > 0$. \end{definition} \noindent It is easy to see that if $E \subset F$ is a subfield, then the characteristic of $E$ is the same as the characteristic of $F$. \begin{example} \label{example-characteristic} The characteristic of $\mathbf{F}_p$ is $p$, and that of $\mathbf{Q}$ is $0$. \end{example} \section{Field extensions} \label{section-extensions} \noindent In general, though, we are interested not so much in fields by themselves but in field {\it extensions}. This is perhaps analogous to studying not rings but {\it algebras} over a fixed ring. The nice thing for fields is that the notion of a field over another field'' just recovers the notion of a field extension, by the next result. \begin{lemma} \label{lemma-field-maps-injective} If $F$ is a field and $R$ is a nonzero ring, then any ring homomorphism $\varphi : F \to R$ is injective. \end{lemma} \begin{proof} Indeed, let $a \in \Ker(\varphi)$ be a nonzero element. Then we have $\varphi(1) = \varphi(a^{-1} a) = \varphi(a^{-1}) \varphi(a) = 0$. Thus $1 = \varphi(1) = 0$ and $R$ is the zero ring. \end{proof} \begin{definition} \label{definition-extension} If $F$ is a field contained in a field $E$, then $E$ is said to be a {\it field extension} of $F$. We shall write $E/F$ to indicate that $E$ is an extension of $F$. \end{definition} \noindent So if $F, F'$ are fields, and $F \to F'$ is any ring-homomorphism, we see by Lemma \ref{lemma-field-maps-injective} that it is injective, and $F'$ can be regarded as an extension of $F$, by a slight abuse of language. Alternatively, a field extension of $F$ is just an $F$-algebra that happens to be a field. This is completely different than the situation for general rings, since a ring homomorphism is not necessarily injective. \medskip\noindent Let $k$ be a field. There is a {\it category} of field extensions of $k$. An object of this category is an extension $E/k$, that is a (necessarily injective) morphism of fields $$k \to E,$$ while a morphism between extensions $E/k$ and $E'/k$ is a $k$-algebra morphism $E \to E'$; alternatively, it is a commutative diagram $$\xymatrix{ E \ar[rr] & & E' \\ & k \ar[ru] \ar[lu] & }$$ The set of morphisms from $E \to E'$ in the category of extensions of $k$ will be denoted by $\Mor_k(E, E')$. \begin{definition} \label{definition-tower} A {\it tower} of fields $E_n/E_{n - 1}/\ldots/E_0$ consists of a sequence of extensions of fields $E_n/E_{n - 1}$, $E_{n - 1}/E_{n - 2}$, $\ldots$, $E_1/E_0$. \end{definition} \noindent Let us give a few examples of field extensions. \begin{example} \label{example-monogenic-extension} Let $k$ be a field, and $P \in k[x]$ an irreducible polynomial. We have seen that $k[x]/(P)$ is a field (Example \ref{example-quotient-polymial-ring}). Since it is also a $k$-algebra in the obvious way, it is an extension of $k$. \end{example} \begin{example} \label{example-field-of-meromorphic-functions-extension-C} If $X$ is a Riemann surface, then the field of meromorphic functions $\mathbf{C}(X)$ (Example \ref{example-field-of-meromorphic-functions}) is an extension field of $\mathbf{C}$, because any element of $\mathbf{C}$ induces a meromorphic --- indeed, holomorphic --- constant function on $X$. \end{example} \noindent Let $F/k$ be a field extension. Let $S \subset F$ be any subset. Then there is a {\it smallest} subextension of $F$ (that is, a subfield of $F$ containing $k$) that contains $S$. To see this, consider the family of subfields of $F$ containing $S$ and $k$, and take their intersection; one checks that this is a field. By a standard argument one shows, in fact, that this is the set of elements of $F$ that can be obtained via a finite number of elementary algebraic operations (addition, multiplication, subtraction, and division) involving elements of $k$ and $S$. \begin{definition} \label{definition-generated-by} Let $k$ be a field. If $F/k$ is an extension of fields and $S \subset F$, we write $k(S)$ for the smallest subfield of $F$ containing $k$ and $S$. We will say that $S$ {\it generates the field extension} $k(S)/k$. If $S = \{\alpha\}$ is a singleton, then we write $k(\alpha)$ instead of $k(\{\alpha\})$. We say $F/k$ is a {\it finitely generated field extension} if there exists a finite subset $S \subset F$ with $F = k(S)$. \end{definition} \noindent For instance, $\mathbf{C}$ is generated by $i$ over $\mathbf{R}$. \begin{exercise} \label{exercise-C-not-countably-generated} Show that $\mathbf{C}$ does not have a countable set of generators over $\mathbf{Q}$. \end{exercise} \noindent Let us now classify extensions generated by one element. \begin{lemma}[Classification of simple extensions] \label{lemma-field-extension-generated-by-one-element} If a field extension $F/k$ is generated by one element, then it is $k$-isomorphic either to the rational function field $k(t)/k$ or to one of the extensions $k[t]/(P)$ for $P \in k[t]$ irreducible. \end{lemma} \noindent We will see that many of the most important cases of field extensions are generated by one element, so this is actually useful. \begin{proof} Let $\alpha \in F$ be such that $F = k(\alpha)$; by assumption, such an $\alpha$ exists. There is a morphism of rings $$k[t] \to F$$ sending the indeterminate $t$ to $\alpha$. The image is a domain, so the kernel is a prime ideal. Thus, it is either $(0)$ or $(P)$ for $P \in k[t]$ irreducible. \medskip\noindent If the kernel is $(P)$ for $P \in k[t]$ irreducible, then the map factors through $k[t]/(P)$, and induces a morphism of fields $k[t]/(P) \to F$. Since the image contains $\alpha$, we see easily that the map is surjective, hence an isomorphism. In this case, $k[t]/(P) \simeq F$. \medskip\noindent If the kernel is trivial, then we have an injection $k[t] \to F$. One may thus define a morphism of the quotient field $k(t)$ into $F$; given a quotient $R(t)/Q(t)$ with $R(t), Q(t) \in k[t]$, we map this to $R(\alpha)/Q(\alpha)$. The hypothesis that $k[t] \to F$ is injective implies that $Q(\alpha) \neq 0$ unless $Q$ is the zero polynomial. The quotient field of $k[t]$ is the rational function field $k(t)$, so we get a morphism $k(t) \to F$ whose image contains $\alpha$. It is thus surjective, hence an isomorphism. \end{proof} \section{Finite extensions} \label{section-finite-extensions} \noindent If $F/E$ is a field extension, then evidently $F$ is also a vector space over $E$ (the scalar action is just multiplication in $F$). \begin{definition} \label{definition-degree} Let $F/E$ be an extension of fields. The dimension of $F$ considered as an $E$-vector space is called the {\it degree} of the extension and is denoted $[F : E]$. If $[F : E]<\infty$ then $F$ is said to be a {\it finite} extension of $E$. \end{definition} \begin{example} \label{example-C-over-R} The field $\mathbf{C}$ is a two dimensional vector space over $\mathbf{R}$ with basis $1, i$. Thus $\mathbf{C}$ is a finite extension of $\mathbf{R}$ of degree 2. \end{example} \begin{lemma} \label{lemma-finite-goes-up} Let $K/E/F$ be a tower of algebraic field extensions. If $K$ is finite over $F$, then $K$ is finite over $E$. \end{lemma} \begin{proof} Direct from the definition. \end{proof} \noindent Let us now consider the degree in the most important special example, that given by Lemma \ref{lemma-field-extension-generated-by-one-element}, in the next two examples. \begin{example}[Degree of a rational function field] \label{example-degree-rational-function-field} If $k$ is any field, then the rational function field $k(t)$ is {\it not} a finite extension. For example the elements $\left\{t^n, n \in \mathbf{Z}\right\}$ are linearly independent over $k$. \medskip\noindent In fact, if $k$ is uncountable, then $k(t)$ is {\it uncountably} dimensional as a $k$-vector space. To show this, we claim that the family of elements $\{1/(t- \alpha), \alpha \in k\} \subset k(t)$ is linearly independent over $k$. A nontrivial relation between them would lead to a contradiction: for instance, if one works over $\mathbf{C}$, then this follows because $\frac{1}{t-\alpha}$, when considered as a meromorphic function on $\mathbf{C}$, has a pole at $\alpha$ and nowhere else. Consequently any sum $\sum c_i \frac{1}{t - \alpha_i}$ for the $c_i \in k^*$, and $\alpha_i \in k$ distinct, would have poles at each of the $\alpha_i$. In particular, it could not be zero. \medskip\noindent Amusingly, this leads to a quick proof of the Hilbert Nullstellensatz over the complex numbers. For a slightly more general result, see Algebra, Theorem \ref{algebra-theorem-uncountable-nullstellensatz}. \end{example} \begin{lemma} \label{lemma-finite-finitely-generated} A finite extension of fields is a finitely generated field extension. The converse is not true. \end{lemma} \begin{proof} Let $F/E$ be a finite extension of fields. Let $\alpha_1, \ldots, \alpha_n$ be a basis of $F$ as a vector space over $E$. Then $F = E(\alpha_1, \ldots, \alpha_n)$ hence $F/E$ is a finitely generated field extension. The converse is not true as follows from Example \ref{example-degree-rational-function-field}. \end{proof} \begin{example}[Degree of a simple algebraic extension] \label{example-degree-simple-algebraic-extension} Consider a monogenic field extension $E/k$ of the form discussed in Example \ref{example-monogenic-extension}. In other words, $E = k[t]/(P)$ for $P \in k[t]$ an irreducible polynomial. Then the degree $[E : k]$ is just the degree $d = \deg(P)$ of the polynomial $P$. Indeed, say \begin{equation} \label{equation-P} P = a_d t^d + a_1 t^{d - 1} + \ldots + a_0. \end{equation} with $a_d \not = 0$. Then the images of $1, t, \ldots, t^{d - 1}$ in $k[t]/(P)$ are linearly independent over $k$, because any relation involving them would have degree strictly smaller than that of $P$, and $P$ is the element of smallest degree in the ideal $(P)$. \medskip\noindent Conversely, the set $S = \{1, t, \ldots, t^{d - 1}\}$ (or more properly their images) spans $k[t]/(P)$ as a vector space. Indeed, we have by (\ref{equation-P}) that $a_d t^d$ lies in the span of $S$. Since $a_d$ is invertible, we see that $t^d$ is in the span of $S$. Similarly, the relation $t P(t) = 0$ shows that the image of $t^{d + 1}$ lies in the span of $\{1, t, \ldots, t^d\}$ --- by what was just shown, thus in the span of $S$. Working upward inductively, we find that the image of $t^n$ for $n \geq d$ lies in the span of $S$. \end{example} \noindent This confirms the observation that $[\mathbf{C}: \mathbf{R}] = 2$, for instance. More generally, if $k$ is a field, and $\alpha \in k$ is not a square, then the irreducible polynomial $x^2 - \alpha \in k[x]$ allows one to construct an extension $k[x]/(x^2 - \alpha)$ of degree two. We shall write this as $k(\sqrt{\alpha})$. Such extensions will be called {\it quadratic,} for obvious reasons. \medskip\noindent The basic fact about the degree is that it is {\it multiplicative in towers.} \begin{lemma}[Multiplicativity] \label{lemma-multiplicativity-degrees} Suppose given a tower of fields $F/E/k$. Then $$[F:k] = [F:E][E:k]$$ \end{lemma} \begin{proof} Let $\alpha_1, \ldots, \alpha_n \in F$ be an $E$-basis for $F$. Let $\beta_1, \ldots, \beta_m \in E$ be a $k$-basis for $E$. Then the claim is that the set of products $\{\alpha_i \beta_j, 1 \leq i \leq n, 1 \leq j \leq m\}$ is a $k$-basis for $F$. Indeed, let us check first that they span $F$ over $k$. \medskip\noindent By assumption, the $\{\alpha_i\}$ span $F$ over $E$. So if $f \in F$, there are $a_i \in E$ with $$f = \sum\nolimits_i a_i \alpha_i,$$ and, for each $i$, we can write $a_i = \sum b_{ij} \beta_j$ for some $b_{ij} \in k$. Putting these together, we find $$f = \sum\nolimits_{i,j} b_{ij} \alpha_i \beta_j,$$ proving that the $\{\alpha_i \beta_j\}$ span $F$ over $k$. \medskip\noindent Suppose now that there existed a nontrivial relation $$\sum\nolimits_{i,j} c_{ij} \alpha_i \beta_j = 0$$ for the $c_{ij} \in k$. In that case, we would have $$\sum\nolimits_i \alpha_i \left( \sum\nolimits_j c_{ij} \beta_j \right) = 0,$$ and the inner terms lie in $E$ as the $\beta_j$ do. Now $E$-linear independence of the $\{\alpha_i\}$ shows that the inner sums are all zero. Then $k$-linear independence of the $\{\beta_j\}$ shows that the $c_{ij}$ all vanish. \end{proof} \noindent We sidetrack to a slightly tangential definition. \begin{definition} \label{definition-number-field} A field $K$ is said to be a {\it number field} if it has characteristic $0$ and the extension $\mathbf{Q} \subset K$ is finite. \end{definition} \noindent Number fields are the basic objects in algebraic number theory. We shall see later that, for the analog of the integers $\mathbf{Z}$ in a number field, something kind of like unique factorization still holds (though strict unique factorization generally does not!). \section{Algebraic extensions} \label{section-algebraic-extensions} \noindent An important class of extensions are those where every element generates a finite extension. \begin{definition} \label{definition-algebraic} Consider a field extension $F/E$. An element $\alpha \in F$ is said to be {\it algebraic} over $E$ if $\alpha$ is the root of some nonzero polynomial with coefficients in $E$. If all elements of $F$ are algebraic then $F$ is said to be an {\it algebraic extension} of $E$. \end{definition} \noindent By Lemma \ref{lemma-field-extension-generated-by-one-element}, the subextension $E(\alpha)$ is isomorphic either to the rational function field $E(t)$ or to a quotient ring $E[t]/(P)$ for $P \in E[t]$ an irreducible polynomial. In the latter case, $\alpha$ is algebraic over $E$ (in fact, the proof of Lemma \ref{lemma-field-extension-generated-by-one-element} shows that we can pick $P$ such that $\alpha$ is a root of $P$); in the former case, it is not. \begin{example} \label{example-C-algebraic-over-R} The field $\mathbf{C}$ is algebraic over $\mathbf{R}$. Namely, if $\alpha = a + ib$ in $\mathbf{C}$, then $\alpha^2 - 2a\alpha + a^2 + b^2 = 0$ is a polynomial equation for $\alpha$ over $\mathbf{R}$. \end{example} \begin{example} \label{example-compact-riemann-surface-is-finite-over-P1} Let $X$ be a compact Riemann surface, and let $f \in \mathbf{C}(X) - \mathbf{C}$ any nonconstant meromorphic function on $X$ (see Example \ref{example-field-of-meromorphic-functions}). Then it is known that $\mathbf{C}(X)$ is algebraic over the subextension $\mathbf{C}(f)$ generated by $f$. We shall not prove this. \end{example} \begin{lemma} \label{lemma-algebraic-goes-up} Let $K/E/F$ be a tower of field extensions. \begin{enumerate} \item If $\alpha \in K$ is algebraic over $F$, then $\alpha$ is algebraic over $E$. \item if $K$ is algebraic over $F$, then $K$ is algebraic over $E$. \end{enumerate} \end{lemma} \begin{proof} This is immediate from the definitions. \end{proof} \noindent We now show that there is a deep connection between finiteness and being algebraic. \begin{lemma} \label{lemma-finite-is-algebraic} A finite extension is algebraic. In fact, an extension $E/k$ is algebraic if and only if every subextension $k(\alpha)/k$ generated by some $\alpha \in E$ is finite. \end{lemma} \noindent In general, it is very false that an algebraic extension is finite. \begin{proof} Let $E/k$ be finite, say of degree $n$. Choose $\alpha \in E$. Then the elements $\{1, \alpha, \ldots, \alpha^n\}$ are linearly dependent over $E$, or we would necessarily have $[E : k] > n$. A relation of linear dependence now gives the desired polynomial that $\alpha$ must satisfy. \medskip\noindent For the last assertion, note that a monogenic extension $k(\alpha)/k$ is finite if and only $\alpha$ is algebraic over $k$, by Examples \ref{example-degree-rational-function-field} and \ref{example-degree-simple-algebraic-extension}. So if $E/k$ is algebraic, then each $k(\alpha)/k$, $\alpha \in E$, is a finite extension, and conversely. \end{proof} \noindent We can extract a lemma of the last proof (really of Examples \ref{example-degree-rational-function-field} and \ref{example-degree-simple-algebraic-extension}): a monogenic extension is finite if and only if it is algebraic. We shall use this observation in the next result. \begin{lemma} \label{lemma-algebraic-finitely-generated} \begin{slogan} A finitely generated algebraic extension is finite. \end{slogan} Let $k$ be a field, and let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be elements of some extension field such that each $\alpha_i$ is algebraic over $k$. Then the extension $k(\alpha_1, \ldots, \alpha_n)/k$ is finite. That is, a finitely generated algebraic extension is finite. \end{lemma} \begin{proof} Indeed, each extension $k(\alpha_{1}, \ldots, \alpha_{i+1})/k(\alpha_1, \ldots, \alpha_{i})$ is generated by one element and algebraic, hence finite. By multiplicativity of degree (Lemma \ref{lemma-multiplicativity-degrees}) we obtain the result. \end{proof} \noindent The set of complex numbers that are algebraic over $\mathbf{Q}$ are simply called the {\it algebraic numbers.} For instance, $\sqrt{2}$ is algebraic, $i$ is algebraic, but $\pi$ is not. It is a basic fact that the algebraic numbers form a field, although it is not obvious how to prove this from the definition that a number is algebraic precisely when it satisfies a nonzero polynomial equation with rational coefficients (e.g. by polynomial equations). \begin{lemma} \label{lemma-algebraic-elements} Let $E/k$ be a field extension. Then the elements of $E$ algebraic over $k$ form a subextension of $E/k$. \end{lemma} \begin{proof} Let $\alpha, \beta \in E$ be algebraic over $k$. Then $k(\alpha, \beta)/k$ is a finite extension by Lemma \ref{lemma-algebraic-finitely-generated}. It follows that $k(\alpha + \beta) \subset k(\alpha, \beta)$ is a finite extension, which implies that $\alpha + \beta$ is algebraic by Lemma \ref{lemma-finite-is-algebraic}. Similarly for the difference, product and quotient of $\alpha$ and $\beta$. \end{proof} \noindent Many nice properties of field extensions, like those of rings, will have the property that they will be preserved by towers and composita. \begin{lemma} \label{lemma-algebraic-permanence} Let $E/k$ and $F/E$ be algebraic extensions of fields. Then $F/k$ is an algebraic extension of fields. \end{lemma} \begin{proof} Choose $\alpha \in F$. Then $\alpha$ is algebraic over $E$. The key observation is that $\alpha$ is algebraic over a finitely generated subextension of $k$. That is, there is a finite set $S \subset E$ such that $\alpha$ is algebraic over $k(S)$: this is clear because being algebraic means that a certain polynomial in $E[x]$ that $\alpha$ satisfies exists, and as $S$ we can take the coefficients of this polynomial. It follows that $\alpha$ is algebraic over $k(S)$. In particular, the extension $k(S, \alpha)/ k(S)$ is finite. Since $S$ is a finite set, and $k(S)/k$ is algebraic, Lemma \ref{lemma-algebraic-finitely-generated} shows that $k(S)/k$ is finite. Using multiplicativity (Lemma \ref{lemma-multiplicativity-degrees}) we find that $k(S,\alpha)/k$ is finite, so $\alpha$ is algebraic over $k$. \end{proof} \noindent The method of proof in the previous argument --- that being algebraic over $E$ was a property that {\it descended} to a finitely generated subextension of $E$ --- is an idea that recurs throughout algebra. It often allows one to reduce general commutative algebra questions to the Noetherian case for example. \begin{lemma} \label{lemma-size-algebraic-extension} Let $E/F$ be an algebraic extension of fields. Then the cardinality $|E|$ of $E$ is at most $\max(\aleph_0, |F|)$. \end{lemma} \begin{proof} Let $S$ be the set of nonconstant polynomials with coefficients in $F$. For every $P \in S$ the set of roots $r(P, E) = \{\alpha \in E \mid P(\alpha) = 0\}$ is finite (details omitted). Moreover, the fact that $E$ is algebraic over $F$ implies that $E = \bigcup_{P \in S} r(P, E)$. It is clear that $S$ has cardinality bounded by $\max(\aleph_0, |F|)$ because the cardinality of a finite product of copies of $F$ has cardinality at most $\max(\aleph_0, |F|)$. Thus so does $E$. \end{proof} \begin{lemma} \label{lemma-subalgebra-algebraic-extension-field} Let $E/F$ be a finite or more generally an algebraic extension of fields. Any subring $F \subset R \subset E$ is a field. \end{lemma} \begin{proof} Let $\alpha \in R$ be nonzero. Then $1, \alpha, \alpha^2, \ldots$ are contained in $R$. By Lemma \ref{lemma-finite-is-algebraic} we find a nontrivial relation $a_0 + a_1 \alpha + \ldots + a_d \alpha^d = 0$. We may assume $a_0 \not = 0$ because if not we can divide the relation by $\alpha$ to decrease $d$. Then we see that $$a_0 = \alpha (- a_1 - \ldots - a_d \alpha^{d - 1})$$ which proves that the inverse of $\alpha$ is the element $a_0^{-1} (- a_1 - \ldots - a_d \alpha^{d - 1})$ of $R$. \end{proof} \begin{lemma} \label{lemma-algebraic-extension-self-map} Let $E/F$ an algebraic extension of fields. Any $F$-algebra map $f : E \to E$ is an automorphism. \end{lemma} \begin{proof} If $E/F$ is finite, then $f : E \to E$ is an $F$-linear injective map (Lemma \ref{lemma-field-maps-injective}) of finite dimensional vector spaces, and hence bijective. In general we still see that $f$ is injective. Let $\alpha \in E$ and let $P \in F[x]$ be a polynomial such that $P(\alpha) = 0$. Let $E' \subset E$ be the subfield of $E$ generated by the roots $\alpha = \alpha_1, \ldots, \alpha_n$ of $P$ in $E$. Then $E'$ is finite over $F$ by Lemma \ref{lemma-algebraic-finitely-generated}. Since $f$ preserves the set of roots, we find that $f|_{E'} : E' \to E'$. Hence $f|_{E'}$ is an isomorphism by the first part of the proof and we conclude that $\alpha$ is in the image of $f$. \end{proof} \section{Minimal polynomials} \label{section-minimal-polynomials} \noindent Let $E/k$ be a field extension, and let $\alpha \in E$ be algebraic over $k$. Then $\alpha$ satisfies a (nontrivial) polynomial equation in $k[x]$. Consider the set of polynomials $P \in k[x]$ such that $P(\alpha) = 0$; by hypothesis, this set does not just contain the zero polynomial. It is easy to see that this set is an {\it ideal.} Indeed, it is the kernel of the map $$k[x] \to E, \quad x \mapsto \alpha$$ Since $k[x]$ is a PID, there is a {\it generator} $P \in k[x]$ of this ideal. If we assume $P$ monic, without loss of generality, then $P$ is uniquely determined. \begin{definition} \label{definition-minimal-polynomial} The polynomial $P$ above is called the {\it minimal polynomial} of $\alpha$ over $k$. \end{definition} \noindent The minimal polynomial has the following characterization: it is the monic polynomial, of smallest degree, that annihilates $\alpha$. Any nonconstant multiple of $P$ will have larger degree, and only multiples of $P$ can annihilate $\alpha$. This explains the name {\it minimal}. \medskip\noindent Clearly the minimal polynomial is {\it irreducible}. This is equivalent to the assertion that the ideal in $k[x]$ consisting of polynomials annihilating $\alpha$ is prime. This follows from the fact that the map $k[x] \to E, x \mapsto \alpha$ is a map into a domain (even a field), so the kernel is a prime ideal. \begin{lemma} \label{lemma-degree-minimal-polynomial} The degree of the minimal polynomial is $[k(\alpha) : k]$. \end{lemma} \begin{proof} This is just a restatement of the argument in Lemma \ref{lemma-field-extension-generated-by-one-element}: the observation is that if $P$ is the minimal polynomial of $\alpha$, then the map $$k[x]/(P) \to k(\alpha), \quad x \mapsto \alpha$$ is an isomorphism as in the aforementioned proof, and we have counted the degree of such an extension (see Example \ref{example-degree-simple-algebraic-extension}). \end{proof} \noindent So the observation of the above proof is that if $\alpha \in E$ is algebraic, then $k(\alpha) \subset E$ is isomorphic to $k[x]/(P)$. \section{Algebraic closure} \label{section-algebraic-closure} \noindent The fundamental theorem of algebra'' states that $\mathbf{C}$ is algebraically closed. A beautiful proof of this result uses Liouville's theorem in complex analysis, we shall give another proof (see Lemma \ref{lemma-C-algebraically-closed}). \begin{definition} \label{definition-algebraically-closed} A field $F$ is said to be {\it algebraically closed} if every algebraic extension $E/F$ is trivial, i.e., $E = F$. \end{definition} \noindent This may not be the definition in every text. Here is the lemma comparing it with the other one. \begin{lemma} \label{lemma-algebraically-closed} Let $F$ be a field. The following are equivalent \begin{enumerate} \item $F$ is algebraically closed, \item every irreducible polynomial over $F$ is linear, \item every nonconstant polynomial over $F$ has a root, \item every nonconstant polynomial over $F$ is a product of linear factors. \end{enumerate} \end{lemma} \begin{proof} If $F$ is algebraically closed, then every irreducible polynomial is linear. Namely, if there exists an irreducible polynomial of degree $> 1$, then this generates a nontrivial finite (hence algebraic) field extension, see Example \ref{example-degree-simple-algebraic-extension}. Thus (1) implies (2). If every irreducible polynomial is linear, then every irreducible polynomial has a root, whence every nonconstant polynomial has a root. Thus (2) implies (3). \medskip\noindent Assume every nonconstant polynomial has a root. Let $P \in F[x]$ be nonconstant. If $P(\alpha) = 0$ with $\alpha \in F$, then we see that $P = (x - \alpha)Q$ for some $Q \in F[x]$ (by division with remainder). Thus we can argue by induction on the degree that any nonconstant polynomial can be written as a product $c \prod (x - \alpha_i)$. \medskip\noindent Finally, suppose that every nonconstant polynomial over $F$ is a product of linear factors. Let $E/F$ be an algebraic extension. Then all the simple subextensions $F(\alpha)/F$ of $E$ are necessarily trivial (because the only irreducible polynomials are linear by assumption). Thus $E = F$. We see that (4) implies (1) and we are done. \end{proof} \noindent Now we want to define a universal'' algebraic extension of a field. Actually, we should be careful: the algebraic closure is {\it not} a universal object. That is, the algebraic closure is not unique up to {\it unique} isomorphism: it is only unique up to isomorphism. But still, it will be very handy, if not functorial. \begin{definition} \label{definition-algebraic-closure} Let $F$ be a field. We say $F$ is {\it algebraically closed} if every algebraic extension $E/F$ is trivial, i.e., $E = F$. An {\it algebraic closure} of $F$ is a field $\overline{F}$ containing $F$ such that: \begin{enumerate} \item $\overline{F}$ is algebraic over $F$. \item $\overline{F}$ is algebraically closed. \end{enumerate} \end{definition} \noindent If $F$ is algebraically closed, then $F$ is its own algebraic closure. We now prove the basic existence result. \begin{theorem} \label{theorem-existence-algebraic-closure} Every field has an algebraic closure. \end{theorem} \noindent The proof will mostly be a red herring to the rest of the chapter. However, we will want to know that it is {\it possible} to embed a field inside an algebraically closed field, and we will often assume it done. \begin{proof} Let $F$ be a field. By Lemma \ref{lemma-size-algebraic-extension} the cardinality of an algebraic extension of $F$ is bounded by $\max(\aleph_0, |F|)$. Choose a set $S$ containing $F$ with $|S| > \max(\aleph_0, |F|)$. Let's consider triples $(E, \sigma_E, \mu_E)$ where \begin{enumerate} \item $E$ is a set with $F \subset E \subset S$, and \item $\sigma_E : E \times E \to E$ and $\mu_E : E \times E \to E$ are maps of sets such that $(E, \sigma_E, \mu_E)$ defines the structure of a field extension of $F$ (in particular $\sigma_E(a, b) = a +_F b$ for $a, b \in F$ and similarly for $\mu_E$), and \item $F \subset E$ is an algebraic field extension. \end{enumerate} The collection of all triples $(E, \sigma_E, \mu_E)$ forms a set $I$. For $i \in I$ we will denote $E_i = (E_i, \sigma_i, \mu_i)$ the corresponding field extension to $F$. We define a partial ordering on $I$ by declaring $i \leq i'$ if and only if $E_i \subset E_{i'}$ (this makes sense as $E_i$ and $E_{i'}$ are subsets of the same set $S$) and we have $\sigma_i = \sigma_{i'}|_{E_i \times E_i}$ and $\mu_i = \mu_{i'}|_{E_i \times E_i}$, in other words, $E_{i'}$ is a field extension of $E_i$. \medskip\noindent Let $T \subset I$ be a totally ordered subset. Then it is clear that $E_T = \bigcup_{i \in T} E_i$ with induced maps $\sigma_T = \bigcup \sigma_i$ and $\mu_T = \bigcup \mu_i$ is another element of $I$. In other words every totally order subset of $I$ has a upper bound in $I$. By Zorn's lemma there exists a maximal element $(E, \sigma_E, \mu_E)$ in $I$. We claim that $E$ is an algebraic closure. Since by definition of $I$ the extension $E/F$ is algebraic, it suffices to show that $E$ is algebraically closed. \medskip\noindent To see this we argue by contradiction. Namely, suppose that $E$ is not algebraically closed. Then there exists an irreducible polynomial $P$ over $E$ of degree $> 1$, see Lemma \ref{lemma-algebraically-closed}. By Lemma \ref{lemma-finite-is-algebraic} we obtain a nontrivial finite extension $E' = E[x]/(P)$. Observe that $E'/F$ is algebraic by Lemma \ref{lemma-algebraic-permanence}. Thus the cardinality of $E'$ is $\leq \max(\aleph_0, |F|)$. By elementary set theory we can extend the given injection $E \subset S$ to an injection $E' \to S$. In other words, we may think of $E'$ as an element of our set $I$ contradicting the maximality of $E$. This contradiction completes the proof. \end{proof} \begin{lemma} \label{lemma-map-into-algebraic-closure} Let $F$ be a field. Let $\overline{F}$ be an algebraic closure of $F$. Let $M/F$ be an algebraic extension. Then there is a morphism of $F$-extensions $M \to \overline{F}$. \end{lemma} \begin{proof} Consider the set $I$ of pairs $(E, \varphi)$ where $F \subset E \subset M$ is a subextension and $\varphi : E \to \overline{F}$ is a morphism of $F$-extensions. We partially order the set $I$ by declaring $(E, \varphi) \leq (E', \varphi')$ if and only if $E \subset E'$ and $\varphi'|_E = \varphi$. If $T = \{(E_t, \varphi_t)\} \subset I$ is a totally ordered subset, then $\bigcup \varphi_t : \bigcup E_t \to \overline{F}$ is an element of $I$. Thus every totally ordered subset of $I$ has an upper bound. By Zorn's lemma there exists a maximal element $(E, \varphi)$ in $I$. We claim that $E = M$, which will finish the proof. If not, then pick $\alpha \in M$, $\alpha \not \in E$. The $\alpha$ is algebraic over $E$, see Lemma \ref{lemma-algebraic-goes-up}. Let $P$ be the minimal polynomial of $\alpha$ over $E$. Let $P^\varphi$ be the image of $P$ by $\varphi$ in $\overline{F}[x]$. Since $\overline{F}$ is algebraically closed there is a root $\beta$ of $P^\varphi$ in $\overline{F}$. Then we can extend $\varphi$ to $\varphi' : E(\alpha) = E[x]/(P) \to \overline{F}$ by mapping $x$ to $\beta$. This contradicts the maximality of $(E, \varphi)$ as desired. \end{proof} \begin{lemma} \label{lemma-algebraic-closures-isomorphic} Any two algebraic closures of a field are isomorphic. \end{lemma} \begin{proof} Let $F$ be a field. If $M$ and $\overline{F}$ are algebraic closures of $F$, then there exists a morphism of $F$-extensions $\varphi : M \to \overline{F}$ by Lemma \ref{lemma-map-into-algebraic-closure}. Now the image $\varphi(M)$ is algebraically closed. On the other hand, the extension $\varphi(M) \subset \overline{F}$ is algebraic by Lemma \ref{lemma-algebraic-goes-up}. Thus $\varphi(M) = \overline{F}$. \end{proof} \section{Relatively prime polynomials} \label{section-relatively-prime} \noindent Let $K$ be an algebraically closed field. Then the ring $K[x]$ has a very simple ideal structure as we saw in Lemma \ref{lemma-algebraically-closed}. In particular, every polynomial $P \in K[x]$ can be written as $$P = c(x - \alpha_1) \ldots (x - \alpha_n),$$ where $c$ is the constant term and the $\alpha_1, \ldots, \alpha_n \in k$ are the roots of $P$ (counted with multiplicity). Clearly, the only irreducible polynomials in $K[x]$ are the linear polynomials $c(x - \alpha)$, $c, \alpha \in K$ (and $c \neq 0$). \begin{definition} \label{definition-relatively-prime} If $k$ is any field, we say that two polynomials in $k[x]$ are {\it relatively prime} if they generate the unit ideal in $k[x]$. \end{definition} \noindent Continuing the discussion above, if $K$ is an algebraically closed field, two polynomials in $K[x]$ are relatively prime if and only if they have no common roots. This follows because the maximal ideals of $K[x]$ are of the form $(x - \alpha)$, $\alpha \in K$. So if $F, G \in K[x]$ have no common root, then $(F, G)$ cannot be contained in any $(x - \alpha)$ (as then they would have a common root at $\alpha$). \medskip\noindent If $k$ is {\it not} algebraically closed, then this still gives information about when two polynomials in $k[x]$ generate the unit ideal. \begin{lemma} \label{lemma-relatively-prime-polynomials} Two polynomials in $k[x]$ are relatively prime precisely when they have no common roots in an algebraic closure $\overline{k}$ of $k$. \end{lemma} \begin{proof} The claim is that any two polynomials $P, Q$ generate $(1)$ in $k[x]$ if and only if they generate $(1)$ in $\overline{k}[x]$. This is a piece of linear algebra: a system of linear equations with coefficients in $k$ has a solution if and only if it has a solution in any extension of $k$. Consequently, we can reduce to the case of an algebraically closed field, in which case the result is clear from what we have already proved. \end{proof} \section{Separable extensions} \label{section-separable-extensions} \noindent In characteristic $p$ something funny happens with irreducible polynomials over fields. We explain this in the following lemma. \begin{lemma} \label{lemma-irreducible-polynomials} Let $F$ be a field. Let $P \in F[x]$ be an irreducible polynomial over $F$. Let $P' = \text{d}P/\text{d}x$ be the derivative of $P$ with respect to $x$. Then one of the following two cases happens \begin{enumerate} \item $P$ and $P'$ are relatively prime, or \item $P'$ is the zero polynomial. \end{enumerate} Then second case can only happen if $F$ has characteristic $p > 0$. In this case $P(x) = Q(x^q)$ where $q = p^f$ is a power of $p$ and $Q \in F[x]$ is an irreducible polynomial such that $Q$ and $Q'$ are relatively prime. \end{lemma} \begin{proof} Note that $P'$ has degree $< \deg(P)$. Hence if $P$ and $P'$ are not relatively prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg(P)$ contradicting the irreducibility of $P$. This proves we have the dichotomy between (1) and (2). \medskip\noindent Assume we are in case (2) and $P = a_d x^d + \ldots + a_0$. Then $P' = da_d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see that this forces $a_d, \ldots, a_1 = 0$ which would mean $P$ is constant a contradiction. Thus we conclude that the characteristic $p$ is positive. In this case the condition $P' = 0$ forces $a_i = 0$ whenever $p \not | i$. In other words, $P(x) = P_1(x^p)$ for some nonconstant polynomial $P_1$. Clearly, $P_1$ is irreducible as well. By induction on the degree we see that $P_1(x) = Q(x^q)$ as in the statement of the lemma, hence $P(x) = Q(x^{pq})$ and the lemma is proved. \end{proof} \begin{definition} \label{definition-separable} Let $F$ be a field. Let $K/F$ be an extension of fields. \begin{enumerate} \item We say an irreducible polynomial $P$ over $F$ is {\it separable} if it is relatively prime to its derivative. \item Given $\alpha \in K$ algebraic over $F$ we say $\alpha$ is {\it separable} over $F$ if its minimal polynomial is separable over $F$. \item If $K$ is an algebraic extension of $F$, we say $K$ is {\it separable}\footnote{For nonalgebraic extensions this definition does not make sense and is not the correct one.} over $F$ if every element of $K$ is separable over $F$. \end{enumerate} \end{definition} \noindent By Lemma \ref{lemma-irreducible-polynomials} in characteristic $0$ every irreducible polynomial is separable, every algebraic element in an extension is separable, and every algebraic extension is separable. \begin{lemma} \label{lemma-separable-goes-up} Let $K/E/F$ be a tower of algebraic field extensions. \begin{enumerate} \item If $\alpha \in K$ is separable over $F$, then $\alpha$ is separable over $E$. \item if $K$ is separable over $F$, then $K$ is separable over $E$. \end{enumerate} \end{lemma} \begin{proof} We will use Lemma \ref{lemma-irreducible-polynomials} without further mention. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $Q$ be the minimal polynomial of $\alpha$ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Then $P' = Q'R + QR'$. Thus if $Q' = 0$, then $Q$ divides $P$ and $P'$ hence $P' = 0$ by the lemma. This proves (1). Part (2) follows immediately from (1) and the definitions. \end{proof} \begin{lemma} \label{lemma-recognize-separable} Let $F$ be a field. An irreducible polynomial $P$ over $F$ is separable if and only if $P$ has pairwise distinct roots in an algebraic closure of $F$. \end{lemma} \begin{proof} Suppose that $\alpha \in F$ is a root of both $P$ and $P'$. Then $P = (x - \alpha)Q$ for some polynomial $Q$. Taking derivatives we obtain $P' = Q + (x - \alpha)Q'$. Thus $\alpha$ is a root of $Q$. Hence we see that if $P$ and $P'$ have a common root, then $P$ does not have pairwise distinct roots. Conversely, if $P$ has a repeated root, i.e., $(x - \alpha)^2$ divides $P$, then $\alpha$ is a root of both $P$ and $P'$. Combined with Lemma \ref{lemma-relatively-prime-polynomials} this proves the lemma. \end{proof} \begin{lemma} \label{lemma-nr-roots-unchanged} Let $F$ be a field and let $\overline{F}$ be an algebraic closure of $F$. Let $p > 0$ be the characteristic of $F$. Let $P$ be a polynomial over $F$. Then the set of roots of $P$ and $P(x^p)$ in $\overline{F}$ have the same cardinality (not counting multiplicity). \end{lemma} \begin{proof} Clearly, $\alpha$ is a root of $P(x^p)$ if and only if $\alpha^p$ is a root of $P$. In other words, the roots of $P(x^p)$ are the roots of $x^p - \beta$, where $\beta$ is a root of $P$. Thus it suffices to show that the map $\overline{F} \to \overline{F}$, $\alpha \mapsto \alpha^p$ is bijective. It is surjective, as $\overline{F}$ is algebraically closed which means that every element has a $p$th root. It is injective because $\alpha^p = \beta^p$ implies $(\alpha - \beta)^p = 0$ because the characteristic is $p$. And of course in a field $x^p = 0$ implies $x = 0$. \end{proof} \noindent Let $F$ be a field and let $P$ be an irreducible polynomial over $F$. Then we know that $P = Q(x^q)$ for some separable irreducible polynomial $Q$ (Lemma \ref{lemma-irreducible-polynomials}) where $q$ is a power of the characteristic $p$ (and if the characteristic is zero, then $q = 1$\footnote{A good convention for this chapter is to set $0^0 = 1$.} and $Q = P$). By Lemma \ref{lemma-nr-roots-unchanged} the number of roots of $P$ and $Q$ in any algebraic closure of $F$ is the same. By Lemma \ref{lemma-recognize-separable} this number is equal to the degree of $Q$. \begin{definition} \label{definition-separable-degree} Let $F$ be a field. Let $P$ be an irreducible polynomial over $F$. The {\it separable degree} of $P$ is the cardinality of the set of roots of $P$ in any algebraic closure of $F$ (see discussion above). Notation $\deg_s(P)$. \end{definition} \noindent The separable degree of $P$ always divides the degree and the quotient is a power of the characteristic. If the characteristic is zero, then $\deg_s(P) = \deg(P)$. \begin{situation} \label{situation-finitely-generated} Here $F$ be a field and $K/F$ is a finite extension generated by elements $\alpha_1, \ldots, \alpha_n \in K$. We set $K_0 = F$ and $$K_i = F(\alpha_1, \ldots, \alpha_i)$$ to obtain a tower of finite extensions $K = K_n / K_{n - 1} / \ldots / K_0 = F$. Denote $P_i$ the minimal polynomial of $\alpha_i$ over $K_{i - 1}$. Finally, we fix an algebraic closure $\overline{F}$ of $F$. \end{situation} \noindent Let $F$, $K$, $\alpha_i$, and $\overline{F}$ be as in Situation \ref{situation-finitely-generated}. Suppose that $\varphi : K \to \overline{F}$ is a morphism of extensions of $F$. Then we obtain maps $\varphi_i : K_i \to \overline{F}$. In particular, we can take the image of $P_i \in K_{i - 1}[x]$ by $\varphi_{i - 1}$ to get a polynomial $P_i^\varphi \in \overline{F}[x]$. \begin{lemma} \label{lemma-count-embeddings} In Situation \ref{situation-finitely-generated} the correspondence $$\Mor_F(K, \overline{F}) \longrightarrow \{(\beta_1, \ldots, \beta_n)\text{ as below}\}, \quad \varphi \longmapsto (\varphi(\alpha_1), \ldots, \varphi(\alpha_n))$$ is a bijection. Here the right hand side is the set of $n$-tuples $(\beta_1, \ldots, \beta_n)$ of elements of $\overline{F}$ such that $\beta_i$ is a root of $P_i^\varphi$. \end{lemma} \begin{proof} Let $(\beta_1, \ldots, \beta_n)$ be an element of the right hand side. We construct a map of fields corresponding to it by induction. Namely, we set $\varphi_0 : K_0 \to \overline{F}$ equal to the given map $K_0 = F \subset \overline{F}$. Having constructed $\varphi_{i - 1} : K_{i - 1} \to \overline{F}$ we observe that $K_i = K_{i - 1}[x]/(P_i)$. Hence we can set $\varphi_i$ equal to the unique map $K_i \to \overline{F}$ inducing $\varphi_{i - 1}$ on $K_{i - 1}$ and mapping $x$ to $\beta_i$. This works precisely as $\beta_i$ is a root of $P_i^\varphi$. Uniqueness implies that the two constructions are mutually inverse. \end{proof} \begin{lemma} \label{lemma-count-embeddings-explicitly} In Situation \ref{situation-finitely-generated} we have $|\Mor_F(K, \overline{F})| = \prod_{i = 1}^n \deg_s(P_i)$. \end{lemma} \begin{proof} This follows immediately from Lemma \ref{lemma-count-embeddings}. Observe that a key ingredient we are tacitly using here is the well-definedness of the separable degree of an irreducible polynomial which was observed just prior to Definition \ref{definition-separable-degree}. \end{proof} \noindent We now use the result above to characterize separable field extensions. \begin{lemma} \label{lemma-separably-generated-separable} Assumptions and notation as in Situation \ref{situation-finitely-generated}. If each $P_i$ is separable, i.e., each $\alpha_i$ is separable over $K_{i - 1}$, then $$|\Mor_F(K, \overline{F})| = [K : F]$$ and the field extension $K/F$ is separable. If one of the $\alpha_i$ is not separable over $K_{i - 1}$, then $|\Mor_F(K, \overline{F})| < [K : F]$. \end{lemma} \begin{proof} If $\alpha_i$ is separable over $K_{i - 1}$ then $\deg_s(P_i) = \deg(P_i) = [K_i : K_{i - 1}]$ (last equality by Lemma \ref{lemma-degree-minimal-polynomial}). By multiplicativity (Lemma \ref{lemma-multiplicativity-degrees}) we have $$[K : F] = \prod [K_i : K_{i - 1}] = \prod \deg(P_i) = \prod \deg_s(P_i) = |\Mor_F(K, \overline{F})|$$ where the last equality is Lemma \ref{lemma-count-embeddings-explicitly}. By the exact same argument we get the strict inequality $|\Mor_F(K, \overline{F})| < [K : F]$ if one of the $\alpha_i$ is not separable over $K_{i - 1}$. \medskip\noindent Finally, assume again that each $\alpha_i$ is separable over $K_{i - 1}$. Let $\gamma = \gamma_1 \in K$ be arbitrary. Then we can find additional elements $\gamma_2, \ldots, \gamma_m$ such that $K = F(\gamma_1, \ldots, \gamma_m)$ (for example we could take $\gamma_2 = \alpha_1, \ldots, \gamma_{n + 1} = \alpha_n$). Then we see by the last part of the lemma (already proven above) that if $\gamma$ is not separable over $F$ we would have the strict inequality $|\Mor_F(K, \overline{F})| < [K : F]$ contradicting the very first part of the lemma (already prove above as well). \end{proof} \begin{lemma} \label{lemma-separable-equality} Let $K/F$ be a finite extension of fields. Let $\overline{F}$ be an algebraic closure of $F$. Then we have $$|\Mor_F(K, \overline{F})| \leq [K : F]$$ with equality if and only if $K$ is separable over $F$. \end{lemma} \begin{proof} This is a corollary of Lemma \ref{lemma-separably-generated-separable}. Namely, since $K/F$ is finite we can find finitely many elements $\alpha_1, \ldots, \alpha_n \in K$ generating $K$ over $F$ (for example we can choose the $\alpha_i$ to be a basis of $K$ over $F$). If $K/F$ is separable, then each $\alpha_i$ is separable over $F(\alpha_1, \ldots, \alpha_{i - 1})$ by Lemma \ref{lemma-separable-goes-up} and we get equality by Lemma \ref{lemma-separably-generated-separable}. On the other hand, if we have equality, then no matter how we choose $\alpha_1, \ldots, \alpha_n$ we get that $\alpha_1$ is separable over $F$ by Lemma \ref{lemma-separably-generated-separable}. Since we can start the sequence with an arbitrary element of $K$ it follows that $K$ is separable over $F$. \end{proof} \begin{lemma} \label{lemma-separable-permanence} Let $E/k$ and $F/E$ be separable algebraic extensions of fields. Then $F/k$ is a separable extension of fields. \end{lemma} \begin{proof} Choose $\alpha \in F$. Then $\alpha$ is separable algebraic over $E$. Let $P = x^d + \sum_{i < d} a_i x^i$ be the minimal polynomial of $\alpha$ over $E$. Each $a_i$ is separable algebraic over $k$. Consider the tower of fields $$k \subset k(a_0) \subset k(a_0, a_1) \subset \ldots \subset k(a_0, \ldots, a_{d - 1}) \subset k(a_0, \ldots, a_{d - 1}, \alpha)$$ Because $a_i$ is separable algebraic over $k$ it is separable algebraic over $k(a_0, \ldots, a_{i - 1})$ by Lemma \ref{lemma-separable-goes-up}. Finally, $\alpha$ is separable algebraic over $k(a_0, \ldots, a_{d - 1})$ because it is a root of $P$ which is irreducible (as it is irreducible over the possibly bigger field $E$) and separable (as it is separable over $E$). Thus $k(a_0, \ldots, a_{d - 1}, \alpha)$ is separable over $k$ by Lemma \ref{lemma-separably-generated-separable} and we conclude that $\alpha$ is separable over $k$ as desired. \end{proof} \begin{lemma} \label{lemma-separable-elements} Let $E/k$ be a field extension. Then the elements of $E$ separable over $k$ form a subextension of $E/k$. \end{lemma} \begin{proof} Let $\alpha, \beta \in E$ be separable over $k$. Then $\beta$ is separable over $k(\alpha)$ by Lemma \ref{lemma-separable-goes-up}. Thus we can apply Lemma \ref{lemma-separable-permanence} to $k(\alpha, \beta)$ to see that $k(\alpha, \beta)$ is separable over $k$. \end{proof} \section{Linear independence of characters} \label{section-independence-characters} \noindent Here is the statement. \begin{lemma} \label{lemma-independence-characters} Let $L$ be a field. Let $G$ be a group. Let $\chi_1, \ldots, \chi_n : G \to L^*$ be pairwise distinct homomorphisms of groups. Then $\chi_1, \ldots, \chi_n$ are $L$-linearly independent: if $\lambda_1, \ldots, \lambda_n \in L$ not all zero, then $\sum \lambda_i\sigma_i(g) \not = 0$ for some $g \in G$. \end{lemma} \begin{proof} If $n = 1$ this is true because $\chi_1(e) = 1$ if $e \in G$ is the neutral element. We prove the result by induction for $n > 1$. Suppose that $\lambda_1, \ldots, \lambda_n \in L$ not all zero. If $\lambda_i = 0$ for some, then we win by induction on $n$. Since we want to show that $\sum \lambda_i\sigma_i(g) \not = 0$ for some $g \in G$ we may after dividing by $-\lambda_n$ assume that $\lambda_n = -1$. Then the only way we get in trouble is if $$\chi_n(g) = \sum\nolimits_{i = 1, \ldots, n - 1} \lambda_i\chi_i(g)$$ for all $g \in G$. Fix $h \in G$. Then we would also get \begin{align*} \chi_n(h)\chi_n(g) & = \chi_n(hg) \\ & = \sum\nolimits_{i = 1, \ldots, n - 1} \lambda_i\chi_i(hg) \\ & = \sum\nolimits_{i = 1, \ldots, n - 1} \lambda_i\chi_i(h) \chi_i(g) \end{align*} In other words, we would get $$\chi_n(g) = \sum\nolimits_{i = 1, \ldots, n - 1} \lambda_i \chi_i(h) \chi_n(h)^{-1}\chi_i(g)$$ Substracting from the previous relation we obtain $$0 = \sum\nolimits_{i = 1, \ldots, n - 1} \lambda_i (1 - \chi_i(h) \chi_n(h)^{-1}) \chi_i(g)$$ for all $g \in G$. Since $\lambda_i \not = 0$ we conclude that $1 - \chi_i(h) \chi_n(h)^{-1} = 0$ for all $i$ by induction. The choice of $h$ above was arbitrary, so we conclude that $\chi_i = \chi_n$ for $i \leq n - 1$ which contradicts the assumption that our characters $\chi_i$ are pairwise distinct. \end{proof} \begin{lemma} \label{lemma-independence-embeddings} Let $K/F$ and $L/F$ be field extensions. Let $\sigma_1, \ldots, \sigma_n : K \to L$ be pairwise distinct morphisms of $F$-extensions. Then $\sigma_1, \ldots, \sigma_n$ are $L$-linearly independent: if $\lambda_1, \ldots, \lambda_n \in L$ not all zero, then $\sum \lambda_i\sigma_i(\alpha) \not = 0$ for some $\alpha \in K$. \end{lemma} \begin{proof} Apply Lemma \ref{lemma-independence-characters} to the restrictions of $\sigma_i$ to the groups of units. \end{proof} \begin{lemma} \label{lemma-finite-separable-tensor-alg-closed} Let $K/F$ and $L/F$ be field extensions with $K/F$ finite separable and $L$ algebraically closed. Then the map $$K \otimes_F L \longrightarrow \prod\nolimits_{\sigma \in \Hom_F(K, L)} L,\quad \alpha \otimes \beta \mapsto (\sigma(\alpha)\beta)_\sigma$$ is an isomorphism of $L$-algebras. \end{lemma} \begin{proof} Choose a basis $\alpha_1, \ldots, \alpha_n$ of $K$ as a vector space over $F$. By Lemma \ref{lemma-separable-equality} (and a tiny omitted argument) the set $\Hom_F(K, L)$ has $n$ elements, say $\sigma_1, \ldots, \sigma_n$. In particular, the two sides have the same dimension $n$ as vector spaces over $L$. Thus if the map is not an isomorphism, then it has a kernel. In other words, there would exist $\mu_j \in L$, $j = 1, \ldots, n$ not all zero, with $\sum \alpha_j \otimes \mu_j$ in the kernel. In other words, $\sum \sigma_i(\alpha_j)\mu_j = 0$ for all $i$. This would mean the $n \times n$ matrix with entries $\sigma_i(\alpha_j)$ is not invertible. Thus we can find $\lambda_1, \ldots, \lambda_n \in L$ not all zero, such that $\sum \lambda_i\sigma_i(\alpha_j) = 0$ for all $j$. Now any element $\alpha \in K$ can be written as $\alpha = \sum \beta_j \alpha_j$ with $\beta_j \in F$ and we would get $$\sum \lambda_i\sigma_i(\alpha) = \sum \lambda_i\sigma_i(\sum \beta_j \alpha_j) = \sum \beta_j \sum \lambda_i\sigma_i(\alpha_j) = 0$$ which contradicts Lemma \ref{lemma-independence-embeddings}. \end{proof} \section{Purely inseparable extensions} \label{section-purely-inseparable} \noindent Purely inseparable extensions are the opposite of the separable extensions defined in the previous section. These extensions only show up in positive characteristic. \begin{definition} \label{definition-purely-inseparable} Let $F$ be a field of characteristic $p > 0$. Let $K/F$ be an extension. \begin{enumerate} \item An element $\alpha \in K$ is {\it purely inseparable} over $F$ if there exists a power $q$ of $p$ such that $\alpha^q \in F$. \item The extension $K/F$ is said to be {\it purely inseparable} if and only if every element of $K$ is purely inseparable over $F$. \end{enumerate} \end{definition} \noindent Observe that a purely inseparable extension is necessarily algebraic. Let $F$ be a field of characteristic $p > 0$. An example of a purely inseparable extension is gotten by adjoining the $p$th root of an element $t \in F$ which does not yet have one. Namely, the lemma below shows that $P = x^p - t$ is irreducible, and hence $$K = F[x]/(P) = F[t^{1/p}]$$ is a field. And $K$ is purely inseparable over $F$ because every element $$a_0 + a_1t^{1/p} + \ldots + a_{p - 1}t^{p - 1/p}, a_i \in F$$ has $p$th power equal to $$(a_0 + a_1t^{1/p} + \ldots + a_{p - 1}t^{p - 1/p})^p = a_0^p + a_1^p t + \ldots + a_{p - 1}^pt^{p - 1} \in F$$ This situation occurs for the field $\mathbf{F}_p(t)$ of rational functions over $\mathbf{F}_p$. \begin{lemma} \label{lemma-take-pth-root} Let $p$ be a prime number. Let $F$ be a field of characteristic $p$. Let $t \in F$ be an element which does not have a $p$th root in $F$. Then the polynomial $x^p - t$ is irreducible over $F$. \end{lemma} \begin{proof} To see this, suppose that we have a factorization $x^p - t = f g$. Taking derivatives we get $f' g + f g' = 0$. Note that neither $f' = 0$ nor $g' = 0$ as the degrees of $f$ and $g$ are smaller than $p$. Moreover, $\deg(f') < \deg(f)$ and $\deg(g') < \deg(g)$. We conclude that $f$ and $g$ have a factor in common. Thus if $x^p - t$ is reducible, then it is of the form $x^p - t = c f^n$ for some irreducible $f$, $c \in F^*$, and $n > 1$. Since $p$ is a prime number this implies $n = p$ and $f$ linear, which would imply $x^p - t$ has a root in $F$. Contradiction. \end{proof} \noindent We will see that taking $p$th roots is a very important operation in characteristic $p$. \begin{lemma} \label{lemma-purely-inseparable-permanence} Let $E/k$ and $F/E$ be purely inseparable extensions of fields. Then $F/k$ is a purely inseparable extension of fields. \end{lemma} \begin{proof} Say the characteristic of $k$ is $p$. Choose $\alpha \in F$. Then $\alpha^q \in E$ for some $p$-power $q$. Whereupon $(\alpha^q)^{q'} \in k$ for some $p$-power $q'$. Hence $\alpha^{qq'} \in k$. \end{proof} \begin{lemma} \label{lemma-purely-inseparable-elements} Let $E/k$ be a field extension. Then the elements of $E$ purely-inseparable over $k$ form a subextension of $E/k$. \end{lemma} \begin{proof} Let $p$ be the characteristic of $k$. Let $\alpha, \beta \in E$ be purely inseparable over $k$. Say $\alpha^q \in k$ and $\beta^{q'} \in k$ for some $p$-powers $q, q'$. If $q''$ is a $p$-power, then $(\alpha + \beta)^{q''} = \alpha^{q''} + \beta^{q''}$. Hence if $q'' \geq q, q'$, then we conclude that $\alpha + \beta$ is purely inseparable over $k$. Similarly for the difference, product and quotient of $\alpha$ and $\beta$. \end{proof} \begin{lemma} \label{lemma-finite-purely-inseparable} Let $E/F$ be a finite purely inseparable field extension of characteristic $p > 0$. Then there exists a sequence of elements $\alpha_1, \ldots, \alpha_n \in E$ such that we obtain a tower of fields $$E = F(\alpha_1, \ldots, \alpha_n) \supset F(\alpha_1, \ldots, \alpha_{n - 1}) \supset \ldots \supset F(\alpha_1) \supset F$$ such that each intermediate extension is of degree $p$ and comes from adjoining a $p$th root. Namely, $\alpha_i^p \in F(\alpha_1, \ldots, \alpha_{i - 1})$ is an element which does not have a $p$th root in $F(\alpha_1, \ldots, \alpha_{i - 1})$ for $i = 1, \ldots, n$. \end{lemma} \begin{proof} By induction on the degree of $E/F$. If the degree of the extension is $1$ then the result is clear (with $n = 0$). If not, then choose $\alpha \in E$, $\alpha \not \in F$. Say $\alpha^{p^r} \in F$ for some $r > 0$. Pick $r$ minimal and replace $\alpha$ by $\alpha^{p^{r - 1}}$. Then $\alpha \not \in F$, but $\alpha^p \in F$. Then $t = \alpha^p$ is not a $p$th power in $F$ (because that would imply $\alpha \in F$, see Lemma \ref{lemma-nr-roots-unchanged} or its proof). Thus $F \subset F(\alpha)$ is a subextension of degree $p$ (Lemma \ref{lemma-take-pth-root}). By induction we find $\alpha_1, \ldots, \alpha_n \in E$ generating $E/F(\alpha)$ satisfying the conclusions of the lemma. The sequence $\alpha, \alpha_1, \ldots, \alpha_n$ does the job for the extension $E/F$. \end{proof} \begin{lemma} \label{lemma-separable-first} \begin{slogan} Any algebraic field extension is uniquely a separable field extension followed by a purely inseparable one. \end{slogan} Let $E/F$ be an algebraic field extension. There exists a unique subextension $F \subset E_{sep} \subset E$ such that $E_{sep}/F$ is separable and $E/E_{sep}$ is purely inseparable. \end{lemma} \begin{proof} If the characteristic is zero we set $E_{sep} = E$. Assume the characteristic if $p > 0$. Let $E_{sep}$ be the set of elements of $E$ which are separable over $F$. This is a subextension by Lemma \ref{lemma-separable-elements} and of course $E_{sep}$ is separable over $F$. Given an $\alpha$ in $E$ there exists a $p$-power $q$ such that $\alpha^q$ is separable over $F$. Namely, $q$ is that power of $p$ such that the minimal polynomial of $\alpha$ is of the form $P(x^q)$ with $P$ separable algebraic, see Lemma \ref{lemma-irreducible-polynomials}. Hence $E/E_{sep}$ is purely inseparable. Uniqueness is clear. \end{proof} \begin{definition} \label{definition-insep-degree} Let $E/F$ be an algebraic field extension. Let $E_{sep}$ be the subextension found in Lemma \ref{lemma-separable-first}. \begin{enumerate} \item The integer $[E_{sep} : F]$ is called the {\it separable degree} of the extension. Notation $[E : F]_s$. \item The integer $[E : E_{sep}]$ is called the {\it inseparable degree}, or the {\it degree of inseparability} of the extension. Notation $[E : F]_i$. \end{enumerate} \end{definition} \noindent Of course in characteristic $0$ we have $[E : F] = [E : F]_s$ and $[E : F]_i = 1$. By multiplicativity (Lemma \ref{lemma-multiplicativity-degrees}) we have $$[E : F] = [E : F]_s [E : F]_i$$ even in case some of these degrees are infinite. In fact, the separable degree and the inseparable degree are multiplicative too (see Lemma \ref{lemma-multiplicativity-all-degrees}). \begin{lemma} \label{lemma-separable-degree} Let $K/F$ be a finite extension. Let $\overline{F}$ be an algebraic closure of $F$. Then $[K : F]_s = |\Mor_F(K, \overline{F})|$. \end{lemma} \begin{proof} We first prove this when $K/F$ is purely inseparable. Namely, we claim that in this case there is a unique map $K \to \overline{F}$. This can be seen by choosing a sequence of elements $\alpha_1, \ldots, \alpha_n \in K$ as in Lemma \ref{lemma-finite-purely-inseparable}. The irreducible polynomial of $\alpha_i$ over $F(\alpha_1, \ldots, \alpha_{i - 1})$ is $x^p - \alpha_i^p$. Applying Lemma \ref{lemma-count-embeddings-explicitly} we see that $|\Mor_F(K, \overline{F})| = 1$. On the other hand, $[K : F]_s = 1$ in this case hence the equality holds. \medskip\noindent Let's return to a general finite extension $K/F$. In this case choose $F \subset K_s \subset K$ as in Lemma \ref{lemma-separable-first}. By Lemma \ref{lemma-separable-equality} we have $|\Mor_F(K_s, \overline{F})| = [K_s : F] = [K : F]_s$. On the other hand, every field map $\sigma' : K_s \to \overline{F}$ extends to a unique field map $\sigma : K \to \overline{F}$ by the result of the previous paragraph. In other words $|\Mor_F(K, \overline{F})| = |\Mor_F(K_s, \overline{F})|$ and the proof is done. \end{proof} \begin{lemma}[Multiplicativity] \label{lemma-multiplicativity-all-degrees} Suppose given a tower of algebraic field extensions $K/E/F$. Then $$[K : F]_s = [K : E]_s [E : F]_s \quad\text{and}\quad [K : F]_i = [K : E]_i [E : F]_i$$ \end{lemma} \begin{proof} We first prove this in case $K$ is finite over $F$. Since we have multiplicativity for the usual degree (by Lemma \ref{lemma-multiplicativity-degrees}) it suffices to prove one of the two formulas. By Lemma \ref{lemma-separable-degree} we have $[K : F]_s = |\Mor_F(K, \overline{F})|$. By the same lemma, given any $\sigma \in \Mor_F(E, \overline{F})$ the number of extensions of $\sigma$ to a map $\tau : K \to \overline{F}$ is $[K : E]_s$. Namely, via $E \cong \sigma(E) \subset \overline{F}$ we can view $\overline{F}$ as an algebraic closure of $E$. Combined with the fact that there are $[E : F]_s = |\Mor_F(E, \overline{F})|$ choices for $\sigma$ we obtain the result. \medskip\noindent If the extensions are infinite one can write $K$ as the union of all finite subextension $F \subset K' \subset K$. For each $K'$ we set $E' = E \cap K'$. Then we have the formulas of the lemma for $K'/E'/F$ by the first paragraph. Since $[K : F]_s = \sup \{[K' : F]_s\}$ and similarly for the other degrees (some details omitted) we obtain the result in general. \end{proof} \section{Normal extensions} \label{section-normal} \noindent Let $P \in F[x]$ be a nonconstant polynomial over a field $F$. We say $P$ {\it splits completely into linear factors over $F$} or {\it splits completely over $F$} if there exist $c \in F^*$, $n \geq 1$, $\alpha_1, \ldots, \alpha_n \in F$ such that $$P = c(x - \alpha_1) \ldots (x - \alpha_n)$$ in $F[x]$. Normal extensions are defined as follows. \begin{definition} \label{definition-normal} Let $E/F$ be an algebraic field extension. We say $E$ is {\it normal} over $F$ if for all $\alpha \in E$ the minimal polynomial $P$ of $\alpha$ over $F$ splits completely into linear factors over $E$. \end{definition} \noindent As in the case of separable extensions, it takes a bit of work to establish the basic properties of this notion. \begin{lemma} \label{lemma-normal-goes-up} Let $K/E/F$ be a tower of algebraic field extensions. If $K$ is normal over $F$, then $K$ is normal over $E$. \end{lemma} \begin{proof} Let $\alpha \in K$. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $Q$ be the minimal polynomial of $\alpha$ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Hence, if $P$ splits completely over $K$, then so does $Q$. \end{proof} \begin{lemma} \label{lemma-intersect-normal} Let $F$ be a field. Let $M/F$ be an algebraic extension. Let $F \subset E_i \subset M$, $i \in I$ be subextensions with $E_i/F$ normal. Then $\bigcap E_i$ is normal over $F$. \end{lemma} \begin{proof} Direct from the definitions. \end{proof} \begin{lemma} \label{lemma-characterize-normal} Let $E/F$ be an algebraic extension of fields. Let $\overline{F}$ be an algebraic closure of $F$. The following are equivalent \begin{enumerate} \item $E$ is normal over $F$, and \item for every pair $\sigma, \sigma' \in \Mor_F(E, \overline{F})$ we have $\sigma(E) = \sigma'(E)$. \end{enumerate} \end{lemma} \begin{proof} Let $\mathcal{P}$ be the set of all minimal polynomials over $F$ of all elements of $E$. Set $$T = \{\beta \in \overline{F} \mid P(\beta) = 0\text{ for some }P \in \mathcal{P}\}$$ It is clear that if $E$ is normal over $F$, then $\sigma(E) = T$ for all $\sigma \in \Mor_F(E, \overline{F})$. Thus we see that (1) implies (2). \medskip\noindent Conversely, assume (2). Pick $\beta \in T$. We can find a corresponding $\alpha \in E$ whose minimal polynomial $P \in \mathcal{P}$ annihilates $\beta$. Because $F(\alpha) = F[x]/(P)$ we can find an element $\sigma_0 \in \Mor_F(F(\alpha), \overline{F})$ mapping $\alpha$ to $\beta$. By Lemma \ref{lemma-map-into-algebraic-closure} we can extend $\sigma_0$ to a $\sigma \in \Mor_F(E, \overline{F})$. Whence we see that $\beta$ is in the common image of all embeddings $\sigma : E \to \overline{F}$. It follows that $\sigma(E) = T$ for any $\sigma$. Fix a $\sigma$. Now let $P \in \mathcal{P}$. Then we can write $$P = (x - \beta_1) \ldots (x - \beta_n)$$ for some $n$ and $\beta_i \in \overline{F}$ by Lemma \ref{lemma-algebraically-closed}. Observe that $\beta_i \in T$. Thus $\beta_i = \sigma(\alpha_i)$ for some $\alpha_i \in E$. Thus $P = (x - \alpha_1) \ldots (x - \alpha_n)$ splits completely over $E$. This finishes the proof. \end{proof} \begin{lemma} \label{lemma-normally-generated} Let $E/F$ be an algebraic extension of fields. If $E$ is generated by $\alpha_i \in E$, $i \in I$ over $F$ and if for each $i$ the minimal polynomial of $\alpha_i$ over $F$ splits completely in $E$, then $E/F$ is normal. \end{lemma} \begin{proof} Let $P_i$ be the minimal polynomial of $\alpha_i$ over $F$. Let $\alpha_i = \alpha_{i, 1}, \alpha_{i, 2}, \ldots, \alpha_{i, d_i}$ be the roots of $P_i$ over $E$. Given two embeddings $\sigma, \sigma' : E \to \overline{F}$ over $F$ we see that $$\{\sigma(\alpha_{i, 1}), \ldots, \sigma(\alpha_{i, d_i})\} = \{\sigma'(\alpha_{i, 1}), \ldots, \sigma'(\alpha_{i, d_i})\}$$ because both sides are equal to the set of roots of $P_i$ in $\overline{F}$. The elements $\alpha_{i, j}$ generate $E$ over $F$ and we find that $\sigma(E) = \sigma'(E)$. Hence $E/F$ is normal by Lemma \ref{lemma-characterize-normal}. \end{proof} \begin{lemma} \label{lemma-lift-maps} Let $L/M/K$ be a tower of algebraic extensions. \begin{enumerate} \item If $M/K$ is normal, then any automorphism $\tau$ of $L/K$ induces an automorphism $\tau|_M : M \to M$. \item If $L/K$ is normal, then $K$-algebra map $\sigma : M \to L$ extends to an automorphism of $L$. \end{enumerate} \end{lemma} \begin{proof} Choose an algebraic closure $\overline{L}$ of $L$ (Theorem \ref{theorem-existence-algebraic-closure}). \medskip\noindent Let $\tau$ be as in (1). Then $\tau(M) = M$ as subfields of $\overline{L}$ by Lemma \ref{lemma-characterize-normal} and hence $\tau|_M : M \to M$ is an automorphism. \medskip\noindent Let $\sigma : M \to L$ be as in (2). By Lemma \ref{lemma-map-into-algebraic-closure} we can extend $\sigma$ to a map $\tau : L \to \overline{L}$, i.e., such that $$\xymatrix{ L \ar[r]_\tau & \overline{L} \\ M \ar[u] \ar[ru]_\sigma & K \ar[l] \ar[u] }$$ is commutative. By Lemma \ref{lemma-characterize-normal} we see that $\tau(L) = L$. Hence $\tau : L \to L$ is an automorphism which extends $\sigma$. \end{proof} \begin{definition} \label{definition-automorphisms} Let $E/F$ be an extension of fields. Then $\text{Aut}(E/F)$ or $\text{Aut}_F(E)$ denotes the automorphism group of $E$ as an object of the category of $F$-extensions. Elements of $\text{Aut}(E/F)$ are called {\it automorphisms of $E$ over $F$} or {\it automorphisms of $E/F$}. \end{definition} \noindent Here is a characterization of normal extensions in terms of automorphisms. \begin{lemma} \label{lemma-normal-and-automorphisms} Let $E/F$ be a finite extension. We have $$|\text{Aut}(E/F)| \leq [E : F]_s$$ with equality if and only if $E$ is normal over $F$. \end{lemma} \begin{proof} Choose an algebraic closure $\overline{F}$ of $F$. Recall that $[E : F]_s = |\Mor_F(E, \overline{F})|$. Pick an element $\sigma_0 \in \Mor_F(E, \overline{F})$. Then the map $$\text{Aut}(E/F) \longrightarrow \Mor_F(E, \overline{F}),\quad \tau \longmapsto \sigma_0 \circ \tau$$ is injective. Thus the inequality. If equality holds, then every $\sigma \in \Mor_F(E, \overline{F})$ is gotten by precomposing $\sigma_0$ by an automorphism. Hence $\sigma(E) = \sigma_0(E)$. Thus $E$ is normal over $F$ by Lemma \ref{lemma-characterize-normal}. \medskip\noindent Conversely, assume that $E/F$ is normal. Then by Lemma \ref{lemma-characterize-normal} we have $\sigma(E) = \sigma_0(E)$ for all $\sigma \in \Mor_F(E, \overline{F})$. Thus we get an automorphism of $E$ over $F$ by setting $\tau = \sigma_0^{-1} \circ \sigma$. Whence the map displayed above is surjective. \end{proof} \begin{lemma} \label{lemma-normal-embeddings-differ-by-aut} Let $L/K$ be an algebraic normal extension of fields. Let $E/K$ be an extension of fields. Then either there is no $K$-embedding from $L$ to $E$ or there is one $\tau : L \to E$ and every other one is of the form $\tau \circ \sigma$ where $\sigma \in \text{Aut}(L/K)$. \end{lemma} \begin{proof} Given $\tau$ replace $L$ by $\tau(L) \subset E$ and apply Lemma \ref{lemma-lift-maps}. \end{proof} \section{Splitting fields} \label{section-splitting-fieds} \noindent The following lemma is a useful tool for constructing normal field extensions. \begin{lemma} \label{lemma-splitting-field} Let $F$ be a field. Let $P \in F[x]$ be a nonconstant polynomial. There exists a smallest field extension $E/F$ such that $P$ splits completely over $E$. Moreover, the field extension $E/F$ is normal and unique up to (nonunique) isomorphism. \end{lemma} \begin{proof} Choose an algebraic closure $\overline{F}$. Then we can write $P = c (x - \beta_1) \ldots (x - \beta_n)$ in $\overline{F}[x]$, see Lemma \ref{lemma-algebraically-closed}. Note that $c \in F^*$. Set $E = F(\beta_1, \ldots, \beta_n)$. Then it is clear that $E$ is minimal with the requirement that $P$ splits completely over $E$. \medskip\noindent Next, let $E'$ be another minimal field extension of $F$ such that $P$ splits completely over $E'$. Write $P = c (x - \alpha_1) \ldots (x - \alpha_n)$ with $c \in F$ and $\alpha_i \in E'$. Again it follows from minimality that $E' = F(\alpha_1, \ldots, \alpha_n)$. Moreover, if we pick any $\sigma : E' \to \overline{F}$ (Lemma \ref{lemma-map-into-algebraic-closure}) then we immediately see that $\sigma(\alpha_i) = \beta_{\tau(i)}$ for some permutation $\tau : \{1, \ldots, n\} \to \{1, \ldots, n\}$. Thus $\sigma(E') = E$. This implies that $E'$ is a normal extension of $F$ by Lemma \ref{lemma-characterize-normal} and that $E \cong E'$ as extensions of $F$ thereby finishing the proof. \end{proof} \begin{definition} \label{definition-splitting-field} Let $F$ be a field. Let $P \in F[x]$ be a nonconstant polynomial. The field extension $E/F$ constructed in Lemma \ref{lemma-splitting-field} is called the {\it splitting field of $P$ over $F$}. \end{definition} \begin{lemma} \label{lemma-normal-closure} \begin{slogan} Existence of normal closure of finite extensions of fields. \end{slogan} Let $E/F$ be a finite extension of fields. There exists a unique smallest finite extension $K/E$ such that $K$ is normal over $F$. \end{lemma} \begin{proof} Choose generators $\alpha_1, \ldots, \alpha_n$ of $E$ over $F$. Let $P_1, \ldots, P_n$ be the minimal polynomials of $\alpha_1, \ldots, \alpha_n$ over $F$. Set $P = P_1 \ldots P_n$. Observe that $(x - \alpha_1) \ldots (x - \alpha_n)$ divides $P$, since each $(x - \alpha_i)$ divides $P_i$. Say $P = (x - \alpha_1) \ldots (x - \alpha_n)Q$. Let $K/E$ be the splitting field of $P$ over $E$. We claim that $K$ is the splitting field of $P$ over $F$ as well (which implies that $K$ is normal over $F$). This is clear because $K/E$ is generated by the roots of $Q$ over $E$ and $E$ is generated by the roots of $(x - \alpha_1) \ldots (x - \alpha_n)$ over $F$, hence $K$ is generated by the roots of $P$ over $F$. \medskip\noindent Uniqueness. Suppose that $K'/E$ is a second smallest extension such that $K'/F$ is normal. Choose an algebraic closure $\overline{F}$ and an embedding $\sigma_0 : E \to \overline{F}$. By Lemma \ref{lemma-map-into-algebraic-closure} we can extend $\sigma_0$ to $\sigma : K \to \overline{F}$ and $\sigma' : K' \to \overline{F}$. By Lemma \ref{lemma-intersect-normal} we see that $\sigma(K) \cap \sigma'(K')$ is normal over $F$. By minimality we conclude that $\sigma(K) = \sigma(K')$. Thus $\sigma \circ (\sigma')^{-1} : K' \to K$ gives an isomorphism of extensions of $E$. \end{proof} \begin{definition} \label{definition-normal-closure} Let $E/F$ be a finite extension of fields. The field extension $K/E$ constructed in Lemma \ref{lemma-normal-closure} is called the {\it normal closure $E$ over $F$}. \end{definition} \noindent One can construct the normal closure inside any given normal extension. \begin{lemma} \label{lemma-normal-closure-inside-normal} Let $L/K$ be an algebraic normal extension. \begin{enumerate} \item If $L/M/K$ is a subextension with $M/K$ finite, then there exists a tower $L/M'/M/K$ with $M'/K$ finite and normal. \item If $L/M'/M/K$ is a tower with $M/K$ normal and $M'/M$ finite, then there exists a tower $L/M''/M'/M/K$ with $M''/M$ finite and $M''/K$ normal. \end{enumerate} \end{lemma} \begin{proof} Proof of (1). Let $M'$ be the smallest subextension of $L/K$ containing $M$ which is normal over $K$. By Lemma \ref{lemma-normal-closure} this is the normal closure of $M/K$ and is finite over $K$. \medskip\noindent Proof of (2). Let $\alpha_1, \ldots, \alpha_n \in M'$ generate $M'$ over $M$. Let $P_1, \ldots, P_n$ be the minimal polynomials of $\alpha_1, \ldots, \alpha_n$ over $K$. Let $\alpha_{i, j}$ be the roots of $P_i$ in $L$. Let $M'' = M(\alpha_{i, j})$. It follows from Lemma \ref{lemma-normally-generated} (applied with the set of generators $M \cup \{\alpha_{i, j}\}$) that $M''$ is normal over $K$. \end{proof} \section{Roots of unity} \label{section-roots-of-1} \noindent Let $F$ be a field. For an integer $n \geq 1$ we set $$\mu_n(F) = \{\zeta \in F \mid \zeta^n = 1\}$$ This is called the {\it group of $n$th roots of unity} or {\it $n$th roots of $1$}. It is an abelian group under multiplication with neutral element given by $1$. Observe that in a field the number of roots of a polynomial of degree $d$ is always at most $d$. Hence we see that $|\mu_n(F)| \leq n$ as it is defined by a polynomial equation of degree $n$. Of course every element of $\mu_n(F)$ has order dividing $n$. Moreover, the subgroups $$\mu_d(F) \subset \mu_n(F),\quad d | n$$ each have at most $d$ elements. This implies that $\mu_n(F)$ is cyclic. \begin{lemma} \label{lemma-cyclic} Let $A$ be an abelian group of exponent dividing $n$ such that $\{x \in A \mid dx = 0\}$ has cardinality at most $d$ for all $d | n$. Then $A$ is cyclic of order dividing $n$. \end{lemma} \begin{proof} The conditions imply that $|A| \leq n$, in particular $A$ is finite. The structure of finite abelian groups shows that $A = \mathbf{Z}/e_1\mathbf{Z} \oplus \ldots \oplus \mathbf{Z}/e_r\mathbf{Z}$ for some integers $1 < e_1 | e_2 | \ldots | e_r$. This would imply that $\{x \in A \mid e_1 x = 0\}$ has cardinality $e_1^r$. Hence $r = 1$. \end{proof} \noindent Applying this to the field $\mathbf{F}_p$ we obtain the celebrated result that the group $(\mathbf{Z}/p\mathbf{Z})^*$ is a cyclic group. More about this in the section on finite fields. \medskip\noindent One more observation is often useful: If $F$ has characteristic $p > 0$, then $\mu_{p^n}(F) = \{1\}$. This is true because raising to the $p$th power is an injective map on fields of characteristic $p$ as we have seen in the proof of Lemma \ref{lemma-nr-roots-unchanged}. (Of course, it also follows from the statement of that lemma itself.) \section{Finite fields} \label{section-finite} \noindent Let $F$ be a finite field. It is clear that $F$ has positive characteristic as we cannot have an injection $\mathbf{Q} \to F$. Say the characteristic of $F$ is $p$. The extension $\mathbf{F}_p \subset F$ is finite. Hence we see that $F$ has $q = p^f$ elements for some $f \geq 1$. \medskip\noindent Let us think about the group of units $F^*$. This is a finite abelian group, so it has some exponent $e$. Then $F^* = \mu_e(F)$ and we see from the discussion in Section \ref{section-roots-of-1} that $F^*$ is a cyclic group of order $q - 1$. (A posteriori it follows that $e = q - 1$ as well.) In particular, if $\alpha \in F^*$ is a generator then it clearly is true that $$F = \mathbf{F}_p(\alpha)$$ In other words, the extension $F/\mathbf{F}_p$ is generated by a single element. Of course, the same thing is true for any extension of finite fields $E/F$ (because $E$ is already generated by a single element over the prime field). \section{Primitive elements} \label{section-primitive-element} \noindent Let $E/F$ be a finite extension of fields. An element $\alpha \in E$ is called a {\it primitive element of $E$ over $F$} if $E = F(\alpha)$. \begin{lemma}[Primitive element] \label{lemma-primitive-element} Let $E/F$ be a finite extension of fields. The following are equivalent \begin{enumerate} \item there exists a primitive element for $E$ over $F$, and \item there are finitely many subextensions $E/K/F$. \end{enumerate} Moreover, (1) and (2) hold if $E/F$ is separable. \end{lemma} \begin{proof} Let $\alpha \in E$ be a primitive element. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $E \subset M$ be a splitting field for $P$ over $E$, so that $P(x) = (x - \alpha)(x - \alpha_2) \ldots (x - \alpha_n)$ over $M$. For ease of notation we set $\alpha_1 = \alpha$. Next, let $E/K/F$ be a subextension. Let $Q$ be the minimal polynomial of $\alpha$ over $K$. Observe that $\deg(Q) = [E : K]$. Writing $Q = x^d + \sum_{i < d} a_i x^i$ we claim that $K$ is equal to $L = F(a_0, \ldots, a_{d - 1})$. Indeed $\alpha$ has degree $d$ over $L$ and $L \subset K$. Hence $[E : L] = [E : K]$ and it follows that $[K : L] = 1$, i.e., $K = L$. Thus it suffices to show there are at most finitely many possibilities for the polynomial $Q$. This is clear because we have a factorization $P = QR$ in $K[x]$ in particular in $E[x]$. Since we have unique factorization in $E[x]$ there are at most finitely many monic factors of $P$ in $E[x]$. \medskip\noindent If $F$ is a finite field (equivalently $E$ is a finite field), then $E/F$ has a primitive element by the discussion in Section \ref{section-finite}. Next, assume $F$ is infinite and there are at most finitely many proper subfields $E/K/F$. List them, say $K_1, \ldots, K_N$. Then each $K_i \subset E$ is a proper sub $F$-vector space. As $F$ is infinite we can find a vector $\alpha \in E$ with $\alpha \not \in K_i$ for all $i$ (a finite union of proper subvector spaces is never a subvector space; details omitted). Then $\alpha$ is a primitive element for $E$ over $F$. \medskip\noindent Having established the equivalence of (1) and (2) we now turn to the final statement of the lemma. Choose an algebraic closure $\overline{F}$ of $F$. Enumerate the elements $\sigma_1, \ldots, \sigma_n \in \Mor_F(E, \overline{F})$. Since $E/F$ is separable we have $n = [E : F]$ by Lemma \ref{lemma-separable-equality}. Note that if $i \not = j$, then $$V_{ij} = \Ker(\sigma_i - \sigma_j : E \longrightarrow \overline{F})$$ is not equal to $E$. Hence arguing as in the preceding paragraph we can find $\alpha \in E$ with $\alpha \not \in V_{ij}$ for all $i \not = j$. It follows that $|\Mor_F(F(\alpha), \overline{F})| \geq n$. On the other hand $[F(\alpha) : F] \leq [E : F]$. Hence equality by Lemma \ref{lemma-separable-equality} and we conclude that $E = F(\alpha)$. \end{proof} \section{Trace and norm} \label{section-trace-pairing} \noindent Let $L/K$ be a finite extension of fields. By Lemma \ref{lemma-vector-space-is-free} we can choose an isomorphism $L \cong K^{\oplus n}$ of $K$-modules. Of course $n = [L : K]$ is the degree of the field extension. Using this isomorphism we get for a $K$-algebra map $$L \longrightarrow \text{Mat}(n \times n, K),\quad \alpha \longmapsto \text{matrix of multiplication by }\alpha$$ Thus given $\alpha \in L$ we can take the trace and the determinant of the corresponding matrix. Of course these quantities are independent of the choice of the basis chosen above. More canonically, simply thinking of $L$ as a finite dimensional $K$-vector space we have $\text{Trace}_K(\alpha : L \to L)$ and the determinant $\text{Det}_K(\alpha : L \to L)$. \begin{definition} \label{definition-trace-norm} Let $L/K$ be a finite extension of fields. For $\alpha \in L$ we define the {\it trace} $\text{Trace}_{L/K}(\alpha) = \text{Trace}_K(\alpha : L \to L)$ and the {\it norm} $\text{Norm}_{L/K}(\alpha) = \text{Det}_K(\alpha : L \to L)$. \end{definition} \noindent It is clear from the definition that $\text{Trace}_{L/K}$ is $K$-linear and satisfies $\text{Trace}_{L/K}(\alpha) = [L : K]\alpha$ for $\alpha \in K$. Similarly $\text{Norm}_{L/K}$ is multiplicative and $\text{Norm}_{L/K}(\alpha) = \alpha^{[L : K]}$ for $\alpha \in K$. This is a special case of the more general construction discussed in Exercises, Exercises \ref{exercises-exercise-trace-det} and \ref{exercises-exercise-trace-det-rings}. \begin{lemma} \label{lemma-characteristic-vs-minimal-polynomial} Let $L/K$ be a finite extension of fields. Let $\alpha \in L$ and let $P$ be the minimal polynomial of $\alpha$ over $K$. Then the characteristic polynomial of the $K$-linear map $\alpha : L \to L$ is equal to $P^e$ with $e \deg(P) = [L : K]$. \end{lemma} \begin{proof} Choose a basis $\beta_1, \ldots, \beta_e$ of $L$ over $K(\alpha)$. Then $e$ satisfies $e \deg(P) = [L : K]$ by Lemmas \ref{lemma-degree-minimal-polynomial} and \ref{lemma-multiplicativity-degrees}. Then we see that $L = \bigoplus K(\alpha) \beta_i$ is a direct sum decomposition into $\alpha$-invariant subspaces hence the characteristic polynomial of $\alpha : L \to L$ is equal to the characteristic polynomial of $\alpha : K(\alpha) \to K(\alpha)$ to the power $e$. \medskip\noindent To finish the proof we may assume that $L = K(\alpha)$. In this case by Cayley-Hamilton we see that $\alpha$ is a root of the characteristic polynomial. And since the characteristic polynomial has the same degree as the minimal polynomial, we find that equality holds. \end{proof} \begin{lemma} \label{lemma-trace-and-norm-from-minimal-polynomial} Let $L/K$ be a finite extension of fields. Let $\alpha \in L$ and let $P = x^d + a_1 x^{d - 1} + \ldots + a_d$ be the minimal polynomial of $\alpha$ over $K$. Then $$\text{Norm}_{L/K}(\alpha) = (-1)^{[L : K]} a_d^e \quad\text{and}\quad \text{Trace}_{L/K}(\alpha) = - e a_1$$ where $e d = [L : K]$. \end{lemma} \begin{proof} Follows immediately from Lemma \ref{lemma-characteristic-vs-minimal-polynomial} and the definitions. \end{proof} \begin{lemma} \label{lemma-trace-and-norm-linear} Let $L/K$ be a finite extension of fields. Let $V$ be a finite dimensional vector space over $L$. Let $\varphi : V \to V$ be an $L$-linear map. Then $$\text{Trace}_K(\varphi : V \to V) = \text{Trace}_{L/K}(\text{Trace}_L(\varphi : V \to V))$$ and $$\text{Det}_K(\varphi : V \to V) = \text{Norm}_{L/K}(\text{Det}_L(\varphi : V \to V))$$ \end{lemma} \begin{proof} Choose an isomorphism $V = L^{\oplus n}$ so that $\varphi$ corresponds to an $n \times n$ matrix. In the case of traces, both sides of the formula are additive in $\varphi$. Hence we can assume that $\varphi$ corresponds to the matrix with exactly one nonzero entry in the $(i, j)$ spot. In this case a direct computation shows both sides are equal. \medskip\noindent In the case of norms both sides are zero if $\varphi$ has a nonzero kernel. Hence we may assume $\varphi$ corresponds to an element of $\text{GL}_n(L)$. Both sides of the formula are multiplicative in $\varphi$. Since every element of $\text{GL}_n(L)$ is a product of elementary matrices we may assume that $\varphi$ either looks like $$E_{12}(\lambda) = \left( \begin{matrix} 1 & \lambda & \ldots \\ 0 & 1 & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right) \quad\text{or}\quad E_1(a) = \left( \begin{matrix} a & 0 & \ldots \\ 0 & 1 & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right)$$ (because we may also permute the basis elements if we like). In both cases the formula is easy to verify by direct computation. \end{proof} \begin{lemma} \label{lemma-trace-and-norm-tower} Let $M/L/K$ be a tower of finite extensions of fields. Then $$\text{Trace}_{M/K} = \text{Trace}_{L/K} \circ \text{Trace}_{M/L} \quad\text{and}\quad \text{Norm}_{M/K} = \text{Norm}_{L/K} \circ \text{Norm}_{M/L}$$ \end{lemma} \begin{proof} Think of $M$ as a vector space over $L$ and apply Lemma \ref{lemma-trace-and-norm-linear}. \end{proof} \noindent The trace pairing is defined using the trace. \begin{definition} \label{definition-trace-pairing} Let $L/K$ be a finite extension of fields. The {\it trace pairing} for $L/K$ is the symmetric $K$-bilinear form $$Q_{L/K} : L \times L \longrightarrow K,\quad (\alpha, \beta) \longmapsto \text{Trace}_{L/K}(\alpha\beta)$$ \end{definition} \noindent It turns out that a finite extension of fields is separable if and only if the trace pairing is nondegenerate. \begin{lemma} \label{lemma-separable-trace-pairing} Let $L/K$ be a finite extension of fields. The following are equivalent: \begin{enumerate} \item $L/K$ is separable, \item $\text{Trace}_{L/K}$ is not identically zero, and \item the trace pairing $Q_{L/K}$ is nondegenerate. \end{enumerate} \end{lemma} \begin{proof} It is clear that (3) implies (2). If (2) holds, then pick $\gamma \in L$ with $\text{Trace}_{L/K}(\gamma) \not = 0$. Then if $\alpha \in L$ is nonzero, we see that $Q_{L/K}(\alpha, \gamma/\alpha) \not = 0$. Hence $Q_{L/K}$ is nondegenerate. This proves the equivalence of (2) and (3). \medskip\noindent Suppose that $K$ has characteristic $p$ and $L = K(\alpha)$ with $\alpha \not \in K$ and $\alpha^p \in K$. Then $\text{Trace}_{L/K}(1) = p = 0$. For $i = 1, \ldots, p - 1$ we see that $x^p - \alpha^{pi}$ is the minimal polynomial for $\alpha^i$ over $K$ and we find $\text{Trace}_{L/K}(\alpha^i) = 0$ by Lemma \ref{lemma-trace-and-norm-from-minimal-polynomial}. Hence for this kind of purely inseparable degree $p$ extension we see that $\text{Trace}_{L/K}$ is identically zero. \medskip\noindent Assume that $L/K$ is not separable. Then there exists a subfield $L/K'/K$ such that $L/K'$ is a purely inseparable degree $p$ extension as in the previous paragraph, see Lemmas \ref{lemma-separable-first} and \ref{lemma-finite-purely-inseparable}. Hence by Lemma \ref{lemma-trace-and-norm-tower} we see that $\text{Trace}_{L/K}$ is identically zero. \medskip\noindent Assume on the other hand that $\text{Trace}_{L/K}$ is not identically zero. Let $L/K'/K$ be a maximal subfield separable over $K$. Then by Lemma \ref{lemma-trace-and-norm-tower} we see that $\text{Trace}_{L/K'}$ is not identically zero. Then we pick $\alpha \in L$ such that $\text{Trace}_{L/K'}(\alpha) \not = 0$. Then by Lemma \ref{lemma-trace-and-norm-from-minimal-polynomial} we see that $\alpha$ is separable over $K'$. If $\alpha \not \in K'$ then $K'$ is not maximal. If $\alpha \in K'$ then Lemma \ref{lemma-trace-and-norm-from-minimal-polynomial} shows that the characteristic of $K$ does not divide $[L : K']$ which implies that $L/K'$ is separable (as the inseparable degree of $L/K'$ is forced to be $1$, see Definition \ref{definition-insep-degree}) and hence trivial. \end{proof} \noindent Let $K$ be a field and let $Q : V \times V \to K$ be a bilinear form on a finite dimensional vector space over $K$. Say $\dim_K(V) = n$. Then $Q$ defines a linear map $Q : V \to V^*$, $v \mapsto Q(v, -)$ where $V^* = \Hom_K(V, K)$ is the dual vector space. Hence a linear map $$\text{Det}(Q) : \wedge^n(V) \longrightarrow \wedge^n(V)^*$$ If we pick a basis element $\omega \in \wedge^n(V)$, then we can write $\text{Det}(Q)(\omega) = \lambda \omega^\wedge$, where $\omega^\wedge$ is the dual basis element in $\wedge^n(V)^*$. If we change our choice of $\omega$ into $c \omega$ for some $c \in K^*$, then $\omega^\wedge$ changes into $c^{-1} \omega^\wedge$ and therefore $\lambda$ changes into $c^2 \lambda$. Thus the class of $\lambda$ in $K/(K^*)^2$ is well defined and is called the {\it discriminant of $Q$}. Unwinding the definitions we see that $$\lambda = \det(Q(v_i, v_j)_{1 \leq i, j \leq n})$$ if $\{v_1, \ldots, v_n\}$ is a basis for $V$ over $K$. Observe that the discriminant is nonzero if and only if $Q$ is nondegenerate. \begin{definition} \label{definition-discriminant} Let $L/K$ be a finite extension of fields. The {\it discriminant of $L/K$} is the discriminant of the trace pairing $Q_{L/K}$. \end{definition} \noindent By the discussion above and Lemma \ref{lemma-separable-trace-pairing} we see that the discriminant is nonzero if and only if $L/K$ is separable. For $a \in K$ we often say the discriminant is $a$'' when it would be more correct to say the discriminant is the class of $a$ in $K/(K^*)^2$. \begin{exercise} \label{exercise-quadratic-discriminant} Let $L/K$ be an extension of degree $2$. Show that exactly one of the following happens \begin{enumerate} \item the discriminant is $0$, the characteristic of $K$ is $2$, and $L/K$ is purely inseparable obtained by taking a square root of an element of $K$, \item the discriminant is $1$, the characteristic of $K$ is $2$, and $L/K$ is separable of degree $2$, \item the discriminant is not a square, the characteristic of $K$ is not $2$, and $L$ is obtained from $K$ by taking the square root of the discriminant. \end{enumerate} \end{exercise} \section{Galois theory} \label{section-galois-theory} \noindent Here is the definition. \begin{definition} \label{definition-galois} A field extension $E/F$ is called {\it Galois} if it is algebraic, separable, and normal. \end{definition} \noindent It turns out that a finite extension is Galois if and only if it has the correct'' number of automorphisms. \begin{lemma} \label{lemma-finite-Galois} Let $E/F$ be a finite extension of fields. Then $E$ is Galois over $F$ if and only if $|\text{Aut}(E/F)| = [E : F]$. \end{lemma} \begin{proof} Assume $|\text{Aut}(E/F)| = [E : F]$. By Lemma \ref{lemma-normal-and-automorphisms} this implies that $E/F$ is separable and normal, hence Galois. Conversely, if $E/F$ is separable then $[E : F] = [E : F]_s$ and if $E/F$ is in addition normal, then Lemma \ref{lemma-normal-and-automorphisms} implies that $|\text{Aut}(E/F)| = [E : F]$. \end{proof} \noindent Motivated by the lemma above we introduce the Galois group as follows. \begin{definition} \label{definition-galois-group} If $E/F$ is a Galois extension, then the group $\text{Aut}(E/F)$ is called the {\it Galois group} and it is denoted $\text{Gal}(E/F)$. \end{definition} \noindent If $L/K$ is an infinite Galois extension, then one should think of the Galois group as a topological group. We will return to this in Section \ref{section-infinite-galois}. \begin{lemma} \label{lemma-galois-goes-up} Let $K/E/F$ be a tower of algebraic field extensions. If $K$ is Galois over $F$, then $K$ is Galois over $E$. \end{lemma} \begin{proof} Combine Lemmas \ref{lemma-normal-goes-up} and \ref{lemma-separable-goes-up}. \end{proof} \noindent Let $G$ be a group acting on a field $K$ (by field automorphisms). We will often use the notation $$K^G = \{x \in K \mid \sigma(x) = x \ \forall \sigma \in G\}$$ and we will call this the {\it fixed field} for the action of $G$ on $K$. \begin{lemma} \label{lemma-galois-over-fixed-field} Let $K$ be a field. Let $G$ be a finite group acting faithfully on $K$. Then the extension $K/K^G$ is Galois, we have $[K : K^G] = |G|$, and the Galois group of the extension is $G$. \end{lemma} \begin{proof} Given $\alpha \in K$ consider the orbit $G \cdot \alpha \subset K$ of $\alpha$ under the group action. Consider the polynomial $$P = \prod\nolimits_{\beta \in G \cdot \alpha} (x - \beta) \in K[x]$$ The key to the whole lemma is that this polynomial is invariant under the action of $G$ and hence has coefficients in $K^G$. Namely, for $\tau \in G$ we have $$P^\sigma = \prod\nolimits_{\beta \in G \cdot \alpha} (x - \tau(\beta)) = \prod\nolimits_{\beta \in G \cdot \alpha} (x - \beta) = P$$ because the map $\beta \mapsto \tau(\beta)$ is a permutation of the orbit $G \cdot \alpha$. Thus $P \in K^G[x]$. Since also $P(\alpha) = 0$ as $\alpha$ is an element of its orbit we conclude that the extension $K/K^G$ is algebraic. Moreover, the minimal polynomial $Q$ of $\alpha$ over $K^G$ divides the polynomial $P$ just constructed. Hence $Q$ is separable (by Lemma \ref{lemma-recognize-separable} for example) and we conclude that $K/K^G$ is separable. Thus $K/K^G$ is Galois. To finish the proof it suffices to show that $[K : K^G] = |G|$ since then $G$ will be the Galois group by Lemma \ref{lemma-finite-Galois}. \medskip\noindent Pick finitely many elements $\alpha_i \in K$, $i = 1, \ldots, n$ such that $\sigma(\alpha_i) = \alpha_i$ for $i = 1, \ldots, n$ implies $\sigma$ is the neutral element of $G$. Set $$L = K^G(\{\sigma(\alpha_i); 1 \leq i \leq n, \sigma \in G\}) \subset K$$ and observe that the action of $G$ on $K$ induces an action of $G$ on $L$. We will show that $L$ has degree $|G|$ over $K^G$. This will finish the proof, since if $L \subset K$ is proper, then we can add an element $\alpha \in K$, $\alpha \not \in L$ to our list of elements $\alpha_1, \ldots, \alpha_n$ without increasing $L$ which is absurd. This reduces us to the case that $K/K^G$ is finite which is treated in the next paragraph. \medskip\noindent Assume $K/K^G$ is finite. By Lemma \ref{lemma-primitive-element} we can find $\alpha \in K$ such that $K = K^G(\alpha)$. By the construction in the first paragraph of this proof we see that $\alpha$ has degree at most $|G|$ over $K$. However, the degree cannot be less than $|G|$ as $G$ acts faithfully on $K^G(\alpha) = L$ by construction and the inequality of Lemma \ref{lemma-normal-and-automorphisms}. \end{proof} \begin{theorem}[Fundamental theorem of Galois theory] \label{theorem-galois-theory} Let $L/K$ be a finite Galois extension with Galois group $G$. Then we have $K = L^G$ and the map $$\{\text{subgroups of }G\} \longrightarrow \{\text{subextensions }K \subset M \subset L\},\quad H \longmapsto L^H$$ is a bijection whose inverse maps $M$ to $\text{Gal}(L/M)$. The normal subgroups $H$ of $G$ correspond exactly to those subextensions $M$ with $M/K$ Galois. \end{theorem} \begin{proof} By Lemma \ref{lemma-galois-goes-up} given a subextension $L/M/K$ the extension $L/M$ is Galois. Of course $L/M$ is also finite (Lemma \ref{lemma-finite-goes-up}). Thus $|\text{Gal}(L/M)| = [L : M]$ by Lemma \ref{lemma-finite-Galois}. Conversely, if $H \subset G$ is a finite subgroup, then $[L : L^H] = |H|$ by Lemma \ref{lemma-galois-over-fixed-field}. It follows formally from these two observations that we obtain a bijective correspondence as in the theorem. \medskip\noindent If $H \subset G$ is normal, then $L^H$ is fixed by the action of $G$ and we obtain a canonical map $G/H \to \text{Aut}(L^H/K)$. This map has to be injective as $\text{Gal}(L/L^H) = H$. Hence $|G/H| = [L^H : K]$ and $L^H$ is Galois by Lemma \ref{lemma-finite-Galois}. \medskip\noindent Conversely, assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma \ref{lemma-lift-maps} we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau|_M \in \text{Gal}(M/K)$. This induces a homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $H$. Thus $H$ is a normal subgroup. \end{proof} \begin{lemma} \label{lemma-ses-galois} Let $L/M/K$ be a tower of fields. Assume $L/K$ and $M/K$ are finite Galois. Then we obtain a short exact sequence $$1 \to \text{Gal}(L/M) \to \text{Gal}(L/K) \to \text{Gal}(M/K) \to 1$$ of finite groups. \end{lemma} \begin{proof} Namely, by Lemma \ref{lemma-lift-maps} we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau|_M \in \text{Gal}(M/K)$ which gives us the homomorphism on the right. The map on the left identifies the left group with the kernel of the right arrow. The sequence is exact because the sizes of the groups work out correctly by multiplicativity of degrees in towers of finite extensions (Lemma \ref{lemma-multiplicativity-degrees}). One can also use Lemma \ref{lemma-lift-maps} directly to see that the map on the right is surjective. \end{proof} \section{Infinite Galois theory} \label{section-infinite-galois} \noindent The Galois group comes with a canonical topology. \begin{lemma} \label{lemma-galois-profinite} Let $E/F$ be a Galois extension. Endow $\text{Gal}(E/F)$ with the coarsest topology such that $$\text{Gal}(E/F) \times E \longrightarrow E$$ is continuous when $E$ is given the discrete topology. Then \begin{enumerate} \item for any topological space $X$ and map $X \to \text{Aut}(E/F)$ such that the action $X \times E \to E$ is continuous the induced map $X \to \text{Gal}(E/F)$ is continuous, \item this topology turns $\text{Gal}(E/F)$ into a profinite topological group. \end{enumerate} \end{lemma} \begin{proof} Throughout this proof we think of $E$ as a discrete topological space. Recall that the compact open topology on the set of self maps $\text{Map}(E, E)$ is the universal topology such that the action $\text{Map}(E, E) \times E \to E$ is continuous. See Topology, Example \ref{topology-example-automorphisms-of-a-set} for a precise statement. The topology of the lemma on $\text{Gal}(E/F)$ is the induced topology coming from the injective map $\text{Gal}(E/F) \to \text{Map}(E, E)$. Hence the universal property (1) follows from the corresponding universal property of the compact open topology. Since the set of invertible self maps $\text{Aut}(E)$ endowed with the compact open topology forms a topological group, see Topology, Example \ref{topology-example-automorphisms-of-a-set}, and since $\text{Gal}(E/F) = \text{Aut}(E/F) \to \text{Map}(E, E)$ factors through $\text{Aut}(E)$ we obtain a topological group. In other words, we are using the injection $$\text{Gal}(E/F) \subset \text{Aut}(E)$$ to endow $\text{Gal}(E/F)$ with the induced structure of a topological group (see Topology, Section \ref{topology-section-topological-groups}) and by construction this is the coarsest structure of a topological group such that the action $\text{Gal}(E/F) \times E \to E$ is continuous. \medskip\noindent To show that $\text{Gal}(E/F)$ is profinite we argue as follows (our argument is necessarily nonstandard because we have defined the topology before showing that the Galois group is an inverse limit of finite groups). By Topology, Lemma \ref{topology-lemma-profinite-group} it suffices to show that the underlying topological space of $\text{Gal}(E/F)$ is profinite. For any subset $S \subset E$ consider the set $$G(S) = \{ f : S \to E \mid \begin{matrix} f(\alpha)\text{ is a root of the minimal polynomial}\\ \text{of }\alpha\text{ over }F\text{ for all }\alpha \in S \end{matrix} \}$$ Since a polynomial has only a finite number of roots we see that $G(S)$ is finite for all $S \subset E$ finite. If $S \subset S'$ then restriction gives a map $G(S') \to G(S)$. Also, observe that if $\alpha \in S \cap F$ and $f \in G(S)$, then $f(\alpha) = \alpha$ because the minimal polynomial is linear in this case. Consider the profinite topological space $$G = \lim_{S \subset E\text{ finite}} G(S)$$ Consider the canonical map $$c : \text{Gal}(E/F) \longrightarrow G,\quad \sigma \longmapsto (\sigma|_S : S \to E)_S$$ This is injective and unwinding the definitions the reader sees the topology on $\text{Gal}(E/F)$ as defined above is the induced topology from $G$. An element $(f_S) \in G$ is in the image of $c$ exactly if (A) $f_S(\alpha) + f_S(\beta) = f_S(\alpha + \beta)$ and (M) $f_S(\alpha)f_S(\beta) = f_S(\alpha\beta)$ whenever this makes sense (i.e., $\alpha, \beta, \alpha + \beta, \alpha\beta \in S$). Namely, this means $\lim f_S : E \to E$ will be an $F$-algebra map and hence an automorphism by Lemma \ref{lemma-algebraic-extension-self-map}. The conditions (A) and (M) for a given triple $(S, \alpha, \beta)$ define a closed subset of $G$ and hence $\text{Gal}(E/F)$ is homeomorphic to a closed subset of a profinite space and therefore profinite itself. \end{proof} \begin{lemma} \label{lemma-galois-infinite} Let $L/M/K$ be a tower of fields. Assume both $L/K$ and $M/K$ are Galois. Then there is a canonical surjective continuous homomorphism $c : \text{Gal}(L/K) \to \text{Gal}(M/K)$. \end{lemma} \begin{proof} By Lemma \ref{lemma-lift-maps} given $\tau : L \to L$ in $\text{Gal}(L/K)$ the restriction $\tau|_M : M \to M$ is an element of $\text{Gal}(M/K)$. This defines the homomorphism $c$. Continuity follows from the universal property of the topology: the action $$\text{Gal}(L/K) \times M \longrightarrow M,\quad (\tau, x) \longmapsto \tau(x) = c(\tau)(x)$$ is continuous as $M \subset L$ and the action $\text{Gal}(L/K) \times L \to L$ is continuous. Hence continuity of $c$ by part (1) of Lemma \ref{lemma-galois-profinite}. Lemma \ref{lemma-lift-maps} also shows that the map is surjective. \end{proof} \noindent Here is a more standard way to think about the Galois group of an infinite Galois extension. \begin{lemma} \label{lemma-infinite-galois-limit} Let $L/K$ be a Galois extension with Galois group $G$. Let $\Lambda$ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda$ corresponds to $L/L_\lambda/K$ with $L_\lambda/K$ finite Galois with Galois group $G_\lambda$. Define a partial ordering on $\Lambda$ by the rule $\lambda \geq \lambda'$ if and only if $L_\lambda \supset L_{\lambda'}$. Then \begin{enumerate} \item $\Lambda$ is a directed partially ordered set, \item $L_\lambda$ is a system of $K$-extensions over $\Lambda$ and $L = \colim L_\lambda$, \item $G_\lambda$ is an inverse system of finite groups over $\Lambda$, the transition maps are surjective, and $$G = \lim_{\lambda \in \Lambda} G_\lambda$$ as a profinite group, and \item each of the projections $G \to G_\lambda$ is continuous and surjective. \end{enumerate} \end{lemma} \begin{proof} Every subfield of $L$ containing $K$ is separable over $K$ (follows immediately from the definition). Let $S \subset L$ be a finite subset. Then $K(S)/K$ is finite and there exists a tower $L/E/K(S)/K$ such that $E/K$ is finite Galois, see Lemma \ref{lemma-normal-closure-inside-normal}. Hence $E = L_\lambda$ for some $\lambda \in \Lambda$. This certainly implies the set $\Lambda$ is not empty. Also, given $\lambda_1, \lambda_2 \in \Lambda$ we can write $L_{\lambda_i} = K(S_i)$ for finite sets $S_1, S_2 \subset L$ (Lemma \ref{lemma-finite-finitely-generated}). Then there exists a $\lambda \in \Lambda$ such that $K(S_1 \cup S_2) \subset L_\lambda$. Hence $\lambda \geq \lambda_1, \lambda_2$ and $\Lambda$ is directed (Categories, Definition \ref{categories-definition-directed-system}). Finally, since every element in $L$ is contained in $L_\lambda$ for some $\lambda \in \Lambda$, it follows from the description of filtered colimits in Categories, Section \ref{categories-section-directed-colimits} that $\colim L_\lambda = L$. \medskip\noindent If $\lambda \geq \lambda'$ in $\Lambda$, then we obtain a canonical surjective map $G_\lambda \to G_{\lambda'}$, $\sigma \mapsto \sigma|_{L_{\lambda'}}$ by Lemma \ref{lemma-ses-galois}. Thus we get an inverse system of finite groups with surjective transition maps. \medskip\noindent Recall that $G = \text{Aut}(L/K)$. By Lemma \ref{lemma-galois-infinite} the restriction $\sigma|_{L_\lambda}$ of a $\sigma \in G$ to $L_\lambda$ is an element of $G_\lambda$. Moreover, this procedure gives a continuous surjection $G \to G_\lambda$. Since the transition mappings in the inverse system of $G_\lambda$ are given by restriction also, it is clear that we obtain a canonical continuous map $$G \longrightarrow \lim_{\lambda \in \Lambda} G_\lambda$$ Continuity by definition of limits in the category of topological groups; recall that these limits commute with the forgetful functor to the categories of sets and topological spaces by Topology, Lemma \ref{topology-lemma-topological-group-limits}. On the other hand, since $L = \colim L_\lambda$ it is clear that any element of the inverse limit (viewed as a set) defines an automorphism of $L$. Thus the map is bijective. Since the topology on both sides is profinite, and since a bijective continuous map of profinite spaces is a homeomorphism (Topology, Lemma \ref{topology-lemma-bijective-map}), the proof is complete. \end{proof} \begin{theorem}[Fundamental theorem of infinite Galois theory] \label{theorem-inifinite-galois-theory} Let $L/K$ be a Galois extension. Let $G = \text{Gal}(L/K)$ be the Galois group viewed as a profinite topological group (Lemma \ref{lemma-galois-profinite}). Then we have $K = L^G$ and the map $$\{\text{closed subgroups of }G\} \longrightarrow \{\text{subextensions }K \subset M \subset L\},\quad H \longmapsto L^H$$ is a bijection whose inverse maps $M$ to $\text{Gal}(L/M)$. The finite subextensions $M$ correspond exactly to the open subgroups $H \subset G$. The normal closed subgroups $H$ of $G$ correspond exactly to subextensions $M$ Galois over $K$. \end{theorem} \begin{proof} We will use the result of finite Galois theory (Theorem \ref{theorem-galois-theory}) without further mention. Let $S \subset L$ be a finite subset. There exists a tower $L/E/K$ such that $K(S) \subset E$ and such that $E/K$ is finite Galois, see Lemma \ref{lemma-normal-closure-inside-normal}. In other words, we see that $L/K$ is the union of its finite Galois subextensions. For such an $E$, by Lemma \ref{lemma-galois-infinite} the map $\text{Gal}(L/K) \to \text{Gal}(E/K)$ is surjective and continuous, i.e., the kernel is open because the topology on $\text{Gal}(E/K)$ is discrete. In particular we see that no element of $M \setminus K$ is fixed by $\text{Gal}(L/K)$ as $E^{\text{Gal}(E/K)} = K$. This proves that $L^G = K$. \medskip\noindent Lemma \ref{lemma-galois-goes-up} given a subextension $L/M/K$ the extension $L/M$ is Galois. It is immediate from the definition of the topology on $G$ that the subgroup $\text{Gal}(L/M)$ is closed. By the above applied to $L/M$ we see that $L^{\text{Gal}(L/M)} = M$ \medskip\noindent Conversely, let $H \subset G$ be a closed subgroup. We claim that $H = \text{Gal}(L/L^H)$. The inclusion $H \subset \text{Gal}(L/L^H)$ is clear. Suppose that $g \in \text{Gal}(L/L^H)$. Let $S \subset L$ be a finite subset. We will show that the open neighbourhood $U_S(g) = \{g' \in G \mid g'(s) = g(s)\}$ of $g$ meets $H$. This implies that $g \in H$ because $H$ is closed. Let $L/E/K$ be a finite Galois subextension containing $K(S)$ as in the first paragraph of the proof and consider the homomorphism $c : \text{Gal}(L/K) \to \text{Gal}(E/K)$. Then $L^H \cap E = E^{c(H)}$. Since $g$ fixes $L^H$ it fixes $E^{c(H)}$ and hence $c(g) \in c(H)$ by finite Galois theory. Pick $h \in H$ with $c(h) = c(g)$. Then $h \in U_S(g)$ as desired. \medskip\noindent At this point we have established the correspondence between closed subgroups and subextensions. \medskip\noindent Assume $H \subset G$ is open. Arguing as above we find that $H$ containes $\text{Gal}(E/K)$ for some large enough finite Galois subextension $E$ and we find that $L^H$ is contained in $E$ whence finite over $K$. Conversely, if $M$ is a finite subextension, then $M$ is generated by a finite subset $S$ and the corresponding subgroup is the open subset $U_S(e)$ where $e \in G$ is the neutral element. \medskip\noindent Assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma \ref{lemma-galois-infinite} there is a surjective continuous homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $\text{Gal}(L/M)$. Thus $\text{Gal}(L/M)$ is a normal closed subgroup. \medskip\noindent Finally, assume $N \subset G$ is normal and closed. For any $L/E/K$ as in the first paragraph of the proof, the image $c(N) \subset \text{Gal}(E/K)$ is a normal subgroup. Hence $L^N = \bigcup E^{c(N)}$ is a union of Galois extensions of $K$ (by finite Galois theory) whence Galois over $K$. \end{proof} \begin{lemma} \label{lemma-ses-infinite-galois} Let $L/M/K$ be a tower of fields. Assume $L/K$ and $M/K$ are Galois. Then we obtain a short exact sequence $$1 \to \text{Gal}(L/M) \to \text{Gal}(L/K) \to \text{Gal}(M/K) \to 1$$ of profinite topological groups. \end{lemma} \begin{proof} This is a reformulation of Lemma \ref{lemma-galois-infinite}. \end{proof} \section{The complex numbers} \label{section-complex-numbers} \noindent The fundamental theorem of algebra states that the field of complex numbers is an algebraically closed field. In this section we discuss this briefly. \medskip\noindent The first remark we'd like to make is that you need to use a little bit of input from calculus in order to prove this. We will use the intuitively clear fact that every odd degree polynomial over the reals has a real root. Namely, let $P(x) = a_{2k + 1} x^{2k + 1} + \ldots + a_0 \in \mathbf{R}[x]$ for some $k \geq 0$ and $a_{2k + 1} \not = 0$. We may and do assume $a_{2k + 1} > 0$. Then for $x \in \mathbf{R}$ very large (positive) we see that $P(x) > 0$ as the term $a_{2k + 1} x^{2k + 1}$ dominates all the other terms. Similarly, if $x \ll 0$, then $P(x) < 0$ by the same reason (and this is where we use that the degree is odd). Hence by the intermediate value theorem there is an $x \in \mathbf{R}$ with $P(x) = 0$. \medskip\noindent A conclusion we can draw from the above is that $\mathbf{R}$ has no nontrivial odd degree field extensions, as elements of such extensions would have odd degree minimal polynomials. \medskip\noindent Next, let $K/\mathbf{R}$ be a finite Galois extension with Galois group $G$. Let $P \subset G$ be a $2$-sylow subgroup. Then $K^P/\mathbf{R}$ is an odd degree extension, hence by the above $K^P = K$, which in turn implies $G = P$. (All of these arguments rely on Galois theory of course.) Thus $G$ is a $2$-group. If $G$ is nontrivial, then we see that $\mathbf{C} \subset K$ as $\mathbf{C}$ is (up to isomorphism) the only degree degree $2$ extension of $\mathbf{R}$. If $G$ has more than $2$ elements we would obtain a quadratic extension of $\mathbf{C}$. This is absurd as every complex number has a square root. \medskip\noindent The conclusion: $\mathbf{C}$ is algebraically closed. Namely, if not then we'd get a nontrivial finite extension $\mathbf{C} \subset K$ which we could assume normal (hence Galois) over $\mathbf{R}$ by Lemma \ref{lemma-normal-closure}. But we've seen above that then $K = \mathbf{C}$. \begin{lemma}[Fundamental theorem of algebra] \label{lemma-C-algebraically-closed} The field $\mathbf{C}$ is algebraically closed. \end{lemma} \begin{proof} See discussion above. \end{proof} \section{Kummer extensions} \label{section-Kummer} \noindent Let $K$ be a field. Let $n \geq 2$ be an integer such that $K$ contains a primitive $n$th root of $1$. Let $a \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root $b$ of the equation $x^n = a$. Then $L/K$ is Galois. If $G = \text{Gal}(L/K)$ is the Galois group, then the map $$G \longrightarrow \mu_n(K),\quad \sigma \longmapsto \sigma(b)/b$$ is an injective homomorphism of groups. In particular, $G$ is cyclic of order dividing $n$ as a subgroup of the cyclic group $\mu_n(K)$. Kummer theory gives a converse. \begin{lemma}[Kummer extensions] \label{lemma-Kummer} Let $K \subset L$ be a Galois extension of fields whose Galois group is $\mathbf{Z}/n\mathbf{Z}$. Assume moreover that the characteristic of $K$ is prime to $n$ and that $K$ contains a primitive $n$th root of $1$. Then $L = K[z]$ with $z^n \in K$. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Artin-Schreier extensions} \label{section-Artin-Schreier} \noindent Let $K$ be a field of characteristic $p > 0$. Let $a \in K$. Let $L$ be an extension of $K$ obtained by adjoining a root $b$ of the equation $x^p - x = a$. Then $L/K$ is Galois. If $G = \text{Gal}(L/K)$ is the Galois group, then the map $$G \longrightarrow \mathbf{Z}/p\mathbf{Z},\quad \sigma \longmapsto \sigma(b) - b$$ is an injective homomorphism of groups. In particular, $G$ is cyclic of order dividing $p$ as a subgroup of $\mathbf{Z}/p\mathbf{Z}$. The theory of Artin-Schreier extensions gives a converse. \begin{lemma}[Artin-Schreier extensions] \label{lemma-Artin-Schreier} Let $K \subset L$ be a Galois extension of fields of characteristic $p > 0$ with Galois group $\mathbf{Z}/p\mathbf{Z}$. Then $L = K[z]$ with $z^p - z \in K$. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Transcendence} \label{section-transcendence} \noindent We recall the standard definitions. \begin{definition} \label{definition-transcendence} Let $k \subset K$ be a field extension. \begin{enumerate} \item A collection of elements $\{x_i\}_{i \in I}$ of $K$ is called {\it algebraically independent} over $k$ if the map $$k[X_i; i\in I] \longrightarrow K$$ which maps $X_i$ to $x_i$ is injective. \item The field of fractions of a polynomial ring $k[x_i; i \in I]$ is denoted $k(x_i; i\in I)$. \item A {\it purely transcendental extension} of $k$ is any field extension $k \subset K$ isomorphic to the field of fractions of a polynomial ring over $k$. \item A {\it transcendence basis} of $K/k$ is a collection of elements $\{x_i\}_{i \in I}$ which are algebraically independent over $k$ and such that the extension $k(x_i; i\in I) \subset K$ is algebraic. \end{enumerate} \end{definition} \begin{example} \label{example-pi-transcendental} The field $\mathbf{Q}(\pi)$ is purely transcendental because $\pi$ isn't the root of a nonzero polynomial with rational coefficients. In particular, $\mathbf{Q}(\pi) \cong \mathbf{Q}(x)$. \end{example} \begin{lemma} \label{lemma-transcendence-degree} Let $E/F$ be a field extension. A transcendence basis of $E$ over $F$ exists. Any two transcendence bases have the same cardinality. \end{lemma} \begin{proof} Let $A$ be an algebraically independent subset of $E$. Let $G$ be a subset of $E$ containing $A$ that generates $E/F$. We claim we can find a transcendence basis $B$ such that $A \subset B \subset G$. To prove this consider the collection of algebraically independent subsets $\mathcal{B}$ whose members are subsets of $G$ that contain $A$. Define a partial ordering on $\mathcal{B}$ using inclusion. Then $\mathcal{B}$ contains at least one element $A$. The union of the elements of a totally ordered subset $T$ of $\mathcal{B}$ is an algebraically independent subset of $E$ over $F$ since any algebraic dependence relation would have occurred in one of the elements of $T$ (since polynomials only involve finitely many variables). The union also contains $A$ and is contained in $G$. By Zorn's lemma, there is a maximal element $B \in \mathcal{B}$. Now we claim $E$ is algebraic over $F(B)$. This is because if it wasn't then there would be an element $f \in G$ transcendental over $F(B)$ since $E(G) = F$. Then $B \cup\{f\}$ wold be algebraically independent contradicting the maximality of $B$. Thus $B$ is our transcendence basis. \medskip\noindent Let $B$ and $B'$ be two transcendence bases. Without loss of generality, we can assume that $|B'| \leq |B|$. Now we divide the proof into two cases: the first case is that $B$ is an infinite set. Then for each $\alpha \in B'$, there is a finite set $B_{\alpha}$ such that $\alpha$ is algebraic over $E(B_{\alpha})$ since any algebraic dependence relation only uses finitely many indeterminates. Then we define $B^* = \bigcup_{\alpha\in B'} B_{\alpha}$. By construction, $B^* \subset B$, but we claim that in fact the two sets are equal. To see this, suppose that they are not equal, say there is an element $\beta \in B \setminus B^*$. We know $\beta$ is algebraic over $E(B')$ which is algebraic over $E(B^*)$. Therefore $\beta$ is algebraic over $E(B^*)$, a contradiction. So $|B| \leq |\bigcup_{\alpha \in B'} B_{\alpha}|$. Now if $B'$ is finite, then so is $B$ so we can assume $B'$ is infinite; this means $$|B| \leq |\bigcup\nolimits_{\alpha \in B'} B_{\alpha}| = |B'|$$ because each $B_\alpha$ is finite and $B'$ is infinite. Therefore in the infinite case, $|B| = |B'|$. \medskip\noindent Now we need to look at the case where $B$ is finite. In this case, $B'$ is also finite, so suppose $B = \{\alpha_1, \ldots, \alpha_n\}$ and $B' = \{\beta_1, \ldots, \beta_m\}$ with $m \leq n$. We perform induction on $m$: if $m = 0$ then $E/F$ is algebraic so $B = \emptyset$ so $n = 0$. If $m > 0$, there is an irreducible polynomial $f \in E[x, y_1, \ldots, y_n]$ such that $f(\beta_1, \alpha_1, \ldots, \alpha_n) = 0$ and such that $x$ occurs in $f$. Since $\beta_1$ is not algebraic over $F$, $f$ must involve some $y_i$ so without loss of generality, assume $f$ uses $y_1$. Let $B^* = \{\beta_1, \alpha_2, \ldots, \alpha_n\}$. We claim that $B^*$ is a basis for $E/F$. To prove this claim, we see that we have a tower of algebraic extensions $$E/ F(B^*, \alpha_1) / F(B^*)$$ since $\alpha_1$ is algebraic over $F(B^*)$. Now we claim that $B^*$ (counting multiplicity of elements) is algebraically independent over $E$ because if it weren't, then there would be an irreducible $g\in E[x, y_2, \ldots, y_n]$ such that $g(\beta_1, \alpha_2, \ldots, \alpha_n) = 0$ which must involve $x$ making $\beta_1$ algebraic over $E(\alpha_2, \ldots, \alpha_n)$ which would make $\alpha_1$ algebraic over $E(\alpha_2, \ldots, \alpha_n)$ which is impossible. So this means that $\{\alpha_2, \ldots, \alpha_n\}$ and $\{\beta_2, \ldots, \beta_m\}$ are bases for $E$ over $F(\beta_1)$ which means by induction, $m = n$. \end{proof} \begin{definition} \label{definition-transcendence-degree} Let $k \subset K$ be a field extension. The {\it transcendence degree} of $K$ over $k$ is the cardinality of a transcendence basis of $K$ over $k$. It is denoted $\text{trdeg}_k(K)$. \end{definition} \begin{lemma} \label{lemma-transcendence-degree-tower} Let $k \subset K \subset L$ be field extensions. Then $$\text{trdeg}_k(L) = \text{trdeg}_K(L) + \text{trdeg}_k(K).$$ \end{lemma} \begin{proof} Choose a transcendence basis $A \subset K$ of $K$ over $k$. Choose a transcendence basis $B \subset L$ of $L$ over $K$. Then it is straightforward to see that $A \cup B$ is a transcendence basis of $L$ over $k$. \end{proof} \begin{example} \label{example-pi-e-transcendental} Consider the field extension $\mathbf{Q}(e, \pi)$ formed by adjoining the numbers $e$ and $\pi$. This field extension has transcendence degree at least $1$ since both $e$ and $\pi$ are transcendental over the rationals. However, this field extension might have transcendence degree $2$ if $e$ and $\pi$ are algebraically independent. Whether or