books/cring/fields.tex

\chapter{Fields and Extensions}

In this chapter, we shall discuss the theory of fields.
Recall that a \textbf{field} is an integral domain for which all non-zero 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. 

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 integral domain can be embedded in a field
(its quotient field), and any \emph{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{Introduction}


Recall once again that:
\begin{definition} 
A \textbf{field} is an integral domain where every non-zero element is
invertible. Alternatively, it is a set $k$, endowed with binary operations of
addition and multiplication, which satisfy the usual axioms of commutativity,
associativity, distributivity, $1$ and $0$ (and $1 \neq 0$!), and additive and
multiplicative inverses.
\end{definition} 

A \textbf{subfield} is a subset closed under these operations: equivalently, it
is a subring that is itself a field.

For a field $k$, we write $k^*$ for the subset $k  \setminus \left\{0\right\}$.
(This generalizes the usual notation \cref{} $R^*$ that refers to the group of
invertible elements in a ring $R$.)

\subsection{Examples}
To get started, let us begin by providing several examples of fields. The reader should
recall (\cref{maximalfield}) that if $R$ is a ring and $I \subset R$ an
ideal, then $R/I$ is a field precisely when $I$ is maximal.

\begin{example}
One of the most familiar examples of a field is the rational
numbers $\mathbb{Q}$.
\end{example}

\begin{example}
If $p$ is a prime number, then $\mathbb{Z}/(p)$ is a field, denoted
$\mathbb{F}_p$. Indeed, $(p)$ is a
maximal ideal in $\mathbb{Z}$. Thus, fields may be finite: $\mathbb{F}_p$
contains $p$ elements.  
\end{example}


\begin{example}[Quotients of the polynomial ring]
In a principal ideal domain, every prime ideal is principal. Now, by
\rref{polyringED}, if $k$ is a field, then the polynomial ring $k[x]$ is a PID. 
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 $\mathbb{C}$
can be constructed as $\mathbb{R}[x]/(x^2 + 1)$.
\end{example} 



\begin{exercise} 
What is $\mathbb{C}[x]/(x^2 + 1)$?
\end{exercise} 

\begin{example}[Quotient fields]
Recall from \cref{quotfld1} that, given an integral domain $A$, there is an
imbedding $A \hookrightarrow K(A)$ into a field $K(A)$ formally constructed as
quotients $a/b, a, b \in A$ (and $b \neq 0$) modulo an evident equivalence
relation.
This is called the \textbf{quotient field.}
The quotient field has the following universal property: given an injection
$\phi: A
\hookrightarrow K$ for a field $K$, there is a unique map $\psi: K(A) \to K$ making
the diagram commutative (i.e. a map of $A$-algebras).
Indeed, it is clear how to define such a map: we set
\[ \psi(a/b) = \phi(a)/\phi(b),  \]
where injectivity of $\phi$ assures that $\phi(b) \neq 0$ if $ b \neq 0$.

If the map is not injective, then such a factorization may not exist. Consider
the imbedding $\mathbb{Z} \to \mathbb{Q}$ into its quotient field, and consider
the map $\mathbb{Z} \to \mathbb{F}_p$: this last map goes from $\mathbb{Z} $
into a field, but it does not factor through $\mathbb{Q}$ (as $p$ is invertible
in $\mathbb{Q}$ and zero in $\mathbb{F}_p$!).
\end{example} 


\begin{example}[Rational function field] \label{monofldext}
\label{rationalfnfld}
If $k$ is a field, then we can consider the field $k(x)$ of \textbf{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 $F, G \in k[x]$ with the obvious
equivalence relation.
\end{example} 


Here is a fancier example of a field.
\begin{example} 
\label{meromorphicfn}
Let $X$ be a Riemann surface.\footnote{Readers not familiar with Riemann
surfaces may ignore this example.} Let $\mathbb{C}(X)$ denote the
set of meromorphic functions on $X$; clearly $\mathbb{C}(X)$ is a ring under
multiplication and addition of functions. It turns out that in fact
$\mathbb{C}(X)$ is a
field; this is because 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 $\mathbb{C}(S^2)$ is the
field of rational functions $\mathbb{C}(z)$. 
\end{example}



One reason fields are so nice from the point of view of most other chapters
in this book is that the theory of $k$-modules (i.e. vector spaces), for $k$ a field, is very simple.
Namely:

\begin{proposition}  \label{vectorspacefree}
If $k$ is a field, then every $k$-module is free.
\end{proposition} 
\begin{proof} 
Indeed, by linear algebra we know that a $k$-module (i.e. vector space) $V$ has a
\emph{basis} $\mathcal{B} \subset V$, which defines an isomorphism from the
free vector space on $\mathcal{B}$ to $V$.
\end{proof} 

\begin{corollary} \label{fieldsemisimple}
Every exact sequence of modules over a field splits.
\end{corollary} 
\begin{proof} 
This follows from \cref{} and \cref{vectorspacefree}, as every vector space is
projective.
\end{proof} 

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 \cref{fieldsemisimple} is a statement about the \emph{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 \emph{abelian category}).

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: \emph{extensions} of fields.

\subsection{The characteristic of a field}
\label{more-fields}

In the category of rings, there is an \emph{initial object} $\mathbb{Z}$: any
ring $R$ has a map from $\mathbb{Z}$ 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.

Let $F$ be a field. As $\mathbb{Z}$ is the initial object of the category of
rings, there is a ring map $f : \mathbb{Z} \to F$, see
\rref{integersinitial}.
The image of this ring map is an integral domain (as a subring of a field)
hence the kernel of $f$ is a prime ideal in $\mathbb{Z}$, see
\rref{primeifdomain}.
Hence the kernel of $f$ is either $(0)$ or $(p)$ for some prime number $p$, see
\rref{integerprimes}.

In the first case we see that $f$ is injective, and in this case
we think of $\mathbb{Z}$ 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 $p, q \in \mathbb{Z}$ with $q \not = 0$.
Hence in this case we may and we do think of $\mathbb{Q}$ as a subring of $F$.
One can easily see that this is the smallest subfield of $F$ in this case.

In the second case, i.e., when $\text{Ker}(f) = (p)$ we see that
$\mathbb{Z}/(p) = \mathbb{F}_p$ is a subring of $F$.  Clearly it is the smallest subfield of $F$.

Arguing in this way we see that every field contains a smallest subfield
which is either $\mathbb{Q}$ or finite equal to $\mathbb{F}_p$ for some
prime number $p$.

\begin{definition}
The \textbf{characteristic} of a field $F$ is $0$ if
$\mathbb{Z} \subset F$, or is a prime $p$ if $p = 0$ in $F$.
The \textbf{prime subfield of $F$} is the smallest subfield of $F$
which is either $\mathbb{Q} \subset F$ if the characteristic is zero, or
$\mathbb{F}_p \subset F$ if the characteristic is $p > 0$.
\end{definition}


It is easy to see that if $E$ is a field containing $k$, then the characteristic of
$E$ is the same as the characteristic of $k$. 

\begin{example} 
The characteristic of $\mathbb{Z}/p$ is $p$, and that of $\mathbb{Q}$ is $0$.
This is obvious from the definitions.
\end{example} 


\section{Field extensions}

\subsection{Preliminaries}

In general, though, we are interested not so much in fields by themselves but
in field \emph{extensions.} This is perhaps analogous to studying not rings
but \emph{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{proposition} \label{fieldinj} If $F$ is a field and $R$ is any ring, then any ring homomorphism $f:F\rightarrow
R$ is either injective or the zero map (in which case $R=0$).
\end{proposition}

\begin{proof} Indeed, $\ker(f)$ is an ideal in
$F$. But there are only two ideals in $F$, namely $(0)$ and $(1)$. If $f$ is identically
zero, then $1=f(1)=0$ in $R$, so $R=0$ too.
\end{proof}

\begin{definition} If $F$ is a field contained in a field $G$, then $G$ is said
to be a \textbf{field extension} of $F$. We shall write $G/F$ to indicate
that  $G$ is an extension of $F$.
\end{definition}

So if $F, F'$ are fields, and $F \to F'$ is any ring-homomorphism, we see by 
\cref{fieldinj} that it is injective,\footnote{The zero ring is not a field!} and $F'$ can be regarded as an extension
of $F$, by a slight abuse of notation. 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. 

Let $k$ be a field. There is a \emph{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 $E/k, E'/k$ is a $k$-algebra morphism $E \to E'$;
alternatively, it is a commutative diagram
\[ \xymatrix{
E \ar[rr] & & E' \\
& k \ar[ru] \ar[lu] &
}.\]


\begin{definition} 
A \textbf{tower} of field extensions $E'/E/k$ consists of an extension $E/k$
and an extension $E'/E$.
\end{definition} 

It is easy to see that any morphism $E \to E'$ in the category of
$k$-extensions gives a tower. 


Let us give a few examples of field extensions.

\begin{example} 
Let $k$ be a field, and $P \in k[x]$ an irreducible polynomial. We have seen
that $k[x]/(P)$ is a field (\rref{monofldext}). Since it is also a $k$-algebra
in the obvious way, it is an extension of $k$.
\end{example} 

\begin{example} 
If $X$ is a Riemann surface, then the field of meromorphic functions
$\mathbb{C}(X)$ (see \cref{meromorphicfn}) is an extension field of
$\mathbb{C}$, because any element of $\mathbb{C}$ induces a
meromorphic---indeed, holomorphic---constant function on $X$.
\end{example} 

Let $F/k$ be a field extension. Let $S \subset F$ be any subset. 
Then there is a  \emph{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 easily checks that this is a field.
It is easy to see, 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} 
If $F/k$ is an extension and $S \subset F$, we write $k(S)$ for the smallest
subextension of $F$ containing $S$.
We will say that $S$ \textbf{generates} the extension $k(S)/k$.
\end{definition} 

For instance, $\mathbb{C}$ is generated by $i$ over $\mathbb{R}$.

\begin{exercise} 
Show that $\mathbb{C}$ does not have a countable set of generators over
$\mathbb{Q}$.
\end{exercise} 


Let us now classify extensions generated by one element.
\begin{proposition}[Simple extensions of a field] \label{fldmono} 
If an extension $F/k$ is generated by one element, then it is $F$ 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{proposition} 

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.

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, $k[t]/(P) \simeq F$.

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}




\subsection{Finite extensions}
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} 
The dimension of $F$
considered as an $E$-vector space is called the \textbf{degree} of the extension and is
denoted $[F:E]$. If $[F:E]<\infty$ then $F$ is said to be a
\textbf{finite} extension.
\end{definition} 

\begin{example} 
$\mathbb{C}$ is obviously a finite extension of $\mathbb{R}$ (of degree 2).
\end{example} 

Let us now consider the degree in the most important special example, that
given by \cref{fldmono}, in the next two examples.

\begin{example}[Degree of a simple transcendental extension]
\label{monodeg1}
If $k$ is any field, then the rational function field $k(t)$ is \emph{not} a
finite extension. The elements $\left\{t^n, n \in \mathbb{Z}\right\}$
are linearly independent over $k$.

In fact, if $k$ is uncountable, then $k(t)$ is \emph{uncountably} dimensional
as a $k$-vector space. To show this, we claim that the family of elements
$\left\{1/(t- \alpha), \alpha \in k\right\} \subset k(t)$ is linearly independent over $k$. A
nontrivial relation between them would lead to a contradiction: for instance,
if one works over $\mathbb{C}$, then this follows because
$\frac{1}{t-\alpha}$, when considered as a meromorphic function on
$\mathbb{C}$, 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.

(Amusingly, this leads
to a quick if suboptimal proof of the Hilbert Nullstellensatz; see \cref{}.)
\end{example} 

\begin{example}[Degree of a simple algebraic extension]
\label{monodeg2}
Consider a monogenic  field extension $E/k$ of the form in
\rref{monofldext}, say $E = k[t]/(P)$ for $P \in k[t]$ an irreducible
polynomial.
Then the degree $[E:k]$ is just the degree $\deg P$. 
Indeed, without loss of generality, we can assume $P$ monic, say
\begin{equation} \label{P} P = t^n + a_1 t^{n-1} + \dots + a_0.\end{equation}
It is then easy to see that the images of $1, t, \dots, t^{n-1}$ 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 $(P)$.

Conversely, the set $S=\left\{1, t, \dots, t^{n-1}\right\}$ (or more
properly their images) spans $k[t]/(P)$ as a vector space.
Indeed, we have by \eqref{P} that $t^n$ lies in the span of $S$.
Similarly, the relation $tP(t)=0$ shows that the image of  $t^{n+1}$ lies in the span of
$\left\{1, t, \dots, t^n\right\}$---by what was just shown, thus in the span of
$S$. Working upward inductively, we find
that the image of $t^M$ for $M  \geq n$ lies in the span of $S$.
\end{example} 

This confirms the observation that $[\mathbb{C}: \mathbb{R}] = 2$, 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
\textbf{quadratic,} for obvious reasons.


The basic fact about the degree is that it is \emph{multiplicative in
towers.}

\begin{proposition}[Multiplicativity]
Suppose given a tower $F/E/k$. Then
\[ [F:k] = [F:E][E:k].  \]
\end{proposition} 
\begin{proof} 
Let $\alpha_1, \dots, \alpha_n \in F$ be an $E$-basis for $F$. Let $\beta_1,
\dots, \beta_m \in E$ 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 $k$.

By assumption, the $\left\{\alpha_i\right\}$ span $F$ over $E$. So if $f \in
F$, there are $a_i \in E$ with 
\[ f = \sum a_i\alpha_i,  \]
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
\[ f = \sum_{i,j} b_{ij}\alpha_i \beta_j,  \]
proving that the $\left\{\alpha_i \beta_j\right\}$ span $F$ over $k$.

Suppose now that there existed a nontrivial relation
\[ \sum_{i,j} c_{ij} \alpha_i \beta_j =0 \]
for the $c_{ij} \in k$. In that case, we would have
\[ \sum_i \alpha_i \left( \sum_j c_{ij} \beta_j \right) =0, \]
and the inner terms lie in $E$ as the $\beta_j$ do. Now $E$-linear independence of
the $\left\{\alpha_i\right\}$ shows that the inner sums are all zero. Then
$k$-linear independence of the $\left\{\beta_j\right\}$ shows that the
$c_{ij}$ all vanish.
\end{proof} 

We sidetrack to a slightly tangential definition:
\begin{definition} 
 A field extensions $K$ of $\mathbb{Q}$ is said to be a \textbf{number field}
if it is a finite extension of $\mathbb{Q}$.
\end{definition} 
Number fields are the basic objects in algebraic number theory. We shall see
later that,
for the analog of the integers $\mathbb{Z}$ in a number field, something kind
of like unique factorization still holds (though strict unique factorization
generally does not!).

\subsection{Algebraic extensions}

Consider a field extension $F/E$.

\begin{definition}
An element $\alpha\in F$ is said to be \textbf{algebraic} over $E$ if
$\alpha$ is the root of some polynomial with coefficients in $E$. If all
elements of $F$ are \textbf{algebraic} then $F$ is said to be an algebraic extension. 
\end{definition}

By \cref{fldmono}, 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, it
satisfies the polynomial $P$!); in the former case, it is not.

\begin{example} 
$\mathbb{C}$ is algebraic over $\mathbb{R}$.
\end{example} 

\begin{example} 
Let $X$ be a compact Riemann surface, and $f \in \mathbb{C}(X) - \mathbb{C}$ any
nonconstant meromorphic function on $X$ (see \cref{meromorphicfn}). Then it is known that
$\mathbb{C}(X)$ is algebraic over the subextension $\mathbb{C}(f)$ generated by
$f$. We shall not prove this.
\end{example} 

We now show that there is a deep connection between finiteness and being
algebraic.
\begin{proposition} \label{finalgebraic}
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{proposition} 
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 
$\left\{1, \alpha, \dots, \alpha^n\right\}$ 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.

For the last assertion, note that a monogenic extension  $k(\alpha)/k$ is
finite if and only $\alpha$ is algebraic over $k$, by \cref{monodeg1} and
\cref{monodeg2}.
So if $E/k$ is algebraic, then each $k(\alpha)/k, \alpha \in E$, is a finite
extension, and conversely.
\end{proof} 



We can extract a corollary of the last proof (really of \cref{monodeg1} and
\cref{monodeg2}): a monogenic extension is finite
if and only if it is algebraic.
We shall use this observation in the next result.

\begin{corollary} \label{fingenalg}
Let $k$ be a field, and let $\alpha_1, \alpha_2, \dots, \alpha_n$ be elements
of some extension field such that each $\alpha_i$ is finite over $k$. Then the
extension $k(\alpha_1, \dots, \alpha_n)/k$ is finite.
That is, a finitely generated algebraic extension is finite.
\end{corollary} 
\begin{proof} 
Indeed, each $k(\alpha_{1}, \dots, \alpha_{i+1})/k(\alpha_1, \dots,
\alpha_{i})$ is monogenic and algebraic, hence finite.
\end{proof} 

The set of complex numbers that are algebraic over $\mathbb{Q}$ are simply
called the \textbf{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{corollary} 
Let $E/k$ be a field extension. Then the elements of $E$ algebraic over $k$
form a field.
\end{corollary} 
\begin{proof} 
Let $\alpha, \beta \in E$ be algebraic over 
$k$. Then $k(\alpha, \beta)/k$ is a finite extension by \cref{fingenalg}. It follows that $k(\alpha
+ \beta) \subset k(\alpha, \beta)$ is a finite extension, which implies that
$\alpha + \beta$ is algebraic by \cref{finalgebraic}.
\end{proof} 


Many nice properties of field extensions, like those of rings, will have the property
that they will be preserved by towers and composita.


\begin{proposition}[Towers]
Let $E/k$ and $F/E$ be algebraic. Then $F/k$ is algebraic.
\end{proposition} 
\begin{proof} 
Choose $\alpha \in F$. Then $\alpha$ is algebraic over $E$. 
The key observation is that $\alpha$ is algebraic over a \emph{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, $k(S,
\alpha)/ k(S)$ is finite. Since $S$ is a finite set, and $k(S)/k$ is algebraic,
\cref{fingenalg} shows that $k(S)/k$ is finite. Together we find that
$k(S,\alpha)/k$ is finite, so $\alpha$ is algebraic over $k$.
\end{proof} 

The method of proof in the previous argument---that being algebraic over $E$ was a
property that \emph{descended} to a finitely generated subextension of $E$---is
an idea that recurs throughout algebra, and will be put to use more generality
in \cref{}.

\subsection{Minimal polynomials}

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 $P(x) \in k[x]$ 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 \emph{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 \emph{generator} $m(x) \in k[x]$ of this ideal. If we assume $m$
monic, without loss of generality, then $m$ is uniquely determined.

\begin{definition} 
$m(x)$ as above is called the \textbf{minimal polynomial} of $\alpha$ over $k$.
\end{definition} 

The minimal polynomial has the following characterization: it is the monic
polynomial, of smallest degree, that annihilates $\alpha$. (Any nonconstant
multiple of $m(x)$ will have larger degree, and only multiples of $m(x)$ can
annihilate $\alpha$.)
This explains the name \emph{minimal.}

Clearly the minimal polynomial is \emph{irreducible.} This is equivalent to the
assertion that the ideal in $k[x]$ consisting of polynomials annihilating
$\alpha$ is prime. But 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{proposition} 
The degree of the minimal polynomial is $[k(\alpha):k]$.
\end{proposition} 
\begin{proof} 
This is just a restatement of the argument in \cref{monofld}: the observation is that if $m(x)$
is the minimal polynomial of $\alpha$, then the map
\[ k[x]/(m(x)) \to k(\alpha), \quad x \mapsto \alpha  \]
is an isomorphism as in the aforementioned proof, and we have counted the degree
of such an extension  (see \cref{monodeg2}).
\end{proof} 

So the observation of the above proof is that if $\alpha \in E$ is algebraic,
then $k(\alpha) \subset E$ is isomorphic to $k[x]/(m(x))$.


\subsection{Algebraic closure}

Now we want to define a ``universal'' algebraic extension of a field. Actually,
we should be careful: the algebraic closure is \emph{not} a universal object.
That is, the algebraic closure is not unique up to \emph{unique} isomorphism:
it is only unique up to isomorphism. But still, it will be very handy, if not
functorial.


\begin{definition} 
Let $F$ be a field. An \textbf{algebraic closure} of $F$ is a field
$\overline{F}$ containing $F$ such that:
\begin{enumerate}[\textbf{AC} 1]
\item $\overline{F} $ is algebraic over $F$.
\item $\overline{F}$ is \textbf{algebraically closed} (that is, every
non-constant polynomial in $\overline{F}[X]$ has a root in $\overline{F}$).
\end{enumerate}
\end{definition} 

The ``fundamental theorem of algebra'' states that $\mathbb{C}$ is
algebraically closed. While the easiest proof of this result uses Liouville's
theorem in complex analysis, we shall give a mostly algebraic proof below
(\cref{}).

We now prove the basic existence result.


\begin{theorem}
Every field has an algebraic closure.
\end{theorem}

The proof will mostly be a red herring to the rest of the chapter. However, we
will want to know that it is \emph{possible} to embed a field inside an
algebraically closed field, and we will often assume it done.
\begin{proof}
Let $ K$ be a field and $ \Sigma$ be the set of all monic irreducibles in $ K[x]$. Let $ A = K[\{x_f : f \in \Sigma\}]$ be the polynomial ring generated by indeterminates $ x_f$, one for each $ f \in \Sigma$. Then let $ \mathfrak{a}$ be the ideal of $ A$ generated by polynomials of the form $ f(x_f)$ for each $ f \in \Sigma$.

\emph{Claim 1}. $ \mathfrak{a}$ is a proper ideal.

\emph{Proof of claim 1}. Suppose $ \mathfrak{a} = (1)$, so there exist finitely many polynomials $ f_i \in \Sigma$ and $ g_i \in A$ such that $ 1 = f_1(x_{f_1}) g_1 + \dotsb + f_k(x_{f_k}) g_k$. Each $ g_i$ uses some finite collection of indeterminates $ V_i \{x_{f_{i_1}}, \dotsc, x_{f_{i_{k_i}}}\}$. This notation is ridiculous, so we simplify it.

We can take the union of all the $ V_i$, together with the indeterminates $ x_{f_1}, \dotsc, x_{f_k}$ to get a larger but still finite set of indeterminates $ V = \{x_{f_1}, \dotsc, x_{f_n}\}$ for some $ n \geq k$ (ordered so that the original $ x_{f_1}, \dotsc, x_{f_k}$ agree the first $ k$ elements of $ V$). Now we can regard each $ g_i$ as a polynomial in this new set of indeterminates $ V$.
Then, we can write $ 1 = f_1(x_{f_1}) g_1 + \dotsb + f_n(x_{f_n}) g_n$ where for each $ i > k$, we let $ g_i = 0$ (so that we've adjoined a few zeroes to the right hand side of the equality).
Finally, we define $ x_i = x_{f_i}$, so that we have
$ 1 = f_1(x_1)g_1(x_1, \dotsc, x_n) + \dotsb + f_n(x_n) g_n(x_1, \dotsc, x_n)$.

Suppose $ n$ is the minimal integer such that there exists an expression of this form, so that
\[ \mathfrak{b} = (f_1(x_1), \dotsc, f_{n-1}(x_{n-1})) \]
is a proper ideal of $ B = K[x_1, \dotsc, x_{n-1}]$, but
\[ (f_1(x_1), \dotsc, f_n(x_n)) \]
is the unit ideal in $ B[x_n]$. Let $ \hat{B} = B/\mathfrak{b}$ (observe that this ring is nonzero). We have a composition of maps
\[ B[x_n] \to \hat{B}[x_n] \to \hat{B}[x_n]/(\widehat{f_n(x_n)}) \]
where the first map is reduction of coefficients modulo $ \mathfrak{b}$, and the second map is the quotient by the principal ideal generated by the image $ \widehat{f_n(x_n)}$ of $ f_n(x_n)$ in $ \hat{B}[x_n]$. We know $ \hat{B}$ is a nonzero ring, so since $ f_n$ is monic, the top coefficient of $ \widehat{f_n(x_n)}$ is still $ 1 \in \hat{B}$. In particular, the top coefficient cannot be nilpotent. Furthermore, since $ f_n$ was irreducible, it is not a constant polynomial, so by the characterization of units in polynomial rings, $ \widehat{f_n(x_n)}$ is not a unit, so it does not generate the unit ideal. Thus the quotient $ \hat{B}[x_n]/(\widehat{f_n(x_n)})$ should not be the zero ring.

On the other hand, observe that each $ f_i(x_i)$ is in the kernel of this composition, so in fact the entire ideal $ (f_1(x_1), \dotsc, f_n(x_n))$ is contained in the kernel. But this ideal is the unit ideal, so all of $ B[x_n]$ is in the kernel of this composition. In particular, $ 1 \in B[x_n]$ is in the kernel, and since ring maps preserve identity, this forces $ 1 = 0$ in $ \hat{B}[x_n]/(\widehat{f_n(x_n)})$, which makes this the the zero ring. This contradicts our previous observation, and proves the claim that $ \mathfrak{a}$ is a proper ideal.

Now, given claim 1, there exists a maximal ideal $ \mathfrak{m}$ of $ A$ containing $ \mathfrak{a}$. Let $ K_1 = A/\mathfrak{m}$. This is an extension field of $ K$ via the inclusion given by
\[ K \to A \to A/\mathfrak{m} \]
(this map is automatically injective as it is a map between fields). Furthermore every $ f \in \Sigma$ has a root in $ K_1$. Specifically, the coset $ x_f + \mathfrak{m}$ in $ A/\mathfrak{m} = K_1$ is a root of $ f$ since
\[ f(x_f + \mathfrak{m}) = f(x_f) + \mathfrak{m} = 0. \]

Inductively, given $ K_n$ for some $ n \geq 1$, repeat the construction with $ K_n$ in place of $ K$ to get an extension field $ K_{n+1}$ of $ K_n$ in which every irreducible $ f \in K_n[x]$ has a root. Let $ L = \bigcup_{n = 1}^{\infty} K_n$.

\emph{Claim 2}. Every $ f \in L[x]$ splits completely into linear factors in $ L$.

\emph{Proof of claim 2}. We induct on the degree of $ f$. In the base case, when $ f$ itself is linear, there is nothing to prove. Inductively, suppose every polynomial in $ L[x]$ of degree less than $ n$ splits completely into linear factors, and suppose
\[ f = a_0 + a_1x + \dotsb + a_nx^n \in L[x] \]
has degree $ n$. Then each $ a_i \in K_{n_i}$ for some $ n_i$, so let $ n = \max n_i$ and regard $ f$ as a polynomial in $ K_n[x]$. If $ f$ is reducible in $ K_n[x]$, then we have a factorization $ f = gh$ with the degree of $ g, h$ strictly less than $ n$. Therefore, inductively, they both split into linear factors in $ L[x]$, so $ f$ must also. On the other hand, if $ f$ is irreducible, then by our construction, it has a root $ a\in K_{n+1}$, so we have $ f = (x - a) g$ for some $ g \in K_{n+1}[x]$ of degree $ n - 1$. Again inductively, we can split $ g$ into linear factors in $ L$, so clearly we can do the same with $ f$ also. This completes the proof of claim 2.

Let $ \bar{K}$ be the set of algebraic elements in $ L$. Clearly $ \bar{K}$ is an algebraic extension of $ K$. If $ f \in \bar{K}[x]$, then we have a factorization of $ f$ in $ L[x]$ into linear factors
\[ f = b(x - a_1)(x - a_2) \dotsb (x - a_n). \]
for $ b \in \bar{K}$ and, a priori, $ a_i \in L$. But each $ a_i$ is a root of $ f$, which means it is algebraic over $ \bar{K}$, which is an algebraic extension of $ K$; so by transitivity of "being algebraic," each $ a_i$ is algebraic over $ K$. So in fact we conclude that $ a_i \in \bar{K}$ already, since $ \bar{K}$ consisted of all elements algebraic over $ K$. Therefore, since $ \bar{K}$ is an algebraic extension of $ K$ such that every $ f \in \bar{K}[x]$ splits into linear factors in $ \bar{K}$, $ \bar{K}$ is the algebraic closure of $ K$.

\end{proof}

\add{two algebraic closures are isomorphic}

Let $K$ be an algebraically closed field. Then the ring $K[x]$ has a very
simple ideal structure.
Since every polynomial $P \in K[x]$ has a root, it follows that there is always
a decomposition (by dividing repeatedly)
\[ P =c (x-\alpha_1)\dots(x-\alpha_n)  ,\]
where $c$ is the constant term and the $\left\{\alpha_i\right\} \subset k$ are the roots
of $P$.
In particular:
\begin{proposition} 
For $K$ algebraically closed, the only irreducible polynomials in $K[x]$ are
the linear polynomials $c(x-\alpha), \ c, \alpha \in K$ (and $c \neq 0$).
\end{proposition} 

In particular, two polynomials in $K[x]$ are \textbf{relatively prime}
(i.e., generate the unit ideal) 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 $\alpha$).


If $k$ is \emph{not} algebraically closed, then this still gives
information about when two polynomials in $k[x]$ generate the unit ideal.

\begin{definition} 
If $k$ is any field, we say that two polynomials in $k[x]$ are
\textbf{relatively prime} if they generate the unit ideal in $k[x]$.
\end{definition} 

\begin{proposition} \label{primepoly}
Two polynomials in $k[x]$ are relatively prime precisely when they
have no common roots in an algebraic closure $\overline{k}$ of $k$.
\end{proposition} 
\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{Separability and normality}


\subsection{Separable extensions}

Throughout, $F \subset K$ is a finite field extension.  We fix once and for
all an algebraic closure $\overline{F}$ for $F$ and an embedding of $F$ in $M$.


\begin{definition}
For an element $\alpha \in K$ with minimal polynomial $q \in F[x]$, we say
$q$ and $\alpha$ are \textbf{separable} if $q$ has distinct roots (in some
algebraic closure $\overline{F}$!), and we say $K$ is
separable if this holds for all $\alpha \in K$.
\end{definition}



By \cref{primepoly}, separability of a polynomial $P \in F[x]$ is equivalent
to $(P, P') = 1$ in $F[x]$.
Indeed, this follows from the fact that $P$ has no multiple roots if and only if $P, P'$ have no
common roots.

\begin{lemma} $q(x) \in F[x]$ is separable if and only if $\gcd(q, q') = 1$,
where $q'$ is the formal derivative of $q$. 
\label{der_poly}
\end{lemma}




\subsection{Purely inseparable extensions}

\begin{definition}
For an element $\alpha \in K$ with minimal polynomial $q$, we say $\alpha$ is \textbf{purely
inseparable} if $q$ has only one root.  We say $K$ is splitting if each $q$
splits in $K$.
\label{def:sepsplit}
\end{definition}


\begin{definition} If $K = F(\alpha)$ for some $\alpha$ with minimal polynomial
$q(x) \in F[x]$, then by \rref{sep_poly}, $q(x) = r(x^{p^d})$, where $p =
\Char{F}$ (or $1$ if $\Char{F} = 0$) and $r$ is separable; in this case we
also denote $\deg_s(K/F) = \deg(r), \deg_i(K/F) = p^d$.  \label{def:prim_sep}
\end{definition}


\section{Galois theory}
\subsection{Definitions}

Throughout, $F \subset K$ is a finite field extension.  We fix once and for
all an algebraic closure $M$ for both and an embedding of $F$ in $M$.  When
necessary, we write $K = F(\alpha_1, \dots, \alpha_n)$, and $K_0 = F, K_i =
F(\alpha_1, \dots, \alpha_i)$, $q_i$ the minimal polynomial of $\alpha_i$ over
$F_{i - 1}$, $Q_i$ that over $F$.

\begin{definition} $\Aut(K/F)$ denotes the group of automorphisms of $K$ which fix
$F$ (pointwise!).  $\Emb(K/F)$ denotes the set of embeddings of $K$ into $M$
respecting the chosen embedding of $F$.
\label{def:gal}
\end{definition}

\begin{definition} By $\deg(K/F)$ we mean the dimension of $K$ as an $F$-vector
space.  We denote $K_s/F$ the set of elements of $K$ whose minimal polynomials
over $F$ have distinct roots; by \rref{sep_subfield} this is a subfield, and
$\deg(K_s/F) = \deg_s(K/F)$ and $\deg(K/K_s) = \deg_i(K/F)$ by definition.
\label{def:sep}
\end{definition}
\subsection{Theorems}
\begin{lemma} If $\Char{F} = 0$ then $K_s = K$.  If $\Char{F} = p > 0$, then for
any irreducible $q(x) \in K[x]$, there is some $d \geq 0$ and polynomial $r(x)
\in K[x]$ such that $q(x) = r(x^{p^d})$, and $r$ is separable and irreducible.
\label{sep_poly}
\end{lemma}

\begin{proof} By formal differentiation, $q'(x)$ has positive degree unless
each exponent is a multiple of $p$; in characteristic zero this never occurs.
If this is not the case, since $q$ is irreducible, it can have no factor in
common with $q'$ and therefore has distinct roots by \rref{der_poly}.

If $p > 0$, let $d$ be the largest integer such that each exponent of $q$ is a
multiple of $p^d$, and define $r$ by the above equation.  Then by
construction, $r$ has at least one exponent which is not a multiple of $p$,
and therefore has distinct roots. \end{proof}

\begin{corollary} In the statement of \rref{sep_poly}, $q$ and $r$ have the same
number of roots.
\label{sep_roots}
\end{corollary}

\begin{proof} $\alpha$ is a root of $q$ if and only if $\alpha^{p^d}$ is a
root of $r$; i.e. the roots of $q$ are the roots of $x^{p^d} - \beta$, where
$\beta$ is a root of $r$.  But if $\alpha$ is one such root, then $(x -
\alpha)^{p^d} = x^{p^d} - \alpha^{p^d} = x^{p^d} - \beta$ since $\Char{K} =
p$, and therefore $\alpha$ is the only root of $x^{p^d} - \beta$. \end{proof}

\begin{lemma} The correspondence which to each $g \in \Emb(K/F)$ assigns the
$n$-tuple $(g(\alpha_1), \dots, g(\alpha_n))$ of elements of $M$ is a
bijection from $\Emb(K/F)$ to the set of tuples of $\beta_i \in M$, such that
$\beta_i$ is a root of $q_i$ over $K(\beta_1, \dots, \beta_{i - 1})$.
\label{emb_roots}
\end{lemma}

\begin{proof} First take $K = F(\alpha) = F[x]/(q)$, in which case the maps $g
\colon K \to M$ over $F$ are identified with the elements $\beta \in M$ such
that $q(\beta) = 0$ (where $g(\alpha) = \beta$).

Now, considering the tower $K = K_n / K_{n - 1} / \dots / K_0 = F$, each
extension of which is primitive, and a given embedding $g$, we define
recursively $g_1 \in \Emb(K_1/F)$ by restriction and subsequent $g_i$ by
identifying $K_{i - 1}$ with its image and restricting $g$ to $K_i$.  By the
above paragraph each $g_i$ corresponds to the image $\beta_i = g_i(\alpha_i)$,
each of which is a root of $q_i$.  Conversely, given such a set of roots of
the $q_i$, we define $g$ recursively by this formula. \end{proof}

\begin{corollary} $|\Emb(K/F)| = \prod_{i = 1}^n \deg_s(q_i)$.
\label{emb_size}
\end{corollary}

\begin{proof} This follows immediately by induction from \rref{emb_roots} by
\rref{sep_roots}. \end{proof}

\begin{lemma} For any $f \in \Emb(K/F)$, the map $\Aut(K/F) \to \Emb(K/F)$ given
by $\sigma \mapsto f \circ \sigma$ is injective.  
\label{aut_inj}
\end{lemma}

\begin{proof} This is immediate from the injectivity of $f$. \end{proof}

\begin{corollary} $\Aut(K/F)$ is finite.
\label{aut_fin}
\end{corollary}

\begin{proof} By \rref{aut_inj}, $\Aut(K/F)$ injects into $\Emb(K/F)$, which by
\rref{emb_size} is finite. \end{proof}

\begin{proposition} The inequality
\begin{equation*}
|\Aut(K/F)| \leq |\Emb(K/F)|
\end{equation*}
is an equality if and only if the $q_i$ all split in $K$.
\label{aut_ineq}
\end{proposition}

\begin{proof} The inequality follows from \rref{aut_inj} and from \rref{aut_fin}.
Since both sets are finite, equality holds if and only if the injection of
\rref{aut_inj} is surjective (for fixed $f \in \Emb(K/F)$).

If surjectivity holds, let $\beta_1, \dots, \beta_n$ be arbitrary roots of
$q_1, \dots, q_n$ in the sense of \rref{emb_roots}, and extract an embedding $g
\colon K \to M$ with $g(\alpha_i) = \beta_i$.  Since the correspondence $f
\mapsto f \circ \sigma$ ($\sigma \in \Aut(K/F)$) is a bijection, there is some
$\sigma$ such that $g = f \circ \sigma$, and therefore $f$ and $g$ have the
same image.  Therefore the image of $K$ in $M$ is canonical, and contains
$\beta_1, \dots, \beta_n$ for any choice thereof.

If the $q_i$ all split, let $g \in \Emb(K/F)$ be arbitrary, so the
$g(\alpha_i)$ are roots of $q_i$ in $M$ as in \rref{emb_roots}.  But the $q_i$
have all their roots in $K$, hence in the image $f(K)$, so $f$ and $g$ again
have the same image, and $f^{-1} \circ g \in \Aut(K/F)$.  Thus $g = f \circ
(f^{-1} \circ g)$ shows that the map of \rref{aut_inj} is surjective.
\end{proof}

\begin{corollary} Define
\begin{equation*}
D(K/F) = \prod_{i = 1}^n \deg_s(K_i/K_{i - 1}).
\end{equation*}
Then the chain of equalities and inequalities
\begin{equation*}
|\Aut(K/F)| \leq |\Emb(K/F)| = D(K/F) \leq \deg(K/F)
\end{equation*}
holds; the first inequality is an equality if and only if each $q_i$ splits in
$K$, and the second if and only if each $q_i$ is separable.
\label{large_aut_ineq}
\end{corollary}

\begin{proof} The statements concerning the first inequality are just
\rref{aut_ineq}; the interior equality is just \rref{emb_size}; the latter
inequality is obvious from the multiplicativity of the degrees of field
extensions; and the deduction for equality follows from the definition of
$\deg_s$. \end{proof}

\begin{corollary} The $q_i$ respectively split and are separable in $K$ if and only
if the $Q_i$ do and are.
\label{absolute_sepsplit}
\end{corollary}

\begin{proof} The ordering of the $\alpha_i$ is irrelevant, so we may take
each $i = 1$ in turn.  Then $Q_1 = q_1$ and if either of the equalities in
\rref{large_aut_ineq} holds then so does the corresponding statement here.
Conversely, clearly each $q_i$ divides $Q_i$, so splitting or separability
for the latter implies that for the former. \end{proof}

\begin{corollary} Let $\alpha \in K$ have minimal polynomial $q$; if the $Q_i$ are
respectively split, separable, and purely inseparable over $F$ then $q$ is as
well.
\label{global_sepsplit}
\end{corollary}

\begin{proof} We may take $\alpha$ as the first element of an alternative
generating set for $K/F$.  The numerical statement of \rref{large_aut_ineq}
does not depend on the particular generating set, hence the conditions given
hold of the set containing $\alpha$ if and only if they hold of the canonical
set ${\alpha_1, \dots, \alpha_n}$.

For purely inseparable, if the $Q_i$ all have only one root then $|\Emb(K/F)|
= 1$ by \rref{large_aut_ineq}, and taking $\alpha$ as the first element of a
generating set as above shows that $q$ must have only one root as well for
this to hold. \end{proof}

\begin{corollary} $K_s$ is a field and $\deg(K_s/F) = D(K/F)$.
\label{sep_subfield}
\end{corollary}

\begin{proof} Assume $\Char{F} = p > 0$, for otherwise $K_s = K$.  Using
\rref{sep_poly}, write each $Q_i = R_i(x^{p^{d_i}})$, and let $\beta_i =
\alpha_i^{p^{d_i}}$.  Then the $\beta_i$ have $R_i$ as minimal polynomials and
the $\alpha_i$ satisfy $s_i = x^{p^{d_i}} - \beta_i$ over $K' = F(\beta_1,
\dots, \beta_n)$.  Therefore the $\alpha_i$ have minimal polynomials over $K'$
dividing the $s_i$ and hence those polynomials have but one distinct root.

By \rref{global_sepsplit}, the elements of $K'$ are separable, and those of
$K'$ purely inseparable over $K'$.  In particular, since these minimal
polynomials divide those over $F$, none of these elements is separable, so $K'
= K_s$.

The numerical statement follows by computation:
\begin{equation*}
\deg(K/K') = \prod_{i = 1}^n p^{d_i}
	= \prod_{i = 1}^n \frac{\deg(K_i/K_{i - 1})}{\deg_s(K_i/K_{i - 1})}
	= \frac{\deg(K/F)}{D(K/F)}. 
	\end{equation*}
\end{proof}

\begin{theorem} The following inequality holds:
\begin{equation*}
|\Aut(K/F)| \leq |\Emb(K/F)| = \deg_s(K/F) \leq \deg(K/F).
\end{equation*}
Equality holds on the left if and only if $K/F$ is splitting; it holds on the
right if and only if $K/F$ is separable.
\label{galois_size}
\end{theorem}

\begin{proof} The numerical statement combines \rref{large_aut_ineq} and
\rref{sep_subfield}.  The deductions combine \rref{absolute_sepsplit} and
\rref{global_sepsplit}. \end{proof}

\subsection{Definitions}

Throughout, we will denote as before $K/F$ a finite field extension, and $G =
\Aut(K/F)$, $H$ a subgroup of $G$.  $L/F$ is a subextension of $K/F$.

\begin{definition} When $K/F$ is separable and splitting, we say it is Galois and
write $G = \Gal(K/F)$, the Galois group of $K$ over $F$.
\label{defn:galois_extension}
\end{definition}

\begin{definition} The fixed field of $H$ is the field $K^H$ of elements fixed by
the action of $H$ on $K$.  Conversely, $G_L$ is the fixing subgroup of $L$,
the subgroup of $G$ whose elements fix $L$.
\label{defn:fixing}
\end{definition}

\subsection{Theorems}

\begin{lemma} A polynomial $q(x) \in K[x]$ which splits in $K$ lies in
$K^H[x]$ if and only if its roots are permuted by the action of $H$.  In this
case, the sets of roots of the irreducible factors of $q$ over $K^H$ are the orbits
of the action of $H$ on the roots of $q$ (counting multiplicity).
\label{root_action}
\end{lemma}

\begin{proof} Since $H$ acts by automorphisms, we have $\sigma q(x) = q(\sigma
x)$ as a functional equation on $K$, so $\sigma$ permutes the roots of $q$.
Conversely, since the coefficients of $\sigma$ are the elementary symmetric
polynomials in its roots, $H$ permuting the roots implies that it fixes the
coefficients.

Clearly $q$ is the product of the polynomials $q_i$ whose roots are the orbits
of the action of $H$ on the roots of $q$, counting multiplicities, so it
suffices to show that these polynomials are defined over $K^H$ and are
irreducible.  Since $H$ acts on the roots of the $q_i$ by construction, the
former is satisfied.  If some $q_i$ factored over $K^H$, its factors would
admit an action of $H$ on their roots by the previous paragraph.  The roots of
$q_i$ are distinct by construction, so its factors do not share roots; hence
the action on the roots of $q_i$ would not be transitive, a contradiction.
\end{proof}

\begin{corollary} Let $q(x) \in K[x]$; if it is irreducible, then $H$ acts
transitively on its roots; conversely, if $q$ is separable and $H$ acts
transitively on its roots, then $q(x) \in K^H[x]$ is irreducible.
\label{sep_irred}
\end{corollary}

\begin{proof} Immediate from \rref{root_action}. \end{proof}

\begin{lemma} If $K/F$ is Galois, so is $K/L$, and $\Gal(K/L) = G_L$..
\label{sub_galois}
\end{lemma}

\begin{proof} $K/F$ Galois means that the minimal polynomial over $F$ of every
element of $K$ is separable and splits in $K$; the minimal polynomials over $L
= K^H$ divide those over $F$, and therefore this is true of $K/L$ as well;
hence $K/L$ is likewise a Galois extension. $\Gal(K/L) = \Aut(K/L)$ consists
of those automorphisms $\sigma$ of $K$ which fix $L$; since $F \subset L$ we
have \emph{a fortiori} that $\sigma$ fixes $F$, hence $\Gal(K/L) \subset G$
and consists of the subgroup which fixes $L$; i.e. $G_L$. \end{proof}

\begin{corollary} If $K/F$ and $L/F$ are Galois, then the action of $G$ on elements of $L$
defines a surjection of $G$ onto $\Gal(L/F)$.  Thus $G_L$ is normal in $G$ and $\Gal(L/F) \cong G/G_L$.  Conversely, if $N \subset G$ is normal, then $K^N/F$ is Galois.
\label{normal}
\end{corollary}

\begin{proof} $L/F$ is splitting, so by \rref{root_action} the elements of $G$
act as endomorphisms (hence automorphisms) of $L/F$, and the kernel of this action is $G_L$.  By
\rref{sub_galois}, we have $G_L = \Gal(K/L)$, so $|G_L| = |\Gal(K/L)| = [K : L] = [K : F] / [L : F]$,
or rearranging and using that $K/F$ is Galois, we get $|G|/|G_L| = [L : F] =
|\Gal(L/F)|$.  Thus the map $G \to \Gal(L/F)$ is surjective and thus the induced map $G/G_L \to
\Gal(L/F)$ is an isomorphism.

Conversely, let $N$ be normal and take $\alpha \in K^N$.  For any conjugate $\beta$ of $\alpha$, we
have $\beta = g(\alpha)$ for some $g \in G$; let $n \in N$.  Then $n(\beta) = (ng)(\alpha) =
g(g^{-1} n g)(\alpha) = g(\alpha) = \beta$, since $g^{-1} n g \in N$ by normality of $N$.  Thus
$\beta \in K^N$, so $K^N$ is splitting, i.e., Galois. \end{proof}

\begin{proposition} If $K/F$ is Galois and $H = G_L$, then $K^H = L$.
\label{fixed_field}
\end{proposition}

\begin{proof} By \rref{sub_galois}, $K/L$ and $K/K^H$ are both Galois.  By
definition, $\Gal(K/L) = G_L = H$; since $H$ fixes $K^H$ we certainly have
$H < \Gal(K/K^H)$, but since $L \subset K^H$ we have \emph{a fortiori} that
$\Gal(K/K^H) < \Gal(K/L) = H$, so $\Gal(K/K^H) = H$ as well.  It follows
from \rref{galois_size} that $\deg(K/L) = |H| = \deg(K/K^H)$, so that $K^H =
L$. \end{proof}

\begin{lemma} If $K$ is a finite field, then $K^\ast$ is cyclic.
\label{fin_cyclic}
\end{lemma}

\begin{proof} $K$ is then a finite extension of $\mathbb{F}_p$ for $p =
\Char{K}$, hence has order $p^n$, $n = \deg(K/\mathbb{F}_p)$.  Thus
$\alpha^{p^n} = \alpha$ for all $\alpha \in K$, since $|K^\ast| = p^n - 1$.
It follows that every element of $K$ is a root of $q_n(x) = x^{p^n} - x$.  For
any $d < n$, the elements of order at most $p^d - 1$ satisfy $q_d(x)$, which has
$p^d$ roots.  It follows that there are at least $p^n(p - 1) > 0$ elements of
order exactly $p^n - 1$, so $K^\ast$ is cyclic. \end{proof}

\begin{corollary} If $K$ is a finite field, then $\Gal(K/F)$ is cyclic, generated by
the Frobenius automorphism.
\label{fin_gal_cyclic}
\end{corollary}

\begin{proof} First take $F = \mathbb{F}_p$.  Then the map $f_i(\alpha) =
\alpha^{p^i}$ is an endomorphism, injective since $K$ is a field, and
surjective since it is finite, hence an automorphism.  Since every $\alpha$
satisfies $\alpha^{p^n} = \alpha$, $f_n = 1$, but by \rref{fin_cyclic}, $f_{n -
1}$ is nontrivial (applied to the generator).  Since $n = \deg(K/F)$, $f =
f_1$ generates $\Gal(K/F)$.

If $F$ is now arbitrary, by \rref{fixed_field} we have $\Gal(K/F) =
\Gal(K/\mathbb{F}_p)_F$, and every subgroup of a cyclic group is cyclic.
\end{proof}

\begin{corollary} If $K$ is finite, $K/F$ is primitive.
\label{fin_prim_elt}
\end{corollary}

\begin{proof} No element of $G$ fixes the generator $\alpha$ of $K^\ast$, so
it cannot lie in any proper subfield.  Therefore $F(\alpha) = K$. \end{proof}

\begin{proposition} If $F$ is infinite and $K/F$ has only finitely many subextensions, then it is
primitive.
\label{gen_prim_elt}
\end{proposition}

\begin{proof} We proceed by induction on the number of generators of $K/F$.

If $K = F(\alpha)$ we are done.  If not, $K = F(\alpha_1, \dots, \alpha_n) =
F(\alpha_1, \dots, \alpha_{n - 1})(\alpha_n) = F(\beta, \alpha_n)$ by
induction, so we may assume $n = 2$.  There are infinitely many subfields
$F(\alpha_1 + t \alpha_2)$, with $t \in F$, hence two of them are equal, say for $t_1$ and
$t_2$.  Thus, $\alpha_1 + t_2 \alpha_2 \in F(\alpha_1 + t_1 \alpha_2)$.  Then
$(t_2 - t_1)\alpha_2 \in F(\alpha_1 + t_1 \alpha_2)$, hence $\alpha_2$ lies in
this field, hence $\alpha_1$ does.  Therefore $K = F(\alpha_1 + t_1
\alpha_2)$. \end{proof}

\begin{corollary} If $K/F$ is separable, it is primitive, and the generator may be
taken to be a linear combination of any finite set of generators of $K/F$.
\label{prim_elt}
\end{corollary}

\begin{proof} We may embed $K/F$ in a Galois extension $M/F$ by adjoining all
the conjugates of its generators.  Subextensions of $K/F$ are as well subextensions
of $K'/F$ and by \rref{fixed_field} the map $H \mapsto (K')^H$ is a surjection
from the subgroups of $G$ to the subextensions of $K'/F$, which are hence
finite in number.  By \rref{fin_prim_elt} we may assume $F$ is infinite.  The
result now follows from \rref{gen_prim_elt}. \end{proof}

\begin{corollary}
 If $K/F$ is Galois and $H \subset G$, then if $L = K^H$, we have $H = G_L$.
 \label{fixing_subgroup}
\end{corollary}

\begin{proof}
 Let $\alpha$ be a primitive element for $K/L$.  The polynomial $\prod_{h \in H} (x - h(\alpha))$ is fixed by $H$, and therefore has coefficients in $L$, so $\alpha$ has $|H|$ conjugate roots over $L$.  But since $\alpha$ is primitive, we have $K = L(\alpha)$, so the minimal polynomial of $\alpha$ has degree $\deg(K/L)$, which is the same as the number of its roots.  Thus $|H| = \deg(K/L)$.  Since $H \subset G_L$ and $|G_L| = \deg(K/L)$, we have equality.
\end{proof}


\begin{theorem} The correspondences $H \mapsto K^H$, $L \mapsto G_L$ define
inclusion-reversing inverse maps between the set of subgroups of $G$ and the
set of subextensions of $K/F$, such that normal subgroups and Galois subfields
correspond.
\label{fundamental_theorem}
\end{theorem}

\begin{proof} This combines \rref{fixed_field}, \rref{fixing_subgroup}, and \rref{normal}.
\end{proof}


\section{Transcendental Extensions}


There is a distinguished type of transcendental extension: those that are
``purely transcendental.'' 
\begin{definition} A field extension $E'/E$ is purely transcendental if it is
obtained by adjoining a set $B$ of algebraically independent elements. A set of
elements is algebraically independent over $E$ if there is no nonzero polynomial$P$
with coefficients in $E$ such
that $P(b_1,b_2,\cdots b_n)=0$ for any finite subset of elements $b_1, \dots,
b_n \in B$.
\end{definition}

\begin{example} The field $\mathbb{Q}(\pi)$ is purely transcendental; in
particular, $\mathbb{Q}(\pi)\cong\mathbb{Q}(x)$ with the isomorphism fixing
$\mathbb{Q}$. \end{example}
Similar to the degree of an algebraic extension, there is a way of keeping
track of the number of algebraically independent generators that are required to
generate a purely transcendental extension.
\begin{definition} Let $E'/E$ be a purely transcendental extension generated by
some set of algebraically independent elements $B$. Then the transcendence
degree $trdeg(E'/E)=\#(B)$ and $B$ is called a transcendence basis for $E'/E$
(we will see later that $trdeg(E'/E)$ is independent of choice of basis).
\end{definition}
In general, let $F/E$ be a field extension, we can always construct an
intermediate extension $F/E'/E$ such that $F/E'$ is algebraic and $E'/E$ is
purely transcendental. Then if $B$ is a transcendence basis for $E'$, it is
also called a transcendence basis for $F$. Similarly, $trdeg(F/E)$ is defined to
be
$trdeg(E'/E)$.
\begin{theorem} Let $F/E$ be a field extension, a transcendence basis exists.
\end{theorem}
\begin{proof} Let $A$ be an algebraically independent subset of $F$. Now pick a
subset $G\subset F$ that generates $F/E$, we can find a transcendence basis
$B$ such that $A\subset B\subset G$. Define a collection of algebraically
independent sets $\mathcal{B}$ whose members are subsets of $G$ that contain
$A$. The set can be partially ordered inclusion and contains at least one
element, $A$. The union of elements of $\mathcal{B}$ is algebraically
independent since any algebraic dependence relation would have occurred in one
of the elements of $\mathcal{B}$ since the polynomial is only allowed to be over
finitely many variables. The union also satisfies $A\subset
\bigcup\mathcal{B}\subset G$ so by Zorn's lemma, there is a maximal element
$B\in\mathcal{B}$. Now we claim $F$ is algebraic over $E(B)$. This is because
if it wasn't then there would be a transcendental element $f\in G$ (since
$E(G)=F$)such that $B\cup\{f\}$ wold be algebraically independent contradicting
the
maximality of $B$. Thus $B$ is our transcendence basis. \end{proof}
Now we prove that the transcendence degree of a field extension is independent
of choice of basis.
\begin{theorem} Let $F/E$ be a field extension. Any two transcendence bases for
$F/E$ have the same cardinality. This shows that the $trdeg(E/F)$ is well
defined. \end{theorem}
\begin{proof} 
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 $E(B^*)$. Therefor $\beta$ is algebraic over $E(B^*)$, a
contradiction. So $\#(B)\leq \sum_{\alpha\in B'} \#(B_{\alpha})$. Now if $B'$ is
finite, then so is $B$ so we can assume $B'$ is infinite; this means
\begin{equation} \#(B)\leq \sum_{\alpha\in B'}\#(B_{\alpha})=\#(\coprod
B_{\alpha})\leq \#(B'\times\mathbb{Z})=\#(B')\end{equation} with the inequality $\#(\coprod
B_{\alpha}) \leq \#(B'\times \mathbb{Z})$ given by the correspondence
$b_{\alpha_i}\mapsto (\alpha,i)\in B'\times \mathbb{Z}$ with $B_\alpha =
\{b_{\alpha_1},b_{\alpha_2}\cdots b_{\alpha_{n_\alpha}}\}$ Therefore in the
infinite case, $\#(B)=\#(B')$.

Now we need to look at the case where $B$ is finite. In this case, $B'$ is also
finite, so suppose $B=\{\alpha_1,\cdots\alpha_n\}$ and
$B'=\{\beta_1,\cdots\beta_m\}$ with $m\leq n$. We perform induction on $m$: if
$m=0$ then $F/E$ is algebraic so $B=\null$ so $n=0$, otherwise there is an
irreducible polynomial $f\in E[x,y_1,\cdots y_n]$ such that
$f(\beta_1,\alpha_1,\cdots \alpha_n) = 0$. Since $\beta_1$ is not algebraic over
$E$, $f$ must involve some $y_i$ so without loss of generality, assume $f$ uses
$y_1$. Let $B^*=\{\beta_1,\alpha_2,\cdots\alpha_n\}$. We claim that $B^*$ is a
basis for $F/E$. To prove this claim, we see that we have a tower of algebraic
extensions $F/E(B^*,\alpha_1)/E(B^*)$ since $\alpha_1$ is algebraic over
$E(B^*)$. 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 $g\in E[x,y_2,\cdots y_n]$ such that
$g(\beta_1,\alpha_2,\cdots\alpha_n)=0$ which must involve $x$ making $\beta_1$
algebraic over $E(\alpha_2,\cdots \alpha_n)$ which would make $\alpha_1$
algebraic over $E(\alpha_2,\cdots \alpha_n)$ which is impossible. So this means
that $\{\alpha_2,\cdots\alpha_n\}$ and $\{\beta_2,\cdots\beta_m\}$ are bases for
$F$ over $E(\beta_1)$ which means by induction, $m=n$. \end{proof}

\begin{example} Consider the field extension $\mathbb{Q}(e,\pi)$ 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 not this is true is
unknown and the problem of determining $trdeg(\mathbb{Q}(e,\pi))$ is an open
problem.\end{example}

\begin{example} let $E$ be a field and $F=E(t)/E$. Then $\{t\}$ is a
transcendence basis since $F=E(t)$. However, $\{t^2\}$ is also a transcendence
basis since $E(t)/E(t^2)$ is algebraic. This illustrates that while we can
always decompose an extension $F/E$ into an algebraic extension $F/E'$ and a
purely transcendental extension $E'/E$, this decomposition is not unique and
depends on choice of transcendence basis. \end{example}

\begin{exercise} If we have a tower of fields $G/F/E$, then $trdeg(G/E)=trdeg(F/E)+trdeg(G/F)$. \end{exercise}

\begin{example} 
Let $X$ be a compact Riemann surface. Then the function field $\mathbb{C}(X)$
(see \cref{meromorphicfn}) has transcendence degree one over $\mathbb{C}$. In
fact, \emph{any} finitely generated extension of $\mathbb{C}$ of transcendence
degree one arises from a Riemann surface. There is even an equivalence of
categories between the category of compact Riemann surfaces and
(non-constant) holomorphic maps 
and the opposite category of finitely generated extensions of $\mathbb{C}$ and
morphisms of $\mathbb{C}$-algebras. See \cite{Fo81}.

There is an algebraic version of the above statement as well. Given an
(irreducible) algebraic curve in projective space over an algebraically
closed field $k$ (e.g. the complex numbers), one can consider its ``field of rational
functions:'' basically, functions that look like quotients of polynomials,
where the denominator does not identically vanish on the curve. 
There is a similar anti-equivalence of categories between smooth projective curves and
non-constant morphisms of curves and finitely generated extensions of $k$ of
transcendence degree one. See \cite{Ha77}.
\end{example} 


\subsection{Linearly Disjoint Field Extensions}
Let $k$ be a field, $K$ and $L$ field extensions of $k$. Suppose also that $K$ and $L$ are embedded in some larger field $\Omega$. 

\begin{definition} The compositum of $K$ and $L$ written $KL$ is $k(K\cup L)=L(K)=K(L)$. 
\end{definition}



\begin{definition} $K$ and $L$ are said to be linearly disjoint over $k$ if the following map is injective:
\begin{equation} \theta: K\otimes_k L\rightarrow KL \end{equation} defined by $x\otimes y\mapsto xy$. 
\end{definition}