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\begin{document}
\centerline{Math 361 lecture for Friday, Week 1}
\bigskip

\centerline{Field extensions}
\bigskip

In the first lecture, in order to think about Pythagorean triples, we extended the field of
rational numbers by adding $i$ to get the field $\Q(i)=\{a+bi: a,b\in \Q\}$.  We
then considered the Gaussian integers sitting inside of that field,
\[
  \Z[i]=\{m+ni: m,n\in\Z\}\subset \Q(i).
\]
We picture the situation as follows, with lines representing the
superset/subset relation:
\begin{center}
  \begin{tikzcd}
    \Q(i)\arrow[r,dash] \arrow[d,dash]& \Z[i]\arrow[d,dash]\\
    \Q\arrow[r,dash]& \Z.
  \end{tikzcd}
\end{center}
To understand Pythagorean triples, we then considered factorization of elements
in $\Z[i]$. Our goal is to generalize this construction: extend a field by
adding some elements not in the field, and then consider a subset that plays
the role of the integers inside that field.  Of special interest with be
factorization properties of these new integers.  
\bigskip

\centerline{\textsc{Extension fields}}
A \emph{field extension} is a pair of fields $K\subseteq L$.  In that case, $L$
is automatically a vector space over~$K$.  It's dimension is denoted
\[
  [L\mathbin{:} K] := \dim_{K}L,
\]
and we might display this like so:
\begin{center}
  \begin{tikzcd}
    L\arrow[d,dash,"{[L\mathbin{:}K]}"]\\
    K.
  \end{tikzcd}
\end{center}
If $[L\mathbin{:}K]<\infty$, we say that $L$ is a \emph{finite field extension
of $K$}.
For example, a basis for $\Q(i)$ over $\Q$ is $\{1,i\}$, and we have
\begin{center}
  \begin{tikzcd}
    \Q(i)\arrow[d,dash,"2"]\\
    \Q.
  \end{tikzcd}
\end{center}
We usually denote a field extension~$K\subseteq L$ by $L/K$. 

\begin{prop}
  Suppose $K,H$ and $L$ are fields with $K\subseteq H\subseteq L$, and suppose
  that $[L\mathbin{:}K]<\infty$.  Then~$[L\mathbin{:}H]<\infty$ and $[H\mathbin{:}K]<\infty$, and
  \[
    [L\mathbin{:}K]=[L\mathbin{:}H][H\mathbin{:}K].
  \]
\end{prop}
\begin{proof}
 Homework.
\end{proof}
\bigskip

\centerline{\textsc{Algebraic elements of a field extension}}

\begin{defn}
  Let $L/K$ be a field extension.  Then $\alpha\in L$ is
  \emph{algebraic} over $K$ if there exists a nonzero polynomial $f\in K[x]$ such
  that $f(\alpha)=0$.
\end{defn}

\begin{eg}
  The numbers $\sqrt{2},i\in \C$ are algebraic over $\Q$ since they are zeros of
  $x^{2}-2$ and $x^{2}-1$, respectively.  The numbers $e,\pi\in\R$ are not
  algebraic over~$\Q$ (although this is not easy to prove).  If~$Q(t)$ denotes
  the field of rational functions in $t$ (whose elements have the
  form~$f/g$ with $f,g\in \Q[t]$ and $g\neq0$), then~$t$ is not algebraic over
  $\Q$.  However, $t$ is algebraic over $\Q(t)$ since is it satisfies the
  polynomial $x-t$ having coefficients in $\Q(t)$.  It is also algebraic
  over~$\Q(t^2)$ (the field of rational functions in $t$ with coefficients in
  $\Q$) since it satisfies the polynomial~$x^2-t\in\Q(t^2)[x]$.  
\end{eg}

\begin{prop} If $L/K$ is a field extension and $\alpha\in L$ is
  algebraic over~$K$, then there exists a unique monic polynomial~$p\in K[x]$ of
  minimal positive degree such that $p(\alpha)=0$.  (A polynomial is
  \emph{monic} if its leading coefficient is~$1$, i.e., if the coefficient of
  its term of highest degree is~$1$.)
\end{prop}
\begin{proof} Let $I$ be the set of $f\in K[x]$ such that $f(\alpha)=0$.
  Then~$I$ is an ideal, and since~$K[x]$ is a PID, there exists $p\in K[x]$ such
  that $I=(p)$. By dividing through by the leading coefficient of~$p$, we may
  assume that $p$ is monic.  If $f$ is any nonzero element of $I$, we may write
 \[
   f = pq
 \]
 for some nonzero $q\in K[x]$.  We have
 \[
   \deg(f)=\deg(pq)=\deg(p)+\deg(q)\geq \deg(p).
 \]
 If $\deg(f)=\deg(p)$, then~$\deg(q)=0$. So $q$ is a nonzero element of~$K$.
 Two polynomials are, by definition, equal if and only if their coefficients are
 equal.  So if~$\deg(f)=\deg(p)$ and~$f$ is monic, it follows that~$f=p$.
\end{proof}

\begin{defn}
  The polynomial $p$ in the above proposition is called the \emph{minimal
  polynomial for $\alpha$ over $K$}.
\end{defn}

\begin{prop}
  Let~$L/K$ be a field extension, and let~$\alpha\in L$ be algebraic over~$K$.
  Let $p$ be a monic polynomial such that~$p(\alpha)=0$.  Then~$p$ is the
  minimal polynomial for $\alpha$ over~$K$ if and only if~$p$ is irreducible.
\end{prop}
\begin{proof} First, suppose that $p$ is the minimal polynomial for~$\alpha$.
  Suppose $p=fg$ with $f,g\in K[x]$.
  Then~$0=p(\alpha)=f(\alpha)g(\alpha)$ implies that~$f(\alpha)=0$ or
  $g(\alpha)=0$.  Without loss of generality, say $f(\alpha)=0$.  Further,
  since~$p$ has positive degree, $p\neq0$, and hence,~$f\neq0$.  So
  since~$f(\alpha)=0$, it follows that~$\deg(f)>0$.\footnote{If~$\deg(f)=0$, then $f$ is
  a constant.  In that case, the only way we could have~$f(\alpha)=0$ is for
  that constant to be~$0$. However, $f\neq0$.} We have
  \[
    \deg(p)=\deg(f)+\deg(g).
  \]
  By minimality of~$p$, we have $\deg(f)=\deg(p)$ and $\deg(g)=0$.  So~$g$ is a
  (nonzero) element of~$K$, and nonzero elements of $K$ are invertible in~$K[x]$, i.e., units.  We
  have shown that if $f$ factors, then one of those factors is a unit.  Thus, $f$
  is irreducible.

  Conversely, suppose that $p$ is irreducible and that~$f\in K[x]$ is the
  minimal polynomial for~$\alpha$ over $K$.  By definition of the minimal
  polynomial, we have $\deg f\leq\deg p$ and $f(\alpha)=0$.  Use the division algorithm to write
  \[
    p = qf+r
  \]
  for some $q,r\in K[x]$ with $\deg r<\deg f$.  Then,
  \[
    0=p(\alpha)=q(\alpha)f(\alpha)+r(\alpha)=r(\alpha).
  \]
  If~$r\neq0$, let $\tilde{r}$ be~$r$ divided by its leading coefficient.  It follows that
  $\tilde{r}$ is a monic polynomial with $\tilde{r}(\alpha)=0$ and
  $\deg(\tilde{r})<\deg(f)$, contradicting the definition of~$f$.  So $r=0$ and
  $p=qf$.  Since~$p$ is irreducible $f\not\in K$, it  must be that~$q$ is a unit
  (i.e., and nonzero element of~$K$).  Since~$p$ and~$f$ are monic, $p=f$.
\end{proof}

Let $L/K$ be a field extension, and let $\alpha\in L$ be algebraic over
$K$.  Let $K[\alpha]$ be the smallest subring of $L$ containing~$\alpha$, and
let $K(\alpha)$ be the smallest subfield of $L$ containing $\alpha$.  Then
\[
  K[\alpha]:= \{f(\alpha):f\in K[x]\},
\]
and 
\[
  K(\alpha) :=\left\{\frac{f(\alpha)}{g(\alpha)}: f,g\in K[x], g(\alpha)\neq0\right\}.
\]

\begin{thm}\label{thm:01fri}
  Let $L/K$ be a field extension.  Then~$\alpha\in L$ is algebraic over~$K$ if
  and only if~$[K(\alpha)\mathbin{:}K]<\infty$.  In this case, $K[\alpha]=K(\alpha)$ and
  $[K(\alpha)\mathbin{:}K]=\deg(p)$ where $p$ is the minimal polynomial for $\alpha$
  over~$K$. 
\end{thm}
\begin{proof}
 First suppose that~$[K(\alpha)\mathbin{:}K]=n<\infty$.
 Then~$1,\alpha,\alpha^{2},\ldots,\alpha^{n}$ are $n+1$ (not necessarily
 distinct) elements in a vector
 space of dimension~$n$, so they are linearly dependent.  This
 means~$\sum_{i=0}^{n} c_i \alpha^{i}=0$ for some $c_{i}\in K$, not all zero.
 Define the polynomial~$f(x)=\sum_{i=0}^{n} c_i x^{i}$.  Then~$f\in K[x]$ and
 $f(\alpha)=0$.  So~$\alpha$ is algebraic over $K$.

 Conversely, suppose that~$\alpha$ is algebraic over~$K$, and
 let~$p=\sum_{i=0}^{n} a_i x^{i}$ be its minimal polynomial.  We first claim
 that~$1,\alpha,\alpha^{2},\ldots,\alpha^{n-1}$ are linearly independent.  If
 not, then there is a nontrivial linear relation~$\sum_{i=0}^{n-1} b_i
 \alpha^{i}$.  Defining~$f= \sum_{i=0}^{n-1} b_i x^{i}$, we have~$f\in
 K[x]$ and~$f(\alpha)=0$.  However, $\deg(f)<\deg(p)=n$, which contradicts the
 minimality of $p$. 

 Next, consider the vector space
 $V=\Span_{K}\{1,\alpha,\alpha^{2},\ldots,\alpha^{n-1}\}$.  We claim that~$V$ is
 a field.  To see this, first note that rearranging~$p(\alpha)=0$, we have
 \[
   \alpha^{n}=-\sum_{i=0}^{n-1} a_i \alpha^{i}.
 \]
 It follows that $V$ is closed under multiplication.  Most of the field
 properties then follow trivially from the fact that $V\subseteq L$ and $L$ is a
 field.  What remains is to show that nonzero elements have inverses.  So let
 $v\in V\setminus\{0\}$ and write~$v=\sum_{i=0}^{n-1} b_{i}\alpha^{i}$ for some
 $b_{i}\in K$.  Define $h=\sum_{i=0}^{n-1} b_i x^{i}\in K[x]$.  So
 $v=h(\alpha)$.  Now~$p$ is irreducible, hence, prime.  So the only prime
 factor~$h$ and $p$ could share is $p$, itself, but~$\deg(h)<\deg(p)$.  Hence,
 $h$ and $p$ are relatively prime, i.e., they share no prime factors.  It
 follows that there are $f,g\in K[x]$ such that
 \[
   fh+gp=1.
 \]
Therefore, 
 \[
  1=f(\alpha)h(\alpha)+g(\alpha)p(\alpha)=g(\alpha)h(\alpha)=f(\alpha)v.
 \]
 Hence~$v$ has the multiplicative inverse~$f(\alpha)\in V$.

 Since~$V$ is a field in~$L$ containing $\alpha$ and $K(\alpha)$ is the smallest
 field in $L$ containing $\alpha$, it follows that $K(\alpha)\subseteq V$.  On
 the other hand, $V\subseteq K[\alpha]$.  In sum,
 \[
   K(\alpha)\subseteq V \subseteq K[\alpha]\subseteq K(\alpha).
 \]
We have shown that $K(\alpha)=K[\alpha]$ of dimension $\deg(p)$. 
\end{proof}

Here is another perspective.  Suppose that $\alpha\in L$ is algebraic over the
subfield~$K$.  Define a mapping
\begin{align*}
  \phi\colon K[x]&\to K[\alpha]\\
  f&\mapsto f(\alpha).
\end{align*}
We have
\[
  \ker(\phi)=\{f\in K[x]: f(\alpha)=0\}.
\]
Since~$K[x]$ is a PID, we can write~$\ker(\phi)=(p)$ for some monic polynomial $p\in
K[x]$.  In fact, $p$ must be the minimal polynomial for $\alpha$.  By the standard isomorphism theorem, we have
\[
  K[x]/(p) \approx K[\alpha].
\]
Since $p$ is irreducible and $K[x]$ is a PID, it follows that $(p)$ is a maximal
ideal.  Hence, $K[x]/(p)$ is a field.  Therefore, $K[\alpha]$ is a field, and it
follows that~$K[\alpha]=K(\alpha)$.

\begin{cor}
  If $[L\mathbin{:} K]<\infty$ and $\alpha\in L$, then $\alpha$ is algebraic
  over~$K$.
\end{cor}
\begin{proof}
  Suppose $[L\mathbin{:} K]<\infty$ and $\alpha\in L$.  Then since $K(\alpha)$
  is a $K$-subvector space of $L$, it follows that $[K(\alpha)\mathbin{:}
  K]<\infty$, and the result follows from Theorem~\ref{thm:01fri}. 
\end{proof}
\end{document}
