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\bec {\Large\bf Splitting Fields and Normal Extensions} \ece
Here we are usually dealing with a ``ground field" $F$ and an extension field $K$.
Throughout we will only consider finite extensions $K:F$. In most examples
$F$ is the field of rational numbers $Q$. While other examples will be considered,
one may well think about $F$ as $Q$ in the first reading.
\begin{define} We say $f(x)$ completely splits into linear factors in $K[x]$ if we
may write
\beq\label{one} f(x) = \prod_{i=1}^r (x-\ah_i)^{m_i} \eeq
\end{define}
\begin{define} We say $K$ is the splitting field of a polynomial $f(x)\in F[x]$
over $F$ if
\ben
\item $f(x)$ completely splits into linear factors in $K[x]$.
\item $K=F(\ah_1,\ldots,\ah_r)$, where the $\ah_i$ are the roots of $f(x)$ in $K$.
That is, $K$ may be generated from $F$ using the roots of $f(x)$ and has nothing more.
\een
\end{define}
Our goal is the following result, but we shall first require preliminaries interesting
in their own right.
\begin{theorem}\label{bigsplit}
Suppose $K\subset C$ is the splitting field of $f(x)\in F[x]$ over $F$.
Suppose $g(x)\in F[x]$ is irreducible (over $F$) and suppose further that there
is an $\beta\in K$ with $g(\beta)=0$. Then $g(x)$ completely splits into linear
factors in $K[x]$.
\end{theorem}
{\tt Remark:} The condition that $K$ is a subfield of the complex numbers isn't totally
necessary but it somewhat simplifies the presentation and this is the main case we shall
use.
\begin{define}
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let $h(x)=h_0+h_1x+\ldots+h_wx^w \in K_1[x]$. Then $\sig h$ is that polynomial
achieved by applying $\sig$ to all of the coefficients. That is,
\[ (\sig h)(x) = \sig(h_0)+\sig(h_1)x+\ldots+ (\sig h_w)x^w \]
We note $(\sig h)(x)\in K_2[x]$.
\end{define}
\begin{theorem}\label{aa1} If $c(x)=a(x)b(x)$ in $K_1[x]$ then $(\sig c)(x)=(\sig a)(x)(\sig b)(x)$ in
$K_2[x]$
\end{theorem}
{\tt Proof:} Immediate.
\begin{theorem}\label{aa2} $p(x)\in K_1[x]$ is irreducible in $K_1$ if and only if
$(\sig p(x))\in K_2[x]$ is irreducible in $K_2$.
\end{theorem}
{\tt Proof:} If $p(x)=a(x)b(x)$ in $K_1[x]$ then $(\sig p)=(\sig a)(\sig b)$ in $K_2[x]$. Conversely we may apply
the isomorphism $\sig^{-1}$ so if $(\sig p)(x)=a(x)b(x)$ in $K_2[x]$, $p(x)=(\sig^{-1}a)(x)(\sig^{-1}b(x))$ in
$K_1[x]$.
\begin{theorem}\label{aa3} Let $f(x)\in F[x]$ be irreducible over $F$.
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let
$f(x)=p_1(x)\cdots p_l(x)$
be the factorization of $f(x)$ into irreducible factors in $K_1[x]$.
Then
$f(x)=(\sig p_1)(x)\cdots (\sig p_l)(x)$
is the factorization of $f(x)$ into irreducible factors in $K_2[x]$.
\end{theorem}
{\tt Proof:} As $f(x)\in F[x]$, $(\sig f)(x)$ is $f(x)$. From Theorem \ref{aa1},
$f(x)=(\sig p_1)(x)\cdots (\sig p_l)(x)$ is a factorization and from Theorem
\ref{aa2} the factors are irreducible in $K_2[x]$.
Now we shall {\em extend} an isomorphism to a larger structure.
\begin{theorem}\label{aa4}{\tt Isomorphism Extension Theorem.}
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let $p(x)\in K_1[x]$ be irreducible and let $\ah$ be a root of $p$.
Let $\beta$ be a root of $(\sig p)(x)$. Then we may extend $\sig$
to an isomorphism $\sig^+ K_1(\ah)\ra K_2(\beta)$ by setting
$\sig^+(\ah)=\beta$.
\end{theorem}
{\tt Proof:} Set $n$ equal the degree of $p(x)$ so we may write
\[ K_1(\ah) = \{ c_0+c_1\ah+\ldots+ c_{n-1}\ah^{n-1}: c_0,\ldots,c_{n-1}\in K_1\} \]
and
\[ K_2(\beta) = \{ d_0+d_1\beta+\ldots+ d_{n-1}\beta^{n-1}: d_0,\ldots,d_{n-1}\in K_2\} \]
Then $\sig^+$ is defined by
\[
\sig(c_0+c_1\ah+\ldots+ c_{n-1}\ah^{n-1})= d_0+d_1\beta+\ldots+ d_{n-1}\beta^{n-1} \]
where $d_j=\sig(c_j)$. Write
$p(x)=x^n-a_{n-1}x^{n-1}-\ldots-a_0$.
To check that $\sig^+$
preserves products we basically (formally, we'd need to write more here)
have to look at
$\ah^n= a_{n-1}\ah^{n-1}+\ldots+a_0$.
We have
$(\sig p)(x)=x^n-b_{n-1}x^{n-1}-\ldots-b_0$.
where $b_j=\sig(a_j)$. So
$\beta^n= b_{n-1}\beta^{n-1}+\ldots+b_0$ and indeed $\sig(\ah^n)=\beta^n$.
{\tt Example:} Let $F=Q$, $\theta=2^{1/3}$, $\omega=e^{2\pi i/3}$, $\eta=\theta\omega$. Define
$\sig: Q(\theta)\ra Q(\eta)$ by $\sig(\theta)=\eta$. Take $p(x)=x^2+x+1$ so that in this
case $\sig p$ is $p$. Take the root $\omega$ for both sides. Now we extend $\sig$
to $\sig^+:Q(\theta,\omega)\ra Q(\eta,\omega)$ by setting $\sig^+(\omega)=\omega$.
Applying Theorem \ref{aa4} repeatedly we get:
\begin{theorem}\label{aa4e}{\tt Full Isomorphism Extension Theorem.}
Let $\sig:K_1\ra K_2$ be an isomorphism over $F$.
Let $f(x)\in K_1[x]$ have complex roots $\ah_1,\ldots,\ah_s$.
Let $\beta_1,\ldots,\beta_s$ denote the complex roots of $(\sig f)(x)\in K_2[x]$.
Then $\sig$ may be extended to an isomorphism
$\sig^{++}:K_1(\ah_1,\ldots,\ah_s)\ra K_2(\beta_1,\ldots,\beta_s)$
\end{theorem}
Proof: Let $p_1(x)\in K_1[x]$ be the irreducible polynomial of $\ah_1$
over $K_1$. Let $\beta_1$ be a root of $(\sig p_1)(x)$. From
Theorem \ref{aa4} we extend $\sig$ to $\sig^+: K_1(\ah_1)\ra K_2(\beta_1)$.
Continue in this fashion extending each of the $a_i$. (When $a_i$ is
already in the field you don't do anything.) At the end you have
a $\sig^{++}$ with domain $K_1(\ah_1,\ldots,\ah_s)$ But the $\sig^{++}(\ah_i)$
must be the roots of $(\sig f)(x)$ so they must be the $\beta_1,\ldots,\beta_s$
giving the desired extension.
Now we prove Theorem \ref{bigsplit}.
Let $\ah_1,\ldots,\ah_r$ denote the complex roots of $f(x)$ and let
$\beta_1=\beta,\beta_2,\ldots,\beta_s$ denote the complex roots of $g(x)$. If the
theorem fails we can assume, without loss of generality, that $\beta_1\in K$ and
$\beta_2\not\in K$. Set $K_1=F(\beta_1)$, $K_2=F(\beta_2)$. As $\beta_1,\beta_2$
have the same minimal polynomial $g(x)$ over $F$ we find an isomorphism
$\sig:K_1\ra K_2$ which preserves $F$ and has $\sig(\beta_1)=\beta_2$.
Now from Theorem \ref{aa4e} we extend $\sig$ to the roots of $f(x)$. As
$f(x)\in F[x]$, $(\sig f)(x)=f(x)$ so both $f(x)$ and $(\sig f)(x)$ have
the roots $\ah_1,\ldots,\ah_r$. So the extended $\sig^{++}$ permutes
these roots. That is, $\sig^{++}$ is an isomorphism from
$K_1(\ah_1,\ldots,\ah_r)$ to $K_2(\ah_1,\ldots,\ah_r)$.
Whats wrong with this? Well, remember that $K_1=F(\beta_1)$ with $\beta_1\in F(\ah_1,\ldots,\ah_r)$
so that $K_1(\ah_1,\ldots,\ah_r)=F(\ah_1,\ldots,\ah_r)$.
But $K_2=F(\beta_2)$ with $\beta_2\not\in F(\ah_1,\ldots,\ah_r)$
and so $K_2(\ah_1,\ldots,\ah_r)= F(\ah_1,\ldots,\ah_r,\beta_2)$ is a nontrivial extension of
$K_1(\ah_1,\ldots,\ah_r)$. But isomorphisms over $F$ preserve dimension over $F$ (a nice exercise!)
and the Tower Theorem would give
$[F(\beta_2,\ah_1,\ldots,\ah_r):F]$ to be strictly bigger than
$[F(\beta_1,\ah_1,\ldots,\ah_r):F]$, a contradiction.
Now that we have proven Theorem \ref{bigsplit} we give an important definition that distinguishes
certain kind of field extensions.
\begin{definition}\label{normal} Suppose $F\subset K$ are subfields of $C$ with $K$ a finite
extension of $F$. We say that the extension
$K:F$ is {\em normal} if the following hold.
\ben
\item\label{y1} There is an $f(x)\in F[x]$ with $K$ the splitting field of $f(x)$ over $F$.
\item\label{y2} {\em Every} $g(x)\in F[x]$ which is irreducible (over $F$) and has a root in $K$
completely splits into linear factors in $K[x]$.
\een
When this occurs we often say that $K$ is a normal extension of $F$.
\end{definition}
\begin{theorem}\label{easy1}
The two conditions in Definition \ref{normal} are equuivalent. That is, either
one implies the other.
\end{theorem}
{\tt Proof:} We've already done the hard part. Theorem \ref{bigsplit} gives that
condition \ref{y1} implies condition \ref{y2}. Now assume condition \ref{y2}.
As $[K:F]$ is finite write $K=F(\ah_1,\ldots,\ah_s)$ for some finite number of
$\ah_1,\ldots,\ah_s$. For each $\ah_i$ let $p_i(x)\in F[x]$ be its irreducible
polynomial over $F$.
\par We claim $K$ is the splitting field of $f(x)$ where we
set $f(x)$ to be the product
$p_1(x)\cdots p_s(x)$. By Condition \ref{y2} {\em all} of the roots of each
$p_i(x)$ are in $F$ and so the extension of $F$ by all of the roots of $f(x)$
(that is, all of the roots of each $p_i(x)$) is still inside of $K$. But the
roots include $\ah_1,\ldots,\ah_s$ so the extension must include $F(\ah_1,\ldots,\ah_s)$
which is all of $K$. That is, the extension of $F$ by all of the roots of $f(x)$
is precisely $K$, giving the claim.
{\tt Some Examples:}
\ben
\item $K=Q(2^{1/3})$ is {\em not} a normal extension of $Q$ as the polynomial $x^3-2\in Q[x]$
(irreducible by Eisenstein's criterion)
has one root in $K$ but its other roots are not in $K$.
\item $K=Q(\sqrt{2},\sqrt{3})$ is a normal extension of $Q$ as the polynomial $(x^2-2)(x^2-3)$
has roots $\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}$ and extending $Q$ by these four roots gives
precisely $K$.
\item $K=Q(2^{1/3},\omega)$ (with $\omega=e^{2\pi i/3}$) is a normal extension of $Q$ as the
polynomial $x^3-2\in Q[x]$ has roots $2^{1/3},2^{1/3}\omega,2^{1/3}\omega^2$ and extending
$Q$ by these three roots gives precisely $K$.
\item Let $\eps=e^{2\pi i/5}$. Then $K=Q(\eps)$ is a normal extension of $Q$ as the polynomial
$(x^5-1)/(x-1)=x^4+x^3+x^2+x+1$ has roots $\eps,\eps^2,\eps^3,\eps^4$ and extending $Q$ by
these four roots gives precisely $K$.
\item Let $K=Q(2^{1/4})$ and $F=Q(2^{1/2})$. Then $K$ is a normal extension of $F$ as the
polynomial $x^2-2^{1/2}\in F[x]$ has roots $2^{1/4},-2^{1/4}$ and extending $F$ by these
two roots gives precisely $K$.
\item Let $K=Q(2^{1/4})$. Then $K$ is {\em not} a normal extension of $Q$ as the polynomial
$x^4-2\in Q[x]$ (irreducible by Eisenstein's criterion) has two roots $2^{1/4},-2^{1/4}$ in
$K$ but the other two roots $2^{1/4}i, -2^{1/4}i$ are not in $K$. (One reason why $2^{1/4}i\not\in K$
is that all elements of $K$ are real.)
\een
{\tt A Cautionary Note:} The last two examples emphasize that when we talk about a normal extension
we are talking about {\em two} fields, that $K$ is normal over $F$. Further, consider the tower
$Q\subset F\subset K$, with $F=Q(2^{1/2})$ and $K=Q(2^{1/4})$. Then $F$ is a normal extension of $Q$
as it is an extension of $Q$ by the two roots of $x^2-2$. We've seen that $K$ is a normal extension
of $F$. But it is {\em not} true (as we just saw) that $K$ is a normal extension of $Q$. That is,
we do {\em not} have a transitive property for normality, just because blip is normal over blop
which is normal over blunk we {\em cannot} deduce that blip is normal over blunk.
While we have to be careful about towers of fields, the following is useful and easy.
\begin{theorem}\label{cc1} {\tt The Middle Normal Theorem}
Let $K\subset L \subset F$ be fields and {\em assume} $F:K$ is a normal field extension.
Then $F:L$ is a normal field extension.
\end{theorem}
{\tt Proof:} From Definition \ref{normal}, condition \ref{y1},
there is an $f(x)\in K[x]$ with $F$ the splitting field of $f(x)$ over $K$. That is, $f$ splits
entirely in $K[x]$ with roots $\ah_1,\ldots,\ah_r\in F$ and $F=K(\ah_1,\ldots,\ah_r)$. But now we can
simply consider $f(x)$ as a polynomial in $L[x]$. It still splits entirely in $K[x]$ with
roots $\ah_1,\ldots,\ah_r$. As $K\subset L$ we have
$K(\ah_1,\ldots,\ah_r)\subset L(\ah_1,\ldots,\ah_r)$ and since $\ah_1,\ldots,\ah_r\in F$ and $L\subset F$,
$L(\ah_1,\ldots,\ah_r)\subset K$ so that $L(\ah_1,\ldots,\ah_r)=F$ and so the $F$ is a normal extension
over $L$ by the same Definition \ref{normal}, condition \ref{y1} and the same $f(x)$.
{\tt Caution:} Under the assumptions of Theorem \ref{cc1} we do {\em not} necessarily have $L:K$ a
normal extension.
Here is a nice property of normal field extensions that say, somehow, that they are nailed down.
\begin{theorem}\label{fixF} Let $K:F$ be a normal field extension. Let $K'$ be a field and $\sig:K\ra K'$
an isomorphism over $F$. (Recall, this means $\sig(c)=c$ for all $c\in F$.) Then $K'=K$.
\end{theorem}
{\tt Proof:} We can write $K=F(\ah_1,\ldots,\ah_s)$. Then $K'=F(\sig(\ah_1),\ldots,\sig(\ah_s))$.
For each $i$, $\ah_i$ and $\sig(\ah_i)$ satisfy the same irreducible polynomial $p_i(x)\in F[x]$.
As $K:F$ is normal this means $\sig(\ah_i)\in K$. Thus $K'\subset K$. Similarly, going backward
with $\sig^{-1}$, $K\subset K'$ and so $K=K'$.
{\tt A Cautionary Note:} Theorem \ref{fixF} does not say that each element of $K$ is fixed by
$\sig$. Indeed, $\sig$ can move around the elements of $K$ but the set of elements remains the
same.
\begin{theorem}\label{biggernomal} Let $K:F$ be a finite field extension. Then there is an
extension $K\subset K^+$ so that $K^+$ is a normal field extension of $F$.
\end{theorem}
{\tt Proof:} As $[K:F]$ is finite we can write $K=F(\ah_1,\ldots,\ah_r)$ for some finite
number of $\ah$'s. Let $p_i(x)$ be the minimal polynomial for $\ah_i$ in $F[x]$. Set
$K^+$ to be the splitting field for the product $f(x)=p_1(x)\cdots p_r(x)$. As a splitting
field it is a normal extension of $F$ and it contains $\ah_1,\ldots,\ah_r$ and therefore $K$.
\end{document}