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\begin{document}

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\parbox[b]{2in}{\tiny
  \href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}: \href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
  \newline \href{http://www.math.toronto.edu/~drorbn/classes/\#2526}{2025-26}: \href{http://drorbn.net/25-347}{MAT 347 Groups, Rings, Fields}:
}~{\Large\bf Galois Theory: The Fundamental Theorem}
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  \qrcode[height=1.2em,level=L,nolink]{drorbn.net/25-347}
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\begin{multicols*}{2}

The statement appearing here, which is a weak version of the full \textbf{fundamental theorem of Galois theory},
is taken from Gallian's book and is meant to match our discussion in class.
The proof is taken from Hungerford's book, except modified to fit our notations and conventions
and simplified as per our weakened requirements.

Here and everywhere below our base field $F$ will be a field of characteristic $0$.

\WSection{Statement}

\textbf{Theorem.}
Let $E$ be a splitting field over $F$.
Then there is a bijective correspondence between the set $\{K:E/K/F\}$
of intermediate field extensions $K$ lying between $F$ and $E$ and the set
$\{H:H<\operatorname{Gal}(E/F)\}$ of subgroups $H$ of the Galois group
$\operatorname{Gal}(E/F)$ of the original extension $E/F$:
\EquationStar{\{K:E/K/F\}\quad\leftrightarrow\quad\{H:H<\operatorname{Gal}(E/F)\}.}

The bijection is given by mapping every intermediate extension $K$
to the subgroup $\operatorname{Gal}(E/K)$ of elements in $\operatorname{Gal}(E/F)$
that preserve $K$,
\EquationStar{\Phi:\quad K\mapsto\operatorname{Gal}(E/K):=\{\phi:E\to E:\phi|_K=I\},}
and reversely, by mapping every subgroup $H$ of $\operatorname{Gal}(E/F)$
to its fixed field $E_H$:
\EquationStar{\Psi:\quad H\mapsto E_H:=\{x\in E:\forall h\in H,\ hx=x\}.}

This correspondence has the following further properties:
\begin{enumerate}
\item It is inclusion-reversing: if $H_1\subset H_2$ then $E_{H_1}\supset E_{H_2}$
and if $K_1\subset K_2$ then $\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_2)$.
\item It is degree/index respecting:
$[E:K]=|\operatorname{Gal}(E/K)|$ and
$[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]$.
\item Splitting fields correspond to normal subgroups:
If $K$ in $E/K/F$ is the splitting field of a polynomial in $F[x]$ then
$\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ and
$\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.
\end{enumerate}

{
  \centering
  \includegraphics[width=\linewidth]{08-401-AllInOne.png}
  %\newline \sl The Fundamental Theorem of Galois Theory, all in one.
}

\WSection{Lemmas}

\WSubsection{Zeros of Irreducible Polynomials}

\textbf{Lemma 1.}
An irreducible polynomial $f$ over a field of characteristic $0$ has no multiple roots.

\textbf{Proof.}
$f$ is irreducible and so its gcd with a lower degree polynomial, $f'$, must be 1. \qed

\WSubsection{Uniqueness of Splitting Fields}

\textbf{Lemma 2.}
Let $\phi:F_1\to F_2$ be an isomorphism of fields,
let $f_1\in F_1[x]$ be a polynomial and let $f_2=\phi(f_1)$,
and let $E_1$ and $E_2$ be splitting fields for $f_1$ and $f_2$
over $F_1$ and $F_2$, respectively.
Then there is an isomorphism $\bar\phi:E_1\to E_2$ (generally not unique)
that extends $\phi$.

\textbf{Proof.}
See the proof of Theorem 20.4 on page 360 of Gallian's book. \qed

\WSubsection{The Primitive Element Theorem}

The celebrated ``Primitive Element Theorem'' is just a lemma for us:

\textbf{Lemma 3.}
Let $a$ and $b$ be algebraic elements of some extension $E$ of $F$.
Then there exists a single element $c$ of $E$ so that $F(a,b)=F(c)$.
(And so by induction, every finite extension of $E$ is ``simple'',
meaning, is generated by a single element, called ``a primitive element'' for that extension).

\textbf{Proof.}
See the proof of Theorem 21.6 on page 375 of Gallian's book. \qed

\WSubsection{Splitting Fields are Absolute}

\textbf{Lemma 4} (compare with Hungerford's Theorem 10.15 on page 355).
If $E$ is a splitting field of some polynomial $f$ over $F$ and some irreducible polynomial
$p\in F[x]$ has a root $v$ in $E$, then $p$ splits in $E$.

\parpic[r]{$\xymatrix@R=2mm@C=4mm{
  & L \ar@{-}[ld] \ar@{-}[dd] \ar@{-}[rd] & \\
  E(v) \ar@{-}[dd] \ar@{=}[rd] \ar@{.>}[rr]^(0.33)\simeq & & E(w) \ar@{-}[dd] \ar@{-}[ld] \\
  & E \ar@{-}[dd] & \\
  F(v) \ar@{-}[rd] \ar@{.>}[rr]^(0.33)\simeq & & F[w] \ar@{-}[ld] \\
  & F & \\
}$}
\textbf{Proof.}
Let $L$ be a splitting field of $p$ over $E$.
We need to show that if $w$ is a root of $p$ in $L$, then $w\in E$
(so all the roots of $p$ are in $E$ and hence $p$ splits in $E$).
Consider the two extensions
\EquationStar{E=E(v)/F(v)\ \text{ and }\ E(w)/F(w).}
The ``smaller fields'' $F(v)$ and $F(w)$ in these two extensions are isomorphic as they both arise
by adding a root of the same irreducible polynomial $p$ to the base field $F$.
The ``larger fields'' $E=E(v)$ and $E(w)$ in these two extensions are both the splitting fields of the same polynomial
$f$ over the respective ``small fields'', as $E$ is obtained from $F$ by adding all the roots of $f$, and so $E(v)$ and $E(w)$ or obtained from $F(v)$ and $F(w)$ by adding those same roots.
Thus by the uniqueness of splitting extensions, the isomorphism between $F(v)$ and $F(w)$ extends to an isomorphism
between $E=E(v)$ and $E(w)$, and in particular these two fields are isomorphic and so
$[E:F]=[E(v):F]=[E(w):F]$.
Since all the degrees involved are finite it follows from the last equality that $E(w)=E$.
Therefore $w\in E$. \qed

%\medskip
%\textbf{Sub-lemma.}
%If $E/F$ is a splitting extension of some polynomial $f\in F[x]$ and $z$ is an element of some larger extension $L$ of $E$,
%then $E(z)/F(z)$ is also a splitting extension of $f$.
%
%\textbf{Proof.}
%Let $u_1,\ldots,u_n$ be all the roots of $f$ in $E$.
%Then they remain roots of $f$ in $E(z)$, and since $f$ completely splits already in $E$,
%these are \emph{all} the roots of $f$ in $E(z)$. So
%$E(z)=F(u_1,\ldots,u_n)(z)=F(z)(u_1,\ldots,u_n)$,
%and $E(z)$ is obtained by adding all the roots of $f$ to $F(z)$. \qed

\textbf{Theorem} A finite extension $E/F$ is a splitting extension iff whenever an irreducible $p\in F[x]$ has a root in $E$, it fully splits in $E$.

\textbf{Proof.} The $\Rightarrow$ side is the previous lemma. For $\Leftarrow$, if $E=F(\{\alpha_i: 1\leq i \leq n\})$, then $f=\prod_i\operatorname{minpoly}_F(\alpha_i)$ fully splits in $E$ and all the $\alpha_i$'s are roots of $f$. So $E$ is a splitting field of $f$. \qed

\WSection{Proof of The Fundamental Theorem}

\WSubsection{The Bijection}

\textbf{Proof of $\Psi\circ\Phi=I$.}
More precisely, we need to show that if $K$ is an intermediate field between $E$ and $F$,
then $E_{\operatorname{Gal}(E/K)}=K$.
The inclusion $E_{\operatorname{Gal}(E/K)}\supset K$ is easy, so we turn to prove the other inclusion.
Let $v\in E-K$ be an element of $E$ which is not in $K$.
We need to show that there is some automorphism $\phi\in\operatorname{Gal}(E/K)$ for which $\phi(v)\neq v$;
if such a $\phi$ exists it follows that $v\not\in E_{\operatorname{Gal}(E/K)}$ and this implies the other inclusion.
So let $p$ be the minimal polynomial of $v$ over $K$.
It is not of degree $1$; if it was, we'd have that $v\in K$ contradicting the choice of $v$.
By lemma 4 and using the fact that $E$ is a splitting extension, we know that $p$ splits in $E$,
so $E$ contains all the roots of $p$.
Over a field of characteristic $0$ irreducible polynomials cannot have multiple roots (lemma 1) and hence $p$ must have at least
one other root; call it $w$.
Since $v$ and $w$ have the same minimal polynomial over $K$, we know that $K(v)$ and $K(w)$ are isomorphic; furthermore,
there is an isomorphism $\phi_0:K(v)\to K(w)$ so that $\phi_0|_K=I$ yet $\phi_0(v)=w$.
But $E$ is a splitting field of some polynomial $f$ over $F$ and hence also over $K(v)$ and over $K(w)$.
By the uniqueness of splitting fields (lemma 2), the isomorphism $\phi_0$ can be extended to an isomorphism $\phi:E\to E$;
i.e., to an automorphism of $E$. but then $\phi|_K=\phi_0|_K=I$ so $\phi\in\operatorname{Gal}(E/K)$,
yet $\phi(v)=w\neq v$, as required. \qed

\textbf{Proof of $\Phi\circ\Psi=I$.}
More precisely we need to show that if $H<\operatorname{Gal}(E/F)$ is a subgroup of the Galois group of $E$ over $F$,
then $H=\operatorname{Gal}(E/E_H)$.
The inclusion $H<\operatorname{Gal}(E/E_H)$ is easy.
Note that $H$ is finite since we've proven previously that Galois groups of finite extensions are finite and hence
$\operatorname{Gal}(E/F)$ is finite.
We will prove the following sequence of inequalities:
\EquationStar{|H|\leq|\operatorname{Gal}(E/E_H)|\leq [E:E_H]\leq |H|}
This sequence and the finiteness of $|H|$ imply that these quantities are all equal and since
$H<\operatorname{Gal}(E/E_H)$ it follows that $H=\operatorname{Gal}(E/E_H)$ as required.

The first inequality above follows immediately from the inclusion $H<\operatorname{Gal}(E/E_H)$.

By the Primitive Element Theorem (Lemma 3) we know that there is some element $u\in E$ so that $E=E_H(u)$.
Let $p$ be the minimal polynomial of $u$ over $E_H$.
Distinct elements of $\operatorname{Gal}(E/E_H)$ map $u$ to distinct roots of $p$, but $p$ has exactly $\deg p$ roots.
Hence $|\operatorname{Gal}(E/E_H)|\leq\deg p=[E:E_H]$, proving the second inequality above.

Let $\sigma_1,\ldots,\sigma_n$ be an enumeration of all the elements of $H$,
let $u_i:=\sigma_iu$ (with $u$ as above), and let $f$ be the polynomial
\EquationStar{f=\prod_{i=1}^n(x-u_i).}
Clearly, $f\in E[x]$.
Furthermore, if $\tau\in H$, then left multiplication by $\tau$ permutes the $\sigma_i$'s (this is always true in groups),
and hence the sequence $(\tau u_i=\tau\sigma u_i)_{i=1}^n$ is a permutation of the sequence $(u_i)_{i=1}^n$, hence
\EquationStar{\tau f=\prod_{i=1}^n(x-\tau u_i)=\prod_{i=1}^n(x-u_i)=f,}
and hence $f\in E_H[x]$.
Clearly $f(u)=0$, so $p|f$, so $[E:E_H]=\deg p\leq \deg f=n=|H|$, proving the third inequality above. \qed

\WSubsection{The Properties}

\textbf{Property 1.}
If $H_1\subset H_2$ then $E_{H_1}\supset E_{H_2}$ and if $K_1\subset K_2$ then
$\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_2)$.
% Likely intended: ... > Gal(E/K_2)

\textbf{Proof of Property 1.}
Easy. \qed

\textbf{Property 2.}
$[E:K]=|\operatorname{Gal}(E/K)|$ and $[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]$.

\textbf{Proof of Property 2.}
If $K=E_H$, then $|\operatorname{Gal}(E/K)|=|\operatorname{Gal}(E/E_H)|=[E:E_H]=[E:K]$
as was shown within the proof of $\Phi\circ\Psi=I$.
But every $K$ is $E_H$ for some $H$, so $|\operatorname{Gal}(E/K)|=[E:K]$ for every $K$ between $E$ and $F$.
The second equality follows from the first and from the multiplicativity of the degree/order/index in towers of extensions
and in towers of groups:
\EquationStar{[K:F] = \frac{[E:F]}{[E:K]} = \frac{|\operatorname{Gal}(E/F)|}{|\operatorname{Gal}(E/K)|}
= [\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)].}
\qed

\textbf{Property 3.}
If $K$ in $E/K/F$ is the splitting field of a polynomial in $F[x]$ then $\operatorname{Gal}(E/K)$ is normal in
$\operatorname{Gal}(E/F)$ and
$\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.

\textbf{Proof of Property 3.}
We will define a surjective (onto) group homomorphism
$\rho:\operatorname{Gal}(E/F)\to\operatorname{Gal}(K/F)$
whose kernel is $\operatorname{Gal}(E/K)$.
This shows that $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ (kernels of homomorphisms are always normal)
and then by the first isomorphism theorem for groups, we'll have that
$\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.

Let $\sigma$ be in $\operatorname{Gal}(E/F)$ and let $u$ be an element of $K$.
Let $p$ be the minimal polynomial of $u$ in $F[x]$.
Since $K$ is a splitting field, lemma 4 implies that $p$ splits in $K[x]$, and hence all the other roots of $p$ are also in $K$.
As $\sigma(u)$ is a root of $p$, it follows that $\sigma(u)\in K$ and hence $\sigma(K)\subset K$.
But since $\sigma$ is an isomorphism, $[\sigma(K):F]=[K:F]$ and hence $\sigma(K)=K$.
Hence the restriction $\sigma|_K$ of $\sigma$ to $K$ is an automorphism of $K$, so we can define
$\rho(\sigma)=\sigma|_K$.

Clearly, $\rho$ is a group homomorphism.
The kernel of $\rho$ is those automorphisms of $E$ whose restriction to $K$ is the identity.
That is, it is $\operatorname{Gal}(E/K)$.
Finally, as $E/F$ is a splitting extension, so is $E/K$.
So every automorphism of $K$ extends to an automorphism of $E$ by the uniqueness statement for splitting extensions (lemma 2).
But this means that $\rho$ is onto. \qed

\end{multicols*}
\end{document} 