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\begin{document}

\parindent 0in
\parbox[b]{1.7in}{\tiny
  \href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}: \href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
  \href{http://www.math.toronto.edu/~drorbn/classes/\#2526}{2025-26}:
  \newline \href{http://drorbn.net/25-347}{MAT 347 Groups, Rings, Fields}:
}~{\Large\bf Galois Theory: The Fundamental Theorem}
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  \null\hfill\sheeturl
  \newline\null\hfill 
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  \qrcode[height=1.2em,level=L,nolink]{drorbn.net/25-347}
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\vspace{-4mm}

\begin{multicols}{2}

In this handout all fields are of characteristic $0$.

\textbf{Theorem} (the fundamental theorem of Galois theory in characteristic $0$).
Let $E$ be a splitting field over a field $F$, where $\operatorname{char}F=0$.
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 $E/F$:
\[ \{K:E/K/F\}\quad\longleftrightarrow\quad\{H:H<\operatorname{Gal}(E/F)\}. \]

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 $K$,
\[ K\mapsto\operatorname{Gal}(E/K):=\{\phi:E\to E:\phi|_K=I\}, \]
and reversely, by mapping every subgroup $H$ of $\operatorname{Gal}(E/F)$
to its fixed field $E_H$:
\[ H\mapsto E_H:=\{x\in E:\forall h\in H,\ hx=x\}. \]

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:
$K$ in $E/K/F$ is the splitting field of a polynomial in $F[x]$ if and only if
$\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ and then
$\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.
\end{enumerate}
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%  %\newline \sl The Fundamental Theorem of Galois Theory, all in one.
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\[ \xymatrix{
  E \ar@{<->}[r] \ar@{<-}[d]^{[E:K]}
    & \{e\}=\Gal(E/E) \ar@{->}[d]_{|H|} \\
  K \ar@{<->}[r] \ar@{<-}[d]^{[K:F]}
    & H=\Gal(E/K) \ar@{->}[d]_{[G:H]} \\
  F \ar@{<->}[r]
    & G=\Gal(E/F)
  }
  \quad\parbox[t]{1.38in}{\vspace{11mm}
  And $K$ is splitting iff $H$ is normal and then $\Gal(K/F)=G/H=\Gal(E/F)/\Gal(E/K)$.
  }
\]

\end{multicols}

\noindent\rule{\linewidth}{1pt}

\begin{multicols}{2}

\textbf{Proof of $E_{\operatorname{Gal}(E/K)}=K$.} Let $K$ be an intermediate field between $E$ and $F$.
The inclusion $E_{\operatorname{Gal}(E/K)}\supset K$ is easy, so we turn to prove the other inclusion.
Let $v\in E\setminus K$ 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 $\phi(v)\neq v$;
this impies that $v\not\in E_{\operatorname{Gal}(E/K)}$ and this implies the other inclusion.
Let $p$ be the minimal polynomial of $v$ over $K$.
It is not of degree $1$ as $v\not\in K$.
$E$ is a splitting extension so we know that $E$ contains all the roots of $p$.
Over a field of characteristic $0$ irreducible polynomials cannot have multiple roots (as the $\gcd$ of an irreducible $p$ with the lower-degree yet non-zero $p'$ must be $1$) 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 via 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, the isomorphism $\phi_0$ can be extended to an isomorphism $\phi:E\to E$;
i.e., to an automorphism of $E$ with $\phi|_K=\phi_0|_K=I$ so $\phi\in\operatorname{Gal}(E/K)$.
Yet $\phi(v)=w\neq v$, as required. \qed

\textbf{Proof of $H=\operatorname{Gal}(E/E_H)$.}
Let $H<\operatorname{Gal}(E/F)$ be a subgroup of the Galois group of $E$ over $F$.
The inclusion $H<\Gal(E/E_H)$ is easy, so it is enough to show that $\Gal(E/E_H)$ is finite and that $|\Gal(E/E_H)|\leq |H|$.

By the Primitive Element Theorem 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 $E_H$. Let $n\coloneqq\deg(p)$. To conclude the proof, we will show that
\[ |\Gal(E/E_H)|\leq n \leq |H|. \]

Distinct elements of $\operatorname{Gal}(E/E_H)$ map $u$ to distinct roots of $p$, but $p$ splits in $E$ and so it has exactly $n$ roots. This proves the first inequality.

For the second inequality, let $f$ be the polynomial
\[ f=\prod_{\sigma\in H}(x-\sigma(u)).\]
Clearly, $f\in E[x]$.
Furthermore, if $\tau\in H$, then the action of $\tau$ permutes the $\sigma(u)$'s, hence $\tau(f)=f$ and hence $f\in E_H[x]$.
Clearly $f(u)=0$, so $p|f$, so $n \leq |H|$.

Seeing that $E=E_H(u)$, we have also proven here that $[E:E_H]=\deg(p)=n=|H|$. \qed

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

\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 shown above.
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:
\begin{multline*}
  [K:F] = \frac{[E:F]}{[E:K]} = \frac{|\operatorname{Gal}(E/F)|}{|\operatorname{Gal}(E/K)|} \\
  = [\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]. \qquad\qed
\end{multline*}

\textbf{Proof of Property 3, $\Rightarrow$.}
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 $p$ 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 $\sigma(K)\subset K$. Similarly $\sigma^{-1}(K)\subset K$ which means $K\subset \sigma(K)$ and hence $\sigma(K)=K$.
Hence the restriction $\sigma|_K$ of $\sigma$ to $K$ is an automorphism of $K$ and 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.
But this means that $\rho$ is onto. \qed

\textbf{Proof of Property 3, $\Leftarrow$.} Suppose $H$ is normal in $G$, $p\in F[x]$ is irreducible, and $u\in E_H$ is a root of $p$. We need to show that $p$ splits in $E_H$. We know that $p$ splits in $E$, so we just need to show that if $v\in E$ is a root of $p$, then $v\in E_H$. Let $\sigma_0\colon F(u)\to F(v)$ be the isomorphism that fixes $F$ and maps $u\to v$. By the uniqueness of splitting fields, it can be extended to an automorphism $\sigma\colon E\to E$. Now for any $h\in H$, $h'\coloneqq\sigma^{-1}h\sigma\in H$ as $H$ is normal, hence $h'u=u$ as $u\in E_H$, and hence $hv = h\sigma u = \sigma\sigma^{-1}h\sigma u = \sigma h'u = \sigma u = v$. So $v\in E_H$, as required. \qed

\end{multicols}

\textbf{The subfields of $\bbQ(x^4-2)$ and the subgroups of $D_8$} (adopted from Gallian's book, page 548).

\[ 
  \def\a#1{\ar@{<-}[#1]|-2}
  \qquad\xymatrix@R=13mm@C=15mm{
    & & \bbQ(x^4-2)=\bbQ(2^{1/4},i) \a{dll} \a{dl} \a{d} \a{dr} \a{drr} & & \\
    \bbQ(2^{1/4}) \a{rd} & \bbQ(2^{1/4}i) \a{d} & \bbQ(2^{1/2},i) \a{ld} \a{d} \a{rd} & \bbQ(2^{1/4}(1-i)) \a{d} & \bbQ(2^{1/4}(1+i)) \a{ld} \\
    & \bbQ(2^{1/2}) \a{rd} & \bbQ(i) \a{d} & \bbQ(2^{1/2}i) \a{ld} & \\
    & & \bbQ & & \\
  }
\]

Below $\alpha:(i,2^{1/4})\mapsto(i,-2^{1/4}i)$ and $\beta:(i,2^{1/4})\mapsto(-i,2^{1/4})$:

\[ 
  \def\a#1{\ar@{->}[#1]|-2}
  \hspace{-4mm}\xymatrix@R=13mm@C=8mm{
    & & \{e\} \a{dll} \a{dl} \a{d} \a{dr} \a{drr} & & \\
    \{e,\beta\} \a{rd} & \{e,\alpha^2\beta\} \a{d} & \{e,\alpha^2\} \a{ld} \a{d} \a{rd} & \{e,\alpha\beta\} \a{d} & \{e,\alpha^3\beta\} \a{ld} \\
    & \{e,\alpha^2,\beta,\alpha^2\beta\} \a{rd} & \{e,\alpha,\alpha^2,\alpha^3\} \a{d} & \{e,\alpha^2,\alpha\beta,\alpha^3\beta\} \a{ld} & \\
    & & D_8=\{e,\alpha,\alpha^2,\alpha^3,\beta,\alpha\beta,\alpha^2\beta,\alpha^3\beta\} & & \\
  }
\]

Which of these subgroups are normal? Which is these extensions are splitting?

\end{document}