08-401/The Fundamental Theorem

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In Preparation

The information below is preliminary and cannot be trusted! (v)

The statement appearing here, which is a weak version of the full 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.

Statement

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 $E/F$:

$\{K:E/K/F\}\quad \leftrightarrow \quad \{H:H<\operatorname {Gal} (E/F)\}$.

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 $K$,

$\Phi :\quad K\mapsto \operatorname {Gal} (E/K):=\{g:E\to E:g|_{K}=I\}$,

and reversely, by mapping every subgroup $H$ of $\operatorname {Gal} (E/F)$ to its fixed field $E_{H}$:

$\Psi :\quad H\mapsto E_{H}:=\{x\in E:\forall h\in H,\ hx=x\}$.

This correspondence has the following further properties:

1. 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 $\operatorname {Gal} (E/K_{1})>\operatorname {Gal} (E/K_{1})$.
2. It is degree/index respecting: $[E:K]=|\operatorname {Gal} (E/K)|$ and $[K:F]=[\operatorname {Gal} (E/F):\operatorname {Gal} (E/K)]$.
3. 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 $\operatorname {Gal} (K/F)\cong \operatorname {Gal} (E/F)/\operatorname {Gal} (E/K)$.

Lemmas

The four lemmas below belong to earlier chapters but we skipped them in class (the last one was also skipped by Gallian).

Zeros of Irreducible Polynomials

Lemma 1. An irreducible polynomial over a field of characteristic 0 has no multiple roots.

Proof. See the proof of Theorem 20.6 on page 362 of Gallian's book. $\Box$

Uniqueness of Splitting Fields

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 a unique isomorphism ${\bar {\phi }}:E_{1}\to E_{2}$ that extends $\phi$.

Proof. See the proof of Theorem 20.4 on page 360 of Gallian's book. $\Box$

The Primitive Element Theorem

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

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).

Proof. See the proof of Theorem 21.6 on page 375 of Gallian's book. $\Box$

Splitting Fields are Good at Splitting

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 $E$.

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

$E=E(v)/F(v)$ 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 $E/F$ is a splitting extension for $f$ and we can use the sub-lemma below. 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 and from $[E(w):F]=[E(w):E][E:F]$ that $[E(w):E]=1$ and therefore $E(w)=E$. Therefore $w\in E$. $\Box$

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 $f$.

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 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(z)} . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box}

Proof of The Fundamental Theorem

The Bijection

Proof of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi\circ\Phi=I} . More precisely, we need to show that if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} is an intermediate field between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} , then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\operatorname{Gal}(E/K)}=K} . The inclusion Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\operatorname{Gal}(E/K)}\supset K} is easy, so we turn to prove the other inclusion. Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v\in E-K} be an element of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} which is not in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} . We need to show that there is some automorphism Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi\in\operatorname{Gal}(E/K)} for which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi(v)\neq v} ; if such a Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} exists it follows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v\not\in E_{\operatorname{Gal}(E/K)}} and this implies the other inclusion. So let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} be the minimal polynomial of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} . It is not of degree 1; if it was, we'd have that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v\in K} contradicting the choice of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} . By lemma 4 and using the fact that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} is a splitting extension, we know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} splits in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} , so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} contains all the roots of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} . Over a field of characteristic 0 irreducible polynomials cannot have multiple roots and hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} must have at least one other root; call it Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w} . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w} have the same minimal polynomial over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} , we know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K(v)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K(w)} are isomorphic; furthermore, there is an isomorphism Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_0:K(v)\to K(w)} so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_0|_K=I} yet Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 $E$. but then $\phi |_{K}=\phi _{0}|_{K}=I$ so $\phi \in \operatorname {Gal} (E/K)$, yet $\phi (v)=w\neq v$, as required. $\Box$

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:

$|H|\leq |\operatorname {Gal} (E/E_{H})|\leq [E:E_{H}]\leq |H|$

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 $u_{i}:=\sigma _{i}u$ (with $u$ as above), and let $f$ be the polynomial

$f=\prod _{i=1}^{n}(x-u_{i})$.

Clearly, $f\in E[x]$. Furthermore, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau\in H} , then left multiplication by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau} permutes the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_i} 's (this is always true in groups), and hence the sequence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\tau u_i=\tau\sigma u_i)_{i=1}^n} is a permutation of the sequence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (u_i)_{i=1}^n} , hence

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau f=\prod_{i=1}^n(x-\tau u_i)=\prod_{i=1}^n(x-u_i)=f} ,

and hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\in E_H[x]} . Clearly Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(u)=0} , so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p|f} , so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [E:E_H]=\deg p\leq \deg f=n=|H|} , proving the third inequality above. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box}

The Properties

Property 1. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_1\subset H_2} then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{H_1}\supset E_{H_2}} and if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_1\subset K_2} then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_1)} .

Proof of Property 1. Easy. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box}

Property 2. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [E:K]=|\operatorname{Gal}(E/K)|} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]} .

Proof of Property 2. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K=E_H} , then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\operatorname{Gal}(E/K)|=|\operatorname{Gal}(E/E_H)|=[E:E_H]=[E:K]} as was shown within the proof of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi\circ\Psi=I} . But every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_H} for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H} , so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\operatorname{Gal}(E/K)|=[E:K]} for every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [K:F] = \frac{[E:F]}{[E:K]} = \frac{|\operatorname{Gal}(E/F)|}{|\operatorname{Gal}(E/K)|} = [\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)].\quad\Box}

Property 3. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E/K/F} is the splitting field of a polynomial in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F[x]} then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/K)} is normal in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/F)} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)} .

Proof of Property 3. We will define a surjective (onto) group homomorphism Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho:\operatorname{Gal}(E/F)\to\operatorname{Gal}(K/F)} whose kernel is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/K)} . This shows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/K)} is normal in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/F)} (kernels of homomorphisms are always normal) and then by the first isomorphism theorem for groups, we'll have that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)} .

Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma} be in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \operatorname{Gal}(E/F)} and let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} be an element of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} . Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} be the minimal polynomial of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F[x]} . Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} is a splitting field, lemma 4 implies that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} splits in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K[x]} , and hence all the other roots of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} are also in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} . As Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma(u)} is a root of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p} , it follows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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. But this means that $\rho$ is onto. $\Box$