Exercise 7.3.13

Proof. Let $F\subset L$ a Galois extension, $F\subset K\subset L$ an intermediate field, $G=\mathrm{Gal}(L∕F),H=\mathrm{Gal}(L∕K)$, $N={\mathrm{Core}}_G(H)$, and $M=L_N$ the fixed field of $N$. We show that $M=L_N$ is the Galois closure of $K$ over $F$.

Since $N\subset H,L_N\supset L_H=K$, so $K$ is a subfield of $L_N$.

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As $N$ is normal in $G$, $M=L_N$ is a Galois extension of $F$.

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Let $M^{\prime }$ an extension of $K$ such that $M^{\prime }$ is Galois over $F$, and suppose first that $M^{\prime }\subset L$. We call $S=\mathrm{Gal}(L∕M^{\prime })$.

As $F\subset M^{\prime }$ is a Galois extension, $S=\mathrm{Gal}(L∕M^{\prime })$ is normal in $G$, and since $K\subset M^{\prime }$, $H=\mathrm{Gal}(L∕K)\supset \mathrm{Gal}(L∕M^{\prime })=S$. So $S$ is a subgroup of $H$, and $S$ is normal in $G$. By exercise 12, $S\subset N={\mathrm{Core}}_G(H)$, thus $M=L_N\subset L_S=M^{\prime }$.

$M=L_N$ is so the smallest intermediate field of the extension $F\subset L$ which contains $K$ and is a Galois extension of $F$.

Let $M_0$ be any Galois closure of $K$ over $F$. As $F\subset M$ is a Galois extension, there exists by proposition 7.1.7 an embedding $\psi$ of $M_0$ in $M$ that is the identity on $K$. Then $K\subset \psi (M_0)\subset M\subset L$, and since $M_0\simeq \psi (M_0)$, $\psi (M_0)$ is a Galois extension of $F$. But $M$ is the smallest intermediate field of the extension $F\subset L$ which contains $K$ and is a Galois extension of $F$, therefore $\psi (M_0)=M$, so $\psi :M_0\to M$ is an isomorphism.

If $M^{″}$ is any extension of $K$ which is Galois over $F$, by the definition of a Galois closure, there exists an field homomorphism $\phi :M_0\to M^{″}$ that is the identity on $K$, so $\phi \circ {\psi }^{-1}$ is an embedding from $M$ to $M^{″}$ that is the identity on $L$, so $M=L_N$ is a Galois closure of $K$.

Note: this exercise shows that there exists always a Galois closure of an intermediate field $K$ of a Galois extension $F\subset L$ that is included in $L$. Moreover it is characterized by the fact that it is the smallest intermediate field of $F\subset L$ containing $K$ that is a Galois extension of $F$. Such a subfield of $L$ is unique (not only up to an isomorphism). □