Exercise 7.1.5

Proof. Let $M,M^{\prime }$ two Galois closures of the separable extension $F\subset L$. By Proposition 7.1.7, there exists a field homomorphism $\phi :M\to M^{\prime }$ that is identity on $L$.

As every field homomorphism, $\phi$ is injective, this is an embedding of $M$ in $M^{\prime }$. Moreover $\phi$ is the identity on $L$, so $\phi$ is a $L$-linear injective application between $M$ and $M^{\prime }$ as $L$-vector spaces, thus $[M:L]\le [M^{\prime }:L]$. Exchanging $M$ and $M^{\prime }$, we prove similarly that $[M^{\prime }:L]\le [M:L]$, thus $[M^{\prime }:L]=[M:L]$. An injective linear application between two same dimensional vector spaces is bijective, thus $\phi$ is bijective. Therefore $\phi$ is a field isomorphism that is the identity on $L$.

The Galois closure of a finite separable extension $F\subset L$ is unique up to an isomorphism that is the identity on $L$. □