Exercise 8.5.2

Proof. Let $L$ the splitting field of $f$ over $F$, then

where ${\alpha }_1=\alpha ,{\alpha }_2,\cdots ,{\alpha }_n$ are the roots of $f$ in $L$.

$\text{∙}$ $L$ is a Galois extension of $F$, since $L$ is the splitting field of a separable polynomial $f\in F[x]$ (Theorem 7.1.1).

$\text{∙}$ Let $M$ be any extension of $F(\alpha )$ such that $M$ is Galois over $F$. As $\alpha$ is a root of the irreducible polynomial $f\in F[x]$, and $\alpha \in M$, where $M$ is a normal extension of $F$, the polynomial $f$ splits completely over $F$ and its distinct roots ${\beta }_1=\alpha ={\alpha }_1,{\beta }_1,\cdots ,{\beta }_n$ are in $M$. Let $L^{\prime }=F({\beta }_1,\cdots ,{\beta }_n)$. $L^{\prime }$ is a splitting field of $f$ over $F$, hence there exists an isomorphism $\phi :L\to L^{\prime }$ which is the identity on $F$, that is an embedding of $L$ in $M$ which is the identity on $F$.

So, by definition, $L$ is a Galois closure of $F\subset F(\alpha )$, unique up to isomorphism. □