Exercise 10.1.7

Proof. As $[F:\mathbb{Q}]$ is a finite extension, and as $\mathbb{Q}$ has characteristic 0, by the Theorem of the Primitive Element (Corollary 5.4.2 (b)), $F=\mathbb{Q}(\alpha )$ for some $\alpha \in F$. Let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

As $ℂ$ is an algebraically closed field, $f$ splits completely over $ℂ$. Let ${\alpha }_1=\alpha ,{\alpha }_2,\dots ,{\alpha }_n$ the roots of $f$ in $ℂ$, and $M=\mathbb{Q}({\alpha }_1,\dots ,{\alpha }_n)$ the splitting field of $f$ in $ℂ$. Then

(a)

$\mathbb{Q}\subset M$ is a Galois extension, since the splitting field of $f$ over $\mathbb{Q}$ is a normal extension of $\mathbb{Q}$, and separable since the characteristic of $\mathbb{Q}$ is 0.

(b)

Let $M\subset M^{\prime }$ by any other extension such that $M^{\prime }$ is a Galois extension over $\mathbb{Q}$. As $\alpha \in M\subset M^{\prime }$ is a root of $f$ and as $M^{\prime }$ is normal, $f$ splits completely over $M^{\prime }$, with roots ${\beta }_1={\alpha }_1,{\beta }_2,\dots ,{\beta }_n,\in M^{\prime }$. Let $M^{″}=\mathbb{Q}({\beta }_1,\dots ,{\beta }_n)$, so $M^{″}$ is a splitting field of $f$ over $\mathbb{Q}$. By the uniqueness of splitting fields (Corollary 5.1.7), there is an isomorphism $\phi :M\to M^{″}$ that is the identity on $L$. Since $M^{″}\subset M^{\prime }$, $\phi$ defines a field homomorphism $\phi :M\to M^{\prime }$.

So the parts (a),(b) of the definition of an Galois closure are satisfied.

Conclusion: there is a field $M$ such that $F\subset M\subset ℂ$ and $M$ is a Galois closure of $F$ over $\mathbb{Q}$. □