Exercise 7.1.6

Proof. $F\subset L$ is a finite extension, so $L=F({\alpha }_1,\dots ,{\alpha }_n)$, where ${\alpha }_i\in L$ is algebraic over $F$, with minimal polynomial $p_i\in F[x]$.

Let $f=p_1\cdots p_n$, and $M=L({\beta }_1,\dots ,{\beta }_m)$ the splitting field of $f$ over $L$, where ${\beta }_1,\dots ,{\beta }_m$ are the roots of $f$ in $M$. As the ${\alpha }_i$ are roots of $p_i$, they are roots of $f$, so $\{{\alpha }_1,\dots ,{\alpha }_n\}\subset \{{\beta }_1,\dots ,{\beta }_m\}$. Therefore

Thus $F({\beta }_1,\dots ,{\beta }_m)$ contains $L$ and ${\beta }_1,\dots ,{\beta }_m$, therefore $M=L({\beta }_1,\dots ,{\beta }_m)\subset F({\beta }_1,\dots ,{\beta }_m)$.

Therefore $M=F({\beta }_1,\dots ,{\beta }_m)$ is the splitting field of $f$ over $F$. Then , by Theorem 5.2.4, the extension $F\subset M$ is normal (and finite), so $M$ satisfies (a).

Let $M^{\prime }\supset L$ any normal extension of $F$. As $F\subset M^{\prime }$ is normal, the $p_i$ splits completely over $M^{\prime }$, thus also $f$. Let ${\gamma }_1,\dots ,{\gamma }_m\in M^{\prime }$ the roots of $f$ in $M^{\prime }$, and $M^{″}=F({\gamma }_1,\dots ,{\gamma }_m)\subset M^{\prime }$. As ${\alpha }_i\in L\subset M^{\prime }$, the ${\alpha }_i$ are roots of $f$ in $M^{\prime }$ : $\{{\alpha }_1,\dots ,{\alpha }_n\}\subset \{{\gamma }_1,\dots ,{\gamma }_m\}$, thus $L=F({\alpha }_1,\dots ,{\alpha }_n)\subset F({\gamma }_1,\dots ,{\gamma }_m)=M^{″}$.

$M^{″}$ and $M$ are so two splitting fields of $f$ over $L$. By the unicity of the splitting field (Corollary 5.1.7), there exist a field isomorphism of $M$ in $M^{″}$ that is identity on $L$. Since $M^{″}\subset M^{\prime }$, we can regard this isomorphism as an injective field homomorphism $\phi :M\to M^{\prime }$. □