Exercise 7.1.2

Proof. By hypothesis, the extension $F\subset L$ is finite, normal and separable. As $F\subset L$ is finite, $L=F({\alpha }_1,\cdots ,{\alpha }_n)$, where ${\alpha }_i\in L$ has $p_i$ as minimal polynomial over $F$. If $q_1,\dots ,q_r$ are the distinct elements in the set $\{p_1,\dots ,p_n\}$, then $f=q_1\cdots q_r$ is a product of monic irreducible distinct polynomials (thus $q_i$ is not associate to $q_j$ if $i\ne j$). As in the text, we know by Lemma 5.3.4 that $f$ is separable.

We show that $L$ is the splitting field of $f$ over $F$.

As $q_j=p_i$ for some $i,1\le i\le n$, is the minimal polynomial of ${\alpha }_i\in L$ over $F$, and as $F\subset L$ is normal, then all the roots of $p_i$ are in $L$, so $q_j$ splits completely over $L$, thus $f={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^r{q}_j$ splits completely over $L$. Write ${\beta }_1,\dots ,{\beta }_m\in L$ the roots of $f$, and $L^{\prime }=F({\beta }_1,\dots ,{\beta }_m)\subset L$ the splitting field of $f$ over $F$. As $F\subset L$, and ${\beta }_1,\dots ,{\beta }_m\in L$, we know that $L^{\prime }\subset L$.

As every ${\alpha }_i,1\le i\le n$ is a root of a polynomial $p_i=q_j$, then ${\alpha }_i$ is a root of $f$, so ${\alpha }_i={\beta }_k$ for some $k,1\le k\le m$, thus ${\alpha }_i\in L^{\prime }$. Consequently $\{{\alpha }_1,\dots ,{\alpha }_n\}\subset \{{\beta }_1,\dots ,{\beta }_m\}$ and

$L=L^{\prime }$ is the splitting field of $f$ over $F$.

Conclusion: if $F\subset L$ is a finite normal separable extension, $L$ is the splitting field of a separable polynomial in $F[x]$. □