subsection*{Exercise 13.4.5}

Let $K$ be a finite extension of $F$. Prove that $K$ is a splitting field over $F$ if and only if every irreducible polynomial in $F[x]$ that has a root in $K$ splits completely in $K[x]$. Use Theorems 8 and 27.

\subsubsection*{Solution}

Let $K$ be a splitting field over $F$ and let some irreducible polynomial $f \in F[x]$ such that $f(\alpha) = 0$ for some $\alpha \in K$. Let $\beta$ be an arbitrary root of $f$. We must show $\beta \in K$. By Theorem 8, the identity map $F \to F$ induces an isomorphism $\phi: F(\alpha) \to F(\beta)$, were $\alpha \mapsto \beta$.

Now the fact that $\alpha \in K$ and $K$ is a splitting field of $f$ over $F$ implies $K(\alpha)$ is a splitting field over $F(\alpha)$ straightforwardly, because $K$ is generated over $F$ by the roots of $f$, so also is $K(\alpha)$ over $F(\alpha)$. It follows $K(\beta)$ is a splitting field over $F(\beta)$. Hence by Theorem 27, we can extend $\phi$ to an isomorphism $\psi: K(\alpha) \to K(\beta)$. But $\alpha \in K$ by assumption, hence $[K(\alpha) : K] = 1$. By the isomorphism $\psi$, we conclude that $[K(\beta) : K] = 1$ and that it must be that $\beta \in K$.

Conversely, if every irreducible polynomial $f \in F[x]$ that has a root in $K$ splits completely in $K[x]$, that is, $K$ contains all other roots of $f$ for every $f$, then by multiplicative closure $K$ is a splitting field over $F$.

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