## A finite field extension K of F is a splitting field if and only if every irreducible polynomial over F has no roots in K or splits over K

Let $K$ be a finite extension of a field $F$. Prove that $K$ is a splitting field over $F$ if and only if every irreducible polynomial over $F$ with a root in $K$ splits over $K$.

We begin with some lemmas.

Lemma 1: Let $K$ be a finite extension of $F$, and let $\alpha$ be algebraic over $F$. If $K$ is a splitting field for a set $S \subseteq F[x]$, then $K(\alpha)$ is a splitting field for $S$ considered as polynomials over $F(\alpha)$. Proof: Certainly the roots of elements of $S$ are in $K(\alpha) \supseteq K$. Now suppose $E$ is a splitting field for $S$ over $F(\alpha)$; in particular, $E$ contains $F$ and the roots of the polynomials in $S$, so $K \subseteq E$. Moreover $\alpha \in E$, so $K(\alpha) \subseteq E$. Hence $K(\alpha)$ is a splitting field for $S$ over $F(\alpha)$. $\square$

Suppose $K$ is a splitting field for some (finite) set $S \subseteq F[x]$. Let $q(x)$ be irreducible over $F$, and suppose $\alpha$ and $\beta$ are roots of $q$ with $\alpha \in K$. By Theorem 8 on page 519 of D&F, $\sigma : \alpha \mapsto \beta$ extends to an isomorphism $\sigma : F(\alpha) \rightarrow F(\beta)$. Now $K(\alpha)$ is the splitting field for $S \subseteq F(\alpha)[x]$, and likewise $K(\beta)$ is the splitting field for $S$ over $F(\beta)$. By Theorem 27 on page 541 of D&F, the isomorphism $\sigma$ extends to an isomorphism $\psi : K(\alpha) \rightarrow K(\beta)$. We can visualize this using the following diagram.

A field diagram

Since $\alpha \in K$, $[K(\alpha):K] = 1$, and so $[K(\beta):K] = 1$, and we have $\beta \in K$. That is, if an irreducible polynomial over $F$ has any roots in $K$, then it splits completely over $K$.

Conversely, suppose $K$ is finite over $F$ and that every irreducible polynomial with a root in $K$ has all roots in $K$. Since $K$ is a finite extension, it is algebraic (see Theorem 17 on page 526 of D&F). Say $K = F(\alpha_1,\ldots,\alpha_n)$, and let $m_i$ be the minimal polynomial of $\alpha_i$ over $F$. By our hypothesis, each $m_i$ splits over $K$. Moreover, any field over which all the $m_i$ split must contain the $\alpha_i$. So in fact $K$ is the splitting field over $F$ of the set of polynomials $m_i$.