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

We begin with some lemmas.

Lemma 1: Let be a finite extension of , and let be algebraic over . If is a splitting field for a set , then is a splitting field for considered as polynomials over . Proof: Certainly the roots of elements of are in . Now suppose is a splitting field for over ; in particular, contains and the roots of the polynomials in , so . Moreover , so . Hence is a splitting field for over .

Suppose is a splitting field for some (finite) set . Let be irreducible over , and suppose and are roots of with . By Theorem 8 on page 519 of D&F, extends to an isomorphism . Now is the splitting field for , and likewise is the splitting field for over . By Theorem 27 on page 541 of D&F, the isomorphism extends to an isomorphism . We can visualize this using the following diagram.

A field diagram

Since , , and so , and we have . That is, if an irreducible polynomial over has any roots in , then it splits completely over .

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

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