# Does the splitting field of an irreducible polynomial contain all extensions over which the polynomial factors?

Say $f$ is an irreducible polynomial with coefficients in a field $F$. Say $f$ is no longer irreducible over some extension $K$ of $f$, i.e. $f$ factors into a product of (irreducible) polynomials having degree $\geq 0$. Does the splitting field $E$ of $f$ over $F$ necessarily contain $K$?

## 1 Answer

The splitting field contains every field extension that is generated by adjoining roots of $f$.

It does, of course, not contain every field extension over which $f$ factors. To see this for example take a proper field extension of the splitting field. Clearly $f$ factors there while it is not contained in the splitting field.

• More properly, the splitting field contains an isomorphic copy of every field extension that is generated by adjoining roots of $f$. – egreg Jan 30 '16 at 20:24
• Yes. Or, which is what I implicitly assumed, we work with a fixed algebraic closure. (Otherwise there is not "the splitting field" to begin with.) – quid Jan 30 '16 at 20:28