Past exam of the number theory course of the University of Oxford / 2010 / A / 1 / ii

An irreducible polynomial is separable if its roots in a splitting field are distinct, equivalently if it is coprime to its formal derivative. For a general nonzero polynomial, the factorwise separability convention means that every irreducible factor is separable. This is the convention under which the printed splitting field implication holds without an extra hypothesis.

If instead separability of an arbitrary polynomial is defined to require all its roots to be distinct, that implication needs the polynomial to be squarefree. For example, $(x-1{)}^2$ has separable splitting extension $K/K$, but a repeated root. These two conventions agree for irreducible polynomials; repeated factors are the entire distinction here.

An algebraic field extension $E/K$ is separable if every $\alpha \in E$ has a separable minimal polynomial over $K$.

For an inseparable example, take $K={\mathbb{F}}_p(t)$ and $f(X)=X^p-t$. Eisenstein's criterion at the prime $t$ in ${\mathbb{F}}_p[t]$, followed by Gauss lemma for polynomials, proves irreducibility over its fraction field. Its derivative is zero, and in a splitting field it is $(X-t^{1/p}{)}^p$, so it is inseparable.

Now let $M/K$ be a separable splitting extension of a nonzero polynomial $f$. Every irreducible factor $g$ of $f$ has a root $\alpha \in M$, and $g$ is a scalar multiple of the minimal polynomial of $\alpha$. Separability of $M/K$ makes this polynomial separable. Therefore every irreducible factor of $f$ is separable, which proves the requested conclusion in the factorwise convention. The squarefree qualification under the alternative convention is explained in part (a).