# Separable Algebras 1

In order to actually get blogging again I think I’m going to do a series on things you should learn in first year graduate algebra but don’t. It will consist mostly of short posts that contain examples, definitions, or tricks that come up for me.

Today we’ll discuss separable algebras. Let ${k}$ be a field (it is important that we allow this to possibly be imperfect). Our algebras will always satisfy the standard hypotheses made in commutative algebra books such as Matsumura, so they will be unital, associative, and commutative. A ${k}$-algebra ${A}$ is called separable if for any field extension ${L/k}$ we have ${A\otimes_k L}$ is reduced (contains no nilpotent elements).

By definition ${A}$ itself must be reduced. The purpose of this post is to show how being reduced over ${k}$ is not enough to guarantee being separable. Now one probably wouldn’t expect this considering you have to be reduced over all base changes to be separable, but it is true that it suffices to check being reduced over ${k}$ if the field is perfect.

We will take as a black box for the purpose of this post that if we choose an algebraic closure ${\overline{k}}$ and define ${k^{1/p}=\{x\in \overline{k} : x^p\in k\}}$ to take all ${p}$-th roots inside this algebraic closure, then it suffices to check that ${A\otimes_k k^{1/p}}$ is reduced. Thus we really only need to check one base change. Now if you work in characteristic ${0}$ all the time, you might be thinking how could we ever pick up nilpotent elements just by tensoring with a field extension?

This is because in characteristic ${0}$, your field is automatically perfect and for any perfect field we have ${k^{1/p}=k}$. Thus we recover the statement that for perfect fields we only have to check that ${A}$ is reduced to see it is separable. Now for our counterexample to the idea that this is all we have to check in general.

Let ${k=\mathbb{F}_p(t)}$. Define our algebra ${\displaystyle A=\frac{k[x]}{(x^p-t)}=k(\eta)}$ where ${\eta^p=t}$. Since ${A}$ is just the field extension of ${k}$ we get by adjoining a ${p}$-th root of ${t}$ it is reduced (fields don’t have nilpotents). But now let’s use our black box to see why the choice of ${k^{1/p}}$ is so important. Note that ${\eta\notin k}$, but we do have ${\eta\in k^{1/p}}$.

Now check ${\displaystyle A\otimes_k k^{1/p}=\frac{k[x]}{(x-\eta)^p}\otimes_k k^{1/p}\simeq \frac{k^{1/p}[x]}{(x-\eta)^p}}$ and since ${\eta\in k^{1/p}}$ we have ${(x-\eta)\in k^{1/p}[x]}$ and it is certainly not zero in ${A\otimes_k k^{1/p}}$ but it does have the property that ${(x-\eta)^p=0}$, i.e. we found a nilpotent element. Thus there are reduced algebras over non-perfect fields that are not separable.