# Separability and the Jacobson Radical

Posted by Jason Polak on 30. October 2016 · Write a comment · Categories: math

Let $F$ be a field. We say that a polynomial $f\in F[x]$ is separable if $f$ has no multiple roots in an algebraic closure of $F$. Let $E/F$ be a finite extension. We say that $E$ is a separable extension of $F$ if every element in $E$ is the root of a separable polynomial with coefficients in $F$. If $F$ is perfect or a field of characteristic zero, then any finite extension of $F$ is separable. On the other hand, the extension of $\F_p((t))$ obtained by adjoining a root of $x^p – t$ is not separable, since its factorisation over an algebraic closure is $(x-t^{1/p})^p$ where $t^{1/p}$ is a $p$-th root of $t$.

On the other hand, given an arbitrary ring $R$, the Jacobson radical of $R$ which I'll denote $J(R)$, is the intersection of all the left maximal ideals of $R$. It turns out that $J(R)$ is also the intersection of all the right maximal ideals of $R$ and is therefore a two-sided ideal of $R$.

What does the Jacobson radical have to do with separable field extensions?!

It turns out that we can rephrase separability using the Jacobson radical. Suppose that $E$ can be obtained from $F$ by adjoining a root of a separable polynomial $f$, so that $E \cong F[x]/f$. If $L/F$ is any extension of fields, we claim that $E\otimes_F L$ has zero Jacobson radical. To see this, we can assume that $L$ splits $f$ because the Jacobson radical can always be computed by base change to a finite extension. More precisely:

Theorem. If $R$ is a $k$-algebra and $K/k$ is an algebraic extension of fields then $J(R) = R\cap J(R\otimes_k K)$.

So, if $L$ splits $f$ then $E\otimes_F L^n$ where $n$ is the degree of $f$, since $f$ is separable. And, of course $L^n$ has zero Jacobson radical. It's easy to see by taking a series of separable extensions that $J(E\otimes_F L) = 0$ whenever $E/F$ is a finite separable extension. To summarize:

Theorem.If $E/F$ is a finite separable extension of fields and $L/F$ is any field extension, then $E\otimes_F L$ has zero Jacobson radical.

What about the converse? If $E/F$ is a finite extension such that $E\otimes_F L$ has zero Jacobson radical, is $E/F$ a separable extension? The answer is yes. To see this, let's suppose that $E/F$ is not separable. Then I claim that $E\otimes_F E$ has nonzero Jacobson radical! To prove this, it suffices to show that $E\otimes_F E$ contains a nonzero nilpotent element. That's because $E\otimes_F E$ is finitely generated as an $F$-algebra, and hence is a Hilbert ring, so the nilradical can actually be computed as the intersection of all the maximal ideals – in other words the Jacobson radical is actually the nilradical.

Anyway, if we can prove that $E\otimes_F E$ contains a nilpotent element when $E$ is obtained from $F$ by adjoining a single purely inseparable element, then we are done, because if $L/F$ is any intermediate extension then $L\otimes_F L\to E\otimes_F E$ is an injection.

So assume $E = F[x]/(x^{p^k} – y)$. Then $E\otimes_F E\cong E[x]/(x-y^{p^{-k}})^{p^k}$. Thus, in the ring $E\otimes_F E$, the element corresponding to $(x- y^{p^{-k}})$ is a nonzero nilpotent element. Therefore, if $E/F$ is not separable, then $E\otimes_F E$ has nonzero Jacobson radical. In summary, we have proved:

Theorem. Let $E/F$ be a finite extension of fields. Then $E$ is separable over $F$ if and only if for each extension $L/F$ of fields, the Jacobson radical of $E\otimes_F L$ is zero.