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Say a $\Bbbk$-algebra is separable if $L\otimes _\Bbbk A$ is reduced for every field extension $L/\Bbbk$, and reduced if its underlying ring is reduced.

Separable always implied reduced, and I found a result saying the converse is true over perfect fields. However, I would like an instructive counterexample of a reduced $\Bbbk$-algebra which is not separable. How can such a thing happen?

  • The basic counter-example is $k=\mathbb{F}_p(t)$ and $A=k[X]/(X^p-t)$. (Note that the polynomial $X^p-t$ is irreducible, and $A$ is a field). Now consider $A\otimes_k A$. This is $k[X,Y]/(X^p-t, Y^p-t)=A[Y]/(X^p-Y^p)=A[Y]/(X-Y)^p$. And so $X-Y$ is nilpotent. – Roland Jul 19 '16 at 10:58
  • @Roland sorry, what is $\mathbb F_p(t)$? The field of rational functions with $\mathbb F_p$ coefficients? –  Jul 19 '16 at 11:03
  • Yes exactly, this is the first example of a non perfect field. – Roland Jul 19 '16 at 11:17

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