TITLE: is tensor square of a reduced ring reduced?
QUESTION [5 upvotes]: Let $R$ be a reduced algebra of finite type over a field $k$ of characteristic 0. Let $S$ be a reduced finite $R$-algebra. Is $S \otimes_R S$ reduced?
(In positive characteristic one can get non-reduced tensor products of reduced algebras even over a field.)
I have failed to find a counterexample so I thought that the statement might be true after all. The question is motivated by the discussion in the comments to this question.
If $S=R[x_1, \ldots, x_n]/I$, what one has to show is that the ideal $I \otimes 1 + 1 \otimes I$ in $S \otimes_R S=R[x_1, \ldots, x_n, y_1, \ldots, y_n]$ is radical. However, I have no good ideas as to how to approach this.

REPLY [13 votes]: No. Let $R = k[x,y]/(y^2-x^3)$. Let $S = k[t]$, with the map $R \to S$ given by $(x,y) \mapsto (t^2, t^3)$. So $S \otimes_R S = k[t,u]/(t^2-u^2, t^3-u^3)$. The element $t-u$ is not zero in the tensor product, because all the relations are in degree $>1$. But $(t-u)^3 = 3 (t-u)(t^2-u^2) - 2 (t^3-u^3)$ is in the ideal.