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](https://mathoverflow.net/questions/88108/reduction-of-a-module).


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.



is tensor square of a reduced ring reduced?
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.


