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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.

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up vote 13 down vote accepted

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.

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Thank you! I wonder if it is a coincidence that this counterexample is not flat. –  Dima Sustretov Feb 13 '12 at 20:21
    
Dear David, is it possible to turn this counterexample into a counterexample where S is of the form $R[x_1,\ldots,x_n]/I$ where I is generated by elements linear in $x_i$-s? E.g. applying the functor $Sym$ to $S$ (seen as a finite module over $R$) we get $R[u,v]/(ux−vy)$. However, I can't easily see if $R[u,v,t,s]/(ux−vy,tx−sy)$ is reduced or not. –  Dima Sustretov Feb 15 '12 at 14:03
    
That's reduced. The ideal has primary decomposition $(ux-vy, tx-sy,us-tv) \cap (x,y)$, and each of those are prime. I don't see any fundamental reason something like this couldn't work, though. –  David Speyer Feb 15 '12 at 14:30
    
I think I see why this can't work. For any point $\mathrm{Spec}(K) \in \mathrm{Spec}(R)$, if the equations are of the form you describe, then the fiber of $\mathrm{Spec}(S)$ over $\mathrm{Spec}(K)$ is a linear space. So the fiber of $\mathrm{Spec}(S \otimes_R S)$ over $\mathrm{Spec}(K)$ is the square of that linear space, and is hence reduced. There should be a theorem that, if $Y$ is reduced, and every fiber of $X \to Y$ is reduced, then $X$ is reduced. I don't know a citation for that, though. –  David Speyer Feb 15 '12 at 14:43
    
This is nice. Concerning the theorem that you mention, is $Y$ over a field or is it a general scheme? –  Dima Sustretov Feb 15 '12 at 15:23

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