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Let $A$ and $B$ be noetherian normal rings and let $f:A\rightarrow B$ be a finite but non-flat ring homomorphism. We can also assume $Spec(A)$ connected if necessary. We put on $B$ the structure of $A$-module given $f$.

Is it true that $B\otimes_A B$ is normal?

The example I need to tackle comes from the following situation: take $A$ normal and noetherian, $K$ its function field, $K\subset L$ a finite extension, $B$ the normalization of $A$ in $L$.

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This is false, even with flatness.

Say $k = \mathbb{F}_p$ for a prime number $p$, $A = k[t]$, $B = k[s]$, and $f(t) = s^p$. Then $A$ and $B$ are noetherian, normal, connected, and $B = A[x]/(x^p - t)$, which is finite flat over $A$. However, $B \otimes_A B = A[x,y]/(x^p - t, y^p - t)$, which carries is nilpotent element $x-y$ as $(x-y)^p = x^p - y^p = 0$.

[The same example, in characteristic $0$, shows that $B \otimes_A B$ need not be a domain.]

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  • $\begingroup$ I am confused. Assume $k$ algebraically closed. Why does your example not contradict EGA IV, 6.14.1 where we take $Y=Spec(A)$, $Y^{\prime}=Spec(B)=X$? $\endgroup$ – jikj Jan 19 '14 at 20:30
  • $\begingroup$ The map $f$ in the answer does not satisfy the hypotheses in EGA (look at the $t=0$ fibre). $\endgroup$ – anonymous Jan 19 '14 at 21:14

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