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Basically, my question is whether this answer is correct. Here is the point. Let $R$ be a ring, and let $A$ and $B$ be $R$-algebras. Suppose that $A$ is regular and $B \otimes_R B$ is regular too. Does it follow that $A \otimes_R B$ is regular?

What if we suppose that $R$ is regular and $B$ a smooth $R$-algebra?

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up vote 2 down vote accepted

I think the answer to your first question is "no" and to the second question is "yes".

Let $R = k[x]$ be a polynomial ring in one variable, $A$ the ring $k[y]$ with the map from $R$ to $A$ given by $x \mapsto y^2$. Let $B = k[x]/(x)$ as an $A$-algebra. Then $R,A,B$ are all regular, $B \otimes_R B = k$ is also regular, but $A \otimes_R B = k[y]/(y^2)$ is not regular.

For the second question, the regularity of $R$ is not necessary. If $B$ is smooth over $R$ then $A \otimes_R B$ is smooth over $A$ which is regular. Since regularity is preserved by smooth base change, $A \otimes_R B$ is also regular.

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So my answer to the linked question was wrong, at least in cases when the ground ring is not a field. –  Tom Goodwillie May 23 '11 at 14:36

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