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?


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.

  • 1
    $\begingroup$ So my answer to the linked question was wrong, at least in cases when the ground ring is not a field. $\endgroup$ – Tom Goodwillie May 23 '11 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.