# Is the $B$-tensor power of flat $A$-modules, $A$-flat?

Let $k$ be an algebraically closed field and $A, B$ be two finitely generated $k$-algebras. Suppose $B$ is flat over $A$. Let $M$ be a finitely generated $B$-module which is flat over $A$. Is it true that for any $n>0$, the tensor power $M \otimes_B M \otimes_B ... \otimes_B M$ flat over $A$? Note here the tensor product is taken as $B$-modules.

Analogously, is the $n$-th exterior power $M \wedge_B M \wedge_B ... \wedge_B M$ obtained as the quotient of the above tensor power, going to be flat over $A$?

I know that the answer is false when instead of tensor power, I just take tensor product (tensor product over $B$ of two $A$-flat modules flat over $A$?).

• Take $M=E\oplus F$ where $E$, $F$ are $A$-flat but $E\otimes_B F$ isn't. – Laurent Moret-Bailly Aug 4 '16 at 15:54
• Remark: in the case $A=B$, the question about the exterior powers has a positive answer by Lazard's theorem. – js21 Feb 13 '17 at 10:27