Let $\phi:A \to B$ be a flat ring homomorphism, $M$ be a $B$-module which is flat when considered as an $A$-module. Is the tensor product $M \otimes_B M \otimes_B ... \otimes_B M$ flat over $A$? If not true in general, is there any known cases of $\phi$ (other than etale morphisms) when it holds true? Furthermore, is the exterior power $\wedge^m M$ flat, wedge taken as $B$ modules (i.e., $M \otimes_B ...\otimes_B M$ modulo equivalence relations) for any $m \in \mathbb{Z}$?
3
-
2$\begingroup$ I think yes: (1) by Lazard's theorem, flat modules are colimits of free modules, (2) your functors commute with colimits and (3) they map free modules to free modules. $\endgroup$– Piotr AchingerCommented Mar 5, 2016 at 10:54
-
$\begingroup$ @Vinteuil you are right, I misunderstood the question. Perhaps it is still true that an $A$-flat $B$-module is a colimit of $A$-free $B$-modules, but it is unclear whether $\bigwedge^m_B$ of an $A$-free $B$-module is $A$-free... $\endgroup$– Piotr AchingerCommented Mar 5, 2016 at 11:55
-
1$\begingroup$ Cross-posted: math.stackexchange.com/questions/1683803/… $\endgroup$– user26857Commented Mar 5, 2016 at 19:50
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Let $A=\mathbb{C}[x^2]$, $B=\mathbb{C}[x^2,x^3]$, and $M=\mathbb{C}[x]$.
Then $B$ and $M$ are free as $A$-modules, but $M\otimes_BM$ (and also $\wedge^2M$) has a one-dimensional $A$-submodule spanned by $x\otimes 1 - 1\otimes x$, and so isn't flat.
answered Mar 5, 2016 at 12:56