# Is the tensor product over $B$ of two flat $A$-modules flat over $A$?

Given a morphism of commutative rings $A\to B$ such that $B$ is a flat $A$-module and given $M$, $N$ two $B$-modules flat as $A$-modules, is the tensor product $M\otimes_B N$ flat over $A$??

The tensor product $M\otimes_A N$ is flat over $A$, the proof is not hard:

Given an exact sequence $0\to X\to Y\to Z\to 0$ of $A$-modules since $M$ is a flat $A$-module the sequence $0\to X\otimes_A M\to Y\otimes_A M\to Z\otimes_A M\to 0$ is exact. Again $N$ is a flat $A$-module, and since the tensor commute we have the statement.

I've tried some similar arguments but without any success, and I can't find a counterexample.

There is a morphism $M\otimes_A N\to M\otimes_B N$ (it is shown in here), but I don't know how to use it.

Not in general, not even if $M$ and $N$ are, in fact $B$-algebras. To see this, let $A = k[t]$ be the ring of polynomials in one indeterminate $t$ over a field $k$, and $B = k[x,y]$ be the ring of polynomials in two indeterminates $x,y$ over $k$ made into an $A$-algebra by mapping $t$ to $x+y$. Let $M = B/(y)$ and $N = B/(x)$ be the quotient rings, emphatically not flat as $B$-modules but still flat as $A$-modules because in fact $M = k[x]$ with $t$ being mapped to $x$ and $N = k[y]$ with $t$ being mapped to $y$ are both even isomorphic to $A$. Yet the tensor product $M \otimes_B N$ is $B / (x,y) = k$ which is not flat as an $A$-module.

This is probably easier to view geometrically: $\mathop{\mathrm{Spec}} M$ and $\mathop{\mathrm{Spec}} N$ are two lines in the plane $\mathop{\mathrm{Spec}} B$ whose intersection is a point: each line maps flatly, and even isomorphically, to the line $\mathop{\mathrm{Spec}} A$, but their intersection (which is their fiber product over the plane) does not map flatly.

Very nice question, though!

Assume that as $B$-modules, you have that $N= B\otimes_A N$: note that this is generally false. There is an obvious map $B\otimes_A N\to N$ that sends $b\otimes n\to bn$. One might be tempted to use the functional inverse $n\to 1\otimes n$, but this is not necessarily $B$-linear, unless say $A\to B$ is onto. If it were true, then the argument goes through. See this. As an example of how $B$ might definitely not be isomorphic to $B\otimes_A B$, take $B$ to be a $k$-algebra with $k$ a field, $k\to B$ the structure map, and say $B$ is finitely dimensional of dimension $>1$. Then $B\otimes_k B$ has dimension strictly greater than $B$.

Now take $C$ an exact sequence. Then $N\otimes_A C$ is exact, and since $B$ is flat, so is $B\otimes_A N\otimes_A C$. Now write $$M\otimes_B N =(M\otimes_A B)\otimes_B (B\otimes_A N)$$

Then $M\otimes_B N\otimes C$ is obtained from $B\otimes_A N\otimes_A C$ by tensoring first over $B$ with $B$, which preserves exactness, and then with $M$ over $A$, which also does.

• I don't understand if you want to say that the answer to the question is positive, but in my opinion it's not. Moreover, for $M=N=B$ I think the equation $M\otimes_B N =(M\otimes_A B)\otimes_B (B\otimes_A N)$ is wrong. Commented Mar 26, 2016 at 0:01
• Your answer is rather confusing: first you make a statement (italicized), then you say it's "generally false" (but not pointing out what assumption would make it true) and you give a counterexample which has all the assumptions made in the original situation, then you seem to use this statement, which you demonstrated to be false. Did you perhaps mean "assume" instead of "note"? Commented Mar 26, 2016 at 0:06
• @user26857 I corrected the incorrect claim, I'll try to rewrite if it is still unclear. The equation uses the initial assumption.
– Pedro
Commented Mar 26, 2016 at 0:06
• @Gro-Tsen I made that statement, and upon realizing it was false, corrected the post.
– Pedro
Commented Mar 26, 2016 at 0:07
• @Gro-Tsen I don't understand how the counterexample has "all the assumptions made in the original situation". If $B$ has dimension $>1$ then $k\to B$ is not surjective.
– Pedro
Commented Mar 26, 2016 at 0:11