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(All rings are commutative)

Let $A$ be a noetherian ring. Let $B$ be a noetherian $A$-algebra (not nessecerily f.g!)

Suppose $M$ and $N$ are finitely generated projective $B$-modules (for my application I can assume that $M=N$ are of rank $1$).

Consider the $B\otimes_A B$-module $M\otimes_A N$. Is it projective?

Note that it is flat: Given a $B\otimes_A B$-module $L$, there is a canonical isomorphism $(M\otimes_A N) \otimes_{B\otimes_A B} L \cong M \otimes_B L \otimes_B N$ which clearly shows that if $M$ and $N$ are flat over $B$ then $M\otimes_A N$ is flat over $B\otimes_A B$, but I am not sure about projectivity. In my application, the ring $B\otimes_A B$ is not noetherian.

Any ideas?

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Isn't it finitely generated ? – user18119 Sep 12 '12 at 17:35
@QiL, I believe it is... why? – the L Sep 12 '12 at 17:42
This seems to be straightforward. Write a finite presentation of $M$ as $B$-module, tensor by $N$ over $A$ and see that $M\otimes_A N$ is cokernel of a map of f.p. $B\otimes_A B$-modules. – user18119 Sep 12 '12 at 21:09
Can you elaborate? how do you deduce projectivity? – the L Sep 12 '12 at 21:24
up vote 1 down vote accepted

Consider an exact sequence of $B$-modules $$ B^n \to B^m \to M\to 0$$ Tensoring by $N$ gives rise to an exact sequence of $B$-modules $$ B^n\otimes_A N\to B^m\otimes_A N\to M\otimes_A N\to 0.$$ The above maps are also $C$-linear, where $C=B\otimes_A B$.

Similarly we have an exact sequence of $C$-modules $$C^{mr}=B^m\otimes_A B^r\to C^{ms}=B^m\otimes_A B^s \to B^m\otimes_A N\to 0$$ which shows that $B^m\otimes_A N$, and similarly, $B^n\otimes_A N$ are finitely presented $C$-modules. Therefore $M\otimes_A N$ is finitely presented over $C$.

Finally notice that a finitely presented module is projective if and only if it is flat.

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Thanks! I actually knew that, I just didn't know that a finitely presented flat module is projective :) – the L Sep 13 '12 at 7:57

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