Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been struggeling for some time with the following problem

Let $k$ be a field and $A$ and $B$ two $k$-algebras. We can then view the tensor product $A\otimes_k B$ as a $k$-algebra by $(a_1\otimes b_1)\cdot (a_2\otimes b_2)=(a_1a_2\otimes b_1b_2)$.

Let $M$ be an $A$-module with $pd_A(M)=m$ and $N$ a $B$-module with $pd_B(N)=n$. Prove that $pd_{A\otimes B}(M\otimes N)\leq m+n$.

I have proven that $M\otimes N$ is in fact a $A\otimes_k B$-module but nothing more. I've tried to find a projective resolution for $M\otimes N$ and I've tried to show that $Ext^{m+n}(M\otimes N,C)\cong 0$ for any $C$ but without luck.

Any help would be greatly appreciated.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Let $P_\bullet$ be a projective resolution of length $m$ (existent because of assumptions) and $Q_\bullet$ be a projective resolution of length $n$. Then you can define a complex $P\otimes Q$ with $(P\otimes Q)_i=\bigoplus_{j+\ell=i}P_j\otimes Q_\ell$. And $P_j\otimes Q_\ell\mapsto P_{j+1}\otimes Q_\ell\oplus P_j\otimes Q_{\ell+1}$ with the second differential attached with a sign $(-1)^j$, as explained e.g. on PlanetMath. Now you can prove that this sequence is indeed exact, either by hand or with the Künneth theorem. Obviously it has length $n+m$.

share|improve this answer
    
Thank you very much. –  M88 Jan 20 '13 at 13:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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