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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.

Norbert
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M88
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1 Answers1

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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 $n+m$.

Julian Kuelshammer
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