Let $M$,$N$ be two modules over commutative ring $R$. Can we say that $pd (M \otimes N )= $ $pd( M) + pd( N)$? Thanks!
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2$\begingroup$ No: $R=\mathbb{Z}$, $M=\mathbb{Z}/2$, $N=\mathbb{Z}/3$. $\endgroup$– Neil StricklandJun 15, 2016 at 16:23
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1$\begingroup$ No. For instance take for $R$ a domain, and $M=N=R/(x)$, for some $x\neq 0$. Then $\mathrm{pd}(M\otimes _RN)=\mathrm{pd}(M)=\mathrm{pd}(N)=1$. $\endgroup$– abxJun 15, 2016 at 16:25
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$\begingroup$ thank you! So do we have some inequality between this two invariant? $\endgroup$– Paulo RossiJun 15, 2016 at 16:53
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4$\begingroup$ I don't think so. In the above examples we have $\mathrm{pd}(M\otimes_R N)<\mathrm{pd}(M)+\mathrm{pd}(N)$, but there are examples where $\mathrm{pd}(M\otimes_R N)$ is infinite: take $R=k[[x,y]]/(x^2-y^3)$, $M=R/(x)$, $N=R/(y)$. Then $M\otimes _RN$ is the residual field, which has infinite projective dimension because $R$ is not regular. $\endgroup$– abxJun 15, 2016 at 17:53
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$\begingroup$ And if we assume that $gldim(R)$ is finite? $\endgroup$– Paulo RossiJun 15, 2016 at 17:59
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1 Answer
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If $M$ and $N$ are Tor-independent (i.e., Tor$_i^R(M,N)=0$ for $i>0$), then you have some positive results. For instance if $R$ is local complete intersection, the three modules are of finite projective dimension (and finite) then you have an affirmative answer as a consequence of Auslander-Buchschaum formula and Hunecke-Wiegand depth formula. See e.g. the expository paper http://arxiv.org/pdf/1302.2170.pdf mainly section 6. This answers less than the above comments, but I cannot comment due to reputation.
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2$\begingroup$ I wanted to add to your answer that, in the Tor independent case, you get $\leq$ without further conditions since the tensor product of projective resolutions is a projective resolution of the tensor product. $\endgroup$ Jun 15, 2016 at 23:11