Suppose that $R$ is a ring, that $M$, $N$ and $L$ are right $R$-modules, and that $N$ is an $R-R$-bimodule.
Is there any formula for $Ext^1 (M\otimes_R N,L)$ or generally for $Ext^n (M\otimes_R N,L)?$
Thanks in advance.
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$\begingroup$ I assume $R$ is meant to be commutative; otherwise you need to be more precise about left/right modules. $\endgroup$– YCorCommented Jan 16, 2018 at 5:23
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$\begingroup$ It seems difficult to answer the question in this generality. You can get nice formulas if, for example, M or N is projective over R. $\endgroup$– the LCommented Jan 16, 2018 at 12:10
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$\begingroup$ @YCor You are right. I consider $R$ is not commutative in general and all modules are right $R$-module. $\endgroup$– MHenryCommented Jan 16, 2018 at 14:18
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$\begingroup$ @theL Could you give a formula for the question when $M$ or $N$ is projective? $\endgroup$– MHenryCommented Jan 16, 2018 at 14:19
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$\begingroup$ If $M$ is an $(R_1,R_2)$-module and $N$ a $(R_2,R_3)$-module then $M\otimes_{R_2}N$ is a $(R_1,R_3)$-module. $M,N$ right $R$-modules is not enough to make sense of $M\otimes_R N$. (And you mean Ext as what?) $\endgroup$– YCorCommented Jan 16, 2018 at 14:21
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I assume from how the question is phrased that the OP is hoping for a kind of Hom-Tensor duality for Ext. The first comment below shows that in general
$$\operatorname{Ext}^n(M\otimes_R N,L) \not \cong \operatorname{Ext}^n(M,\operatorname{Hom}_R(N,L))$$
Instead, what you have are the well-known Ext-Tor relations. This makes sense: Ext is a left derived functor of Hom, so should play nicely with Tor, the right derived functor of $\otimes$.
In this answer, I'll show that Hom-Tensor duality of the sort that fails above for Ext holds for $\operatorname{Ext}_{dw}$, the subgroup of Ext for chain complexes consisting of short exact sequences that are degreewise split. This $\operatorname{Ext}_{dw}$ can be computed via the internal $\operatorname{Hom}$ in chain complexes. Let $S^n(M)$ be the chain complex with $M$ in degree $n$, 0 everywhere else, and all differentials 0. The Ext of chain complexes agrees with the Ext of modules, when they sit inside chain complexes via $S^0(-)$. Hence, $\operatorname{Ext}_{dw}(S^0(A),S^0(B))$ is also a subgroup of the usual $\operatorname{Ext}(A,B)$. First, note that: $$\operatorname{Ext}_{dw}^n(X,Y) \cong H_n \operatorname{Hom}(X,Y) \cong \operatorname{Ch}(R)(X,\Sigma^{-n}Y)/\sim, $$ where $\sim$ is chain homotopy equivalence, see Lemma 2.1 of this paper.
Let's apply this to your situation, letting $\otimes$ be the monoidal product in $\operatorname{Ch}(R)$: $$ \begin{aligned} \operatorname{Ext}_{dw}^n(S^0(M\otimes_R N),S^0(L)) &\cong H_n(\operatorname{Hom}(S^0(M\otimes_R N),S^0(L))) \\ &\cong H_n(\operatorname{Hom}(S^0(M)\otimes S^0(N),S^0(L))\\ &\cong H_n(\operatorname{Hom}(S^0(M),\operatorname{Hom}(S^0(N),S^0(L)))) \\ &\cong \operatorname{Ext}_{dw}^n(S^0(M),\operatorname{Hom}(S^0(N),S^0(L)))\\ &\cong \operatorname{Ext}_{dw}^n(S^0(M),S^0(\operatorname{Hom}_R(N,L)). \end{aligned} $$ Here I use that chain complex maps from $S^0(N)$ to $S^0(L)$ are the same as $R$-module maps from $N$ to $L$, since $$\operatorname{Hom}(X,Y)\cong \prod_i \operatorname{Hom}_R(X_i,Y_{n+i}),$$ and for $S^0(N)$ and $S^0(L)$ the only place this product doesn't vanish is $\operatorname{Hom}_R(N,L)$. We conclude that:
$$\operatorname{Ext}_{dw}^n(M\otimes_R N,L) \cong \operatorname{Ext}_{dw}^n(M,\operatorname{Hom}_R(N,L))$$
where modules sit inside chain complexes via the functor $S^0(-)$.
answered Jan 16, 2018 at 14:50
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$\begingroup$ I do not think this is correct. If you let $(R,\pi)$ to be a dvr, $M=L=R$ and $N=R/\pi R$, $n=1$, then, the left side is $N$ and the right side is zero. $\endgroup$– MohanCommented Jan 16, 2018 at 15:09
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$\begingroup$ That's a good example, but I can't see anything wrong in the proof. Now I'm curious to see how this can be resolved. $\endgroup$ Commented Jan 16, 2018 at 15:40
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$\begingroup$ When using Lemma 2.1, aren't you making the assumption that (i) Ext_{dw}^1(X,\Sigma^{-n-1}Y) = \Ext_{dw}^n(X,Y) (for whatever the latter means, since it is not defined in the cited paper) (ii) That Ext_{dw}^n(M,N) = \Ext_R^n(M,N). $\endgroup$ Commented Jan 16, 2018 at 15:51
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$\begingroup$ Indeed, Section 5 of the cited paper tells you that Ext^n_R(M,N) = Ch(R)(P_,S^n(N))\~, where P_ is a projective resolution of M $\endgroup$ Commented Jan 16, 2018 at 15:53
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$\begingroup$ Hi Drew. I was thinking of $Ext^1$ as the group of split short exact sequences, and $Ext_{dw}^1$ as the group of degreewise split short exact sequences of chain complexes. In this form, it seems to me that when the chain complexes have the form $S^0(-)$ then these groups agree. Am I wrong? $\endgroup$ Commented Jan 16, 2018 at 15:59