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Let $M, N, P$ be modules over a commutative ring $R$. The above identity is true for $R$ a field:

Since the RHS $\cong$ (by passing to double dual) $$\operatorname{Hom}(N^* \otimes P^*, M^*)\cong\operatorname{Hom}(P^*,\operatorname{Hom}(N^*,M^*)) \cong P^{**} \otimes\operatorname{Hom}(M,N)\cong P \otimes\operatorname{Hom}(M,N).$$

In a couple examples that I worked out, it held true but the steps do not hold true if $N$ and $M$ have torsion. Is the identity true?

edit: I had originally made a mistake when writing my question. I have changed the question title to make it correct. Originally I had written $\operatorname{Hom}_R(M,N) \otimes_R P=\operatorname{Hom}_R(M\otimes P,N)$ which as Qiaochu Yuan remarks is ludicrous (for the vector space case $RHS= P^*\otimes\operatorname{Hom}(M,N)$). Other than cosmetic changes the body of the text remain unchanged.

user26857
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user062295
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    In your title the LHS is covariantly functorial in $P$ but the RHS is contravariantly functorial. Also, you seem to be assuming that your modules are dualizable; over a field this means they're finite-dimensional and in general it means they're finitely generated projective. – Qiaochu Yuan Apr 05 '16 at 03:04
  • Sorry I had written the right steps but the wrong question. I meant to write – user062295 Apr 05 '16 at 22:17
  • I meant to write $Hom(M,N) \otimes_R P =Hom_R(M,N \otimes P)$ – user062295 Apr 05 '16 at 22:17
  • Related: http://math.stackexchange.com/questions/81429/hom-and-tensor-with-a-flat-module – user26857 Apr 05 '16 at 22:25
  • In the special case in which $M=P$ and $N=R$, what you want is that $P^*\otimes_RP$ be the same as $\hom_R(P,P)$, and that is not true even over a field when $P$ is not finitely generated. – Mariano Suárez-Álvarez Apr 05 '16 at 22:29
  • If, on the other hand, $M$ is a torsion module over a domain $R$, $P=M$ and $N=R$, then $P^=0$ so that $P^\otimes P=0$ yet $\hom(P,P)$ is not zero. You can take $M$ finitely generated here. – Mariano Suárez-Álvarez Apr 05 '16 at 22:31

3 Answers3

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As explained here, this does not hold in general, but it does hold if any of the following sets of hypotheses holds:

  • $P$ is finitely presented projective,
  • $M$ is finitely presented and $P$ is flat,
  • $P$ is finitely presented and $M$ is projective.

As an example of the second case, $P$ can be a localization of the base ring, and you get that if $M$ is finitely presented then hom commutes with localization.

Your argument can be made to work under the additional assumption of dualizability, which for modules is equivalent to being finitely presented projective; see for example this blog post for details. But you assume in your argument that $M, N, P$ are all dualizable, and as mentioned above you actually only need to assume that $P$ is dualizable, although this doesn't cover all cases of interest.

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Qiaochu Yuan
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It is true if $P$ is finitely generated projective, for instance. In this case you have $P\oplus Q=R^n$ and the statement boils down to prove $$\def\H{\operatorname{Hom}_R} \H(M,N)\otimes R^n\cong\H(M,N\otimes_RR^n) $$ which is almost obvious.

If $R$ is a field, $P$ finitely generated projective means $P$ is finite dimensional.

In general you can only say there's a natural map $$ \H(M,N)\otimes_RP\to\H(M,N\otimes_RP) $$ given by the bilinear map $$ \H(M,N)\times P\to\H(M,N\otimes_RP), \qquad (f,p)\mapsto \widetilde{(f,p)} $$ where $$ \widetilde{(f,p)}\colon m\mapsto f(m)\otimes p $$

egreg
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  • I the first part you said that it is enough to prove for $R^n$. That I got. From there how it follows for $P$ where $P$ satisfies $P \oplus Q=R^n$. – budi Jul 24 '16 at 04:36
  • @AnupamAh Hom commutes with finite direct sums – egreg Jul 24 '16 at 05:48
  • For R^n I was able to prove. From this how can I prove for P where P satisfies $P \oplus Q=R^n$. If it is true for $R^n$ then $Hom(L,M) \otimes (P\oplus Q)$ isomorphic to $Hom(L, M \otimes (P \oplus Q)$. From this How can I conclude that $Hom(L,M) \otimes P$ isomorphic to $Hom(L,M \otimes P)$ – budi Jul 24 '16 at 06:00
  • @AnupamAh Consider the appropriate map diagrams and see that they consist of isomorphisms. – egreg Jul 24 '16 at 19:44
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I am going to put Prof. Suarez-Alvarez's comment as an answer.

Right so it is not true in the case for infinite dimensional vector spaces because I need to use double duality to get the identity $Hom(V,W)=V^* \otimes W$. In more detail $Hom(V,W) = Hom (V, Hom(W^*,R))=Hom(V \otimes W^* ,R)=(V\otimes W^*)^*=V^* \otimes W$. I used this identity in the second isomorphism in my question.

Nor is it true when $M=\mathbb{Z}/2$, $N=\mathbb{Z}$, $P=\mathbb{Z}/2$, for LHS=$Hom(\mathbb{Z}/2,\mathbb{Z})\otimes \mathbb{Z}/2=0\otimes \mathbb{Z}/2=0$ and $RHS=Hom(\mathbb{Z}/2,\mathbb{Z}/2)=(\mathbb{Z/2})^4$.

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