# Duals and Tensor products

Let $A$ be a commutative ring with a unit element. Let $M$ and $N$ be $A$-modules. Let $M^v$ and $N^v$ be the dual modules. In general, do we have $M^v \otimes N^v \cong (M\otimes N)^v$? It is definitely true when M and N are free. I believe (though haven't worked out the details) that it is true when M and N are projective. Both Atiyah & Macdonald and Lang don't say anything on the matter.

I came up with this question while studying $Pic(A)$: defined as the group of isomorphism classes of projective modules of rank 1. If I and J are projective $A$-modules of rank 1, then the fact that $Pic(A)$ is a group immediately implies $I^v \otimes J^v \cong (I\otimes J)^v$, as both are the inverse of $I\otimes J$.

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According to your profile you are an undergraduate but you already express yourself very clearly, jkramerm: +1. – Georges Elencwajg Feb 23 '11 at 8:16
Thanks for the kind words but to be honest, Georges, I have already graduated college so I am not an undergraduate anymore. I took a year off, and will start my graduate studies next fall. – jkramerm Feb 24 '11 at 6:30

In general, we have a map $\mu:M^\vee\otimes N^\vee\to (M\otimes N)^\vee$ given by $\mu(\phi\otimes\psi)(\sum_i m_i\otimes n_i)=\sum_i\phi(m_i)\psi(n_i)$; this is presumably what mephisto is referring to, and it is an isomorphism in the case that he mentions. If the ring $A$ is a principal ideal domain and $M$ and $N$ are both finitely generated then the map is again an isomorphism, for rather uninteresting reasons, because the functor $M\mapsto M^\vee$ kills all torsion modules.

Now let $M$ be the free module over $A$ with basis $\{e_n\}_{n\in\mathbb{N}}$ and take $N=M$ and define $\xi:M\otimes M\to A$ by $\xi(e_i\otimes e_i)=1$ and $\xi(e_i\otimes e_j)=0$ for $j\neq i$. I claim that this is not in the image of $\mu$. Indeed, if $\zeta=\mu(\sum_{i=1}^r\phi_i\otimes\psi_i)$ then the matrix $(\zeta(e_i\otimes e_j))_{i,j=0}^k$ has rank at most $r$ for all $k$, and it is clear that $\xi$ does not have this property.

For another interesting example, take $A=k[x,y]/(xy)$ and $M=A/x$ and $N=A/y$. Multiplication by $y$ gives a map $\phi:M\to A$, so $\phi\in M^\vee$. Clearly $x\phi=0$ so multiplication by $\phi$ gives a well-defined map $M=A/x\to M^\vee$, which is an isomorphism. Similarly we have an isomorphism $N\to N^\vee$, so $M^\vee\otimes N^\vee\simeq M\otimes N=A/(x,y)=k$. On the other hand, $(M\otimes N)^\vee=\text{Hom}_A(A/(x,y),A)=0$. Thus, we have a case where $\mu:M^\vee\otimes N^\vee\to(M\otimes N)^\vee$ is not injective.

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+1 for the free but non-f.g example. – Hailong Dao Feb 22 '11 at 19:40
+1 for the f.g. but non-free example. – Georges Elencwajg Feb 23 '11 at 8:46
Beautiful answer. – Ben Lim Jan 16 '13 at 13:53

In addition to what has been said earlier, I would like to add that if you derive all the functors involved then the formula will be true in a pretty big generality. To be more precise, the derived functor of $(-)^\vee$ is $RHom(-,A)$, while the derived functor of $\otimes$ is $\otimes^L$. So, the correct formula is $$RHom(M,A) \otimes^L RHom(N,A) \cong RHom(M\otimes^L N,A).$$ Now if for example $M$ is projective and finitely generated then $RHom(M,A) = M^\vee$, and the tensor product with $M^\vee$ has no higher (actually lower) derived functors, so the LHS gives $M^\vee \otimes RHom(N,A)$. In particular, its cohomology in degree 0 is $M^\vee\otimes N^\vee$. On the other hand, in the RHS you just have $RHom(M\otimes N,A)$ and its cohomology in degree 0 is $(M\otimes N)^\vee$. So, you see that in this case the derived formula implies the usual one. Also, in the general situation you can write down two spectral sequences, one for the LHS and the other for the RHS, which would converge to the same object, and those can be considered as a replacement for the original equality.

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I'm not sure about the convergence of these spectral sequences in general. You are OK if $M$ and $N$ admit resolutions of finite length by finitely generated projective modules, which holds automatically if $A$ is a regular local ring and $M$ and $N$ are finitely generated, for example. If $A$ is not Noetherian then it can happen that $M$ is finitely generated but that it has no resolution by finitely generated projectives. I suspect that convergence fails in this case. – Neil Strickland Feb 22 '11 at 22:38
This is absolutely right. The convergence is a separate issue. – Sasha Feb 23 '11 at 9:57

This is not true in general.

The main problem is that tensor product can create torsion and co-torsion and reflexive modules have neither. (The dual of a finitely generated module is reflexive, that is, isomorphic to its own double dual).

Even if you assume that $M_1$ and $N_1$ are both reflexive over an integral domain, $M_1\otimes N_1$ may have torsion and this gives a counterexample since reflexive modules are duals. Let $M=M_1^\vee$ and $N=N_1^\vee$. Then $M^\vee\otimes N^\vee=M_1\otimes N_1$ has torsion while $(M\otimes N)^\vee$ being a dual does not.

Notice that in this case the map $$M^\vee\otimes N^\vee \to (M\otimes N)^\vee$$ is not injective. An interesting feature of this example is that $M$ and $N$ are torsion-free.

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This answer doesn't seem to be right, unless it is an answer to a slightly different question than the one actually asked. Why should there be a map $M\otimes N\to M^\vee\otimes N^\vee$? – Neil Strickland Feb 22 '11 at 13:06
Neil (and those two who voted it up :): That was a typo and that statement was not used in the answer. Cheers. – Sándor Kovács Feb 22 '11 at 16:48
p.s.: I removed that part from the answer, so just to make it clear, I meant these maps: $M\otimes N \to M^{\vee\vee}\otimes N^{\vee\vee}$ and $M\otimes N \to (M\otimes N)^{\vee\vee}$. – Sándor Kovács Feb 22 '11 at 17:25
+1 for this answer, I think torsion created by tensor product is the most conceptual explanation. – Hailong Dao Feb 22 '11 at 18:58

For those interested in general statements, here is a summary of assumptions under which the canonical morphisms of $A$-modules below are isomorphisms:

If $P$ is finitely generated projective: $$P\stackrel{\sim }{\to} \left( P^{\vee }\right)^{\vee} \quad$$

A module $P$ is finitely generated projective iff the following canonical map is an isomorphism

$$\quad P^\vee \otimes P \stackrel{\sim }{\to} End(P)$$

If $P$ or $P'$ is finitely generated projective

$$P^\vee \otimes P' \stackrel{\sim }{\to} Hom(P,P')$$

If both $P$ and $P'$ or both $P$ and $M$ or both $P'$ and $M'$are finitely generated projective $$Hom(P,M) \otimes Hom(P',M') \stackrel{\sim }{\to} Hom(P\otimes P',M\otimes M')$$

In particular for $P$ or $P'$ finitely generated projective

$$P^\vee \otimes P'^\vee \stackrel{\sim }{\to} (P \otimes P')^\vee$$

(this is mephisto's answer to the OP's question)

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It's true if one of the modules is finitely generated and projective. The canonical map from left to right is obviously an isomorphism when M is free of finite rank, and hence when M is a direct summand of a module that is free of finite rank. [Edit: Removed second sentence, which was irrelevant.]

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Dear Sandor, I think mephisto means the natural map $M^{\vee} \otimes N^{\vee} \to (M\otimes N)^{\vee}$ (given by sending $\phi\otimes \psi$ to the map $m\otimes n \mapsto \phi(m)\psi(n)$). Regards, Matt – Emerton Feb 22 '11 at 14:52
Oops. Silly me. Sorry. :( – Sándor Kovács Feb 22 '11 at 17:23

I may as well add the following. One can always consider the dual map:

$$(M^\vee\otimes N^\vee)^{\vee} \leftarrow ((M\otimes N)^\vee)^{\vee}.$$

I think that under moderate hypotheses (maybe just requiring the modules to be finitely generated is ok, but maybe one needs more: it's certainly ok if additionally the ambient ring is normal) this is an isomorphism.

Here's a brief discussion as to why. Both sides of the equality are reflexive because they are both duals of something. In the normal ring case, the statement reduces to codimension 1 (because reflexive modules in a normal ring are determined in codimension 1, let me know if you need references). Then you are done because the ring becomes a DVR (and thus a PID) and Neil's original answer gives you an isomorphism even without the dual.

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