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Let $M$ be an module over the commutative ring $R$. I'd like to ask do we have the following isomorphism?

$$Hom_R(\wedge^n_RM,R)\simeq \wedge^n_R Hom_R(M,R)$$

We can obviously see it's true for the case $M=R^m$ is a free $R$-module. But I don't know in general.

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    $\begingroup$ I strongly suspect that it's false even for infinite-dimensional vector spaces over fields. Did you check that? $\endgroup$
    – Alexander Shamov
    Commented Sep 15, 2014 at 10:31
  • $\begingroup$ @Alexader: it's flase for finite-dimesional vector spaces. So I would ask: $Hom_R(\wedge ^n_RN,R)\simeq \wedge^n_R Hom_R(N,R)$ ? $\endgroup$
    – Hoang
    Commented Sep 15, 2014 at 10:51
  • $\begingroup$ This does not pass the test of checking the case of free modules of finite rank. If $N$ has rank $a$ and $M$ has rank $b$, with $n \leq a$ then the left side has rank $\binom{a}{n}b$ and the right side has rank $\binom{ab}{n}$. While these are equal if $n = 1$ or $b = 1$, they are certainly not equal for most choices of $a$, $b$, and $n$. $\endgroup$
    – KConrad
    Commented Sep 15, 2014 at 10:52
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    $\begingroup$ @DinhVanHoang: if you want to change your question, you should edit it so people see what you want to ask in the question box rather than leave changes to your question in the comment section. $\endgroup$
    – KConrad
    Commented Sep 15, 2014 at 11:01
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    $\begingroup$ @DinhVanHoang: My comment still applies. For infinite-dimensional vector spaces, for instance, $\mathrm{Hom}(\wedge^2 M, R)$ consists of all skew-symmetric matrices, while $\wedge^2 \mathrm{Hom}(M, R)$ only contains those of finite rank. $\endgroup$
    – Alexander Shamov
    Commented Sep 15, 2014 at 11:16

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