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Sep 15, 2014 at 12:15 comment added Vít Tuček The hom space you are asking about is actually just the dual of $M$. So you can rephrase your question as to whether taking a dual commutes with taking an exterior power. See a related question where the normal tensor power is dealt with: mathoverflow.net/questions/56255/duals-and-tensor-products
Sep 15, 2014 at 11:16 comment added Alexander Shamov @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.
Sep 15, 2014 at 11:08 history edited Hoang CC BY-SA 3.0
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Sep 15, 2014 at 11:01 comment added KConrad @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.
Sep 15, 2014 at 10:52 comment added KConrad 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 $n$.
Sep 15, 2014 at 10:51 comment added Hoang @Alexader: it's flase for finite-dimesional vector spaces. So I would ask: $Hom_R(\wedge ^n_RN,R)\simeq \wedge^n_R Hom_R(N,R)$ ?
Sep 15, 2014 at 10:31 comment added Alexander Shamov I strongly suspect that it's false even for infinite-dimensional vector spaces over fields. Did you check that?
Sep 15, 2014 at 10:31 review First posts
Sep 15, 2014 at 10:40
Sep 15, 2014 at 10:28 history asked Hoang CC BY-SA 3.0