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Let $M$ and $N$ are two $R$-module. Prove or disprove: $Hom_R(M,N) \simeq M^\ast \otimes_R N$.

Approach: Define a bi-linear map $\phi$ : $M^\ast \times N \longrightarrow Hom_R(M,N)$ by $(f,m)\mapsto f_m$
where $ f_m : M \longrightarrow N $ is defined by $f_m (n) = f(m) n, \forall n\in N$.

Hence this $\phi$ gives a linear map $M^\ast \otimes N \longrightarrow Hom_R(M,N)$.

But I can't able to cook up any map from $ Hom_R(M,N) \longrightarrow M^\ast \otimes N$ which is inverse of $\phi$.

Hence my guess is that the statement is wrong. Am I going to right direction? What is the minimum condition needed so that the isomorphism holds. Is it $N$ has to be free over $R$?

Sandipan Das
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    Here is a possible approach to this problem: Find an $R$-module $M$ with $M^\ast = 0$. Take $N = M$. What happens? – Smiley1000 Feb 18 '25 at 07:05

1 Answers1

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I believe a sufficient condition is that $M$ be projective and finitely generated over $R$, in which case your isomorphism is even natural in $N$.

Edit: The reason we can do it in this case is basically that projective modules have a "dual basis" of $\text{Hom}_R(M,R)$ (A module is projective iff it has a projective basis)

Since $M$ is finitely generated, $M\cong m_1R+...+ m_nR$, we can choose a finite dual "basis" $f_1,..,f_n$ such that for all $x\in M$, $x = \sum_{i=1}^nf_i(x)m_i$, and for all $f\in M^*$, $f = \sum_{i=1}^nf(m_i)f_i$.

Then you can define a map $\text{Hom}_R(M,N)\rightarrow M^*\otimes_R N$ by $\phi\mapsto \sum_{i=1}^nf_i\otimes \phi(m_i)$ and check all the things you need to check.

Wyatt Kuehster
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