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I am going over some counterexamples for the the isomorphism $M^{*} \otimes M \rightarrow Hom_R( M,M)$. In particular I have been trying to understand what happens if you remove the various restrictions on the module conditions on $M$ necessary guarantee the bijectivity of the isomorphism. The strongest condition required that I know of $M$ be finitely generated projective so the natural question in search of a counterexample is formulated below:

Let $R$ be the ring $\mathbb{Z} / \mathbb{4Z}$ and let $I = 2 \mathbb{Z} / 4 \mathbb{Z}$ be an ideal of $R$. Consider the quotient $R/I$ as an $R$-module $M$.

Now let $\theta : Hom_R(M,R) \otimes M \rightarrow Hom_R(M,M)$ be the canonical mapping. That is the mapping given by $\theta(f\otimes m)(x)=f(x) m$ for all $f\in Hom_R(M,R)$, all $x \in M$ and all $m\in M$.

How do we show that $\theta$ is not one to one or onto?

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I think you want $f(x)m$ there, instead of $f(x) \otimes m$. Minor point! –  Dylan Moreland Oct 2 '11 at 3:02
    
@DylanMoreland Thanks for the help. Is it true that $R \otimes M \cong M$ ?(I think this is where I got confused in the notation). –  user7980 Oct 2 '11 at 3:05

1 Answer 1

up vote 4 down vote accepted

I think all you need to do is write things out.

M has 2 elements: 0+2Z and 1+2Z.

Hom(M,R) has 2 elements: the zero and 1+2Z → 2+4Z.

Hom(M,M) has 2 elements: zero and the identity.

Hom(M,R) ⊗ M has 2 elements: zero and (1+2Z→2+4Z)⊗(1+2Z).

θ( (1+2Z→2+4Z)⊗(1+2Z) )( 1+2Z ) = (2+4Z)(1+2Z) = 2+2Z = 0.

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