Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 5 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.