3
$\begingroup$

I've been trying to find concrete examples to prove that these statements aren't true:

Let $B$ be an $R$-module and $(A_{j})$ a family of $R$-modules then:

$ \mathrm{Hom}\left ( \prod A_{j} ,B\right )\cong \prod\mathrm{Hom}\left ( A_{j} ,B\right )$

$ \mathrm{Hom}\left ( \prod A_{j} ,B\right )\cong \ \bigoplus\mathrm{Hom}\left ( A_{j} ,B\right )$

For the first statement, I know that if we change the direct product with direct sum in the left side, then it is true. What I've been trying to do is to find an example where the left side is different from zero and when computing the right side, the direct product of Hom's, we get zeros, but I have had no success.

Specifically, I've been struggling to find an example where the left side is nonzero.

Any idea in both cases would be appreciated.

  • 1
    A direct product of abelian groups is zero iff all factors are zero.2012-10-22

2 Answers 2

2

In order to pursue your strategy, you have to take into account Mariano's comment. So perhaps take $A_n = \mathbb Z/n$ ($n \geq 1$) and take $B = \mathbb Q$. Then certainly $\mathrm{Hom}(A_n,B) = 0$ for all $n$.

What about $\mathrm{Hom}(\prod A_n, B)$?

  • 0
    @MattE I could not link the two notions of "torsion-free" for $(1,1,...)$ and injectiveness for $\mathbb Q$ to find a nonzero element of $\mathrm{Hom}(\prod A_n, B)$. Would you please explain more?2016-12-18
2

Let $R$ be a field, $J$ be countable infinite set, $A_j=R$ for all $j\in J$ and $B=R$.Then the first isomorphism does not hold: this follows from the fact that the dual of infinite dimensional vector space has strictly larger dimension.

The same choices also provide an counter-example for the second isomorphism :-)

  • 0
    If we take your choices, then the direct sum on the right would be isomorphic to the direct sum of $R$'s. But, why the direct product on the left is not so? Thanks!2016-12-18