Let $R$ be a commutative ring with unit and let $M$ be a $R$-module.

It holds that $\text{Hom}_R(R^n,M)\cong \prod_{i=1}^n\text{Hom}_R(R,M)$, right?

How could we prove this?

Do we have to define a function and show that it is an homomorphism and bijective?

Do we maybe define the following function?

$$\phi : \text{Hom}_R(R^n,M)\rightarrow \prod_{i=1}^n\text{Hom}_R(R,M)=\text{Hom}_R(R,M)\times \text{Hom}_R(R,M)\times \dots \times \text{Hom}_R(R,M)$$

with $f\mapsto (f_1, f_2, \dots , f_n)$, where $f, f_i\in \text{Hom}_R(R,M)$, for $1\leq i \leq n$

If this mapping is correct:

To show that $\phi$ is an homomorphism, we do the following?

$$\phi (f+g)=(f_1+g_1, f_2+g_2, \dots , f_n+g_n) \\ =(f_1, f_2, \dots , f_n)+(g_1, g_2, \dots , g_n)=\phi (f)+\phi (g) \\ \phi (af)=(af_1, af_2, \dots , af_n)=a(f_1, f_2, \dots , f_n)=a\phi (f)$$

Is this correct?

How can we show that the mapping is bijective?

  • 1
    $\begingroup$ The question is a special case of the isomorphism $\text{Hom}_R(\bigoplus_{i\in I} M_i, N) \cong \prod_{i\in I} \text{Hom}_R(M_i, N)$ where $M_i, N$ are $R$-modules. $\endgroup$ – user348338 Jun 23 '16 at 17:04
  • $\begingroup$ It's not clear to me how you're defining the homomorphism, but yes, that's one way to prove it. $\endgroup$ – Callus - Reinstate Monica Jun 23 '16 at 17:08

I will prove that given a family $\{M_j\}_{j \in J}$ of $R$-modules then $Hom(\oplus M_{j},N) \cong \prod Hom(M_{j},N)$ for any given $R$-module. Then your question follows by setting $M_{j}=R$ for all $\thinspace$ $j \in J$.

Assume that $i_{j} : M_{j} \rightarrow \oplus M_{j}$ is the canonical inclusion. Now define the above homomorphism $\phi: Hom(\oplus M_{j},N) \rightarrow \prod Hom(M_{j},N)$ to be given by $\phi(f)= (f \circ i_{j})_{j \in J}$. In fact what we achieve by doing this, is that since $f \in Hom(\oplus M_{j},N) $ the composition $f \circ i_{j}$ is always a map in $Hom(M_{j},N)$, $\forall j \in J$, so the image of $f$ under $\phi$ is in fact always in $\prod Hom(M_{j},N)$. So, $\phi$ defined as above is the isomorphism you are looking for (prove the latter, isn't difficult now).

  • $\begingroup$ Shouldn't the codomain be the direct product $\prod_{j\in J}\text{Hom}_R(M_j, N)$ ? $\endgroup$ – user348338 Jun 23 '16 at 17:19
  • $\begingroup$ it is a product.. by $(f \circ i_{j})$ I mean the sequence for all $j \in J$ $\endgroup$ – user321268 Jun 23 '16 at 17:21
  • $\begingroup$ @SpamIAm you are right, as I perceive this isn't our problem in that question. $\endgroup$ – user321268 Jun 23 '16 at 17:23
  • $\begingroup$ @SpamIAm, Finite products and finite coproducts of modules are the same. $\endgroup$ – user348338 Jun 23 '16 at 17:23
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    $\begingroup$ @SpamIAm You mentioned: ``These happen to be the same in the category of modules, but I think this bears mentioning." This is true for finite products/coproducts, but not the same for arbitrary products/coproducts. $\endgroup$ – user348338 Jun 23 '16 at 17:29

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