5

We want to show that $Hom_R(R^n,M) \cong M^n$ for $n\in\Bbb Z_{\ge0}$

I have already shown that $Hom_R(R,M) \cong M$ by letting $f:Hom_R(R,M)\rightarrow M$ given by $f(\phi) = \phi(1)$.

I showed that $f$ is bijective and is a group homomorphism, thus $Hom_R(R,M) \cong M$.

It seems too easy to define the same function for $Hom_R(R^n,M) \cong M^n$, I must be doing something wrong. Injectivity and surjectivity were both straightforward, and showing its a homomorphism seemed to go okay... Can you just use the same $f$? How can you show these are isomorphic?

Moya
  • 5,248
shakebeforeopening
  • 83
  • In my opinion, Hanul Jeon's answer is the best way to see this. However, your approach is fine, and maybe this comment will help explain why: $R^{n}$ is a free $R$-module with the usual standard basis ${e_{1}, \ldots, e_{n}} = {(1, 0, \ldots, 0), \ldots, (0, \ldots, 0, 1)}$, so any $R$-module homomorphism from $R$ to $M$ is determined by the values of $e_{1}, \ldots, e_{n}$. Hence, to give an $R$-module homomorphism $R^{n} \to M$ is to specify an $n$-tuple in $M^{n}$, and vice versa. – Alex Wertheim Jan 27 '16 at 00:45

1 Answers1

2

In general, following theorem holds: $$\operatorname{Hom}_R\left(\bigoplus_{i\in I}A_i, B \right)\cong \prod_{i\in I}\operatorname{Hom}_R(A_i,B)$$

Let define a map $\phi$ from the former to the latter defined as $$\phi(f) := \langle f\circ \iota_i\rangle_{i\in I}$$ (where $\iota_i:A_i\to \bigoplus_{i\in I} A_i$ be the canonical insertion) and define map $\psi$ from the latter to the former as $$\psi(\langle g_i\rangle_{i\in I})(a) = \sum_{i\in I}g_i(a_i).$$ Above sum ranges for all $i\in I$. Each $g_i$ is a morphism from $A_i$ to $B$ and $a = \langle a_i\rangle_{i\in I} \in \bigoplus_{i \in I} A_i$. $a_i=0$ but finitely many so $\psi(\langle g_i\rangle_{i\in I})$ is well-defined. You can check that $\phi$ and $\psi$ are homomorphism and inverses each other.

You can find the finite version of above theorem, and its proof is essentially same.

Hanul Jeon
  • 27,376
  • Is this because my mistake was not understanding what elements in $M^n$ look like? Are they n-tuples? – shakebeforeopening Dec 15 '15 at 05:44
  • @shakebeforeopening I can't suspect what your mistake comes from, but I certainly can say that the elements in $M^n$ are $n$-tuples whose components is in $M$. – Hanul Jeon Dec 15 '15 at 05:48