MHB Are Hom_R(R, M) and M Isomorphic?

[mathmari]

Hey!

Let $R$ be a commutative ring and $M$ be a $R$-module.
I want to show that $\text{Hom}_R(R,M)\cong M$.

I have done the following:

We consider the mapping $\phi : \text{Hom}_R(R,M)\rightarrow M$ with $f\mapsto f(1_R)$.

Let $f,g\in \text{Hom}_R(R,M)$.
We have that
$$\Phi (f+g)=(g+g)(1_R)=f(1_R)+g(1_R)=\Phi (f)+\Phi (g) \\ \Phi (af)=(af)(1_R)=af(1_R)=a\phi (f)$$
So, $\Phi$ is an homomorphism.

Let $\Phi (f)=\Phi (g)$. Then $f(1_R)=g(1_R)$.
So, $f(r)=rf(1_R)=rg(1_R)=g(r), \forall r\in R$.
Therefore, $\phi$ is 1-1.

For each $y\in M$ we define $f$ as follows:
$f: R\rightarrow M$ with $f(r)=ry$
So, for each $y\in M$ we have that $\Phi (f)=f(1_R)=y$.
Therefgore, $\Phi$ is onto.

Is everything correct? (Wondering)

 

[Euge]

It's all correct! (Nod)

 

[mathmari]

Great! (Smile)

I have also an other question... Does it stand that $$\text{Hom}_R(R^n,M)\cong \prod_{I=0}^n \text{Hom}_R(R,M)$$ ? (Wondering)

 

[Euge]

Almost. You need to change $I = 0$ to $I = 1$.

I will use the symbol $\operatorname{Hom}_R(R,M)^n$ to represent $\prod_{i = 1}^n \operatorname{Hom}_R(R,M)$. For $1 \le j \le n$, let $e_j\in R^n$ be the element with $1$ in the $j$th coordinate and zeros elsewhere; let $i_j : R \to R^n$ be the $j$th inclusion mapping $r \mapsto re_j$. Since the inclusion mappings are $R$-linear, there is an $R$-linear mapping

$$\Phi : \operatorname{Hom}_R(R^n,M) \to \operatorname{Hom}_R(R,M)^n$$

given by $\Phi(f) = (f\circ i_1,\ldots, f\circ i_n)$. I claim that the mapping

$$\Psi : \operatorname{Hom}_R(R,M)^n \to \operatorname{Hom}_R(R^n,M)$$

defined by the equation $\Psi(g_1,\ldots, g_n) = g_1\circ \pi_1 + \cdots +g_n\circ \pi_n$ is the inverse of $\Psi$. Here, $\pi_i : R^n \to R$ ($1 \le i \le n$) is the projection mapping $\pi_i(r_1,\ldots, r_n) = r_i$. It's important to note that $\pi_i \circ i_j$ is zero for $i \neq j$ and $\operatorname{id}_R$ for $i = j$. Indeed, for all $r\in R$, $(\pi_i \circ i_j)(r) = \pi_i(re_j) = r\delta_{ij}$, which is equal to $0$ if $i \neq j$ and $r$ when $i = j$. Now

$$\Phi(\Psi(g_1,\ldots, g_n)) = \Phi(g_1 + \cdots + g_n) = \left(\sum_{k = 1}^n g_k \circ i_1 \circ \pi_k, \ldots, \sum_{k = 1}^n g_k \circ i_n \circ \pi_k\right) = \left(\sum_{k = 1}^n g_k\delta_{1k},\ldots, \sum_{k = 1}^n g_n \delta_{nk}\right) = (g_1,\ldots, g_n)$$

and

$$\Psi(\Phi(f)) = \Psi(f\circ i_1,\ldots, f\circ i_n) = f\circ i_1\circ \pi_1 + \cdots + f\circ i_n \circ \pi_n = f\circ (i_1 \circ \pi_1 + \cdots + i_n \circ \pi_n) = f$$

where the last identity follows from the fact that $i_1 \circ \pi_1 + \cdots + i_n \circ \pi_n$ is the identity on $R$:

$$(i_1 \circ \pi_1 + \cdots + i_n \circ \pi_n)(r_1,\ldots r_n) = i_1(r_1) + \cdots + i_n(r_n) = r_1 e_1 + \cdots + r_n e_n = (r_1,\ldots, r_n)$$

Therefore $\Psi$ is the inverse of $\Phi$ and $\Phi$ is a bijection. Since $\Phi$ is also $R$-linear, it is an $R$-module isomorphism.