Let $R$ be a ring, and let $V$ denote the $R$-Module $R$. Determine all module homomorphisms (need clarification of some concepts)

Question

Let $R$ be a ring, and let $V$ denote the $R$-module $R$. Determine all module homomorphisms $\varphi : V → V$.

I'm having an issue understanding the structure of this module. The answer is that all homomorphisms are isomorphic to $R$.

Can someone explain how you think about that intuitively? The module itself I suppose is a ring, and it's also over the same ring? How do you imagine that structure?

Furthermore, the homomorphisms being isomorphic to $R$ doesn't make sense to me. The homomorphism is a MAPPING, how is it isomorphic to a ring?

Answer 1

It's very simple : for any free $R$-module $L$, i.e. any $R$-module isomorphic to $R^{(I)}$ for some set $I$, and any $R$-module $M$, one has $$ \operatorname{Hom}_R(L,M)\simeq M^I. $$ Let $\mathcal B=(e_i)_{i\in I}$ a basis for $L$. As a homomorphism $u\colon L\longrightarrow M$ is defined in a unique way by the images of the $e_i$s, the isomorphism maps $u\in \operatorname{Hom}_R(L,M)$ to $(u(e_i))_{i\in I}\in M^I$.

Answer 2

The module structure is very simple : addition in $V$ is simply the addition in $R$, and the scalar multiplication is given by the product in $R$.

If we assume that $R$ has a multiplicative identity $1$, then every $R$-module homomorphism must satisfy $$\varphi(x)=\varphi (x\cdot 1)=x\cdot \phi(1);$$ and $\varphi(1)$ can be any element of $R$; indeed, if you define $\phi(x)=x\cdot \alpha$, then you can easily check that $$\varphi(x+y)=(x+y)\cdot\alpha=x\cdot\alpha +y\cdot\alpha=\varphi(x)+\varphi(y)$$ and $$\varphi(r\cdot x)=(r\cdot x)\cdot \alpha=r\cdot (x\cdot \alpha)=r\cdot \varphi(x),$$ so that $\varphi$ is an $R$-module homomorphism.

Thus every homomorphism of $V$ is of the form $\phi(x)=x\cdot \alpha$. You can easily add $R$-module homomorphisms $V\to V$ (pointwise), and you can also compose them; these operations, seen as a sum and product, give to $Hom_R(V,V)$ the structure of a ring. Then you can check that $$\Phi:R\to Hom_R(V,V) : \alpha\mapsto[x\mapsto x\cdot \alpha]$$ is injective and surjective, and that it is almost a ring homomorphism. I say almost because it actually is an anti-homomorphism : it preserves operation, but it only preserves the product up to order, in the sense that $$\Phi(\alpha\cdot \beta)=\Phi(\beta)\circ \Phi(\alpha).$$ Of course, if $R$ is commutative, then $\Phi$ is really an homomorphism. Moreover, I've considered $R$ as a left module; you can also consider it as a right module, and use a similar construction (just switch $x$ and $\alpha$) to obtain a (true) isomorphism of rings between $R$ and $Hom_R(V,V)$.

Answer 3

The set of all such homomorphisms, $\operatorname{Hom}_R(V,V)$, is a ring in its own right, with the operation being the composition of mappings. So when the answer says that "$\operatorname{Hom}_R(V,V)$ is isomorphic to $R$", I'm sure they mean that this is an isomorphism of rings. And that is actually true, if the ring $R$ has an identity. (Probably somewhere in this chapter of your textbook there's a note saying that this chapter assumes all rings to have an identity.)

Here's the main idea. Let $1_R\in R$ be the identity element of $R$. Assume $\varphi:V → V$ is such a homomorphism. Then it's completely determined by the image of $\varphi$ on $1_R$. More specifically, if $\varphi(1_R)=r$, then for any $x\in R$ we have $\varphi(x)=\varphi(x\cdot1_R)=x\varphi(1_R)=xr$. Then we can define a mapping from $\operatorname{Hom}_R(V,V)$ to $R$ via $\varphi\mapsto\varphi(1_R)$. In other words, we define a mapping $F:\operatorname{Hom}_R(V,V)\to R$ via $F(\varphi)=\varphi(1_R)$ for each $\varphi\in R$. Then it's pretty straightforward to show that this mapping $F$ is an isomorphism.