What is interpretation of follwing lemma?

Question

I had come across following technical lemma as excersise

It states that

Let R be ring with identity M is abelian group. Prove that M is R Module if and only If there is a 1-preserving ring morphism $f:R\to A$ where A=set of Endomorphism between M.

I can prove this lemma .

But is this any significance of above lemma?

Please tell me .

Any help will be appreciated

Answer 1

I suppose you saw the following definition.

Let $R$ be a unital ring and let $M,+$ be an abelian group. $M$ is called an $R$-module if there exists a action $R\times M\to M: (r,m)\mapsto r\cdot m$ such that

• $1\cdot m=m$,
• $(\lambda+\mu)\cdot m = \lambda\cdot m+\mu\cdot m$,
• $\lambda\cdot(m+m')=\lambda \cdot m+\lambda \cdot m'$,
• $(\lambda\mu)\cdot m=\lambda \cdot (\mu\cdot m)$.

The lemma above asks you to show that the above definition coincides in some sense with the following one:

An abelian group $M,+$ is called an $R$-representation if there exists unital ring morphism $\phi\colon R\to \text{End}(M)$.

So let's prove the lemma:

Assume that $M$ is an $R$-module. Define a map $\phi\colon R\to \text{End}(M):r\mapsto \phi_r$ where $\phi_r\colon M\to M:m \mapsto r\cdot m$. You can verify yourself that $\phi$ is a well-defined unital ring morphism.

Conversely, assume that $M$ is an $R$-representation. Define $\cdot\colon R\times M\to M: (r,m)\mapsto \phi(r)(m)$ and verify that this satisfies the properties of an $R$-module.