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A representation of a ring $R$ is an abelian group $M$ together with a ring homomorphism $\rho:R\rightarrow\text{End}(M).$ I know that there is a bijection between $$A=\{\sigma:R\times M\rightarrow M|(M,\sigma)\text{ is a left $R$-module}\}$$ and $$B=\{\rho:R\rightarrow\text{End}(M)|(M,\rho)\text{ is a representation of $R$}\}.$$

What I don't know is a form of this existing bijection. What does it do?

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  • $\begingroup$ Then how do you know it is there? Anyway, it is quite straightforward to describe once you write up the definitions of each set. $\endgroup$ – Tobias Kildetoft Oct 22 '16 at 22:19
  • $\begingroup$ It was a remark in my course. What do you mean by writing up the definitions? Maybe we can take $f:A\rightarrow B,~\sigma\mapsto f(\sigma)$ with $f(\sigma)(r)=\sigma(r,\cdot)$. I just wonder why $f(\sigma)(r)$ is linear. And what is the inverse of $f$? $\endgroup$ – user369147 Oct 22 '16 at 22:26
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Given $\sigma\in A$, there is a corresponding $\rho\in B$ defined by $\rho(r)(m)=\sigma(r,m)$. The facts that $\rho(r)$ is linear and $\rho$ is a ring-homomorphism follow from the axioms of a left $R$-module (the left distributive axiom says exactly that $\rho(r)$ is linear, and the right distributive, associative, and unit axioms say $\rho$ is a ring-homomorphism).

Conversely, given $\rho\in B$, you can define define $\sigma\in A$ by $\sigma(r,m)=\rho(r)(m)$. The fact that $\sigma$ is a left $R$-module structure follows from the fact that each $\rho(r)$ is linear and $\rho$ is a ring-homomorphism, by just reversing the argument from before.

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