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Is there a difference between proving that $\langle \text{End}(M), +, \circ \rangle$ is a ring and proving that $\langle \text{End}_{R}(A) = \text{Hom}_{R}(A,A),+,\circ\rangle$ is a ring, where $M$ is an abelian group and $\text{End}(M)=\{ h \mid h: M \rightarrow M $ is a group homomorphism$\}$, while $\text{End}_{R}(A) = \text{Hom}_{R}(A,A)$ is the group of all the $R$-module homomorphisms from $A$ to $A$.

In my opinion only difference is proving that:the sum of $R$-module homomorphisms is an $R$-module homomorphism, the composition of $R$-module homomorphisms is an $R$-module homomorphism, the identity map of $A$ is an $R$-module homomorphism, $(-f)$ is an $R$-module homomorphism. Am I right?

Thanks!

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  • $\begingroup$ How can you speak of $R$-module endomorphisms of $A$ if it is only an abelian group? Probably you mean that $A$ is an $R$-module, and by $\operatorname{End}(A)$ you mean group endomorphisms, $\operatorname{End}_{\mathbb{Z}}(A)$? $\endgroup$ – lisyarus Nov 3 '17 at 8:25
  • $\begingroup$ I will edit my question sorry @lisyarus $\endgroup$ – user426277 Nov 3 '17 at 8:27
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An $R$-module homomorphism is an abelian group homomorphism that commutes with scalars from $R$:

$$\forall r \in R, m \in M: f(r\cdot m)=r\cdot f(m)$$

When you prove that something is a ring, you have to prove that it is closed under addition and multiplication, so for $\operatorname{End}_R(M)$ you will have to check that the result of adding or multiplying endomorphisms is still an endomorphism, that is, is an abelian group homomorphism and commutes with scalars from $R$.

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