How do we understand the left and right module structure of Endomorphism ring I am reading some materials about noncommuative algebra which made me very confused.
Let $f: A\to B$ be a injective ring homomorphism between two rings, $A$, $B$ are not necessarily commutative. Then the author writes that:

Let $r:B\to E=\mathrm{End}_{A}(B)$ denote the canonical inclusion $b\to r_{b}$.

I do not understand the sentence. Since we have map $i$, we could realize $B$ as a two side module of $A$. What is the meaning of $\mathrm{End}_{A}(B)$ here? It means left $A$-module homomorphism or right $A$-module homomorphism or two side $A$ homomorphism?. Also what is $r_{b}$? Does it means $r_{b}(b’)=bb’$ or $r_{b}(b’)=b’b$? I am very confused. Can help me? Thanks a lot. By the way, could you recommend some references with details about this? The book which I am reading does not give any details.
 A: Typically, if not specified, "module" is taken to mean "left module".  Indeed, sometimes people don't even work with right modules at all and instead think of a right module over a ring $A$ as a left module over the ring $A^{op}$ (which is $A$ with the order of its multiplication reversed).  This gives a unified way of treating both left and right modules together so that you don't need to worry about confusing them notationally.
So, in your quote, $B$ is presumably being considered as a left $A$-module, so $\operatorname{End}_A(B)$ refers to the left $A$-module homomorphisms $B\to B$.  This is confirmed by the choice of notation $r_b$: here $r$ stands for "right multiplication", so $r_b$ is the map that takes an element and multiplies it on the right by $b$.  That is, $r_b(b')=b'b$.  It is common to use $r$ or $\rho$ to denote right multiplication in this way while left multiplication would instead be denoted $l$ or $\lambda$.  Right multiplication commutes with left multiplication (this is just what associativity says: first multiplying by something on the right and then multiplying by something else on the left is the same as doing it in the opposite order), so right multiplication gives a homomorphism of left modules.  Explicitly, $r_b$ is a left $A$-module homomorphism since if $a\in A$ then $r_b(ab)=(ab)b'=a(bb')=ar_b(b')$.
