Show that for every ring $(R,+,\cdot)$, there is an abelian group, $(A,+)$, such that $R$ is isomorphic to a subring of $(\operatorname{End}(A),+,\circ)$.
$(\operatorname{End}(A),+,\circ)$ is the set of homomorphisms of $A$ that form a ring under function addition and composition.
I am thinking to let $\operatorname{End}(A)$ be the group such that $A$ is the abelian group $(R,+)$ and create a ring homomorphism from $(R,+,\cdot)$ into $\operatorname{End}((R,+))$.
Thoughts?
Best Answer
You are correct, (madame or) sir.
This is essentially the ring-theoretic analogue of Cayley's Theorem for groups.
Also, this issue (as a question) came up a while back on Math Overflow.
Added: I had missed that the explicit definition of the map was not contained in the OP's question. A natural ring embedding from $R$ to $\operatorname{End}(R,+)$ is
$r \mapsto \bullet r: (x \in R \mapsto xr)$.
[Or possibly $r \mapsto r \bullet: (x \in R \mapsto rx)$, depending upon your conventions on composition.]