# Every Ring is Isomorphic to a Subring of an Endomorphism Ring of an Abelian Group

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?

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Yes, that works. – Qiaochu Yuan Aug 12 '12 at 22:09

## 2 Answers

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.]

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What homomorphism can I use from $(R,+,\cdot)$ to $End((R,+))$? – neelp Aug 13 '12 at 1:09
@neelp: what options do you have? A ring has two operations. One of them is addition. And the other one is... – Qiaochu Yuan Aug 13 '12 at 1:12

I think instead of using $(R,+)$ as the candidate abelian group, you should use $$A = \oplus_{a\in R}A_a$$ where $A_a$ is the cyclic group generated by the element $a$.

Let $f \in End(A)$. Then we can think of $f$ as $$\oplus_{a\in R}f_a$$ where $f_a \in End(A_a)$.

Finally, we can define a map $\phi: R \rightarrow End(A)$ such that for $b \in R$ $$\phi(b) = \oplus_{a\in R}f_a$$ where $f_a$ is the trivial map if $a \neq b$; and $f_a$ is the identity map if $a = b$.

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Why don't you want to use $(R,+)$? – Pete L. Clark Aug 12 '12 at 22:55
I cannot see an obvious map from $R$ into $End(R)$ – Hongshan Li Aug 13 '12 at 0:35
@Hongshan: how about $r : x \mapsto rx$? – Qiaochu Yuan Aug 13 '12 at 1:12
@Qiaochu Yuan: of course. Thanks) – Hongshan Li Aug 13 '12 at 3:07