Cayley's Theorem for Composition Rings?

Question

A composition ring is a commutative ring $(R,+,\cdot)$ endowed with an additional binary operation $\circ$ satisfying the following properties for all $f,g,h\in R$:

1. $(f+g)\circ h=f\circ h +g\circ h$
2. $(f\cdot g)\circ h=(f\circ h)\cdot(g\circ h)$
3. $(f\circ g)\circ h=f\circ(g\circ h)$

Now if $End(R)$ is the set of all functions from $R$ to $R$, it is a composition ring when endowed with the operations of pointwise addition, pointwise multiplication, and composition of functions.

My question is, is there a composition ring isomorphism between $R$ and some composition subring of $End(R)$? And if so, can we say something more? That is, rather than all functions from $R$ to $R$ can we restrict ourselves to some nicer set of functions?

Answer 1

I found an answer to my question in, of all places, another question on Mathematics.SE. This is true as long as $R$ has a right identity $I$ with respect to $\circ$. Let $\phi:R\rightarrow End(R)$ by defined by $\phi(a)(x)=a\circ x$. Then for all $a,b,x\in R$, we have:

1. $\phi(a+b)(x)=(a+b)\circ x=a\circ x +b\circ x=\phi(a)(x)+\phi(b)(x)=(\phi(a)+\phi(b))(x)$
2. $\phi(a\cdot b)(x)=(a\cdot b)\circ x=(a\circ x)\cdot(b\circ x)=(\phi(a)(x))\cdot(\phi(b)(x))=(\phi(a)\cdot\phi(b))(x)$
3. $\phi(a \circ b)(x)=(a\circ b)\circ x = a\circ(b\circ x)=\phi(a)(\phi(b)(x))=(\phi(a)\circ\phi(b))(x)$

Thus $\phi$ is a composition ring homormorphism. If $\phi(a)=\phi(b)$, then $\phi(a)(I)=\phi(b)(I)$, so $a \circ I=b\circ I$. But $I$ is a right identity with respect to $\circ$, so $a \circ I=a$ and $b\circ I=b$, and thus $a=b$. Therefore, $\phi$ is an injective composition ring homomorphism, which implies that there is a composition ring isomorphism between $R$ and a composition subring of $End(R)$.