# Defining a ring homomorphism from $R$ to $End(R)$

What is a natural ring homomorphism from a ring $R$ to the ring of endomorphisms of $R$, i.e., of group homomorphisms under addition and composition. The map that I've seen defined was send $r$ to $f_r$ where $f_r$ is defined by $f_r (a) = r+a$, but I don't understand how $f_r$ is even a group homomorphism of $R$.

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It might be better to define $f_r$ by $f_r(a) = ra$. –  Dylan Moreland Mar 2 '12 at 3:36

This sounds like you're learning the first bit about module theory. In particular, $R$ is itself an $R$-module, and the ring homomorphism $R \to End(R)$ provides the multiplication.

Perhaps that's off the mark. That's okay. Suppose $r_1, r_2$ are in $R$. Then $f_{r_1} f_{r_2} (a) = f_{r_1} ( a + r_1) = (a + r_1) + r_2 = a + (r_1 + r_2) = f_{r_1 + r_2}$. That's interesting. $f_{r_1 r_2}(a) = a + r_1 r_2 \not =$ anything useful, really, unless your multiplication is my addition.

But if instead, $f_r (a) = ra$, then $f_{r_1} f_{r_2} (a) = f_{r_1}(r_2 a) = r_1 r_2 a = f_{r_1 r_2} (a)$ and $f_{r_1 + r_2} = f_{r_1} + f_{r_2}$ by distributivity and the same direct work. This is a ring homomorphism, and in fact it's in line with the standard idea of $R$ being an $R$-module - it acts on itself by left multiplication.

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Do you know what it means for the map $f_r$ to be a group homomorphism? You have to check that addition in $R$ turns into composition in $\mathrm{End}(R)$ under $f$, i.e. check that $f_{r+s}=f_r\circ f_s$.

As Dylan points out, the multiplication map $m_r(a):=ra$ satisfies ring homomorphism properties; check that the following hold for all $r,s,a\in R$ (it comes down to parroting ring axioms on $R$):

$$(m_r+m_s)(a)=m_{r+s}(a)$$ $$(m_r\circ m_s)(a)=m_{r\cdot s}(a)$$

This means that $+$ and $\circ$ in $\mathrm{End}(R)$ on the LHS is turned into $+$ and $\cdot$ on the RHS.

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