Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $R$.

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

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.