Suppose that $R$ is a commutative ring with unity, $a \in R$, and $\varphi(r) = ar$ defines a ring homomorphism $\varphi$. Prove that $a$ is idempotent, i.e. that $a = a^{2}$.

This is exercise 15 from chapter 17 of A First Course in Abstract Algebra: Rings, Groups and Fields, Second Edition by Marlow Anderson and Todd Feil.

Here is my attempt at a solution.


By definition of $\varphi$, $\varphi(a^{2}) = aa^{2} = a^{3}$. But since $\varphi$ is also a ring homomorphism, $\varphi(a^{2}) = \varphi(aa) = \varphi(a)\varphi(a) = aaaa = a^{4}$.

Now, $a^{3} = a^{4}$ $\dots$

This is as far as I get. I really want to perform multiplicative cancellation and get the answer from here, but I do not know if $a$ is the additive identity and I do not know if $R$ is an integral domain, so that really isn't an option.

I'm not sure how to proceed.

I believe the homomorphism's multiplication preserving property will be important. This suggests that there is some product $xy$ such that when $\varphi$ is applied, it will take on two values with different terms that will then provide the equality $a = a^{2}$ after some algebraic manipulation.


Note that $\varphi(1_R)=a$. But also $\varphi(1_R)=\varphi(1_R\cdot1_R)=a\cdot a=a^2$.


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