Working with Idempotent elements of a Ring Homomorphism

Question

Let $R, S$ be rings with unities $1_R$ and $1_S$ respectively, and let $\theta \colon R \rightarrow S$ be a ring homomorphism. Prove that $\theta(1_R)^2 = \theta(1_R)$. And if $1_S$ and $0_S$ are the only idempotents of $S$, and $\theta$ is not the zero homomorphism, prove that $\theta(3_R) = 3_S$, where $3_S = 1_S + 1_S + 1_S$.

Answer 1

From the other part, note that because $\theta$ is a ring homomorphism then:

$$\theta(1_R)^2=\theta(1_R)\cdot\theta(1_R)=\theta(1_R^2)=\theta(1_R)$$

For the other part, the problem says that $\theta$ is a non-zero ring homomorphism AND tAAVQhe only units of $S$ are $1_S$ and $0_S$. Note from this that $\theta(1_R)$ is an unit of $S$ because of the first part, so the only values could take $\theta(1_R)$ are $0_S$ or $1_S$. If $\theta(1_R)=0_S$, the ring homomorphism would be the zero homomorphism (contradiction to the hypothesis), so $\theta(1_R)=1_S$, then: $$\theta(3_R)=\theta(1_R+1_R+1_R)=\theta(1_R)+\theta(1_R)+\theta(1_R)=1_S+1_S+1_S=3_S$$

The reason of why the ring homomorphism would be the zero homomorphism if $\theta(1_R)=0_S$ is because if you consider any element $x\in R$, $$\theta(r)=\theta(1\cdot r)=\theta(1)\cdot\theta(r)=0_S\theta(r)=0_S$$

So any element evaluated in $\theta$ will let you the $0_S$ if $\theta(1_R)=0_S$