Let $R$ and $S$ be commutative rings and $f : R \to S$ is a ring homomorphism. Show that if R is simple then either $f$ is injective or that $f(a) = 0$ for every $a \in R$.
I have that since $R$ is a simple ring, then we must have that the two sided ideals of $R$ are either $\{0\}$ and $R$. If we consider the ideal $I = \{x \in R | f(x) = 0\}$, since $f$ is a ring homomorphism, then $I$ is a two sided ideal of $R$ or $\{0\}$. Then if $I$ is $R$, $f(a)=0 \; \forall \; a \in R$. Also, $I = \{0\}$ where the kernel of the homomorphism is $0$ $\implies f$ is injective.
I was wondering if my proof is sufficient.All help is appreciated!