Why is $f(a) \neq 0$ if $a \neq 0$ with $f$ being a ring homomorphism?

Question

I tried to prove it via contradiction, but I still need $f$ to be injective. Here is what I tried:

$$ f(a) = 0 = f(0), $$ thus $a = 0$ (only if $f$ is injective), which leads to a contradiction. How do I generalize this for all ring homomorphisms?