# Ring Homomorphism Counterexample

Suppose you have the following conditons: $$f(x+y) = f(x) + f(y), f(xy) = f(x)f(y)$$ and $$f(1) = 0$$, for all $$x, y$$. Would these conditions be sufficient to form a ring homomorphism always or is there a counterexample? I've been trying to show that you automatically get $$f(1) = 1$$ from $$f(xy) = f(x)f(y)$$, but I can't seem to find a counterexample that eliminates the $$f(1) = 0$$ case. Any help would be greatly appreciated.

• Consider the implications of $f(x)=f(1x)=f(1)f(x)$ if $f(1)=0$. – Robert Shore Feb 6 at 2:33
• Ahh so would this only work if f(x) = 0? – Sanjoy Kundu Feb 6 at 2:34
• Yes. If $f(1)=0$, then $f$ is the trivial homomorphism $f=0$. – Robert Shore Feb 6 at 2:53