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.

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

When you say that f(1)(1-f(1))=0 implies f(1)=0 or f(1)=1 you are implicitly assuming that ab=0 implies a=0 or b=0. This is not true in every ring. The rings for which ab=0 implies a=0 or b=0 are called "integral domains" or just "domains" for short. For non-trivial ring homomorphisms, the unity-preserving axiom is necessary to include.


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