If R and S are commutative rings and f is a homomorphism from the additive group of R to the additive group of S with f(1) = 1, give an example of f such that f(ab) it not equal to f(a)f(b) for some a,b in R. I.e., give an example of f as a ring homomorphism except the axiom of preservation of multiplication is violated.
Context: I have tried using maps like Z->Z, however whatever definition I choose, I always either violate preservation of addition or preservation of multiplicative identity. Even maps like Z-> Z[x] I couldn't get to work because it seems (at least in the cases I dealt with) you have to be able to factor things out of the polynomial to preserve addition but then that breaks preservation of the multiplicative identity.
The background for this question is that I thought of this while reading a textbook for an algebra course at a major research university in the United States.
How about this:
$$ R=\Bbb Z[x], S=\Bbb Z $$ $$ f: R \to S $$ $$ f(a) = a \;\forall a \in \Bbb Z, f(x^n) = n \;\forall n > 1 $$
and extend $f$ to the whole $R$ by preserving addition. Then $f$ is an homomorphism between the additive groups, but $2=f(x^2) \ne f(x)f(x) = 1\cdot1 = 1 $.