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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.

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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 $.

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  • $\begingroup$ This seems correct. Quick questions, 1. How did you know that you could extend f to the whole of R by preserving addition? 2. In your answer, f is ultimately defined by f($\sum_{i=0}^{n} a_ix^i)$ = ($\sum_{i=1}^{n} a_ii$) + $a_0$, correct? $\endgroup$
    – Coriolanus
    Commented Apr 22, 2017 at 1:58
  • $\begingroup$ 1. Well, I didn't actually prove it but I'm defining $f$ in a basis of $R$. Elements of $R$ are of the form $\sum_{i=0}^n a_ix^i$, as you noted, so if you let $f(\sum_{i=0}^n a_ix^i) = \sum_{i=0}^n a_if(x^i)$ then it MUST preserve addition. 2. Correct $\endgroup$
    – cronos2
    Commented Apr 22, 2017 at 12:49
  • $\begingroup$ When defining f in a basis of R, how did you know you'd get something of the form $f(\sum_{i=0}^{n} a_ix^i) = \sum_{i=0}^{n}a_if(x^i)$? I'm just trying to understand your thinking during the solution process. $\endgroup$
    – Coriolanus
    Commented Apr 23, 2017 at 12:56
  • $\begingroup$ I guess it's a common practice in algebra. In exercises you will usually be given a map and asked to prove it is an homomorphism, checking the suitable properties. In problems and proofs, however, you will usually want to construct such a map that does something very algebraic, in terms of structure and with a very natural definition. It was the case with this function, but I may also happen in cyclic groups, defining the image of the generator; or in field extensions where $f: F(\alpha)/F \to F(\beta)/F, f(x) = x \forall x \in F, f(\alpha) = \beta$, etc. $\endgroup$
    – cronos2
    Commented Apr 24, 2017 at 11:20
  • $\begingroup$ So, rather than knowing beforehand that $f$ would act in that way, I forced it to behave like that, defining it in a basis and extending its definition with the $R$ and $S$ operations. Note that this implies there must be some relation between both structures' operations so this extension makes any sense. Hope this helps. $\endgroup$
    – cronos2
    Commented Apr 24, 2017 at 11:23

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