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Let $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(x+y)=f(x)+f(y), \forall x,y \in \mathbb Z$ and $f(xy)=f(x)f(y), \forall x,y \in \mathbb Z$. I need to prove that either $f=I_{\mathbb Z}$ is the identity function or $f(x)=0, \forall x \in \mathbb Z$.

Supose $f(x) \neq 0, \forall x \in \mathbb Z$, Then I should be able to show that $f(x)=x$, but I didn't manage to do it. Please, give-me a hint. Thank you!

Walter r
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    Well, if $\neg (\forall x \in \mathbb Z f(x) = 0)$, then $\exists x \in \mathbb Z f(x) \ne 0$. Then what are $f(x + x)$ and $f(x * x)$? – qaphla Aug 22 '14 at 17:22
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    You can't assume $f(x)\neq 0$ for all $x$, you need to assume $f(x)\neq 0$ for some $x$. – Thomas Andrews Aug 22 '14 at 17:23

3 Answers3

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Use the fact that $f$ is additive to show that $f(x) = x f(1)$. (It may be easier to prove it for positive integers first.) The fact that $f(xy) = f(x) f(y)$ then forces $f(1)$ to be $0$ or $1$.

anomaly
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Suppose $x \neq 0$ implies $f(x) \neq 0$. Since $f(x \cdot 1)=f(x)$ and $f(x \cdot 1)=f(x)f(1)$. For $x \neq 0$ we have that $f(1)=1$. The addition property can be used to complete the proof that $f$ is the identity map on $\mathbb{Z}$ in the case $f$ is not the zero map.

After that you need to show that if $f$ is not the identity map, then it is the zero map.

M A Pelto
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$f(0)=f(0+0)=f(0)+f(0)$ implying that: $$f(0)=0$$ and consequently: $$f(n)+f(-n)=f(0)=0$$

$f(1)=f(1.1)=f(1)^2$ implying that: $$f(1)\in\{0,1\}$$

If $n$ is a positive integer then: $$f(n)=f(1+\cdots+1)=f(1)+\cdots+f(1)=nf(1)$$

drhab
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