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Let $R$ be a ring, $\mathbb{Q}$ be the field of rational numbers, and $f,g: R \to \mathbb{Q}$ be homomorphisms such that $f(n)=g(n)$ for all $n\in \mathbb{Z}$, where $\mathbb{Z}$ is the ring of integers, show that $f=g$.

(Hint: first show $f(1/n)= g(1/n)$ for all $n \in \mathbb{Z}$)

egreg
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Manal Alzoubi
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  • If the domain of $f$ is an arbitrary ring $R,$ what is, say, $f(2)?$ – mfl Nov 10 '14 at 22:08
  • @mfl: For people who consider "rings" to have identity, $f(2)$ is unambiguously $f(1_R+_R1_R)$. – hmakholm left over Monica Nov 10 '14 at 22:21
  • This is almost certainly a miscopying of this question: http://math.stackexchange.com/questions/63018/showing-two-ring-homomorphisms-that-agree-on-the-integers-must-agree-on-the-rati?rq=1 . These kind of errors are common in questions like these where the asker just typed what was in a book/assignment and obviously did not put any thought into the question. – PVAL-inactive Nov 10 '14 at 22:43

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It is not always true. For example if the ring $R$ is $\mathbb Z[X]$, then $f$ and $g$ could be the evaluation homomorphism for $X=0$ and $X=1$ respectively. They satisfy $f(n)=g(n)$ for $n\in\mathbb Z$, but $f(X)\ne g(X)$.

The hint doesn't even begin to work here, because in general $1/n$ is not a member of the ring and so $f(1/n)$ and $g(1/n)$ are not even undefined.

hmakholm left over Monica
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