Is there a possibilty to define a (nontrivial) ring structure on $\mathbb{Z}$ (or $\mathbb{Q}, \mathbb{R}$) other than the usual so that $\mathbb{Z}$ (or $\mathbb{Q}, \mathbb{R}$) with that structure becomes an integral domain (or even UFD, PID or euclidean) but not a field?
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$\mathbb Z$ is not a field. What do you mean by 'other'? Non-isomorphic perhaps? – drhab Aug 06 '14 at 09:46
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My memories of ring algebra is very foggy so I don't dare put this as an answer. My intuition is that such a ring structure does not exist. Assume it exists and denotes $Z$ this ring structure (i.e. a countable integral domain that is not a field and not isomorph to $\mathbb Z$). Then $Z$ has a field of fractions denoted $Q$. By construction, $Q$ has characteristic $0$ and the subring ${\mathbb Z}1_Q$ is isomorphic to $\mathbb Z$. But this subring ${\mathbb Z}1_Q$ is just $Z$, hence the contradiction. – Taladris Aug 06 '14 at 09:58
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Yes, of course.
Let $E$ denote your favourite countable Euclidean domain that is not a field, say $\mathbb{Z}[i].$ Since $E$ and $\mathbb{Z}$ are both countable, there is a bijection $f : \mathbb{Z} \rightarrow E$. Now define a binary operation $+_*$ on $\mathbb{Z}$ by writing
$$\forall x,y \in \mathbb{Z}(x+_*y = f^{-1}(f(x)+_E f(y)))$$
Similarly for all the other ring operations / constants. You'll end up with an ring structure on $\mathbb{Z}$ that is isomorphic to $E$.
goblin GONE
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