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A ring homomorphism maps units to units.

I was wondering if it implies that it maps non units to non units. I tried to find a counter example because I think the answer should be no but couldn't find one. Any help is appreciated.

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    $\begingroup$ Well, what if one ring is a field and the other isn't? $\endgroup$
    – Zhen Lin
    Jun 15 '14 at 8:13
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    $\begingroup$ What ring homomorphisms have you looked at? Off the top of my head, I think you have to try hard to find examples that aren't counterexamples: I think the only two classes that are "easy" to find are homomorphisms from a field, and the embedding from a ring into the polynomial ring over it. $\endgroup$
    – user14972
    Jun 15 '14 at 8:19
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Every integral domain is embedded in a 'field of fractions' of its elements, so any such non-field will have a natural (injective) homomorphism to a field which fails this property.

An easy example is $f:\mathbb{Z} \to \mathbb{Q}$ with $f(m) := m$

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    $\begingroup$ This is not true. The ring has to be at least a domain for this to have any chance of being true, and then it is true. (And that assuming the ring is commutative. Without that hypothesis, things are quite more delicate...) $\endgroup$ Jun 15 '14 at 8:23
  • $\begingroup$ Your edit does not fix this. As I wrote, the ring has to be a domain for the map into its field of fractions to be injective (and the «field of fraction» of a non-domain has to be defined with care...) $\endgroup$ Jun 15 '14 at 8:25
  • $\begingroup$ There you go :-) $\endgroup$ Jun 15 '14 at 8:29
  • $\begingroup$ @MarianoSuárez-Alvarez Ah yes, I imagine you're quite correct. It's been a long time since I did the construction, and I forget at which point in the course we started assuming these things. $\endgroup$ Jun 15 '14 at 8:29
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Can you describe the ring morphisms $\mathbb C[x]\to\mathbb C$?

Can you describe the ring morphisms $\mathbb Z\to\mathbb Z/2\mathbb Z$?

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