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When I looked at the wikipedia page about ring homomorphisms (here) I noticed the following statement:

Let $f:R\to S$ be a ring homomorphism.

If $R$ is a field, $S$ is not the zero ring, and $\operatorname{Im}(f)$ is not the zero ring then $f$ is injective.

My question is why the fact that $\operatorname{Im}(f)$ is not the zero ring is required as I don't see where this would be used in a proof.

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    $\begingroup$ Maybe $f(x)=0$ is a ring homomorphism? $\endgroup$ – SmileyCraft Jan 10 at 16:14
  • $\begingroup$ but isn't f(1) = 1 required for f to be a ring morphism? $\endgroup$ – oneguy Jan 10 at 16:19
  • $\begingroup$ Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring. $\endgroup$ – SmileyCraft Jan 10 at 16:21
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This is just an error in the Wikipedia page; the assumption that the image is not the zero ring is unnecessary. If ring homomorphisms are not defined to be unital (that is, $f(1)=1$), then it is necessary, since otherwise $f$ could just map everything to $0$. However, the Wikipedia page does define ring homomorphisms to be unital, so this is not an issue.

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  • $\begingroup$ The zero homomorphism to the trivial ring $\{0\}$ is unital. I think that's what the authors had in mind. $\endgroup$ – rschwieb Jan 10 at 16:27
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    $\begingroup$ But the statement also includes the requirement that $S$ is not the zero ring. $\endgroup$ – Eric Wofsey Jan 10 at 16:31
  • $\begingroup$ Oh that part is superfluous. Got crossed up. $\endgroup$ – rschwieb Jan 10 at 17:01

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