# Ring homomorphism $f$ from field to non-trivial ring where $\operatorname{Im}(f)$ is not the zero ring

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.

• Maybe $f(x)=0$ is a ring homomorphism? – SmileyCraft Jan 10 at 16:14
• but isn't f(1) = 1 required for f to be a ring morphism? – oneguy Jan 10 at 16:19
• Well then indeed $im(f)$ is automatically not the zero ring if both $R$ and $S$ are not the zero ring. – SmileyCraft Jan 10 at 16:21

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