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If we drop the need for 1 to map to 1 in a ring homomorphism, do we recover it as a consequence of the other parts of the definition. If does the fact 1 does not map to 1 have any major consequences on the image of the homorphism?

  • Well, many algebraists use the term "ring homomorphism" as a map that is multiplicative and additive. So, it is required to take $0$ to $0$, but $1$ is not necessarily mapped to $1$ in this convention. They use, I think, the term "unitary ring homomorphism" for a homomorphism that maps $1$ to $1$. More interesting things happen when you don't map $1$ to $1$. For example, if $\phi:R\to S$ is a ring homomorphism, then $\ker \phi\hookrightarrow R$ is a ring homomorphism. In this sense, an ideal is a subring, but if you use the "unitary" definition, an ideal is not a subring. –  Oct 10 '18 at 17:57
  • Maybe relevant: https://math.stackexchange.com/questions/671823/the-infinite-direct-sum-in-the-category-ring. Changing the definition has repercussions in the "web" of rings. –  Oct 10 '18 at 18:16
  • To give a concrete example, consider the function of sets $i:n\in\mathbb Z\mapsto nx\in\mathbb Z[x]$ and right-compose it with the quotient map $\eta: \mathbb Z[x]\to\mathbb Z[x]/(x^2-x)$ to obtain $\eta\circ i:\mathbb Z\to\mathbb Z[x]/(x^2-x)$, which is both additive and multiplicative but does not take $1_\mathbb Z$ to $1_{\mathbb Z[x]/(x^2-x)}$. – Rafi Oct 10 '18 at 18:18

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A multiplicative map $\phi: R \to S$ must send $1_R$ to an idempotent of $S$ because $\phi(1_R)^2=\phi(1_R^2)=\phi(1_R)$.

If $S$ only has trivial idempotents, $0_S$ and $1_S$, then $\phi$ is either the zero map or $\phi(1_R)=1_S$.

lhf
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