Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). A ring homomorphism from $R$ to $S$ is a function $f:R\to S$ such that, for all $x,y\in R$,

• (RH1) $f(x+y)=f(x)+f(y)$;
• (RH2) $f(xy)=f(x)f(y)$;
• (RH3) $f(1_R)=1_S$.

A ring homomorphism from a ring to itself is a ring endomorphism; a bijective ring homomorphism is a ring isomorphism.

(RH3) is imposed, not derived, and the reason is exactly the reason Monoid homomorphism and group homomorphism treats monoids and groups differently. Condition (RH1) says that $f$ is a homomorphism of the additive groups $(R,+,0_R)\to (S,+,0_S)$, and for groups preservation of the identity is free: it follows from (RH1) by cancellation (A group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(g^{-1})=f(g{)}^{-1}$, and $f(g^n)=f(g{)}^n$ for every $n\in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed). Condition (RH2) says only that $f$ is a homomorphism of the multiplicative semigroups, and $(R,\cdot ,1_R)$ is a monoid that need not be a group (Semigroup and monoid); for monoids preservation of the identity does not follow and must be assumed, which is precisely why Monoid homomorphism and group homomorphism puts the clause $f(e)=e^{\mathrm{\prime }}$ into the definition of a monoid homomorphism and leaves it out of the definition of a group homomorphism (Left identity, right identity, and two-sided identity for a binary operation).

So a ring homomorphism is: a homomorphism of additive groups that is also a homomorphism of multiplicative monoids. The clause is not redundant: the companion page exhibits a map $\mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$ satisfying (RH1) and (RH2) and failing (RH3).

• What follows automatically, and what does not. $f(0_R)=0_S$, $f(-x)=-f(x)$ and $f(mx)=mf(x)$ for every integer $m$ all follow from (RH1) alone, and units are carried to units once (RH3) is available; these are A ring homomorphism satisfies $f(0)=0$, $f(-a)=-f(a)$ and $f(ma)=mf(a)$ for $m\in \mathbb{Z}$, carries units to units, and has a subring as its image; composites of ring homomorphisms are ring homomorphisms. What does not follow from (RH1) and (RH2) is (RH3) itself.

• Between fields there is no difference from the published notion. A ring homomorphism between fields is exactly a field homomorphism in the sense of Field homomorphism and embedding, and every such map is injective; that is A ring homomorphism between fields is a field homomorphism in the published sense, and every such map is injective.

• Kernels, quotient rings and the isomorphism theorems are not defined on this page and nothing here uses them.