I was looking at the definition under Wikipedia, which states that for arbitrary rings $R,S$, a ring homomorphism $f:R\to S$ must satisfy $f(1)=1.$ Here, I assume they mean $1$ as the multiplicative identity. Certainly then, this implies the zero map is not a ring homomorphism?
This seems somehow intuitionally false; that is, we would want the zero map to be a ring homomorphism, as it is a group homomorphism between groups, a continuous function between reals, a smooth function between manifolds, etc. Could someone help explain why the zero map in the category of rings seems to be an exception to this pattern?
Best Answer
It's a matter of convention.
For many authors, "rings" are required to be unital rings (have a multiplicative unit); when rings are required to be unital, it makes sense to require the homomorphisms to be unital as well (viewing rings as general algebras with two binary, $+$ and $\times$, one unary, $-$, and two nullary operations, $0$ and $1$, homomorphisms are required to respect all the operations). This is the convention followed, for example, by Lam in his A First Course in Noncommutative Rings (to give a highly regarded, professional ring theorist example of someone who would agree with Wikipedia).
Another way to justify this is to recall that a monoid homomorphism is not merely a semigroup homomorphism between monoids: if $M$ and $N$ are monoids, a monoid homomorphism is a map $f\colon M\to N$ such that $f(ab)=f(a)f(b)$ and $f(e_M) = e_N$ holds. Thus, for instance, the map $(\mathbb{N},\times)\to (\mathbb{N}\times\mathbb{N},\cdot)$, where $\mathbb{N}$ are the nonnegative integers under multiplciation, and $\cdot$ is the coordinatewise multiplication, given by $a\mapsto (a,0)$, is not a monoid homomorphism, even though it is a semigroup homomorphism. (In a sense, it is a "happy accident" that any semigroup homomorphism between groups is also a group homomorphism; but it should really be defined as requiring that it map inverses to inverses and the identity to the identity).
If you view a ring as a set that has a structure of an abelian group under $+$ and a monoid under $\times$, with the two structures connected via the distributive laws, then it makes sense to require the homomorphisms between rings to simultaneously be group homomorphisms of the additive structure, and monoid homomorphisms of the multiplicative structure... and this requires the homomorphisms to map $1$ to $1$.
Under these requirements, the only time that the zero map can be a ring homomorphisms $\zeta\colon R\to S$ is when $S=\{0\}$ is the trivial ring.
For other authors, rings are not required to be unital; when the rings are not required to be unital, you certainly cannot expect homomorphisms to be unital. In that case, the zero map is always a homomorphism between two rings. This is the convention followed, for example, by ring theorists who do radical theory.