Is the zero map (between two arbitrary rings) a ring homomorphism?
Solution 1:
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.
Solution 2:
The (usual) theory of rings has 5 symbols: $0, 1, +, -, \cdot$. A homomorphism has to preserve all of these symbols:
- $\phi(0) = 0$
- $\phi(1) = 1$
- $\phi(-a) = -\phi(a)$
- $\phi(a + b) = \phi(a) + \phi(b)$
- $\phi(a \cdot b) = \phi(a) \cdot \phi(b)$
In the same way, the usual theory of (multiplicative) groups has 3 symbols: $1, \cdot, {}^{-1}$. A group homomorphism has to preserve all three symbols:
- $\phi(1) = 1$
- $\phi(x^{-1}) = \phi(x)^{-1}$
- $\phi(x \cdot y) = \phi(x) \cdot \phi(y)$
You are confused because, in the case of groups, the third property implies the other two. So when one speaks of group homomorphisms, they tend to focus on just the third property.