Is the zero map (between two arbitrary rings) a ring homomorphism? 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?
 A: 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.
A: A variety in the sense of a universal algebra is just a bunch of algebraic structures equipped with functions of certain arities and certain relations. For example, the variety of monoids has two functions, one of arity zero, a constant $1$, and one of arity two, a binary operation $\cdot$, satisfying the relations $1 \cdot m = m \cdot 1 = m$ and $m \cdot (n \cdot k) = (m \cdot n) \cdot k$. A homomorphism between two objects of a variety is a map preserving all the functions. For example, a homomorphism between two monoids $M,N$ is a map $f : M \to N$ of the underlying sets such that $f(1)=1$ and $f(m \cdot n) = f(m) \cdot f(n)$ for all $m,n \in M$. This is not a matter of convention as many might think; it follows the general paradigm that homomorphisms preserve the whole structure.
In fact, with this notion, every variety constitutes a category. You always have to remember in which category you are! When you consider a monoid $M$ as a semigroup $U(M)$, i.e. you forget the unit $1$, you have got a different object. You may regard $U(-)$ as the forgetful functor from the category of monoids to the category of semigroups. It is important to keep in mind that this is not the identity; although many authors etc. treat it as such. This causes many confusions. But when you keep in mind that $U$ takes you into another category, it is clear as crystal: A homomorphism $U(M) \to U(N)$ is by definition a map of the underlying sets which preserves $\cdot$, whereas a homomorphism $M \to N$ is by definition a map of the underlying sets which preserves $\cdot$ and $1$.
Now the same story of rings and rngs (a rng is an abelian group with a distributive and associative multiplication; so roughly a ring without the requirement of a unit). By abuse of notation, let us denote the forgetful functor from rings to rngs again by $U$. Let $R,S$ be rings. Then, by definition, a homomorphism $f : R \to S$ is a map preserving all the structure, i.e. $f(0)=0$, $f(x+y)=f(x)+f(y)$, $f(-x)=-f(x)$, $f(x \cdot y) = f(x) \cdot f(y)$, $f(1)=1$ for all $x,y \in R$. Now the condition $f(x+y)=f(x)+f(y)$ for all $x,y \in R$ already implies $f(0)=0$ and $f(-x)=-f(x)$ for all $x \in R$. Many authors are motivated by this to drop these two conditions from the definition, which in my opinion is a bad idea. It is just a lemma that you only have to check that $+$ is is preserved; the correct definition is just a special case of the more general one in universal algebra, which you should even keep in mind when you don't study universal algebra because algebraic structures never come alone. Of course, this discussion already applies to the variety of groups, see Hurkyl's answer. In contrast to that definition of homomorphism of rings, a homomorphism of rngs is a map preserving $0,+,-,\cdot$ (there is no way to talk about $1$). In particular, a homomorphism $R \to S$ yields a homomorphism $U(R) \to U(S)$, but not vice versa.
Now it is easy to answer the question whether the zero map is a homomorphism. Namely, it depends on the category in which you are working. If a homomorphism of rings $f : R \to S$ happens to be the zero map, we get $1 = f(1) = 0$, thus $S = 0$. So yes, there is a zero homomorphism, but only when the target ring is trivial. In fact, something stronger is true: When $R = 0$, then also $S = 0$. One says that the zero ring is a strict initial object. In contrast to that, for all rngs $P,Q$, the zero map $P \to Q$ is a homomorphism.
By the way, it is quite interesting for rings $R,S$ to study homomorphisms $U(R) \to U(S)$; these correspond to a decomposition $S \cong S_1 \times S_2$ and a homomorphism $R \to S_1$  (Hint: The image of $1$ is idempotent). See also the answers at MO/34332, especially the one by James Borger is very enlightening.
A: 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.
A: If  $\phi : R \to R'$ is a ring homomorphism, then $\phi(1_R)$ is unity in $\phi(R)$.
