Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
I'll post my comment again: the zero map is a ring homomorphism iff the target is the zero ring, in which $1 = 0$. –  Dylan Moreland Jul 1 '12 at 23:34
@Dylan: Using Wikipedia's definition, it seems required to have the multiplicative identity in $R$ to be mapped to the multiplicative identity in $S$, not necessarily the multiplicative identity in the image of $R$, i.e., the zero ring. –  Riem Jul 1 '12 at 23:36
Also, I don't think there's a well-defined notion of "zero maps" between manifolds. Perhaps "linear transformation between vector spaces" would be a better example. –  Adam Saltz Jul 1 '12 at 23:49
You can't add two ring homomorphisms to get another one (unlike continuous real-valued functions or linear transformations between vector spaces where you can), so leaving out the zero map is not that strange. I'm sure this was mentioned in an earlier question on the same topic, but I can't seem to find it right now. –  Ted Jul 2 '12 at 7:28
show 3 more comments

4 Answers

up vote 5 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
@GenericHuman: $e=\phi(e)=\phi(x\cdot x^{-1})=\phi(x)\cdot\phi(x^{-1})$ and the same logic applies for $\phi(x^{-1})\cdot\phi(x)=e$. Hence, $\phi(x^{-1})=\phi(x)^{-1}.$ –  Riem Jul 2 '12 at 2:14
I don't know what I was thinking, sorry for the noise :-) –  Generic Human Jul 2 '12 at 2:15
@Hurkyl: The third property only implies $\phi(1)=1$ when the target group is non-trivial, and the same applies for ring homomorphisms. That is, suppose we have unital rings and a ring homomorphism where $\phi(1)\neq1$. Then $\phi(a)=\phi(a\cdot1)=\phi(a)\cdot\phi(1)\neq\phi(a)$. Hence, the $\phi(1)=1$ requirement is also implicit in the case of (non-trivial) rings. So then the question is: Why include $\phi(1)=1$ for ring homomorphisms, but not $\phi(1)=1$ for group homomorphisms? It seems adding the $\phi(1)=1$ requirement is made to specifically exclude the zero map. –  Riem Jul 2 '12 at 2:19
@Riem: Also, I don't understand your comment about the trivial group. The argument that $\phi(1) = 1$ is implied by the third property and the group axioms works for every group. A one-line version is $$1 = \phi(1) \phi(1)^{-1} = \phi(1 \cdot 1) \phi(1)^{-1} = \phi(1) \phi(1) \phi(1)^{-1} = \phi(1).$$ And we additionally have the fact that for every function from a group to the trivial group, we have $\phi(1) = 1$ (because $1$ is the only element of the target group). –  Hurkyl Jul 2 '12 at 2:57
Indeed; I think it's better to consider the case of monoids; if we want rings to be monoids under multiplication, then we must want ring homomorphisms to be monoid homomorphisms when restricted to the multiplicative structure. –  Arturo Magidin Jul 2 '12 at 3:57
show 2 more comments

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.

share|improve this answer
What may be a matter of convention is whether the objects called "rings" are required to have a unity or not. The nomenclature "rngs" is, although amusing, not universally accepted (or even necessarily accepted by a majority of algebraists). I certainly agree that if you require your rings to have a unity, then you must require your homomorphisms to preserve it. –  Arturo Magidin Jul 2 '12 at 16:09
add comment

If $\phi : R \to R'$ is a ring homomorphism, then $\phi(1_R)$ is unity in $\phi(R)$.

share|improve this answer
This only explains Dylan's comment: "the zero map is a ring homomorphism iff the target is the zero ring, in which $1=0$." But why doesn't the conventional definition of a ring homomorphism include the zero map in general? In my opinion at least, it would only make sense for this to be the case. –  Riem Jul 2 '12 at 0:23
Well, that's news to me. I thought you could have the zero homomorphism between any two rings. proofwiki.org/wiki/Definition:Zero_Homomorphism –  Vectk Jul 2 '12 at 0:31
@Lloyd: Notice that proofwiki.org/wiki/Definition:Ring_(Abstract_Algebra) takes the somewhat unusual approach of leaving 1 out of the definition of a ring, so the zero map is a homomorphism. –  Ben Millwood Jul 2 '12 at 3:15
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.