# How many homomorphisms are there of $\Bbb Z \times \Bbb Z\times \Bbb Z$ into $\Bbb Z$?

I've tried looking around for an explanation to this problem, but I've been having trouble finding a clear solution that specifically focuses on this question:

How many homomorphisms are there of $\Bbb Z \times \Bbb Z\times \Bbb Z$ into $\Bbb Z$?

I'm studying abstract algebra and we have just begun discussing rings. Thank you in advance for your help. I would appreciate it if you could give me step by step help for this problem. This is not for homework- I just need to figure out how to do this problem to prepare for my exam. Thanks!

Edit: Here is a post that addressed the question, but did not answer it fully: Describe all ring homomorphisms

• Are you regarding $\mathbb{Z}^3$ as a module, or as a ring? – goblin Apr 5 '16 at 3:24
• See the above edit- the question comes from a series of questions about ring homomorphisms, but this part of the question is left kind of ambiguous. I guess that's part of my question, too. – asoiaf_obsession_91 Apr 5 '16 at 3:32
• A ring, not a module. – asoiaf_obsession_91 Apr 5 '16 at 3:37
• One should never forget that a $\Bbb Z$-linear map $f:\Bbb Z^n\to\Bbb Z^m$ is given by a matrix of size $m\times n$ with integer coordinates, and uniquely so. – Pedro Tamaroff Apr 5 '16 at 4:00

## 2 Answers

Assuming the question is about ring homomorphisms, there are $3$. Here is how to see this very explicitly.

Write $e_1 = (1, 0, 0), e_2 = (0, 1, 0), e_3 = (0, 0, 1)$. These three elements form a complete set of orthogonal idempotents. This means that they satisfy $e_i e_j = \delta_{ij} e_j$ and $\sum e_i = 1$, or more explicitly

$$e_i^2 = e_i, e_i e_j = 0 \text{ if } i \neq j, e_1 + e_2 + e_3 = 1.$$

The significance of these conditions for us is that they must be preserved by ring homomorphisms. Hence if $f : \mathbb{Z}^3 \to \mathbb{Z}$ is a ring homomorphism, $f(e_i)$ is a complete set of orthogonal idempotents in $\mathbb{Z}$. It's clear that $f$ is determined by the $f(e_i)$.

But $\mathbb{Z}$ only has two idempotents, namely $0$ and $1$. The condition that $\sum f(e_i) = 1$ means exactly one of the $f(e_i)$ is equal to $1$, and then orthogonality is automatic. Hence there are $3$ homomorphisms corresponding to whether $f(e_1), f(e_2)$, or $f(e_3)$ is $1$; these are the $3$ projection maps $\mathbb{Z}^3 \to \mathbb{Z}$.

More generally, homomorphisms $\mathbb{Z}^n \to \mathbb{Z}^m$ correspond to functions $m \to n$.

The algebraic significance of complete sets of orthogonal idempotents in a commutative ring $R$ is that they correspond to decompositions of $R$ as a finite product $\prod_i R e_i$ of rings. The geometric significance is that they correspond to decompositions of $\text{Spec } R$ into a finite disjoint union $\coprod_i \text{Spec } R e_i$ of affine schemes. Here $\text{Spec } \mathbb{Z}^n$ decomposes into $n$ copies of $\text{Spec } \mathbb{Z}$.

• Good answer! $\;$ – goblin Apr 5 '16 at 3:44
• Thank you! I think this is what I was looking for. – asoiaf_obsession_91 Apr 5 '16 at 3:54
• The condition that $\sum f(e_i)=1$ alone actually tells you exactly one of the $f(e_i)$ is $1$. – Eric Wofsey Apr 5 '16 at 3:57
• @Eric: whoops. Yes, of course. – Qiaochu Yuan Apr 5 '16 at 5:09
• I don't quite understand the $Re_i$ thing in the last paragraph. Does this just mean $\prod_{i \in I} R$? – goblin Apr 5 '16 at 13:11

(I'm assuming that the question is about module homomorphisms, not ring homomorphisms.)

Adam is right. In more detail:

Definition/Proposition.

Suppose we're given

• a commutative ring with unity, call it $R$,
• an $R$-module, call it $X$,
• a finite set $S$.

Then there's a natural bijective correspondence between:

• homomorphisms $X \leftarrow R^S$ and
• functions $X \leftarrow S.$

given as follows:

Firstly, assign to each $s \in S$ a vector $e_s \in R^S$ writing by $e_s(s') = [s=s']$, where $[]$ is the Iverson bracket.

$(\Rightarrow)$ Given a homomorphism $h : X \leftarrow R^S$, we get a function $F(h) : X \leftarrow S$ defined as follows: $$F(h)(s) = h(e_s).$$

$(\Leftarrow)$ Given a function $f : X \leftarrow S$, we get a homomorphism $H(f) : X\leftarrow R^S$ by writing: $$H(f)(a) = \sum_{s \in S} a(s)f(s)$$

(The above sum makes sense because $S$ is finite.)

To prove this:

Step 0. Check that $H(f)$ really is a homomorphism.

Step 1. Check that these processes are inverse to each other. That is:

$$H(F(h)) = h, \qquad F(H(f)) = f$$

Step 2. (Optional). If you know some category theory, you can also go ahead and try to prove that $H$ and $F$ are natural transformations.

• Although there are moral reasons to write down arrows in functions the way you do, there are social reasons one would like to avoid it. Is it really a notation that carries a significant advantage? – Pedro Tamaroff Apr 5 '16 at 4:00
• @PedroTamaroff, I really haven't decided yet. Sometimes I flip back and forth between them. The thing that made me use the backward $R \leftarrow S$ here is the agreement with the $R^S$ notation. But anyway, I don't really care about the social reasons, or tradition. As a general rule, I think we take these kinds of things altogether too seriously. What I want is to settle on a truly optimal system of conventions to hand to the next generation. – goblin Apr 5 '16 at 13:08