TITLE: Why is Zero not composite?
QUESTION [7 upvotes]: I saw this question:

Vidya wonders why numbers one and zero are neither composite numbers nor prime numbers. Could you please explain to Vidya what can happen if they are either composite numbers or prime numbers? Use an example to punch your points.

Apart from the fact that I cannot comprehend how to answer, I am also unable to understand how zero is not composite.
A number $n$ is composite if $k\mid n$, where $1<k<n$. $k\mid n$ if $kx=n$. Since, $0k=0$ for all $k$, shouldn't $0$ be some sort of "hyper-composite"?
Is it? If not, why not?

REPLY [2 votes]: In more general stuctures, rings, we define units, irreducible elements and prime elements. (What follows is slightly modified quotes from Wolfram MathWorld.)
Unit

A unit in a ring is an element $u$ such that there exists $u^{-1}$ where $u \cdot u^{-1}=1$.

Irreducible Element

An element $a$ of a ring which is nonzero, not a unit, and whose only divisors are the trivial ones (i.e., the units and the products ua, where u is a unit). Equivalently, an element a is irreducible if the only possible decompositions of a into the product of two factors are of the form $a=u^{-1}ua$, 
  where $u^{-1}$ is the multiplicative inverse of $u$.

Prime Element

A nonzero and noninvertible element a of a ring $\mathbf R$ ... characterized by the condition that whenever $a$ divides a product in $\mathbf R$, $a$ divides one of the factors. The prime elements of $\mathbb Z$ are the prime numbers, $\mathbf P$.

Comments

In an integral domain, every prime element is irreducible, but the converse holds only in unique factorization domains. The ring $\mathbb Z[i \sqrt 5]$, where $i$ is the imaginary unit, is not a unique factorization domain, and there the element $2$ is irreducible, but not prime, since $2$ divides the product 
  $(1-i \sqrt 5)(1+i \sqrt 5)=6$, but it does not divide any of the factors.

In the ring of integers, all irreducible elements are also prime elements. 
Note that prime and irreducible elements cannot be zero or units.
So why omit zero?
Note that $0$ meets the definition of a prime number since, if $0$ divides $ab$, then zero divides $a$ or zero divides $b$.
But zero is not irreducible since every number divides $0$.
So if we want our prime numbers to also be irreducible, we must require that they not be equal to $0$.