Why the integers form an integral domain?

Question

This is the definition of integral domain

A ring that is commutative under multiplication, has a multiplicative identity element, and has no divisors of 0.

0 is an integer. 0 can divide any integers. Namely, 0 has infinitely many divisors. So why integers form an integral domain?

Answer 1

Yes, zero does in fact have infinitely many divisors. By using the term $\textit{zero divisors}$, they mean non-zero elements $x$ and $y$ so that $xy=0.$ If the only time that $xy=0$ is when either $x$ or $y$ is zero, then the ring has no zero divisors. This is true in $\mathbb{Z}$.

For an example of a ring with zero divisors, take the integers modulo six. Neither $2$ nor $3$ equals zero there, yet $2 \times 3 \equiv 0$ modulo six.

Answer 2

First note that if $m<n$ are natural numbers then there exists a positive natural number $d$ such that $n=m+d$. Indeed, the set $A=\{x\in\mathbb{N}:n\leqslant m+x\}$ is nonempty and so by the well-ordering principle of natural numbers, it has the least element $d$. Clearly $d$ is positive. We know that every positive natural number is the successor of a natural number. So $d=y^{+}=y+1$ for some $y\in\mathbb{N}$. But $y\notin A$ and so $m+y<n$. This shows that $m+d=m+y+1\leqslant n$. Hence, $m+d=n$.

We know that the ring of integers $\mathbb{Z}$ is indeed the Grothendieck ring of the semiring of natural numbers $\mathbb{N}$. More precisely, we first define an equivalence relation over the set $\mathbb{N}\times\mathbb{N}$ as $(m,n)\sim(m',n')$ if $m+n'=m'+n$. We denote the equivalence class containing of an ordered pair $(m,n)$ simply by $[m,n]$. We also denote by $\mathbb{Z}=\{[m,n]: m,n\in\mathbb{N}\}$ the set of all equivalence classes obtained by this relation. The set $\mathbb{Z}$ by the addition $[m,n]+[m',n']=[m+m',n+n']$ is an Abelian group. Indeed, $[0,0]$ is the additive identity (it is called the zero integer and is often denoted by $0$). The additive inverse of each $[m,n]\in\mathbb{Z}$ is the element $[n,m]$. For each positive natural number $m$ ($0<m\in\mathbb{N}$), the element $[m,0]\in\mathbb{Z}$ is called a positive integer (and it is often denoted by $+m$ or simply by $m$). Its additive inverse, namely $[0,m]$, is called a negative integer (and it is often denoted by $-m$). Then it is clear that $[m,n]=m-n$ for all $m,n\in\mathbb{N}$. The Additive group of integers $\mathbb{Z}$ with the multiplication $[m,n]\cdot[m',n']=[mm'+nn',mn'+m'n]$ is a commutative ring. The element $[1,0]$ is the multiplicative identity of this ring.

Now we show that $\mathbb{Z}$ is an integral domain. Clearly $\mathbb{Z}$ is a nonzero ring. In fact, the map $\mathbb{N}\rightarrow\mathbb{Z}$ given by $m\mapsto[m,0]$ is an injective morphism of semirings. Now assume $x=[m,n]$ and $y=[m',n']$ are two integers with $xy=0$ where $m,n,m',n'\in\mathbb{N}$. If $x\neq0$ then $m\neq n$ and so $m<n$ or $n<m$. If $m<n$ then there exists a positive natural number $d$ such that $n=m+d$. This yields that $x=[m,m+d]=[0,d]$. Now from $xy=0$ we get that $dm'=dn'$. But $d>0$ and so $m'=n'$. This shows that $y=0$. We will have the same if $n<m$.