# Integral domains and characteristics

Today we shall talk about the characteristic of an integral domain, concentrating mainly on misconceptions and important points.

An integral domain is a commutative ring with the property that if $a\neq 0$ and $b\neq 0$, then $ab\neq 0$. Hence, if $ab=0$, then $a=0$ or $b=0$ (or both).

The characteristic of an integral domain is the lowest positive integer $c$ such that $\underbrace{1+1+\dots +1}_{\text{ c times}}=0$.

Let $a\in R$. Then $\underbrace{a+a+\dots +a}_{\text{ c times}}=a\underbrace{(1+1+\dots +1)}_{\text{ c times}}=0$. This is because $a.0=0$.

If $\underbrace{a+a+\dots +a}_{\text{ d, then we have $a\underbrace{(1+1+\dots +1)}_{\text{ d times}}=0$. This is obvious for $a=0$. If $a\neq 0$, then this implies $\underbrace{1+1+\dots +1}_{\text{ d times}}=0$, which contradicts the fact that $c$ is the lowest positive integer such that $1$ added $c$ times to itself is equal to $0$. Hence, if $c$ is the characteristic of the integral domain $D$, then it is the lowest positive integer such that any non-zero member of $D$, added $c$ times to itself, gives $0$. No member of $D$ can be added a lower number of times to itself to give $0$.

Sometimes $\underbrace{a+a+\dots +a}_{\text{ c times}}$ is written as $ca$. One should remember that this has nothing to do the multiplication operator in the ring. In other words, this does not imply that $\underbrace{a+a+\dots +a}_{\text{ c times}}=c.a$, where $c$ is a member of the domain. In fact, $c$ does NOT have to be a member of the domain. It is just an arbitrary positive integer.

Now on to an important point: something that is not emphasized, but should be. Any expression of the form $\underbrace{\underbrace{a+a+\dots +a}_{\text{m times}}+\underbrace{a+a+\dots +a}_{\text{m times}}+\dots +\underbrace{a+a+\dots +a}_{\text{m times}}}_{\text{n times}}=\underbrace{(a+a+\dots +a)}_{\text{m times}}(\underbrace{1+1+\dots +1}_{\text{n times}})$.

Now use this knowledge to prove that the characteristic of an integral domain, if finite, has to be $0$ or prime.