Integral domain

by Patrick Stevens Jul 28 2016

An integral domain is a ring where the only way to express zero as a product is by having zero as one of the terms.

[summary: An integral domain is a ring in which the only way to make $~$0$~$ as a product is to multiply $~$0$~$ by something. For instance, in an integral domain like [48l $~$\mathbb{Z}$~$], $~$2 \times 3$~$ is not equal to $~$0$~$ because neither $~$2$~$ nor $~$3$~$ is.]

[summary(Technical): An integral domain is a ring in which $~$ab=0$~$ implies $~$a=0$~$ or $~$b=0$~$. (We exclude the ring with one element: that is conventionally not considered an integral domain.)]

In keeping with ring theory as the attempt to isolate each individual property of [48l $~$\mathbb{Z}$~$] and work out how the properties interplay with each other, we define the notion of integral domain to capture the fact that if $~$a \times b = 0$~$ then $~$a=0$~$ or $~$b=0$~$. That is, an integral domain is one which has no "zero divisors": $~$0$~$ cannot be nontrivially expressed as a product. (For uninteresting reasons, we also exclude the ring with one element, in which $~$0=1$~$, from being an integral domain.)


%%hidden(Show solution): Suppose $~$ab = 0$~$, but $~$a \not = 0$~$. We wish to show that $~$b=0$~$.

Since we are working in a field, $~$a$~$ has an inverse $~$a^{-1}$~$; multiply both sides by $~$a^{-1}$~$ to obtain $~$a^{-1} a b = 0 \times a^{-1}$~$. Simplifying, we obtain $~$b = 0$~$. %%


The reason we care about integral domains is because they are precisely the rings in which we may cancel products: if $~$a \not = 0$~$ and $~$ab = ac$~$ then $~$b=c$~$. %%hidden(Proof): Indeed, if $~$ab = ac$~$ then $~$ab-ac = 0$~$ so $~$a(b-c) = 0$~$, and hence (in an integral domain) $~$a=0$~$ or $~$b=c$~$.

Moreover, if we are not in an integral domain, say $~$r s = 0$~$ but $~$r, s \not = 0$~$. Then $~$rs = r \times 0$~$, but $~$s \not = 0$~$, so we can't cancel the $~$r$~$ from both sides. %%

Finite integral domains

If a ring $~$R$~$ is both finite and an integral domain, then it is a field. The proof is an exercise. %%hidden(Show solution): Given $~$r \in R$~$, we wish to find a multiplicative inverse.

Since there are only finitely many elements of the ring, consider $~$S = \{ ar : a \in R\}$~$. This set is a subset of $~$R$~$, because the multiplication of $~$R$~$ is closed. Moreover, every element is distinct, because if $~$ar = br$~$ then we can cancel the $~$r$~$ (because we are in an integral domain), so $~$a = b$~$.

Since there are $~$|R|$~$-many elements of the subset $~$S$~$ (where $~$| \cdot |$~$ refers to the Cardinality), and since $~$R$~$ is finite, $~$S$~$ must in fact be $~$R$~$ itself.

Therefore in particular $~$1 \in S$~$, so $~$1 = ar$~$ for some $~$a$~$. %%