Let

*R* be a

commutative integral domain. We say that

*R* is a

principal ideal domain (or PID) if every

ideal
of

*R* is generated by a single element (ie. has the form

*aR*).

It is easy to give examples of PIDs because of the next result.

**Theorem** Let *R* be a commutative integral domain
which is a Euclidean ring. Then *R* is a PID.

Before we prove this let us just observe that (after the Euclidean ring
writeup) this tells us that the ring of integers, the polynomial
ring over a field and the Gaussian integers are all PIDs.

*Proof of the theorem*.
Let *I* be an ideal of *R*. If I={0} then there is nothing to
prove, so we can assume that *I* contains a nonzero element.
Amongst all the nonzero elemsnts of *I* choose one, *a* say,
with the smallest norm. We will see that *a* generates *I*.

Let *b* be an element of *I*. Since *R* is a
ER (with norm *d*)
there exist *q,r* in *R* such that
*b=aq+r* and such that either *r=0* or
*d(r)<d(a)*. Since *I* is an ideal and *a,b* are elements
of *I* it follows that *r* is too. If *r* is nonzero
then its norm is strictly smaller than that of *a* (contradicting
the way we chose *a*) so if must be that *r=0*. In other
words *b=ar*. This proves that *I=aR*, as required.

**Lemma**
If *R* is a PID and *a* is irreducible in *R*
then *a* is prime.

*Proof*
Suppose *p|ab*. If *p|b*
we are done so suppose that *p* doesn't divide *b*. Consider
the ideal *I=pR+bR*. Since *R* is a PID there exists
*c* in *R* so that *I=cR*. Thus *p* is a multiple of
*c* and so is *b*. Since *p* is irreducible it must be that
either it is an associate of *c* or *c* is a unit. In the former
case this says that *p* divides *b* (contradicting our assumption).
In the latter case we can write
*1=pr+bs*. Thus *a=apr+abs*. Clearly then we see that
*p|a*, as needed.