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.