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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.

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