## Principal Ideal Domains

Previously, I defined a principal ideal domain to be a ring in which every ideal is principal (note to newcomers: the best way to get the hang of these terms is to browse the category “mathematics”). Now I’m going to show that every PID is a Dedekind domain with unique factorization, and vice versa.

Recall that a unique factorization domain is defined by two properties: every irreducible element is prime, and every ascending chain of principal ideals, (a1), (a2), (a3)… where (a(n)) is a subset of (a(n+1)), terminates, i.e. for some n, (a(m)) = (a(n)) for all m > n.

Now, a principal ideal domain is Noetherian. A Noetherian domain is one where every ideal has finitely many generators, and in a principal ideal domain, every ideal has in fact just one generator. So every ascending chain of ideals, principal or not, terminates.

Also, if an ideal (b) contains (a), then we can write a as br for some r in the ring, so b divides a. If a is irreducible, and if (b) contains (a), then either b is associated to a, and (a) = (b), or b is a unit, and (b) is the entire ring; therefore, the ideal (a) is maximal. Hence (a) is prime, which means that a is prime. This proves that every PID is a UFD.

In addition, in a PID, if (a) is prime, then a is prime, which means that a is irreducible, and (a) is maximal. This proves that a PID is Noetherian and satisfies the condition that every prime ideal is maximal.

To show that every PID is integrally closed, I’ll show that every UFD is integrally closed. If c/d is an element of the fraction field of a UFD, and c/d is integral over the ring, then it’s the root of a polynomial of the form x^n + a(n-1)x^(n-1) + … + a2x^2 + a1x + a0.

Now, we have (c/d)^n + a(n-1)(c/d)^(n-1) + … + a2(c/d)^2 + a1(c/d) + a0 = 0. Multiplying both sides by d^n, we get c^n + a(n-1)cd^(n-1) + … + a2d^(n-2)c^2 + a1d^(n-1)c + a0d^n = 0.

Every irreducible factor of d divides a(n-1)cd^(n-1) + … + a2d^(n-2)c^2 + a1d^(n-1)c + a0d^n, so it must divide c^n. Since we’re in a UFD, that factor is prime, so it divides c. We can then cancel out that factor from c/d and look at c1/d1 instead, where cd1 = dc1. By a similar consideration, c1 and d1 have an irreducible factor in common, so canceling, we get c2/d2. Since d has only a finite number of irreducible factors, say n, d(n) will be a unit, and c/d = c(n)/d(n) is an element of the ring.

This shows that if c/d is an element of the fraction field of the ring that is integral over the ring, it’s in the ring. In other words, the ring is integrally closed. We can do this with every UFD, so we get PID –> UFD –> integrally closed.

The other direction, (UFD and Dedekind) –> PID, has two proofs, an annoyingly long one from first principles, and a shorter one from algebraic number theory. The short one is that if I is an ideal of a Dedekind domain, then we can look at its inverse, the fractional ideal I^(-1). By the definition of a fractional ideal, we can find some r such that rI^(-1) is an integral ideal.

We have (rI^(-1))I = (r). r uniquely factors into irreducibles, p1p2…p(n), so we have (r) = (p1)(p2)…(p(n)). Every irreducible is prime, so all of these elements are prime; in particular, all of these ideals are prime. Since factorization into ideals is unique, I is a product of a subset of these prime ideals. But the product of principal ideals is principal, since for example (a)(b) = (ab), so I is itself principal.

Next time, I’m finally going to show that all of these weird creations – PIDs, UFDs, and the likes – really exist by giving algebraic examples.