CHARACTERIZATION of PIDS and NOETHERIAN RINGS with RESPECT to PRIME IDEALS 1. Introduction Most Algebra Textbooks Have an Exerci

CHARACTERIZATION OF PIDS AND NOETHERIAN RINGS WITH RESPECT TO PRIME IDEALS

CIHAN˙ BAHRAN

1. Introduction Most algebra textbooks have an exercise to show that if every prime ideal of a commutative ring R is finitely generated, then R is Noetherian. That is if prime ideals of R are finitely generated, then every ideal of R is finitely generated. An analogue of this statement holds for PIDs: If every prime ideal of an integral domain R is principal, then R is a PID. These are nice and memorable results, but the arguments to prove them are not easy to remember (at least for me). This paper will always remember them even if I don’t. The argument I will use for the Noetherian rings is from Paolo Aluffi’s Alge- bra: Chapter 0 (exercise V.3.15) and the argument for PIDs is from Dummit & Foote’s Abstract Algebra (exercise 8.2.6).

2. Let’s go disco We will use Zorn’s lemma for both proofs. We start with a a lemma heading to this direction.

Lemma 1. Let R be a ring and let (Iλ) be a chain of ideals with respect to λ∈Λ S inclusion (i.e. for every λ1, λ2 ∈ Λ; either λ1 ⊆ λ2 or λ2 ⊆ λ1). Let I = λ∈Λ Iλ. Then we have:

(1) If none of Iλ are principal, then I is not principal. (2) If none of Iλ are ﬁnitely generated, then I is not ﬁnitely generated.

Proof. For (1), suppose to the contrary I = (a), for some a ∈ R. Then a ∈ Iλ for some λ. But then I = (a) ⊆ Iλ. It follows Iλ = I = (a), a contradiction. We employ a similar strategy for (2). Suppose I = (a1, . . . , an). Since (Iλ)λ∈Λ is a chain, some Iλ contains all a1, . . . , an. It follows Iλ = I = (a1, . . . , an), a contradiction. A straightforward application of Zorn’s lemma gives the following: Proposition 2. Let R be a ring. Let

P = {I E R : I is not principal} F = {I E R : I is not ﬁnitely generated} Then, (1) If P= 6 ∅, P has a maximal element (with respect to inclusion). (2) If F 6= ∅, F has a maximal element (with respect to inclusion). 1 2 CIHAN˙ BAHRAN

We will show that the maximal elements (w.r.t ⊆) of P and F are necessarily prime. This gives us what we need. Proposition 3. Let R be an integral domain that is not a PID. Then R has a non-principal prime ideal. Proof. By Proposition 2, R has a maximal non-principal ideal I. Suppose I is not prime. Then there exist x, y ∈ R such that xy ∈ I but x∈ / I , y∈ / I. Let Ix = I + (x) and Iy = I + (y). Since x∈ / I, I is strictly contained in Ix. Therefore Ix must be a principal ideal, say (α). Let J = {r ∈ R : rIx ⊆ I}. Note that J is an ideal of R that contains Iy. So J is also principal, say J = (β). Take any a ∈ I. Since I ⊆ Ix = (α), a = rα for some r ∈ R. As rIx = r(α) ⊆ I, r is in J. Thus a ∈ IxJ. As a ∈ I was arbitrary, we have I ⊆ IxJ. On the other hand, by the very definition of J we have Ix ⊆ J. Thus I = IxJ = (α)(β) = (αβ), a contradiction to I being non-principal. The argument for the Noetherian case is similar but a bit more involved. Proposition 4. Let R be a non-Noetherian commutative ring. Then R has a prime ideal which is not finitely generated. Proof. By Proposition 2, R has a maximal non-finitely-generated ideal I. Note that R/I is a Noetherian ring. (This is because every nonzero ideal of R/I is of the form J/I for some ideal J of R which strictly contains I. Therefore J must be finitely generated, hence J/I is finitely generated.) Suppose I is not a prime ideal. So there exist x, y ∈ R such that xy ∈ I but x∈ / I, y∈ / I. Let J1 = I + (x) and J2 = I + (y). Here we have J1J2 ⊆ I, I ( J1 and I ( J2. Consequently J1 and J2 are finitely generated ideals of R. Consider the R-module J1/J1J2 and its submodule I/J1J2. Since IJ1 ⊆ J1J2, I annihilates J1/J1J2. Therefore J1/J1J2 has also an R/I-module structure. Now since J1 is a finitely generated ideal of R, J1/J1J2 is a finitely generated R-module, hence also a finitely generated R/I-module. Finitely generated modules over Noetherian rings are Noetherian modules, thus J1/J1J2 is a Noetherian R/I-module. So its submodule I/J1J2 is finitely generated as an R/I-module, hence as an R-module. But J1J2 is also clearly a finitely generated R-module, which forces I to be finitely generated, a contradiction.

Department of Mathematics, Bilkent University, Bilkent, Ankara 06800, Turkey