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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 acommutative ring R is finitely generated, then R is Noetherian. That is if primeideals ofR are finitely generated, then every ideal ofR is finitely generated. An

analogue of this statement holds for PIDs: If every prime ideal of an integral domainR 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 ifI dont.

The argument I will use for the Noetherian rings is from Paolo Aluffis Alge-bra: Chapter 0(exercise V.3.15) and the argument for PIDs is from Dummit & FootesAbstract Algebra (exercise 8.2.6).

2. Lets go disco

We will use Zorns lemma for both proofs. We start with a a lemma heading tothis direction.

Lemma 1. Let R be a ring and let (I) be a chain of ideals with respect toinclusion (i.e. for every1, 2 ; either1 2 or2 1). LetI=

I.

Then we have:

(1) If none ofI are principal, thenI is not principal.(2) If none ofI are finitely generated, thenI is not finitely generated.

Proof. For (1), suppose to the contrary I= (a), for some a R. Then a I forsome. But thenI= (a) I. It followsI = I= (a), a contradiction.

We employ a similar strategy for (2). SupposeI= (a1, . . . , an). Since (I)is a chain, some I contains all a1, . . . , an. It follows I = I = (a1, . . . , an), acontradiction.

A straightforward application of Zorns lemma gives the following:

Proposition 2. LetR be a ring. Let

P={I R: I is not principal}

F={I R: I is not finitely generated}

Then,

(1) IfP =, Phas a maximal element (with respect to inclusion).(2) IfF =, Fhas a maximal element (with respect to inclusion).

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2 CIHAN BAHRAN

We will show that the maximal elements (w.r.t ) ofP and F are necessarilyprime. This gives us what we need.

Proposition 3. Let R be an integral domain that is not a PID. Then R has anon-principal prime ideal.

Proof. By Proposition 2, R has a maximal non-principal ideal I. Suppose I is notprime.

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, Iis strictly contained in Ix. ThereforeIx must bea principal ideal, say (). Let J={r R : rIx I}. Note that Jis an ideal ofRthat containsIy. SoJis also principal, say J= ().

Take any a I. SinceIIx= (), a = r for some r R. As rIx= r() I,r is in J. Thus a IxJ. As a Iwas arbitrary, we have I IxJ. On the otherhand, by the very definition ofJ we have Ix J. Thus I=IxJ= ()() = (),a contradiction toI being non-principal.

The argument for the Noetherian case is similar but a bit more involved.

Proposition 4. LetR be a non-Noetherian commutative ring. ThenR has a primeideal which is not finitely generated.

Proof. By Proposition 2, Rhas a maximal non-finitely-generated idealI. Note thatR/Iis a Noetherian ring. (This is because every nonzero ideal ofR/Iis of the formJ/I for some ideal J ofR which strictly contains I. ThereforeJmust be finitelygenerated, henceJ /I is finitely generated.)

Suppose I is not a prime ideal. So there existx, y R such that xy I butx /I, y / I. Let J1 = I+ (x) and J2 = I+ (y). Here we haveJ1J2 I, I J1andI J2. Consequently J1 andJ2 are finitely generated ideals ofR.

Consider the R-module J1/J1J2 and its submodule I/J1J2. Since IJ1 J1J2,I annihilates J1/J1J2. ThereforeJ1/J1J2 has also an R/I-module structure. NowsinceJ1 is a finitely generated ideal ofR,J1/J1J2is a finitely generatedR-module,hence also a finitely generatedR/I-module.

Finitely generated modules over Noetherian rings are Noetherian modules, thusJ1/J1J2 is a Noetherian R/I-module. So its submodule I/J1J2 is finitely gener-ated as an R/I-module, hence as an R-module. ButJ1J2 is also clearly a finitelygeneratedR-module, which forces I to be finitely generated, a contradiction.

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