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 λ2Λ S inclusion (i.e. for every λ1; λ2 2 Λ; either λ1 ⊆ λ2 or λ2 ⊆ λ1). Let I = λ2Λ 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 2 R. Then a 2 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)λ2Λ 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 = fI E R : I is not principalg F = fI E R : I is not finitely generatedg 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 2 R such that xy 2 I but x2 = I , y2 = I. Let Ix = I + (x) and Iy = I + (y). Since x2 = I, I is strictly contained in Ix. Therefore Ix must be a principal ideal, say (α). Let J = fr 2 R : rIx ⊆ Ig. Note that J is an ideal of R that contains Iy. So J is also principal, say J = (β). Take any a 2 I. Since I ⊆ Ix = (α), a = rα for some r 2 R. As rIx = r(α) ⊆ I, r is in J. Thus a 2 IxJ. As a 2 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 2 R such that xy 2 I but x2 = I, y2 = 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 gener- ated 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.