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In this question all rings are commutative with identity.

Consider the following well-known statement:

(*) Let $R$ be a ring and $S$ a multiplicatively closed subset of $R$. Suppose $I$ is an ideal of $R$ maximal among those not meeting $S$. Then $I$ is prime.

There are two easy proofs of this:

(1) The direct method, i.e. a proof along the lines "Suppose every ideal properly containing $I$ meets $S$. Let $ab \in I$. Suppose $a, b \notin I$. Then $(I, a)$ meets $S$, i.e. $s = x + at$ for some $s \in S$, $x \in I$, $t \in R$. Similarly $s' = y + bt'$. Then $ss' = (x + at)(y + bt') = xy + xbt' + yat + abtt' \in S \cap I$, whence I meets $S$.

(2) Construct the ring $S^{-1}R$ and establish the description of all ideals of $R$. The result follows from the fact that $R/I \rightarrow S^{-1}R/S^{-1}I$ is an injection.

I find proof (2) much more appealing, because it is (a) far more informative, and (b) easier to remember [although in this case making up a new proof is easy enough].

Now on to the "meat" of my question: it seems to me that there are many more statements similar to (*), i.e. following this pattern:

(**) Let $I$ be an ideal maximal among those not satisfying property $P$. Then $I$ is prime.

Famous examples of such properties $P$ are "principal" and "finitely generated".

Are there proofs of these (and similar) facts which are "instructive", in the sense of being similar to proof (2) above rather than (1)?

Tom Bachmann
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  • Are you sure that (**) is true for $P$ being "principal"? – George Lowther Sep 20 '11 at 22:44
  • Consider the ring $R$ of analytic functions on the complex plane, and $I$ the ideal of $f\in R$ vanishing at $0$ and $1$. This is not prime, as $X.(1-X)\in I$ but $X,1-X\notin I$. But it is maximal among nonprincipal ideals (the only larger ideals being $R$, $(X)$ and $(1-X)$. – George Lowther Sep 20 '11 at 22:46
  • You can look at the ring $R/I$ to reduce to the case where $I=0$, so you reduce to proving to something of the form "If every nonzero ideal of $R$ satisfies property $P$ then $R$ is an integral domain." – George Lowther Sep 20 '11 at 22:52
  • But $I$ is principal, generated by $X(1-X)$, isn't it? – Tom Bachmann Sep 20 '11 at 23:25
  • Sorry, yes, it is principal. – George Lowther Sep 21 '11 at 01:55

1 Answers1

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Yes, for some very interesting general viewpoints see Lam and Reyes: Oka and Ako Ideal Families in Commutative Rings, and Anderson; Dobbs; and Zafrullah: Some applications of Zorn's lemma in algebra.

Bill Dubuque
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  • This seems more like a very streamlined proof of type (1) to me than a type (2) proof. On the other hand on this level of generality the proof certainly is istructive, so thanks for your answer. – Tom Bachmann Sep 21 '11 at 12:05
  • @Tom It's difficult to respond not knowing what precisely that "this" refers to. – Bill Dubuque Sep 21 '11 at 14:57
  • I'm referring to to the paper by Lam and Reyes, i.e. to proving the sort of statement my question is about by exhibiting a general principle (namely: "if F is an Ako or Oka family, then every maximal wrt not meeting F is prime"). To me this is "essentially the same" as the proof (1) I gave, just "executed much more professionally" (i.e. we isolate the particular properties common to many families that make the proofs work, and then formulate a general result that has all particular statements of interest as easy corollaries). – Tom Bachmann Sep 24 '11 at 12:39
  • My (surely somewhat illusional) hope was that maybe there are proofs that tell more about why the results are true (I know this is not a well-defined notion...). Perhaps as in proof (2), we might construct an auxiliary ring $R'$ where $I$ corresponds to an obviously prime ideal $I'$, and then relate the two quotients $R/I$ and $R'/I'$. – Tom Bachmann Sep 24 '11 at 12:42
  • Your first link rotted. – rschwieb Apr 10 '19 at 15:28