# “Instructive” proof of “If I is maximal among ideals not …, then I is prime”

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)?

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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 1:55

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