# An ideal that is maximal among non-finitely generated ideals is prime.

I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over.

Let $R$ be a commutative ring with identity and let $U$ be an ideal that is maximal among non-finitely generated ideals of $R$. I wish to show that $U$ is a prime ideal.

Assume that $U$ is not prime. Let $x, y\not\in U$ be such that $xy\in U$. $U$ is contained in a maximal ideal $M$ and $xy\in M$, so either $x$ or $y$ is in $M$; assume $x\in M$. The condition $U\subset M$ then implies that there is a ring homomorphism $$\varphi: R/M\to R/U$$

Since $R/M$ is a field, $\varphi$ is injective. Hence, $\varphi(x)\in U$. This is a contradiction, so $U$ must be prime.

The thing that worries me is that I never explicitly used the hypothesis that $U$ was not finitely generated or the result that $M$ must be finitely generated.

-
Do you want the map to go the other way? You can't "un-mod out" :) – Dylan Moreland May 19 '12 at 1:06
Ah, you're right; the map's not well-defined this way. I'll have to try to see what falls out of the reverse map being a surjection now, if that's the right way to go about this problem. – Connor May 19 '12 at 1:09
You also meant, I suppose, "R/M is a field", not U – DonAntonio May 19 '12 at 1:21
Correct as well. I've changed it, though it's not of much importance now. – Connor May 19 '12 at 1:22
@acoustician By the way a corollary of this result is a result due to I.S. Cohen that if every prime ideal in a ring $A$ is finitely generated then $A$ is Noetherian. – user38268 May 19 '12 at 4:47

I propose the following: suppose $\,U\,$ is not prime, thus there exist $$\,x,y\in R \text{ such that }x,y\notin U\,,\,xy\in U\,.$$ Define now $B:=U+\langle y\rangle$.
By maximality of $\,U\,$ we have that $\,B\,$ is f.g., say $$\,B=\Bigl\langle u_i+r_iy\,,\,1\leq i\leq k\,,\,k\in\mathbb{N}\,\,;\,\,u_i\in U\,,\,r_i\in R\Bigr\rangle$$ and let now $$U_y:=\{s\in R\,\,;\,\,sy\in U\}.$$ (1) Check that $\,U_y\,$ is a proper ideal in $\,R\,$.
(2) Show that $\,U_y\,$ is f.g.
Put $\,U_y=\langle s_1,\ldots,s_m\rangle\,$, and take $\,u\in U\Longrightarrow\,\exists v_1,\ldots,v_k, t_1,\ldots,t_k\in R\,\,s.t.$$u=\sum_{n=1}^kv_nu_n+\sum_{n=1}^kt_nr_ny.$$ (3) Show that$\displaystyle{\sum_{n=1}^kt_nr_n}\in U_y.$(4) Putting$\,\Omega:=\{u_i,\ldots,u_k,ys_1,\ldots,ys_m\}\,$, derive the contradiction$\,U=\langle\Omega\rangle$. - I don't know how excited you might be about deeper results along these lines, and the question is completely answered already, but I can't resist mentioning some deeper results. It turns out there are a lot of results of "maximal-implies-prime" flavor (like "ideal maximal among non-principal ideals", "ideal maximal among non-countably generated", "maximal among point annihilators of a module", "ideal maximal among ideals disjoint from a multiplicative set"). For a long time, these were proven on an ad hoc basis, but Lam and Reyes managed to get them all (and apparently new ones!) in one fell swoop. In another paper their approach is used to generalize some classical results of Kaplansky and Cohen about properties of prime ideals propagating to all ideals. They are really fantastic papers, that I think anyone would enjoy. Here are several related papers by Reyes at arXiv: A prime ideal principle for two-sided ideals Noncommutative generalizations of theorems of Cohen and Kaplansky A one-sided Prime Ideal Principle for noncommutative rings Enjoy! - Thanks for the extra results! I'm actually a student at UCSD, so this is a pretty cool to discover. – Connor May 19 '12 at 2:26 @acoustician See also my answer here. – Bill Dubuque May 19 '12 at 2:54 +1: I was about to start looking for links to the Lam-Reyes paper. – Arturo Magidin May 19 '12 at 3:46 The links seem to be broken. – Leullame Jan 13 at 22:11 @Leullame Thanks: I was able to restore some with arXiv, but I could only find three of four. Perhaps I had two versions of the same thing previously... – rschwieb Jan 14 at 14:13 Can't we modify the answer Zev gave to the question to which Bill linked in his above comment? Let$x,y \notin U$. We want$xy \notin U$. Notice that$U + \langle x \rangle, U+ \langle y \rangle$are ideals properly containing$U$and are not contained in$\Sigma = \lbrace I \subset R \; | \; I\;\text{infinitely generated} \rbrace.$Thus,$U + \langle x \rangle, U+\langle y \rangle$are finitely generated, which implies$(U+\langle x \rangle) + (U+\langle y \rangle) = U+\langle x \rangle + \langle y \rangle$is finitely generated. But$U+\langle xy \rangle \subset U+\langle x \rangle + \langle y \rangle$, which implies$U+\langle xy \rangle$is finitely generated and is thus not contained in$\Sigma$. More importantly, we then know that$U \subsetneqq U+\langle xy\rangle$and so$xy \notin U$, which is what we wanted. - In the line just before the last one above you seem to be implying that an ideal contained in a f.g. generated ideal is itself f.g....I don't think this is true in general, but even if it were then the proof above would stop at$\,U\leq\,U+\langle x\rangle\,$, and the ideal in the right is f.g. If you deduce that$\,U+\langle xy\rangle\,$is f.g. because of another reason that I can't see I'd appreciate if you could tell which one. – DonAntonio May 19 '12 at 18:09 Yeah I don't think this is right either; wouldn't U+<xy>=U, so that it's still infinitely generated? – Connor May 19 '12 at 19:52 @DonAntonio You're absolutely right (I believe the polynomial ring over$\mathbb{Z}$in infinitely many indeterminates gives some examples of this phenomenon). I should've seen this error. Although, I wonder if there's a quick way to conclude that$U+\langle xy \rangle\$ is f.g., which would maintain the essence of this proof. Otherwise, I think you hit the nail on the head with your answer. – Derek Allums May 20 '12 at 1:44