# When to use Zorn's Lemma

I was looking at an exercise this morning which I was able to reduce to showing that the nilradical is the the intersection of the prime ideals in a ring -- a fact I remembered was true, but which I tried for a while to prove without success. Bucking my usual tendency to let something like that ruin the rest of my day for research, I dug out my copy of Atiyah and MacDonald and looked up the answer. (The idea is to assume that some non-nilpotent element f lies in every prime ideal, apply Zorn's lemma to the ideals which contain no power of f ordered by inclusion, and then show that the upper bound is prime.)

My reaction to this was something along the lines of, "Ah, I never would have got that, because I never would have tried using Zorn's lemma!" Upon further reflection, I realized that this indicated a serious weakness in my ability to do commutative algebra.

I'm perfectly comfortable using Zorn's lemma for something like showing that an arbitrary vector space has a basis, but when I look at a question like this I'm just not seeing the connection. I know that this doesn't really have a definite answer, but I was hoping that someone would be able to point out some kind of connection that would improve my intuition for when Zorn's lemma might be effective.

EDIT: Thanks to everyone for all the answers. They are all helpful and I had a hard time choosing!

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I'd say: try to use Zorn's lemma any time you have to build a ‘maximal’ object but have no canonical way of doing so. But I agree that its use is not quite so obvious in the exercise you quote... – Zhen Lin Sep 3 '11 at 15:43
Eisenbud remarks in his book that ideals maximal with respect to some property tend to be prime, and gives a few examples. This particular proof might be made more transparent via localization. – Dylan Moreland Sep 3 '11 at 15:50
A related question: How to use Zorn’s lemma, in Gowers's blog. – Pierre-Yves Gaillard Sep 3 '11 at 16:20
"My supervision partner solved the problem using Zorn’s lemma, which we had been told about in a lecture, and I just sat there in disbelief because it hadn’t even remotely occurred to me that Zorn’s lemma might be useful." Good to at least know I'm in the same boat as a future Fields medalist... – Daniel McLaury Sep 3 '11 at 17:49
Good question. Even, I was thinking about asking this here. You also use $\text{Zorn's lemma}$ for proving that a ring with $1$ has a maximal ideal. – user9413 Sep 3 '11 at 17:58

As Dylan Moreland hints at in a comment above, one way to think about your specific question (on nilradicals), which is very commutative algebraic in spirit, is to first localize your ring $A$ at the non-nilpotent element $f$. The problem then amounts to proving that the non-zero ring $A_f$ admits a prime ideal, and this follows from Zorn's lemma: any non-zero ring (with identity) has a maximal (and hence prime) ideal.

This result on the existence of maximal ideals is the standard use of Zorn's lemma in commutative algebra, akin to the existence of bases in linear algebra. If you would like to strengthen your commutative algebra, the solution is perhaps not so much to find a wider range of situations in which to apply Zorn's lemma, but rather to practice applying standard tricks such as localization, so as to find ways to put yourself into situations where this standard application of Zorn's lemma can be used.

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Perhaps this is an overly on-the-nose answer to what is really a more general question, but here goes:

You are given a non-nilpotent element $x$ of a commutative ring $R$ and need to find a prime ideal $\mathfrak{p}$ of $R$ such that $x \notin \mathfrak{p}$. So in particular you need to find a prime ideal of $R$! But without some form Zorn's Lemma / Axiom of Choice there are nonzero commutative rings with no prime ideals whatsoever. Indeed, the Boolean Prime Ideal Theorem is the assertion that every nonzero Boolean ring has a prime ideal, and this is well known to be strictly weaker than Zorn's Lemma / Axiom of Choice but not derivable from standard ZF set theory. See this answer by Chris Phan to a closely related MO question for more information on this point.

Here is a sketch of a better, more general answer: it is often the case that any ideal in a commutative ring which is maximal with respect to some given property $P$ is necessarily prime. Hand in hand with this goes the fact that ideals maximal with respect to $P$ exist, which is usually verified using Zorn's Lemma. The above is a special case of this where the maximality is with respect to exclusion of a certain multiplicatively closed subset. For more on this I highly recommend that you read about the Prime Ideal Principle of Lam and Reyes.

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An exposition of the result of Lam-Reyes can be found in 1.3 of people.fas.harvard.edu/~amathew/chspec.pdf, incidentally. To avoid making this look like pure advertising, let me point out some applications (shown there): one can show using the Lam-Reyes method that an ideal maximal with respect to being not principal is necessarily prime. So to check that a ring is a PIR, one just has to check on prime ideals. Another example is the fact that noetherianness can be checked, again, simply on prime ideals. I have no idea where these results have actually been used, but they are fun! – Akhil Mathew Sep 4 '11 at 3:11
@Akhil: when your product is free, I see no reason for qualms about advertising it! – Pete L. Clark Sep 4 '11 at 3:24
@Akhil: In your exposition, in the definition of an Oka family (Definition 1.15), shouldn't the condition be that $R \in F$ rather than $R \not \in F$? When you apply the Proposition 1.18, you don't want to get $R$ as a maximal not-in-$F$ ideal, so you need $R \in F$. Furthermore, in the examples you give (principal ideals, finitely generated ideals) $R$ is in the family. – Ted Sep 4 '11 at 3:58
@Ted: Thanks for pointing that out. I had made a mistake earlier in this section, which someone else corrected, but I apparently didn't correct it completely then. I've just done so. – Akhil Mathew Sep 4 '11 at 13:45

IMO, you rarely set out to do a problem and think "Oh, I'll use Zorn's lemma! If I can only figure out how...." Instead, the opportunity comes up naturally, and you just have to notice it.

For example, the proof that every vector space has a basis; everyone learns in their first linear algebra course the algorithm to prove a finite-dimensional vector space has a basis:

2. If your current set is a basis, you're done
3. Pick some element not in the span of your current set
5. Go to step 2

Since the vector space is finite dimensional, this algorithm stops and you get a basis.

Now, to prove that all vector spaces have bases, you don't have to do anything new -- you already know the key idea, and in this case you merely have to notice that this algorithm still terminates (but after transfinitely many steps).

Of course, I tend to invoke the well-ordering theorem in my proofs rather than Zorn's Lemma -- as you might guess from my example -- but the same idea can apply.

For the problem in your opening post, the key idea seems to be

Choose an ideal that doesn't contain a power of $x$

And to notice that any particular violation of primeness can be corrected by picking a bigger ideal. It's at this point you notice "Hey, Zorn's lemma lets me do that!" and invoke it to complete your proof.

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