# A confusion about Axiom of Choice and existence of maximal ideals.

The proof of the theorem of statement that every ring has a maximal ideal uses zorn's lemma or the axiom of choice.Now, the defintion of ring as well as the definition of maximal ideal don't depend on the axiom of choice.Is that true? So, given a ring I should be able to check whether it has maximal ideal or not without his axiom of choice. Am I right? So, how can the truth of this statement depend on Axiom of Choice? What is wrong with my assumptions?

To make my question more concrete, assuming that axiom of choice is false, can any one give me a ring which has no maximal ideal? How can a maximal ideal (assuming AoC is correct) turn into a non-maximal ideal just by assuming that AoC is false ?

PS. By rings I mean rings with identity.

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Enter @Asaf Karagila – Git Gud Feb 28 '13 at 17:34
LOL${}{}{}{}{}$ – Asaf Karagila Feb 28 '13 at 17:35
The definition of a family of sets as well as the definition of a choice function also don't depend on the axiom of choice... – Qiaochu Yuan Feb 28 '13 at 18:09
@Downvoter can you please explain the downvote? – Mohan Mar 2 '13 at 13:20

The axiom of choice is in fact equivalent to the assertion that every commutative unital ring has a maximal ideal.

Since the negation of the axiom of choice is as non-constructive as the axiom of choice itself, we can only say that there exists a commutative unital ring without a maximal ideal when the axiom of choice fails. The sets which are involved in this process are non well-orderable.

It is important to understand that much like the axiom of choice only assures that certain objects exist; its negation has a very similar intangible action: it only assures that some sets cannot be well-ordered, some partial orders in which every chain is bounded will not have maximal elements, and some families of non-empty sets do not have a choice function. We have no means of knowing where these objects are without further assumptions.

It is consistent that the axiom of choice holds for every set you have ever dreamed using, but then fails acutely. In that universe you cannot imagine to actually find that set which cannot be well-ordered, or that ring without a maximal ideal, and so on. But it is also consistent that the axiom of choice fails "nearby" and the counterexamples appear in relatively familiar sets (objects related to, or defined from the real numbers for example).

The best way to actually "find" a ring without a maximal ideal is to follow the proof of how the existence of maximal ideals imply the axiom of choice. These proofs often consists of taking a family of non-empty sets, defining some ring (or whatever) and using the maximal ideal to prove the existence of a choice function. Therefore starting with a family of non-empty sets which does not have a choice function guarantees that the process fails and that the ring defined in such proof will not have any maximal ideals.

Some point that came up in my comment exchange with Trevor Wilson under his answer, is that the axioms are syntactical. They allow us to write proofs. It seems to me, upon re-reading this question that you think something like:

Take a universe of ZFC, $R$ is a unital ring and $I$ is a maximal ideal. Now just don't assume that AC holds. Now we can't prove that $I$ is a maximal ideal.

That's false for two main reasons:

1. Once you fixed a universe of sets, the axiom of choice is either true or false in that universe. Even if you don't assume it, it has a truth value, and since we began from a universe of ZFC this truth value is indeed true. We just might not be able to write a proof from ZF that $I$ is maximal, but this is something that is still true in that universe of sets.

So changing the assumptions does not necessarily mean that you have changed your universe of sets.

2. If you have a definition for a ring (e.g. $\Bbb{R^R}$ with pointwise addition and multiplication) then the actual underlying set, and more importantly its subsets, may change between one universe of set theory and another. So for example we cannot prove from ZF that $\Bbb{R^R}$ has a maximal ideal because there are universes of set theory where this is false. But the underlying set $\Bbb{R^R}$ is very different between those universes, and again more importantly, its power set is different.

So to sum up all my answer (with its two additions), simply removing the assumption that the axiom of choice holds will not falsify the axiom of choice. It might be that you just won't be able to write a proof that there exists a maximal ideal in every unital ring.

If, however, you allow yourself to change the universe of set theory then it is possible that a certain ring will "lose" its maximal ideals, simply because we removed those sets from the universe (and in some cases did a whole revamp of the universe altogether).

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Thank you very much. I get it now. – Mohan Feb 28 '13 at 18:05
My pleasure!${}$ – Asaf Karagila Feb 28 '13 at 18:06

How can a maximal ideal (assuming AoC is correct) turn into a non-maximal ideal just by assuming that AoC is false ?

Changing your set of axioms does not change sets (nor does it leave them the same, for that matter.) It only makes sense to discuss the properties of sets once we have fixed axioms. Your question is analogous to the question "how can the pair $\{x,y\}$ disappear just by assuming that the Axiom of Pairing is false?" which also would make no sense. In both cases you have an axiom telling you that $\exists S\,P(S)$ is true where $P$ is some logical property of sets, and if you didn't have the axiom, then you couldn't prove the existence of such an $S$.

The issue with the Axiom of Choice is admittedly more subtle because it postulates the existence of sets without also giving a concrete description of those sets in the way that the Axiom of Pairing does. Indeed, it is impossible to give such a concrete description when you are dealing with infinite sets.

There is no general way to consider infinite sets as having an existence that is independent of one's choice of axioms for set theory. We are finite beings so we cannot just "write down" the set and "check" whether it has a given property. When we say something like "AC implies that the ring $R$ has a maximal ideal," it would be a mistake to believe that the ring $R$—and the collection of all ideals of $R$—already exist as concrete objects accessible to our senses, independently of our choice of axioms. If this were the case then indeed it would just be a matter of "checking", but it isn't the case.

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But definition of a maximal ideal doesn't use AoC.Given a ring and an ideal I can check whether that ideal is maximal or not without using AoC.So, if even if we assume that AoC is false, that ideal should remain maximal because while checking it I didn't use AoC. Right? Is there any flaw in my reasoning? – Mohan Feb 28 '13 at 17:56
@Mohan: The definition of a well-order doesn't use the axiom of choice either; but it impossible to prove that every set can be given such order without the axiom of choice. If your ring is a particular concrete set then you cannot make the maximal ideals disappear. If your ring is a set which is defined from some parameters "All the functions from $\Bbb R$ to itself" then by assuming the axiom of choice fails in a particular manner you have changed the contents of that collection. It is not the same ring anymore. – Asaf Karagila Feb 28 '13 at 18:00
@Mohan How do you 'check' whether an ideal is maximal? Its maximality is asserting the non-existence of a certain object (a larger ideal containing that ideal); in the case of a finite ring obviously you can check by simply iterating all of the subsets of the ring, but when you have an infinite ring how do you propose to 'check' maximality? – Steven Stadnicki Feb 28 '13 at 18:18
@Asaf I don't think it makes sense to say "you have changed the contents of that collection." Formally speaking, infinite sets do not have an existence outside of the context of a fixed set of axioms. Changing the axioms just changes what you can prove about sets. – Trevor Wilson Feb 28 '13 at 18:19
@ Steven For example in a pid I can prove that every maximal ideal is generated by an irreducible element. – Mohan Feb 28 '13 at 18:21

Yes, the existence of maximal ideals in nontrivial commutative rings can be formulated in ZF, but it cannot be proven in ZF! Actually in ZF this assertion is equivalent to the axiom of choice. See A primrose path from Krull to Zorn. The existence of prime ideals is weaker, it is equivalent to the ultrafilter principle.

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Yes, the definition of maximal ideal does not depend on the axiom of choice.

Even assuming the axiom of choice is true, there are rings (necessarily without identity) which do not have maximal ideals. There is a link to an example of such a ring in this solution: http://math.stackexchange.com/a/258076/29335

Since it turns out that the AC is in fact equivalent to all rings with identity having a maximal ideal, assuming the negation of the AC would allow us to conclude that there is a ring with identity which does not have a maximal ideal.

In either case, there are examples of rngs without maximal ideals.

It sounds a little like you're thinking that "if you assume the negation of the AC, then maximal ideals don't exist" but that is not the right negation. The right negation is: "There is a ring with identity which doesn't have a maximal ideal." No matter if you assume the AC or its negation, finite rings always will have maximal ideals, for example.

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"there is a ring with identity which does not have a maximal ideal" do we know what is that ring? – Mohan Feb 28 '13 at 17:42
@Mohan No, it's as Asaf says: the proof of equivalence does not give us a construction. – rschwieb Feb 28 '13 at 17:45

The issue is does there exist any proof of the existence of a maximal ideal in a ring without the use of Zorn's lemma or some other equivalent of the axiom of choice. I am not aware of any.

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