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An algebraic structure such as a group, ring, field, etc. is usually defined to be a set with some operations satisfying certain properties. I am curious what, if anything, goes wrong when the underlying collection of elements is not a set. I don't have any particular example in mind, but I know some exist (in particular the surreals). I'm just asking out of curiosity and I'm not sure where one would look up such a thing -- I am not a set theorist or logician by any stretch.

For a concrete question: what are the most elementary algebra results (on the level of Dummit and Foote, say) which can fail for groups or rings with "too many elements"? Perhaps things like the existence of maximal ideals in commutative rings?

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With which foundation of mathematics do you want to work? ZF? NBG? ZF + universes? – Martin Brandenburg Sep 27 '10 at 15:57
For example when you work in ZF and talk about classes, you cannot quantify over classes and notions such as "maximal ideal" become undefined. But when you think in universes, well, then everything goes through one level higher. – Martin Brandenburg Sep 27 '10 at 15:58

In the current mainstream view of set theory (Zermelo-Fraenkel set theory), the only objects that we can talk about (quantify over) are sets. Informally, a class is a family (collection?) of sets that is defined by a formula $\phi(x,y_1,\dots,y_n)$ with sets $b_1,\dots,b_n$ as parameters. In this case we would denote the class by $\{a:\phi(a,b_1,\dots,b_n)\}$.

Simple example: Let $b$ be a set and consider the class $\{a:b\in a\}$, the class of all sets that contain $b$. Here $\phi(x,y)$ is $y\in x$ and the parameter is $b$.

Now, while it can happen that a certain class is a ring (with the addition and multiplication also being classes), we have no way of speaking about all such classes (in the language of set theory), since we cannot quantify over the formulas of our language within in the language.

Given a fixed formula $\phi(x,y_1,\dots,y_n)$, we can talk about all classes of the form $\{a:\phi(a,b_1,\dots,b_n)\}$ with $b_1,\dots,b_n$ sets by quantifying over the parameters $b_1,\dots,b_n$, but we cannot talk about all rings that are classes, since they will not all have a uniform representation as a class in this sense.

This is the main problem. I believe that all statements about rings that concern the internal arithmetic laws of the ring still go through for rings that are classes, but you get problems with properties that require speaking about the relation of the ring in question to other rings (universal properties etc, but also subrings and ideals as mentioned by Martin Brandenburg in a comment.).

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Nevertheless, the following well-formed meta-theorem may be seen as a generalization of Krull's theorem: Let $R$ be a class ring. Then there is a class ideal $I$ of $R$ such that for every $x \in R - I$, we have $R = I + Rx$. I don't know if this is true. We have somehow to track back the formulas in the proof of the usual Krull's theorem, in order to produce from every formula defining $R$, another formula defining $I$ ... – Martin Brandenburg Sep 27 '10 at 21:36
Yes Martin, this is indeed what you have to do. You cannot prove the theorem for all class rings at once, but you can start with a specific class ring and prove the theorem. The proof, if you succeed, will probably be uniform, i.e., you will have to use the formula defining the ring in several places, but this formula can be replaced by any other formula that defines a ring. So, what you end up with is a scheme of proofs that encompasses all class rings. – Stefan Geschke Sep 28 '10 at 7:08

One issue that might come up is that some standard constructions won't make sense, typically those that involve looking at equivalence classes. For example, if you have a set group and you take a quotient by a subgroup, you're resulting object is a collection of equivalence classes, in this case a set of sets. But when trying to do the same thing with a class group, you might end up with a class of classes, or a set of classes, which is obviously no good. Scott's trick is usually provides a suitable work-around for this problem: when some of your equivalence classes are proper classes, use the elements of minimal (von Neumann) rank in the equivalence class in place of the entire equivalence class.

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Let $R$ be a direct product of $n$ copies of a field, where $n$ is a proper class. $R$-modules seem to work OK but EXT between $R$-modules is not a set. It may not be the most elementary but it is fascinating as it follows that the category of $R$-modules is abelian category that does not admit derived category...

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If we allow objects of a category to be proper classes, why not to allow Hom sets between them to be proper classes as well? – Qfwfq Sep 27 '10 at 15:59
What is up, Doc? Which objects are you talking about? $R$-modules are sets!! In particular, $R$ is not $R$-module :-)) Progenerators gone with the wind... – Bugs Bunny Sep 27 '10 at 16:05
What I want to say is that allowing only certain entities to be "large" and not others may be "unnatural". Do you really want $R$ not to be an $R$-module? [b.t.w. the usual definition of module $M$ over $R$ involve both a set $M$ and a set $R$; if you're up to big-ify $R$, why not to big-ify $M$ as well?] – Qfwfq Sep 27 '10 at 16:49

Depending on your set theory, it's reasonable to think that an axiom of choice holds for sets, but not for larger collections. Since the axiom of choice is very important to algebra (for example, it is equivalent to the statement that every ring has a maximal ideal), lots of things can go wrong.

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@Martin Brandenburg: Absolutely, and moreover I tend to work in some sort of "universe" theory in which choice goes all the way up. But that's why I said "depending on your set theory". – Theo Johnson-Freyd Oct 1 '10 at 15:46

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