Groups and rings which are not sets - MathOverflow

Groups and rings which are not sets

2010-09-27T15:16:39Z

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?

Answer by Bugs Bunny for Groups and rings which are not sets

2010-09-27T15:29:22Z

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...

Answer by Stefan Geschke for Groups and rings which are not sets

2010-09-27T16:10:33Z

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.).

Answer by Theo Johnson-Freyd for Groups and rings which are not sets

2010-09-28T02:16:24Z

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.

Answer by Amit Kumar Gupta for Groups and rings which are not sets

2010-10-01T03:04:11Z

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.