4

Probably this is very elementary but I don't have enough background to answer that question. In particular I am struggling with the following statement: "the collection of all finitely generated projective modules over a ring $R$ is not a set, but the collection of isomorphism classes of this kind of modules is a set". So, how can one prove that? Which is a procedure to decide whether something is a set or not?

Thanks in advance.

MBL
  • 972
  • 1
    Presumably, over a fixed ring $R$? What axioms/definitions are you using for set? For fixed $R$, you can certainly create a set, $\mathcal S$, of $R$-modules such that $M$ is a finitely-generated projective $R$-module if and only if it is isomorphic to some element of $\mathcal S$. That's not the same as saying that the equivalence classes under isomorphisms even comprises a set. – Thomas Andrews Dec 02 '11 at 21:27
  • @ThomasAndrews: Yes, I edited the post to fix the ring $R$. And about the axioms, ZFC would be enough... – MBL Dec 02 '11 at 21:36
  • The collection of isomorphism classes isn't a set either. That is, within ZFC you cannot construct such a thing. – André Nicolas Dec 02 '11 at 22:17

4 Answers4

7

Depends on your set theory. I will work in Zermelo-Fraenkel-with Choice below (ZFC).

What is really at issue is the problem of "proper classes".

Remember that a "module over the ring $R$" is really an ordered triple, $(M,+,\cdot)$, where $M$ is a set, $+$ is a binary operation on $M$ (that is, $+$ is a function $M\times M\to M$) and $\cdot$ is a function $\cdot\colon R\to M$, which satisfies a bunch of properties.

Technically speaking, then, if $\bullet\neq\star$, then the module $(\{\bullet\},+,\cdot)$, where $\bullet+\bullet = \bullet$ and $a\cdot\bullet=\bullet$ for all $a\in R$; and the module $(\{\star\},\oplus,\odot)$ where $\star\oplus\star=\star$ and $a\odot\star=\star$ for all $a\in R$, are different modules, because they are different as ordered triples.

That means that, because the collection of all singletons is not a set, but we can make each singleton set into a finite (hence finitely generated) $R$-module, the collection of all finitely generated $R$-modules cannot be a set. (See the Axiom of Replacement).

However, let $X$ be a set whose cardinality is the larger of $|R|$ and $\aleph_0$. Every finitely generated $R$-module is isomorphic to an $R$-module with underlying set contained in $X$ (a finitely generated $R$-module is a quotient of $R^n$ for some $n$, and the cardinality is always bounded by $\max\{\aleph_0,|R|\}$; now biject it with some subset of $X$ and use transport of structure).

Since $X$ is a set, the collection of all subsets of $X$ is a set (Axiom of the Power Set). And from this, using the Axiom of the Power Set and the Axiom Schema of Specification, one can show that the collection of all function $A\times A\to A$ over all subsets $A$ of $X$, and of all functions $R\times A\to A$ over all subsets $A$ of $X$, are also sets. And from there, the collection of all triples of the form $(A,+,\cdot)$ where $A\subseteq X$, $+\colon A\times A\to A$ is a function, and $\cdot R\times A\to A$ is a function, is also a set. And so the collection of all $R$-modules with underlying set contained in $X$ is a set.

So the collection of all $R$-modules that are both finitely generated and have underlying set contained in $X$ is also a set.

Now we can identify each equivalence class of finitely generated modules over $R$ with a particular element of this latter collection (which is actually a set), and so we conclude that the collection of equivalence classes of finitely generated modules over $R$ is actually a set (however, see Asaf's caveat below). With this in place, what we can do formally is the following: given any finitely generated module over $R$, we know that it is isomorphic to (at least) one of the finitely generated modules in our set (those whose underlying set is a subset of $X$). So we can "pretend" that we have all finitely generated modules "on hand" by working only with those whose underlying set is contained in $X$, because the kind of properties that we are interested in are invariant under isomorphism (they only depend on the isomorphism type of the module, not on how exactly we write down the module, or the names of the elements). So, "morally", we have all finitely generated modules on hand. It's not really a set of equivalence classes (see Asaf's answer), but we do have at least one representative of any class.

And so we can breathe a sigh of relief at being guaranteed that one could, if one were really stickler for details, work entirely within set theory properly, and then promptly forget about it and speak glibly about "all $R$-modules" as if that made sense in set theory, because with a lot of work we can make it make sense.

Community
  • 1
Arturo Magidin
  • 398,050
  • I thank you very much for your detailed answer. Your final paragraph is also great! – MBL Dec 02 '11 at 22:52
5

This is more of a technical comment which is too long to fit into the comment box

In ZFC this statement cannot be true.

Let $P$ be the collection of all projective $R$-modules, and let $\sim$ be the isomorphism relationship: $M\sim N$ if and only if $M$ is isomorphic to $N$ as $R$-modules.

Suppose that $P/\sim$ is the collection of all equivalence classes, if it is a set then $\bigcup(P/\sim)$ by the Axiom of Union. However this set is exactly $P$ which we already assumed that is not a set.

There is a nice way of solving this problem:

There exists a set of representatives, that is a set $S$ such that every projective $R$-module which is finitely generated is isomorphic to a member of $S$.

Note the difference, this is not the collection of isomorphism classes.

Asaf Karagila
  • 393,674
  • Nice! So whenever I read "the set of classes of isomorphism of______" (and I read this kind of things very often and in different books), should I interpret the sentence in that way? (Provided we are in ZFC). – MBL Dec 02 '11 at 22:41
  • This is why I like working in NGB set theory-and frankly,no one's given me a good reason yet why we shouldn't. Someone needs to tell most mathematicians I ask the question that "because" isn't a reason.Hopefully set theorists can give a real answer. – Mathemagician1234 Dec 02 '11 at 22:51
  • Because it is familiar. And perhaps because NGB and relatives require consistent type distinctions that are too easy to forget about. – André Nicolas Dec 03 '11 at 01:29
  • @Andre I'm not sure for strictly mathematical purposes,the type complications that have been notorious in set theoretic models with proper classes are as huge a problem as philosophers have made them out to be.As long as those models are isomorphic to ZFC for "small" enough collections,I think setting those type distinctions is probably manageable.This is important for those of us sympathetic to a consistent foundation for mathematics that handles category theory without a sidewise wink-wink. Frankly,that's what MacLane-Grothendeick Universes sound like to me.But we're getting WAY off topic. – Mathemagician1234 Dec 03 '11 at 03:32
  • 2
    @Mathemagician1234: In NBG classes still cannot be elements of a collection. Furthermore, the question remains the same. The collection is still not a set, and the collection of isomorphism classes is not even a class. You should have to interpret this as you do in ZFC. – Asaf Karagila Dec 03 '11 at 08:16
  • @MBL: Not always, for example, the set of finite dimensional vector spaces over $\mathbb R$ is of course not a set, but if you require that the underlying set is fixed (which is fine since the cardinality of $\mathbb R^n$ is still only continuum) then it is really a set of possible spaces, and a set of isomorphism classes. Furthermore, there is a "trick" to have a set representative for every isomorphism class, in which case the equivalence classes collection is a set, but its union is not all the projective $R$-modules like you may expect normally from equivalence classes. – Asaf Karagila Dec 03 '11 at 08:19
  • @Asaf To apply the axiom of union as you have done, wouldn't each isomorphism class have to be a set? But this is not the case, because, for example, even the isomorphism class of trivial modules is not a set. I admit that I know very little set theory, so I have probably missed something elementary here. I just don't properly understand why the collection of isomorphism classes of all finitely generated project modules does not form a set. Can you put me on the right track? – Henri Johnston Jul 30 '19 at 11:39
  • @Henri: Note that I said that you can apply Union if $P/{\sim}$ is a set. In that case, all of its elements are already sets as well (since elements of sets are sets). – Asaf Karagila Jul 30 '19 at 11:40
  • @Asaf Ah, I see - that elements of sets are sets was the crucial point I was missing. Thanks! – Henri Johnston Jul 30 '19 at 11:44
  • @Henri: Yes, it's a bit subtle if you're not used to it. But for the statement $x\in y$ to even make sense, $x$ and $y$ have to be sets. We abuse this notation when $y$ is a class sometimes to mean $\varphi(x)$ if $\varphi$ defines $y$. But we simply cannot write $\varphi(\psi)$ if $x$ is given by $\psi$. That is not a valid statement in the language of set theory, the most you can do is write $\exists z(\forall u(u\in z\leftrightarrow\psi(z))\land\varphi(z))$. But that's exactly saying that $x$ is a set again. – Asaf Karagila Jul 30 '19 at 11:46
1

The thing to do - generally and not always - is either find a set that exists which contains it (then by axiom schema we're home) or show it can't exist by contradiction [usually the structure will be too large to be a set like if the set of all groups existed, what is it's free group?] and etc.

Adam
  • 1,978
0

The group of permutations of any set is a group, so what about the set of all permutations of "set of all groups"? If there's a "set of all groups", then things like this involve us in Russell's paradox or the like.

It's things like that that can tell you when a class is "too big to be a set".

Michael Hardy
  • 1