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In functional analysis, there are many examples of things that "go wrong" in the nonseparable setting. For instance, my favorite version of the spectral theorem only works for operators on a separable Hilbert space. Within C${}^*$-algebra there are many examples of nice results that require separability. (Dixmier's problem: is every prime C${}^*$-algebra primitive? Yes for separable C*-algebras, no in general.)

I wondered whether there is a similar phenomenon in pure algebra. Are there good examples of results that hold for countable groups, countable dimensional vector spaces, etc., but fail in general?

One example I know about is Whitehead's problem, which has a positive solution for countable abelian groups, but is independent of ZFC in general.

Martin Sleziak
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Nik Weaver
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    If you a countable dimensional $\mathbb C$-algebra then the endomorphism algebra of a simple module is $\mathbb C$ but this is not true for uncountable dimensional ones – Benjamin Steinberg May 09 '21 at 20:01
  • It's also true that countable dimensional Leavitt path algebras and more generally Hausdorff etale groupoid algebras over fields are prime iff primitive but this is false in the uncountable case. – Benjamin Steinberg May 09 '21 at 20:04
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    That said my feeling is that in algebra I think finitely generated versus non finitely generated is really the bigger dividing line. – Benjamin Steinberg May 09 '21 at 20:06
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    @NikWeaver What's your "favorite version of the spectral theorem"? – André Henriques May 09 '21 at 20:08
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    @AndréHenriques: Let $A$ be a bounded self-adjoint operator on a separable Hilbert space. Then there is a measurable bundle $X$ of Hilbert spaces over ${\rm sp}(A)$ and an isomorphism $H\cong L^2({\rm sp}(A), X)$ which takes $A$ to multiplication by $x$. – Nik Weaver May 09 '21 at 20:17
  • It fails in the nonseparable case because one doesn't have a good notion of "measurable Hilbert bundle". E.g. take $A$ to be multiplication by $x$ on $L^2[0,1]$ direct sum multiplication by $x$ on the nonseparable space $l^2[0,1]$. – Nik Weaver May 09 '21 at 20:18
  • @NikWeaver: But that just seems a (perhaps unnecessarily) fancy version of the theorem on spectral representations ($A$ is unitarily equivalent to multiplication by the variable in $\bigoplus L^2(\sigma(A),\rho_{\alpha})$), which works on any Hilbert space. – Christian Remling May 09 '21 at 20:20
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    @ChristianRemling sure, but I like the bundle version better because it's canonical --- there are lots of ways to split $H$ up as $\bigoplus L^2(\sigma(A), \rho_\alpha)$. – Nik Weaver May 09 '21 at 20:23
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    A keyword for many examples might be "$\omega$-categorical". For instance, the property "any non-atomic Boolean algebra is free" is true for countable Boolean algebras but not in general (the countable case is a restatement [thru Stone duality] of the fact that any nonempty metrizable totally disconnected compact space is homeomorphic to the Cantor set). – YCor May 09 '21 at 20:35
  • Would posets count as algebra, or is that getting more towards "abstract set theory"? – Yemon Choi May 09 '21 at 20:43
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    Speaking of $\omega$-categoricity, of course in algebra there is also the opposite phenomenon - that some things are nice in the uncountable case but go wrong in the countable case: vector spaces of the same cardinality over the same (countable) field being isomorphic, algebraically closed fields of the same cardinality and characteristic being isomorphic, etc. – Arno Fehm May 09 '21 at 20:46
  • @NikWeaver: You do have an ordered spectral representation when $H$ is separable, but that's probably just rephrasing your original comment. – Christian Remling May 09 '21 at 21:13
  • @YemonChoi I don't know if they count as algebra but I'd be interested in that kind of example too. – Nik Weaver May 09 '21 at 21:52
  • On the other hand, things can be nicer in the uncountable case: if $F$ is a characteristic zero field of cardinality equal to that of the reals, and $F$ is algebraically closed, then $F$ is isomorphic to the complexes. On the other hand, there are countably infinitely many isomorphism classes of countable characteristic zero algebraically closed fields (isomorphism depends only on transcendence degree). More generally, in characteristic zero, for any uncountable cardinal, there exists a unique algebraically closed field of that cardinality, up to isomorphism. – David Handelman May 09 '21 at 22:04
  • @YCor I believe you're using "non-atomic Boolean algebra" to mean what I'd call an atomless Boolean algebra, but linguistically it should mean "a Boolean algebra which is not an atomic Boolean algebra", which could have some atoms. – bof May 10 '21 at 01:53
  • @bof thanks, indeed, I wasn't aware of the nuance. So I mean atomless, i.e. without atoms. An atom in a Boolean algebra is a nonzero element $x$ such that for no nonzero $y,z$ we have $yz=0$ and $y+z=x$. – YCor May 10 '21 at 05:04
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    @David Handelman all that you wrote is true in characteristic $p$ too: for each prime $p$, uncountable algebraically closed fields of characteristic $p$ are (by Zorn’s Lemma) determined up to field isomorphism by their cardinality and this is false for countable algebraically closed fields of characteristic $p$. In general, an algebraically closed field is determined up to field isomorphism by its characteristic and its transcendence degree over its prime subfield ($\mathbf Q$ or $\mathbf F_p$). Its transcendence degree equals its cardinality for uncountable algebraically closed fields. – KConrad May 10 '21 at 12:36

8 Answers8

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Let $G$ be an abelian group.

The statement

If every subgroup of $G$ of finite rank is $\mathbf{Z}$-free, then $G$ is $\mathbf{Z}$-free.

is a theorem for $G$ countable, but false in general ($\mathbf{Z}^X$ for infinite $X$ is a counterexample).

[Recall that the rank, or $\mathbf{Q}$-rank of an abelian group $A$ is the maximal number of $\mathbf{Z}$-free elements, or equivalently the dimension over $\mathbf{Q}$ of $A\otimes_\mathbf{Z}\mathbf{Q}$. For instance $\mathbf{Q}$ and $\mathbf{Z}[1/n]$ have rank 1.]

YCor
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  • Excellent, just the sort of thing I was looking for. – Nik Weaver May 09 '21 at 20:26
  • This is not too surprising, since "countable" is the next thing after "finite". What about the obvious generalization to cardinal numbers $\lambda$? If $G$ is an abelian group with $\lambda$ elements and every subgroup of rank $< \lambda$ is $\mathbb{Z}$-free, does it follow that $G$ is free? Or is $\mathbb{Z}^{\mathbb{N}}$ with $\lambda=2^{\aleph_0}$ a counterexample (perhaps assuming $\mathrm{CH}$)? Is it true for some cardinals $> \aleph_0$ at least? – Martin Brandenburg May 11 '21 at 18:41
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    @MartinBrandenburg It's true for singular cardinals, such as $\aleph_{\omega}$, but false for $\aleph_n$ ($n$ finite and greater than $0$). I don't think it's known for precisely which $\lambda$ it is true. – Jeremy Rickard May 12 '21 at 18:14
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Countable torsion abelian groups are better behaved than uncountable ones. For example, Kaplansky’s “test problems”

  1. If $G$ and $H$ are isomorphic to direct summands of each other, is $G\cong H$?

  2. If $G\oplus G\cong H\oplus H$, is $G\cong H$?

have positive answers for countable torsion abelian groups, but not for uncountable torsion abelian groups.

Another, slightly more obscure, example involving abelian groups (not just torsion groups this time) is that the answer to

  1. If $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$, is $G\cong G\oplus\mathbb{Z}$?

is positive for countable abelian groups $G$, but not for uncountable abelian groups.

Jeremy Rickard
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  • By the way this translates into the same statements for profinite abelian groups (by Pontryagin duality). – YCor May 09 '21 at 22:20
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    @YCor But Pontryagin duality sends countability to a subtler property - having countably many open subgroups, I guess. – Will Sawin May 10 '21 at 02:22
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    @WillSawin Pontryagin duality sends countable discrete groups onto metrizable compact groups. – YCor May 10 '21 at 05:01
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In rings: Let $R$ be a ring where idempotents lift modulo the Jacobson radical $J(R)$. Any countable set of orthogonal idempotents in $R/J(R)$ lifts to an orthogonal set of idempotents; but this fails for uncountable sets.

In combinatorics of words: Vaughan Pratt's "crossword problem" is another example. Suppose you have a language consisting of words of length $\kappa$ in the letters $\{0,1\}$, satisfying the following three conditions: (1) The all $0$'s word, and the all $1$'s word, belong to your language. (2) For any two distinct positions in $\kappa$, there is a word with $1$ in the first position and $0$ in the second. (3) If you fill in a $\kappa\times\kappa$ crossword, so that every row and every column is a word in your language, then the main diagonal is also a word.

Pratt asked: Must your language contain all possible words? The answer is yes if $\kappa\leq \aleph_0$, and no otherwise.

In the converse direction:

In algebras: Amitsur's theorem says that if $R$ is a nil algebra over an uncountable field, then the polynomial ring $R[X]$ is nil as well. Agata showed that this fails for nil algebras over countable fields.

Pace Nielsen
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  • "In comonoids" is an off-putting way of introducing the crossword problem -- could you say instead "in combinatorial matrix theory"? –  May 10 '21 at 03:05
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    @MattF. But the usual algebraic operations, of matrix addition and multiplication, have nothing to do with the structure. – Pace Nielsen May 10 '21 at 14:32
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    In the Mathematics Subject Classification (https://mathscinet.ams.org/msnhtml/msc2020.pdf), I would have categorized Pratt's problem as 05Bxx, Designs and configurations and specifically 05B20, Combinatorial aspects of matrices. I also see it categorized as 68Rxx Discrete mathematics in relation to computer science, specifically 68R15 Combinatorics on words -- in your paper, which you might have referenced! (https://arxiv.org/abs/1504.07310) Some combination of those subject words might be more suggestive and appealing. –  May 10 '21 at 19:53
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    @MattF. Go ahead and change it as you see fit. – Pace Nielsen May 10 '21 at 20:03
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If $A$ is a countable dimensional $\mathbb C$-algebra, then the endomorphism algebra of a simple $A$-module is isomorphic to $\mathbb C$. But $\mathbb C(t)$ is an uncountable dimension $\mathbb C$-algebra and since it is a field, its regular module is simple. But $\mathrm{End}_{\mathbb C(t)}(\mathbb C(t))\cong \mathbb C(t)$.

The real issue here is that there are no countable dimension division algebras over $\mathbb C$ besides itself.

Benjamin Steinberg
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    So eventually the point is that every field extension of an algebraically closed field of infinite cardinal $\alpha$ has dimension $\ge\alpha$. – YCor May 09 '21 at 20:39
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    Yes. It is really about the algebra having smaller dimension than the cardinality of the field – Benjamin Steinberg May 09 '21 at 20:46
  • This is also known as the failure of the Schur's lemma in uncountable dimensions - see this with a proof in the countable case and a counter-proof in the uncountable case. – Student May 12 '21 at 03:22
  • @Student this is the same counterexample I have in my answer but I didn't put the proof – Benjamin Steinberg May 12 '21 at 12:04
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Since you mentioned Whitehead's problem here is another interesting independence example.

For a group $G$ define its dual $G^\ast$ to be $\mathrm{Hom}(G,\Bbb Z)$. Like in the vector spaces case we get a canonical evaluation homomorphism $j\colon G\to G^{\ast\ast}$ given by $g\mapsto(f\mapsto f(g))$ and we call a group reflexive if $j$ is an isomorphism.

Now let $G$ be free abelian, must it be reflexive? The answer is provably positive for all "small" free abelian groups in $\mathsf{ZFC}$, in fact it is positive for all free abelian groups iff there is no measurable cardinal.

Alessandro Codenotti
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    More precisely the answer is true for a given free abelian group, of cardinal $\alpha$, iff $\alpha<\kappa$, where $\kappa$ is the smallest measurable cardinal (this being viewed as void [=true] condition if $\kappa$ doesn't exist). – YCor May 10 '21 at 05:00
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Certainly countability plays a role when measures are involved. Galois theory has a few such examples. For example, by a result of Jarden, for a countable field $K$, the set of $\sigma$ in the absolute Galois group $G_{K(t)}$ of the rational function field $K(t)$ for which the fixed field ${\rm Fix}(\sigma)$ is pseudo-algebraically closed (i.e. every geometrically irreducible variety over it has a dense set of rational points) has measure 1 with respect to the Haar measure on $G_{K(t)}$. Jarden-Shelah gave an example of an uncountable field $K$ where the set of $\sigma\in G_{K(t)}$ with ${\rm Fix}(\sigma)$ pseudo-algebraically closed is non-measurable.

Arno Fehm
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  • Nonmeasurable, what a twist! Do we know if this set can have a well-defined measure smaller than 1? Maybe even equal to 0? – Wojowu May 11 '21 at 14:30
  • @Wojowu: Indeed, these are very natural questions, but as far as I am aware it's not known. The standard proof applied to a field of cardinality $\kappa$ just gives that the set of such $\sigma$ is an intersection of $\kappa$ many sets of measure 1 (one for each geometrically irreducible variety). – Arno Fehm May 11 '21 at 17:44
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Separable manifolds (which I'll always assume to be Hausdorff) have partitions of unit (= are paracompact). This implies many nice properties.

E.g., if $V\to M$ is a vector bundle and $f,g:X\to M$ are two homotopic maps, then $f^*V\cong g^*V$.

All this fails for non-separable manifolds, as illustrated by the example of the long line $L$, whose tangent bundle is non-trivial but nevertheless there exists a (smooth) vector bundle over $L\times [0,1]$ whose restriction to $L\times \{0\}$ is the tangent bundle of $L$, and whose restriction to $L\times \{1\}$ is trivial.

André Henriques
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Relating to the prime versus primitive issues, it was shown by Abrams, Bell and Rangaswamy that a Leavitt path algebra over a field defined by a countable digraph is prime if and only if it is primitive but this is not true for uncountable graphs.

More generally, I showed that if $\mathscr G$ is a locally compact and totally disconnected second countable Hausdorff etale groupoid, then its convolution algebra over a field is prime if and only if it is primitive but this again fails in the non-second countable case.

Benjamin Steinberg
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