Uncountable counterexamples in algebra

Question

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.

Answer 1

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.]

Answer 2

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

3. 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.

Answer 3

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.

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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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.