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comment Which compact topological spaces are homeomorphic to their ultrapower?
Maybe I'm understanding your definitions wrong, but doesn't any nonprincipal ultrapower in the category of spaces have the indiscrete topology? The quotient map $\prod X\to \prod_\mu X$ is surjective on any nonempty open set (because an open set depends on only finitely many coordinates, and $\mu$ is nonprincipal).
Mar
25
comment Uninteresting questions with interesting answers
I have to say I am shocked that this question has received so many upvotes and has not yet been closed. This question is very subjective and MO is not for collecting bits of mathematical trivia. This question could perhaps have some value if it were geared toward understanding how it is that uninteresting questions have interesting answers (see Timothy Chow's answer, for instance), but in its current form I don't see the point of it and it is attracting a lot of dubious answers. Pietro Majer's comment is also illustrative of the arbitrariness of this question.
Mar
24
comment Uncountable chain of prime ideals in an arbitrary direct product of rings
I'm pretty sure that for a countable product of rings of size $\leq\mathfrak{c}$, the dimension is always either $0$ or $ded(\mathfrak{c})$ (the latter number is the largest cardinality of a chain in $P(\mathfrak{c})$; under CH it is $2^{\mathfrak{c}}$). For instance, in my example above, you can take not just sets of constant sequences but arbitrary subsets of the ultrapower of the set of variables. With a countably infinite variable set, this ultrapower has cardinality $\mathfrak{c}$.
Mar
24
comment Uncountable chain of prime ideals in an arbitrary direct product of rings
Actually, I believe I can say affirmatively that in the latter example, there are chains of primes in the product that are isomorphic to any ordering of the set of variables. Once you've passed to an ultraproduct and modded out the nilpotents, the ideal generated by any subset of the constant sequences corresponding to the variables will be prime. So a countable product of zero-dimensional rings can have arbitrarily large dimension, no matter how you define it.
Mar
24
comment Uncountable chain of prime ideals in an arbitrary direct product of rings
For your second question, there are multiple ways you might define an infinite Krull dimension (do you just care about cardinality of chains, or do you want the length of well-ordered chains, or something else?). I don't know of an upper bound for the dimension of a countable product of finite-dimensional rings under any of these definitions, though. For instance, suppose $A_n$ is of the form $k[x_i]/(x_i^n)$, where $\{x_i\}$ is a very large infinite set of variables. It seems plausible to me that the product might have at least as many prime ideals as there are variables.
Mar
24
comment Uncountable chain of prime ideals in an arbitrary direct product of rings
For your first question, there is always a surjection onto the set of ultrafilters of the index set, given by looking at elements which are $1$ or $0$ on every coordinate. This is a bijection iff each factor is local and the product is zero-dimensional (iff there is an integer $n$ such that every non-unit $x$ in a factor satisfies $x^n=0$).
Mar
23
revised Uncountable chain of prime ideals in an arbitrary direct product of rings
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Mar
23
answered Uncountable chain of prime ideals in an arbitrary direct product of rings
Mar
21
revised Is $\{0,1\}^\omega$ the order-preserving image of $\{0,1\}^\omega$ modulo some finiteness relation?
added 232 characters in body
Mar
21
answered Is $\{0,1\}^\omega$ the order-preserving image of $\{0,1\}^\omega$ modulo some finiteness relation?
Mar
20
comment Can every tromino (including those with gaps) tile the plane?
Note that this is problem is invariant under the action of $GL_n(\mathbb{Z})$, so if it holds for one linearly independent omino, it holds for all of them. But classifying exactly when a non-independent omino tiles seems nontrivial.
Mar
20
answered For a partition of $\mathbb{R}$ into countably infinite sets, must there be an almost-disjoint family of $2^{\frak c}$ many selectors?
Mar
20
awarded  Enlightened
Mar
19
awarded  Nice Answer
Mar
19
revised How many subsets of $[0,1)$ are there modulo null sets?
added 201 characters in body
Mar
19
answered How many subsets of $[0,1)$ are there modulo null sets?
Mar
19
awarded  Nice Answer
Mar
18
awarded  Nice Answer
Mar
17
comment Automorphisms of $\mathbb{C}$ and meromorphic functions
Is there any meromorphic function at all from which $\mathbb{R}$ is definable? It seems plausible to me that any set which is definable from a meromorphic function must be either countable or cocountable, but I haven't thought about it much.
Mar
17
comment Two H-space structures on S^3 and [X,S^3] different as groups for each: Explicit Example?
This group is computed for the standard $H$-space structure here.