Unanswered Questions

156
votes
0answers
7k views

A short proof for $\dim(R[T])=\dim(R)+1$?

If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial ...
140
votes
0answers
4k views

Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
123
votes
0answers
4k views

The Ring Game on $K[x,y,z]$

I recently read about the Ring Game on Mathoverflow, and have been trying to determine winning strategies for each player on various rings. The game has two players and begins with a commutative ...
107
votes
0answers
4k views

Identification of a curious function

During computation of some Shapley values (details below), I encountered the following function: $$ f\left(\sum_{k \geq 0} 2^{-p_k}\right) = \sum_{k \geq 0} \frac{1}{(p_k+1)\binom{p_k}{k}}, $$ where ...
104
votes
1answer
6k views

A variation of Fermat's little theorem

Fermat's little theorem states that for $n$ prime, $$ a^n \equiv a \pmod{n}. $$ The values of $n$ for which this holds are the primes and the Carmichael numbers. If we modify the congruence ...
73
votes
0answers
2k views

What properties of busy beaver numbers are computable?

The busy beaver function $\text{BB}(n)$ describes the maximum number of steps that an $n$-state Turing machine can execute before it halts (assuming it halts at all). It is not a computable function ...
62
votes
0answers
4k views

Grothendieck 's question - any update?

I was reading Barry Mazur's biography and come across this part: Grothendieck was exceptionally patient with me, for when we first met I knew next to nothing about algebra. In one of his first ...
52
votes
0answers
1k views

Is there a homology theory that counts connected components of a space?

It is well-known that the generators of the zeroth singular homology group $H_0(X)$ of a space $X$ correspond to the path components of $X$. I have recently learned that for Čech homology the ...
50
votes
0answers
1k views

Differential forms on fuzzy manifolds

This post will take a bit to set up properly, but it is an easy read (and most likely easy to answer); in any event, please bear with me. Question In the usual setting of open subsets of ...
49
votes
1answer
933 views

Is there a categorical definition of submetry?

(Updated to include effective epimorphism.) This question is prompted by the recent discussion of why analysts don't use category theory. It demonstrates what happens when an analyst tries to use ...
48
votes
0answers
922 views

Does there exist a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?

Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ ...
45
votes
0answers
2k views

Continuous projections in $\ell_1$ with norm $>1$

I was trying to find papers and articles about non-contractive continuous projections in $\ell_1(S)$ where $S$ is an arbitrary set. If it is not studied yet, I would like to know results for the case ...
44
votes
0answers
2k views

Is this similarity to the Fourier transform of the von Mangoldt function real?

Mathematica knows that the logarithm of $n$ is: $$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$ The von Mangoldt function should then be: ...
42
votes
0answers
2k views

$n!$ is never a perfect square if $n\geq2$. Is there a proof of this that doesn't use Chebyshev's theorem?

If $n\geq2$, then $n!$ is not a perfect square. The proof of this follows easily from Chebyshev's theorem, which states that for any positive integer $n$ there exists a prime strictly between $n$ and ...
36
votes
0answers
395 views

Regular way to fill a $1\times1$ square with $\frac{1}{n}\times\frac{1}{n+1}$ rectangles?

The series $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1$$ suggests it might be possible to tile a $1\times1$ square with nonrepeated rectangles of the form $\frac{1}{n}\times\frac{1}{n+1}$. Is there a ...

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