Unanswered Questions

80
votes
0answers
3k 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 ...
69
votes
0answers
2k 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 ...
64
votes
0answers
2k 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 ...
53
votes
0answers
1k 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 ...
50
votes
0answers
1k views

How prove this matrix inequality $\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$?

Question: let matrices $A,B,C\in M_{n}(C)$ be Hermitian and Positive definite matrices, such that:$$A+B+C=I_{n}$$ Show that: $$\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$$ ...
44
votes
0answers
4k 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 ...
41
votes
0answers
733 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 ...
40
votes
0answers
3k 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 ...
38
votes
0answers
884 views

Why are asymptotically one half of the integer compositions gap-free?

This is a longish post about something that has been haunting me for a while about a kind of restricted composition, namely gap-free and complete compositions. First, I will define the terms that are ...
37
votes
0answers
750 views

Is the sequence $(a_n)$ defined by $a_n=\tan{a_{n-1}}$, $a_0=1$, dense in $\Bbb{R}$?

Let $a_0=1,a_n=\tan{a_{n-1}}$. Then is $\{a_n\}_{n=0}^\infty$ dense in $\Bbb{R}$? I've drawn a map of this dynamical system and it seems that the sequence is dense on $\Bbb{R}$.
35
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 ...
34
votes
0answers
798 views

Extending the result $\int_{0}^{\infty} \left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} \right) \, dx = \frac{4}{\pi} $

While generalizing this result, I succeeded in proving that for $\alpha > 0$, $\beta < 1$ and $1 < 2\alpha + \beta < 3$, we have \begin{align*} &\int_{0}^{\infty} \left[ \left( ...
31
votes
0answers
722 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 ...
30
votes
0answers
1k 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: ...
29
votes
0answers
649 views

Applying Freyd-Mitchell's embedding theorem on large categories

One commonly reads that the Freyd-Mitchell's embedding theorem allows proof by diagram chasing in any abelian category. This is not immediately clear, since only small abelian categories can be ...

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