Unanswered Questions

179
votes
0answers
8k 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 ...
158
votes
0answers
5k 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 ...
130
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 ...
79
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 ...
74
votes
0answers
5k 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 ...
67
votes
0answers
2k 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, $f(S)\subset Y$ ...
55
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 ...
49
votes
0answers
981 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 ...
46
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 ...
45
votes
0answers
812 views

Gross-Zagier formulae outside of number theory

(Edit: I have asked this question on MO.) The Gross-Zagier formula and various variations of it form the starting point in most of the existing results towards the Birch and Swinnerton-Dyer ...
42
votes
0answers
674 views

Geometric interpretation of the Riemann-Roch for curves

Let $X$ be a smooth projective curve of genus $g\geq2$ over an algebraically closed field $k$ and denote by $K$ a canonical divisor. I have some clues about the geometrical interpretation of the ...
41
votes
0answers
1k views

The log gamma integral $\int_{0}^{z} \log \Gamma (x) \ \mathrm dx$

One way to evaluate $ \displaystyle\int_{0}^{z} \log \Gamma(x) \ \mathrm dx $ is in terms of the Barnes G-function. $$ \int_{0}^{z} \log \Gamma(x) \ \mathrm dx = \frac{z}{2} \log (2 \pi) + ...
37
votes
0answers
1k views

What does it take to divide by $2$?

Theorem 1 [ZFC, classical logic]: If $A,B$ are sets such that $\textbf{2}\times A\cong \textbf{2}\times B$, then $A\cong B$. That's because the axiom of choice allows for the definition of ...
37
votes
0answers
890 views

Generalization of Liouville's theorem

As proposed in this answer, I wonder if the answer to following question is known. Let $E = E_0$ be the set of elementary functions. For each $i > 0$, inductively define $E_i$ to be the closure ...
36
votes
0answers
2k views

Limit of sequence of growing matrices

Let $$ H=\left(\begin{array}{cccc} 0 & 1/2 & 0 & 1/2 \\ 1/2 & 0 & 1/2 & 0 \\ 1/2 & 0 & 0 & 1/2\\ 0 & 1/2 & 1/2 & 0 \end{array}\right), $$ ...

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