57 votes
Accepted

Clausen's modified Hodge Conjecture

It's a bit of a long story, but I can at least give the idea. Let $X$ be a smooth projective variety over $\mathbb{C}$. The Hodge conjecture says that for all $p\geq 0$, the cycle class map $$Ch^p(X)...
Dustin Clausen's user avatar
15 votes

Does this conic have a rational point?

The answer is no. If there was, we could assume that $X,Y,Z$ are in $\mathbb Q[u,v]$ and are coprime (since that ring is a UFD). Setting $v=0$ we get $X(u,0)^2+uY(u,0)^2=0$ in $\mathbb Q[u]$, which (...
Wojowu's user avatar
  • 25.9k
12 votes
Accepted

Definition of locally symmetric space of reductive groups

There is a very natural, intrinsic definition of a "symmetric space", as a manifold (Riemannian or Hermitian) with an extra symmetry of a certain prescribed type. It is then a theorem, not a ...
David Loeffler's user avatar
10 votes
Accepted

Definition of modular curve associated to $\Gamma(N)$

This is a subtle issue (which has come up before on this site several times, see e.g. is the modular curve X(N) defined over Q? for a related question). Your $S(N)$ is naturally a scheme over $\mathbb{...
David Loeffler's user avatar
8 votes
Accepted

Étale group schemes and specialization

If $k \to \ell$ is any regular field extension (i.e. $k$ is algebraically closed in $\ell$), then $X(k) \to X(\ell)$ is a bijection when $X \to \operatorname{Spec} k$ is étale. Indeed, it suffices to ...
R. van Dobben de Bruyn's user avatar
6 votes
Accepted

Is the value of the power series at 0.1 transcendental?

This question is likely open. We can tell whether $f(1/10)$ is rational or irrational (by asking whether $a_n$ is eventually periodic); in this case, definitely irrational. Can we have an algebraic ...
Gerald Edgar's user avatar
  • 39.4k
6 votes
Accepted

Faithful representations of integral models

Yes, there exists a closed immersion $\mathcal{G}\to \mathrm{GL}_n$ over $\mathbb{Z}$. This is folklore. For a proof, see for example Proposition 3 of arXiv:2012.05708v3
anon's user avatar
  • 96
5 votes

The notion of morphisms between two moduli problems in Katz-Mazur

That's correct. We would like this to give a morphism between the underlying stacks, i.e. the functor that takes $S$ to the groupoid of pairs of an elliptic curve $E/S$ and an element of $\mathcal P_i(...
Will Sawin's user avatar
  • 126k
4 votes

Is there a method to check if two sections of an elliptic surface are dependent over the endomorphism ring or not?

Let $\mathcal O=\mathbb Z[\tau]=\mathbb Z+\mathbb Z\tau$. Then you're asking if the four points $$ P_0(t),\; \tau\bigl(P_0(t)\bigr),\; P_1(t),\; \tau\bigl(P_1(t)\bigr) $$ are $\mathbb Z$-linearly ...
Joe Silverman's user avatar
4 votes
Accepted

Flat scheme-theoretic closure

Explicitly, lets let $R = \mathbb{C}[x]_{(x)}$ so $K = \mathbb{C}(x)$. You can then let $C_K = {\bf P}^1_K$ and $C_R = {\bf P}^1_R$. $C_R$ has a chart that looks like $\mathbb{C}[x]_(x)[y]$. This ...
Karl Schwede's user avatar
  • 19.9k
4 votes

Zariski dense in abelian scheme

How about if $A = E_1\times E_2$ with $E_i$ non isotrivial elliptic curves and $s: S \to P\times 0$. Then your $\mathbb Z\cdot X$ is contained in $E_1\times 0$ and so is not Zariski dense. In fact, ...
Asvin's user avatar
  • 7,302
4 votes

Rational points on genus 3 curves defined by short equations

I have looked a bit at the first equation. It has (at least) seven rational points (as a projective curve). The differences of these points generate a free abelian group of rank three in the Mordell-...
Michael Stoll's user avatar
2 votes

Is the value of the power series at 0.1 transcendental?

For the UPDATE, allowing coefficients $a_n < M$ for a fixed $M$. Then there are examples with $f(1/10)$ rational. Let's do this. Define a sequence $(a_n)$ of coefficients as follows: Start with ...
Gerald Edgar's user avatar
  • 39.4k
1 vote

Symmetric powers, localisation and Frobenius

I want to come back to this, and provide a sketch of the Dold-Puppe argument so people who don't speak German (such as myself) don't have to try and sift through the linked paper to try and find the ...
Logan Hyslop's user avatar

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