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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
43
votes
Accepted
From Zeta Functions to Curves
To determine which potential zeta functions are actual zeta functions of curves is very difficult. The zeta function of a curve is determined by the zeta function of its Jacobian so one could instead …
6
votes
On the field of invariants of a finite group
It seems likely (and may be known) that it does depend on the permutation representation. However, it is a subtle question as the stable isomorphism class (of any faithful permutation, or in fact arbi …
25
votes
Accepted
How does Tate verify his own conjecture for the Fermat hypersurface?
I don't know how Tate did it but here is one way. Let $\zeta$ be such that
$\zeta^{q+1}=-1$ and put $a_j=(0\colon\cdots\colon1\colon\zeta:\cdots\colon0)$,
$j=0,\ldots,i$ with the $1$ in coordinate $2 …
20
votes
extensions, abelian varieties, $\mathbb{G}_m$
Using the Kummer exact sequence
$0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we
get a long exact sequence
$$
0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm …
18
votes
About isogeny theorem for elliptic curves
If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by
the twist by a locally free rank $1$ module over the endomorphism ring of one of
them. This is true for all abelian variet …
6
votes
Relation between l-adic and l'-adic geometric monodromy
As $\pi_1(B)$ is the same over $\overline{\mathbb Q}$ as over $\mathbb C$ we may
embed $\overline{\mathbb Q}$ in $\mathbb C$ and then extend scalars to $\mathbb
C$ to reduce to the case when $\mathbb …
3
votes
Accepted
Q-lattices and commensurability, any insight into the definition and intuition?
The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have
$\Lambda_1,\Lambda_2\subseteq X$ and then we have $\Lambda_1,\Lambda_2\subseteq
\Lambda_1+\Lambda_2\subseteq X$. This mean …
6
votes
Is the tangent space functor from PD formal groups to Lie algebras an equivalence?
I think that this MR0277590 (43 #3323)
André, M.
Hopf algebras with divided powers.
J. Algebra 18 1971 19--50 may be relevant. It says that a graded commutative divided power Hopf algebra is the co-en …
51
votes
Accepted
Separable and algebraic closures?
Geometrically there is a very big difference between separable and algebraic
closures (in the only case where there is a difference at all, i.e., in positive
characteristic $p$). Technically, this com …
7
votes
Accepted
sub-tori of a torus, generated by 1-dimensional subgroup
The key to solving both problems is the use of the following two facts: 1) Any
closed subgroup of $T^n$ is the intersection of the kernels of characters of
$T^n$, i.e., continuous group homomorphisms …
4
votes
Positive solutions of linear Diophantine equations
Just some comments that are well-known in the theory of toric varieties (and no
doubt to other areas as well). What we are asked to determine is membership in a
finitely generated submonoid $\Gamma$ o …
16
votes
Accepted
Finiteness property of automorphism scheme
I wanted to add some things to the comments I had already made but the list of
comments have become very large and the comments I have already made are
becoming more and more difficult to follow so I' …
1
vote
Smooth linear algebraic groups over the dual numbers
I don't understand why the usual proof over a field base doesn't work over a
(local) artinian base $R$ for a flat finite type group scheme over $R$: Let $A$
be the affine algebra of $G$. Take any basi …
9
votes
The etale fundamental group of a field
Also for the étale fundamental group there is in fact always some universal
cover. However, in the abstract way that Grothendieck formulated the theory of
coverings a universal cover would only exist …
7
votes
solutions to equation mod a prime
If you rewrite the equation as $b^2=-a^2/(a^2+1)$ as Will did you can continue as follows: The equation can be rewritten as
$$\left(\frac{b}{a}\right)^2 = -(a^{-2}+1)$$
(excepting $a=b=0$). Putting $x …