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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
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 …
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' …
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 …