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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
19
votes
Accepted
Is there a notion of integration over the algebraic numbers?
You can definitely talk about integration on $\overline{\mathbb{Q}}$. The question is "with respect to what measure?" The reason integration theory works so well over a completion of $\mathbb{Q}$ is t …
12
votes
1
answer
591
views
Hecke-module structure implicit in definition of automorphic forms in Borel-Jacquet's Corval...
Let $G$ be a connected reductive group over a number field $F$, $G_\infty=\prod_{v\mid\infty} G(F_v)$, $\mathbf{A}$ the adèles of $F$, $\mathbf{A}_f$ the finite adèles of $F$. Fix a maximal compact su …
11
votes
1
answer
2k
views
Where in the literature does the anticyclotomic $\mathbf{Z}_p$-extension of an imaginary qua...
If $K$ is an imaginary quadratic field, then the $\mathbf{Z}_p$-rank of $K$ is $2$, meaning that the Galois group of the compositum of all the $\mathbf{Z}_p$-extensions of $K$ in an algebraic closure …
7
votes
Avoiding Minkowski's theorem in algebraic number theory.
Yes, there is a way to get the finiteness of class number as well as the S-unit theorem, avoiding using Minkowski's theorem explicitly. However, it doesn't give you the Minkowski bound. The idea is to …
7
votes
Analog to the Chinese Remainder Theorem in groups other than Z_n.
What you know as the Chinese remainder theorem for the abelian group $\mathbb{Z}/n\mathbb{Z}$ (which you probably don't want to call a ``simple group" unless $n$ is prime, as this term has a technical …
7
votes
Accepted
Relation between partially computable function and complex function
Any complex-valued function on $\mathbb{N}$ can be extended to an entire function, so the answer is "yes." This follows from Theorem 15.13 of Rudin's Real and Complex Analysis, which states that for a …
6
votes
Accepted
Place stabilizers for the absolute Galois Group
If $p$ is a rational prime, then choosing a prime $v$ of $\overline{\mathbf{Q}}$ lying over $p$ amounts to choosing an embedding $i:\overline{\mathbf{Q}}\hookrightarrow\overline{\mathbf{Q}}_p$. This g …
3
votes
Accepted
Pontryagin dual
Let $\Lambda=\mathbf{Z}_p[[T]]$. The group $M[p]$ is dual to $M^\vee/pM^\vee$. So if $M[p]$ is finite, then $M^\vee/pM^\vee$ is finite, and the form of Nakayama's lemma applicable to profinite modules …
3
votes
Accepted
Weil group of a local field, small notational problem
If $G$ is any profinite group, and $a\in\widehat{\mathbf{Z}}$, then for any sequence $(a_n)$ of integers converging in $\widehat{\mathbf{Z}}$ to $a$, the sequence $(g^{a_n})$ converges in $g$ to an el …