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Algebraic number fields, Algebraic integers, Arithmetic Geometry, Elliptic Curves, Function fields, Local fields, Arithmetic groups, Automorphic forms, zeta functions, $L$-functions, Quadratic forms, Quaternion algebras, Homogenous forms, Class groups, Units, Galois theory, Group cohomology, Étale cohomology, Motives, Class field theory, Iwasawa theory, Modular curves, Shimura varieties, Jacobian varieties, Moduli spaces
9
votes
"Violent" field isomorphisms and $p$-adic "degrees" of complex numbers
(Assume AC throughout) For any transcendental $\tau\in\mathbb C$ and any $\tau'\in\overline{\mathbb Q_p}$ transcendental over $\mathbb Q$ there is an isomorphism $\sigma$ satisfying $\sigma(\tau)=\tau …
15
votes
Accepted
Absolute values of non-unit algebraic integers at all infinite places
No. Let $K=\mathbb Q(\sqrt{6})$ and let $\alpha=2+\sqrt{6}\in O_K$. All units of $O_K$ are given by $u=\pm v^n$, where $v=5+2\sqrt{6}$ is the fundamental unit. The key here is that $\alpha$ has relati …
4
votes
Fiberwise isomorphism of number rings $\mathcal{O}_E\otimes \mathbb{F}_p\cong \mathcal{O}_F\...
The entirety of the factorization of primes in a number field $K$ is encoded in its Dedekind zeta function. Therefore if two number fields $E,F$ have coinciding Dedekind zeta functions, the rings $\ma …
18
votes
Accepted
Geometric series in algebraic number fields
This holds precisely for elements which aren't roots of unity. Indeed, by the product formula we have $\prod_v|\alpha|_v=1$, where the product runs over all (finite and infinite) primes. This shows th …
11
votes
Size of a generator of the unit group in a cyclic cubic field
There will not be a smallest one. As you say, $O_K^\times$ has rank $2$, meaning it has two multiplicatively independent generators $u_1,u_2$, which we may take to be positive. This is equivalent to s …
3
votes
Accepted
$K_v(a^{1/m}) /K_v$ is unramified if only if $v(a)≡0 \pmod m$
If $v(a)\not\equiv 0\pmod m$, ramification is easy: just consider the valuation of the element $a^{1/m}$.
The converse is a little subtler than you make it seem, and depending on how exactly you phras …
19
votes
Has Fermat's Last Theorem per se been used?
Recall that around 1977 Mazur has completely classified the possible torsion groups of elliptic curves over $\mathbb Q$. A few years prior, Kubert has worked on this problem and has established a numb …
1
vote
Hilbert class field tower
Posting as an answer to get out of the unanswered list.
Each field $H^i_K$ in the class field tower of $K$ is Galois over $K$. Indeed, let $\sigma$ be any automorphism of the algebraic closure of $K$ …
12
votes
Accepted
Rationality of field embeddings
The answer is yes. Suppose $S$ is nonempty. Write $K=\mathbb Q(\alpha)$ (using the primitive element theorem). Applying your assumption to $\alpha^n$ for all $n\in\mathbb N$ we get that all sums $\sum …
11
votes
3
answers
1k
views
Is every group an ideal class group of a number field?
The inverse Galois problem asks whether every finite group appears as the Galois group of some finite extension of $\mathbb Q$. I was wondering to what extent the analogous problem for ideal class gro …
9
votes
A ring map from algebraic integers to algebraic closure of $\mathbb F_p$
The other answer is excellent and provides a lot more context, but if you are just after the existence statement, then there is a much more straightforward argument.
Since $p$ is not invertible in $\o …
4
votes
Accepted
What's the class group of $\mathbb{Q}^{\mathrm{ab}}$?
From Armand Brumer, The class group of all cyclotomic integers:
As an abelian group, $\mathrm{Pic}(O_\infty)$ is isomorphic to a countable direct sum of copies of $\mathbb Q/\mathbb Z$.
Here $O_\inf …
7
votes
Need for Drinfeld modules compared to elliptic curves over function field
I am not entirely sure what the best answer to this question would look like, but let me elaborate a little on the explicit class field theory point.
Most broadly speaking, the reason why elliptic cur …
7
votes
Does the equation $x^2+x=a$ have an integer solution?
$a$ is an integer of the form $x^2+x$ iff $4a+1$ is a perfect square. We have the following result:
Theorem: Integer $N$ is a perfect square iff it is a square modulo every prime.
Proof: This is tru …
6
votes
Accepted
Completion of infinite degree extension of perfectoid fields is perfectoid?
I'm assuming you mean infinite algebraic extensions, as otherwise there is no standard way of completing them.
Let $K$ be a perfectoid field, let $L$ be an infinite algebraic extension. Then $L$ admit …