Search Results

No, such a field does not exist, since the Galois group of $k_{tr}/k_{nr}$ embeds into the multiplicative group of the residue field of $k_{nr}$, which your hypothesis implies to be finite.

The answer to the question as stated is "no". The original Cohen—Lenstra—Martinet heuristics say nothing about the $2$-Sylow subgroup of the class group of a quadratic field. These heuristics were lat …

A necessary and sufficient condition is that $a$, $b$ be linearly independent in $K^{\times}/K^{\times 2}$ and the Galois group of $K$ cyclically permute the classes of $a$, $b$, and $ab$ in $K^{\time …

This is a fun question, and I had already been thinking of making some comments on this in the question you link to. Apologies in advance for the long post. Actually, the quadratic subfield of $\mathb …

The cyclotomic field $K=\mathbb{Q}(\zeta_{23})$ contains the quadratic field $F=\mathbb{Q}(\sqrt{-23})$, and $F$ has class number $3$ (it is the first quadratic field, when those are ordered by absolu …

I am going to assume that by "algebra" you simply mean a ring. The answer is "no", in general. For example $\mathbb{Z}/5\mathbb{Z}$ is not the unit group of a ring. Indeed, suppose it was the unit gro …

Such $X$ do indeed exist, and are explicitly constructed in I.Z. Ruzsa, Solving a linear equation in a set of integers, Acta Arith., LXV.3 (1993), pp. 259-282, Theorem 7.5. The whole paper is devoted …

Why is the map ${\rm Gal}(FL/L)\to {\rm Gal}(F/K)$ injective? Elements of ${\rm Gal}(FL/L)$ are automorphisms of $FL$ that act trivially on $L$. To be in the kernel of the above map means to also act …

In addition to Felipe's reference, you can also have a look at Tom Fisher's papers https://www.dpmms.cam.ac.uk/~taf1000/papers/congr7and11.html and https://www.dpmms.cam.ac.uk/~taf1000/papers/congr9.h …

No, the set $S_f$ is empty for lots of forms. For example, if $a$ is not a square modulo $b$, then $ax^2+by^2$ is never a square (consider the equation modulo $b$). Thus, $S_f$ is empty for $f(x,y)=6x …

The simplest generalisation to abelian surfaces is, I believe, the statement (which is a theorem, not a conjecture) that the Igusa invariants of an abelian surface with CM by $K$ generate an abelian u …

No, not necessarily. For example, you can take $K\neq \mathbb{Q}$ to be totally real, and take $L$ to be a quadratic CM extension of $K$ whose unit group is equal to that of $K$ (e.g. make sure that $ …

You would usually want the principal Arakelov divisors, i.e. those of the form $(\sum_{\mathfrak{p}}{\rm ord}_{\mathfrak{p}}(a), \sum_\sigma -\log|\sigma(a)|)$ for $a\in F^\times$, to be cocompact in …

Use the Chinese remainder theorem to construct infinitely many $f_i$ such that $f_i$ is Eisenstein at one prime, and $f_i+g$ is Eisenstein at another.

There might well be an elementary construction of infinitely many points (which I cannot think of right now), but in any case, I think that there are experts out there who expect there to be infinitel …