Algebraic class field theory tells us that $\text{Gal}(\mathbb{Q}^{ab}/\mathbb{Q})$ is isomorphic to the group of connected components of the quotient $\mathbb{Q}^{\times}\backslash \mathbb{A}_{\mathbb{Q}}^{\times}\cong \prod_p \mathbb{Z_p}^{\times}\times \mathbb{R}_{>0}$, where $\mathbb{A}_{\mathbb{Q}}$ is the ring of adèles of $\mathbb{Q}$. It’s then said that the group of connected components is given by $\prod_p \mathbb{Z_p}^{\times}$, how can I see this? Thank you very […]

Given the Ulam spiral with center $c = 41$ and the numbers in a clockwise direction, we have, $$\begin{array}{cccccc} \color{red}{61}&62&63&64&\to\\ 60&\color{red}{47}&48&49&50\\ 59&46&\color{red}{\small{c=\,}41}&42&51\\ 58&45&44&\color{red}{43}&52\\ 57&56&55&54&\color{red}{53}&\downarrow\\ \end{array}$$ The main diagonal is defined by Euler’s polynomial $F(n) = n^2+n+41$, and yields distinct primes for 40 consecutive $n = 0\,\text{to}\,39$. If we let $c = 3527$ as in this […]

I tried to solve the exercise VIII.XX in Number Fields by Marcus. It asks to find the Hilbert class field of $Q(\sqrt m)$ for $m=-6,-10,-21,-30$. And the emphasis of this question is on the first two. Attempt to solve the first case. I knew that the class number of $K=Q(\sqrt{-6})$ is $2$, so its Hilbert […]

I got this doubt after going through tables given by Henri Cohen: Advanced topics in Computational number theory on “Hilbert Class Field of Imaginary quadratic field”. pg no.539-542, sec 12.1 and a Book by D A Cox on Primes of the form $x^2+n y^2$. In Cox Book, Hilbert Class field of $K=Q(\sqrt{-14})$ is computed as […]

As the classical book of David Cox argues, Assume the conditions are satisfied and $p$ can be represented as $x^2 + ny^2$. What would be a way to find solutions $(p,x,y)$ efficiently? Ideally, one would have one algorithm that works for all $n$.

Is the Hilbert class field of $K=\mathbb Q\left(\sqrt{-39}\right)$ is equal to $K\left(\sqrt{-39},\frac{\sqrt{1+\sqrt{13}}}{2}\right)$

Consider $\mathbb Q_p^{\text{ur}}$ the maximal unramified extension of the p-adic numbers. Suppose that on $\mathbb Q_p$ we have the usual absolute value that extends $|\frac{a}{b}|_p=\frac{1}{p^{v_p(a)-v_p(b)}}$ to $\mathbb Q_p$. Now it is known that $|\cdot|_p$ extends uniquely to $\mathbb Q_p^{\text{ur}}$. Is this absolute value discrete w.r.t. $\mathbb Q_p^{\text{ur}}$ ? The relevant definitions can be found here […]

Let $p$ be a prime, $n\geq 1$, $\zeta=\zeta_{p^n}$ a primitive $p^n$th root of unity, $L$ a number field, and $\wp$ a prime ideal of the ring of integers of $L$ lying above $p$. Suppose that $L(\zeta)$ is a non-trivial extension of $L$. Is $L(\zeta)/L$ necessarily ramified at $\wp$? I think so, but how do you […]

Let $K$ be a number field and consider the Arithmentic complex $\Gamma_{Ar}(1)^\bullet$ be defined by $$\begin{array} A\Bbb R^{r_1+r_2} & \stackrel{\Sigma}{\longrightarrow} & \Bbb R \\ \uparrow{l(\cdot)=\Pi\log |\sigma_i(\cdot)|} & & \uparrow{-\Sigma \log|\mathfrak p_i|^{-z_i}} \\ K^* & \stackrel{\operatorname{val}}{\longrightarrow} & \oplus_\mathfrak p\Bbb Z \end{array} $$ The left arrow going up is just the same map that appears in the […]

It is a classical result that every extension of $\mathbb{Q}$ is ramified. Put differently: there are no unramified extensions of $\mathbb{Q}$. The classical proof follows from the following two statements: (a) The only number field having discriminant $1$ is $\mathbb{Q}$ itself. (b) A prime number $p$ ramifies in a number field $K$ if and only […]

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