# Legendre symbols as multiplicative homomorphisms in number fields

$\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}$

Suppose we have a finite set of rational primes $B=\{p_1,\ldots,p_k\}$, and $V=\{ x\in\mathbb{Q}^*~|~x\text{ contains only primes in B} \}$. So for example $B=\{2,3,5,7,11\}$, and $x=-\frac{700}{9}$. Elements of $V$ are those that don't contain primes outside of $B$.

Now consider an odd prime $q\not\in B$. The Legendre symbol over $q$ is a well defined homomorphism from $V$ to $\{\pm 1\}$ \begin{align} \legendre{\cdot}{q}:&V\longrightarrow \{\pm 1\}\\ &x\longmapsto \legendre{x}{q} \end{align} Is it true that every element of the space $\operatorname{Hom}(V,\{\pm1\})$ is of that form, i.e. it's a Legendre symbol for a given $q$?

As far as I see it, we have

$$V\cong\mathbb{Z}/2\oplus\bigoplus_{p\in B}\mathbb{Z}$$ where the $\mathbb{Z}/2$ is for the contribution of $-1$, and $\mathbb{Z}$ is for the contribution of $p$ in the factorization of elements of $V$. So to find a map from $V$ to $\{\pm1\}$ we need to find, for every given element $(a_0,\ldots,a_k)\in\{\pm1\}^{k+1}$ a $q$ such that $\legendre{-1}{q}=a_0$ and $\legendre{p_i}{q}=a_i$ for $1\leq i\leq k$. Because of the law of quadratic reciprocity, and because we know that $\legendre{-1}{q}$ and $\legendre{2}{q}$ are equivalent to a congruence relation of $q$ modulo 4/8, this is essentially equivalent to a congruence relation modulo $N=8\prod_{p\in B}p$. Given the Dirichlet arithmetic progression theorem, the primes are equidistributed (by Dirichlet and natural density) over the residue classes ${\mathbb{Z}/N}^*$, so there should exist a prime $q$ for every homomorphism $\chi:V\longrightarrow \{\pm 1\}$.

Now, i finally arrive to my questions.

The first is - can this be proved without the quadratic reciprocity law? Both the existence of a Legendre symbol for every character $\chi\in\operatorname{Hom}(V,\{\pm1\})$, and the equidistribution over the primes.

The second question is - in what way can this be extended to rings of integers $\mathcal{O}$ of number fields $\mathbb{L}\supseteq\mathbb{Q}$? We can still define $B$ as a set of prime ideals of $\mathcal{O}$, and we can use first grade prime ideals $\mathfrak{q}$, the isomorphism $\mathcal{O}/\mathfrak{q}\cong \mathbb{F}_q$ and the map \begin{align} \legendre{\cdot}{\mathfrak{q}}:&V\longrightarrow\{\pm1\}\\ &x\longmapsto \legendre{x \bmod \mathfrak{q}}{q} \end{align} as our Legendre symbols. The problem is that I don't know if there are quadratic reciprocity laws for a general $\mathcal{O}$, and I didn't find anything about the distribution of primes with regards to units of $\mathcal{O}$. There should be a way to connect this to the Chebotarev density theorem, but i just don't see it.

• So i thought about it a little, and I'd like to ask - how one i deduce the character $\left(\frac{\cdot}{\mathfrak{q}}\right)$ induced by $\mathfrak{q}$ given it's Artin symbol $\left(\frac{\mathbb{L}/\mathbb{Q}}{\mathfrak{q}}\right)$ ? – Kolja Nov 9 '16 at 16:07