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Let $\chi$ be a Dirichlet character. This is,

$$\chi: (\mathbb{Z}/m\mathbb{Z})^* \to S^1$$

I'm trying to understand the relation between those characters and fiel extensions.

For example, the principal (trivial) character $\chi_0$ mod $1$ is naturally associated to $\mathbb{Q}$, in the sense that $L(s,\chi_0)=\zeta(s)$.

In particular I'm interested in the following. Let $K$ be a quadratic extensions. We know that:

$$\zeta_K(s)=\zeta(s)L(s,\chi_K)$$

Let's call quadratic characters to those who appear that way. There should be a bijection between quadratic characters and quadratic extensions (right?).

Are all non-principal characters quadratic?

But perhaps the most natural question is,

Given a Dirichlet character $\chi$, how do we get a field extension $K$ from it? Is $K$ quadratic always? Is this process unique?

Thanks for any information. I'd appreciate references also.

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    $\begingroup$ The correspondence between quadratic fields and quadratic Dirichlet characters, that is, those characters such that $\chi(n)=\pm 1$ for $n$ coprime to the conductor of $\chi$, is the Quadratic Reciprocity Law. $\endgroup$ Aug 22, 2016 at 17:15
  • $\begingroup$ Notice also that $L(s,\chi_0)\neq \zeta(s)$. $\endgroup$
    – PITTALUGA
    Aug 22, 2016 at 17:19
  • $\begingroup$ @ÁngelValencia Thanks, but what about the rest of characters? $\endgroup$ Aug 22, 2016 at 17:23
  • $\begingroup$ @PITTALUGA I ment mod 1. $\endgroup$ Aug 22, 2016 at 17:26
  • $\begingroup$ In general, Dirichlet characters are associated to characters of abelian extensions of $\Bbb Q$ via Class Field Theory. Check the relation, for example, at Bump et. al., An Introduction to the Langlands Program, Thm 5.2, p. 13. The book Cyclotomic Fields by L. Washington is more explicit. $\endgroup$ Aug 22, 2016 at 17:29

1 Answer 1

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This is probably not a complete answer to your question but here it is anyways.

Usually, one calls a Dirichlet character $\chi$ quadratic, if it has order 2 in the dual of $(\mathbb{Z}/m\mathbb{Z})^\ast$, i.e. if $\chi^2 = 1$ and $\chi$ is not principal. Notice that every qudratic character takes only the values $\pm 1$ and $-1$ at least once. Equivalently, a quadratic characater is a real and non-principal character.

Next, call an integer $D$ a quadratic discriminant, if it arises as the discriminant of a quadratic number field. Equivalently, $D$ is a quadratic discrimant, if $D$ is $\neq 1$, square-free and $D \equiv 1 \mod{4}$ or if $D = 4d$ for some square-free integer $d \equiv 2$ or $3 \mod 4$.

Attached to every quadratic discriminant $D$ is a special quadratic character, called the Kronecker symbol, sometimes denoted by $\chi_D(n)$ or $\left( \frac{D}{n} \right)$ where $n$ is the argument. One can define it by requiring:

  1. $\chi_D$ is completely multiplicative.
  2. $\chi_D(0) = 0$ and $\chi_D(1) = 1$.
  3. For every odd prime $p$, $\chi_D(p)$ is equal to the Legendre symbol mod $p$.
  4. $\chi_D(2) = \begin{cases}0 & \text{ if } D \equiv 2 \pmod{2}\\ 1 &\text{ if } D \equiv 1 \pmod{8}\\ -1 &\text{ if } D \equiv 5 \pmod{8} \end{cases}$.
  5. $\chi_D(-1) = \begin{cases}1 & \text{ if } D > 0 \\ -1 & \text{ if } D < 0 \end{cases}$.

One has then the following Theorem, whose proof uses a lot of the theory of Dirichlet characters and Gauss' quadratic reciprocity law.

Theorem. For every quadratic discrimant $D$, the Kronecker symbol $\chi_D$ defines a primitive quadratic character mod $|D|$. Conversely, every primitve quadratic character is given by a Kronecker symbol and hence every quadratic character is induced by some Kronecker symbol.

In my opinion, Montgomery's and Vaughan's ''Multiplicative number theory'' or Davenport's classical ''Multiplicative number theory'' is a good reference for this.

Furthermore, the following formula holds for quadratic fields $K$:

$$\zeta_K(s) = \zeta(s)L(s, \chi_D).$$

In your notation $\chi_D = \chi_K$.

The answer to your first question is no with both your and 'my' defintion of quadratic.

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  • $\begingroup$ Thanks, that helps a lot! I'll go through it more carefully, and give a look to the books you reference. $\endgroup$ Aug 22, 2016 at 19:54

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