0
$\begingroup$

Let we have $K= \mathbb{Q}(\sqrt d)$ . For a rational prime $p$, we have the following cases if we consider its ramification in $O_K$:

$p$ is ramified if $p| \Delta$.

$p$ splits if $(\cfrac{d}{p}) = 1$

$p$ is inert if $(\cfrac{d}{p}) = -1$

Now let us define a character modulo $\Delta$, we know that $\Delta$ is $d$ if $d \equiv 1 \mod 4$ and $4d$ if $d \equiv 3 \mod 4$ .

$\chi(p) = 0$ if $p$ is ramified.

$\chi(p)=1$ if $p$ splits if $(\cfrac{d}{p}) = 1$

$\chi(p) = -1$ if p is inert if $(\cfrac{d}{p}) = -1$

From now on I have some problems understanding $\chi(.)$, the rest is my thinking:

$\chi(p)$ is actually $(\cfrac{d}{p}) $. I am not sure what this symbol is. I think that it is not Legendre symbol. Probably it is Kronoecker symbol because otherwise $(\cfrac{d}{2})$ is meaningless.

For example when we have $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{5}]$ what is the corresponding $L-$series $L(\chi,s)$ ?

P.s. I do not know how to prove that $\chi(.)$ is a character. To add, they are $8$ periodic and $5$ periodic since their discriminants are $8$ and $5$ respectively.

$\endgroup$
  • $\begingroup$ Yes, I think it is the Legendre symbol. In the case of $2$, however you should use $(-1)^{\frac{p^2-1}{8}}$ which determines whether $2$ a quadratic residue. $\endgroup$ – Rene Schipperus Feb 24 '18 at 22:26
  • $\begingroup$ What does it come from? $\endgroup$ – Ninja Feb 24 '18 at 22:36
  • $\begingroup$ I dont understand what you mean, the Legendre symbol ? $\endgroup$ – Rene Schipperus Feb 24 '18 at 22:38
  • $\begingroup$ We have $(2|p) = (-1)^{p^2 -1}/8$ but we are looking for $\chi(2) = (8 | 2)$ so I am confused. $\endgroup$ – Ninja Feb 24 '18 at 22:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.