# Proving that a Dirichlet character $\chi$ must be the Kronecker symbol $\chi_D = \left ( \frac{D}{\cdot} \right )$

I'm stuck trying to prove that a particular Dirichlet Character $\chi$ must be the Kronecker symbol $\chi_D = \left ( \frac{D}{\cdot}\right )$. The context is the following.

Let $d$ be a square-free integer and define $D$ to be

$D = \begin{cases} |d| & \text{if$d \equiv 1 \pmod{4}$}\\ 4|d| & \text{if$d \equiv 2, 3 \pmod{4}$} \end{cases}$

Now let $H = \left \{ [i] \in (\mathbb{Z}/D \mathbb{Z})^\times \mid \chi_D (i) = \left ( \frac{D}{i}\right ) = 1 \right \}$ and define X to be the set of Dirichlet characters

$$X = \{ \chi \in \widehat{(\mathbb{Z}/D \mathbb{Z})^\times} \mid \ \chi|_H \equiv 1 \}$$

That is, $X$ is the set of Dirichlet characters on $(\mathbb{Z}/D \mathbb{Z})^\times$ that are trivial when restricted to $H$.

My Problem

Then I would like to prove that $X = \{ \chi_0 , \chi_D \}$ where $\chi_0$ is the trivial character and $\chi_D$ is the character defined by the Kronecker symbol.

My Attempt

So what I tried is taking an arbitrary nontrivial character $\chi \in X$ and I want to show that $\chi = \chi_D$.

By definition of $X$ and $H$, we know already that $\chi(i) = 1$ for any $i \in H$ and since for any $i \in H$ we have $\chi_D (i) = 1$ then this shows that $\chi |_H = \chi_D |_H$. Thus it remains to show that the two characters agree in the set $(\mathbb{Z}/D \mathbb{Z})^\times \setminus H$.

Now, if $i \in (\mathbb{Z}/D \mathbb{Z})^\times \setminus H$ then that means that $\chi_D (i) = -1$. Therefore it must be proved that $\chi(i) = -1$ also.

But at this point I'm stuck and I don't know what to do. I realize that I'm not using properties of the Kronecker symbol and that might be the key, but I don't know how to proceed. I would really appreciate any help with this problem.

Thank you for any help.

## 1 Answer

Do you know that the Kronecker symbol actually defines a character (i.e. a homomorphism)? If not, this is the key property that you need. If yes, then you know that $H$ is a subgroup of $(\mathbb Z/D)^{\times}$, and it might help to regard both $\chi_D$ and $\chi$ (using your notation) as characters on the quotient $(\mathbb Z/D)^{\times}/H$. What is the order of this group? How many non-trivial characters does it admit?

• Dear Matt, thank you very much. It took me some time to answer back because I was writing all the details, but I think I got it. – Adrián Barquero Oct 15 '12 at 3:30
• If I'm not mistaken, $H$ is the kernel of $\chi_D$, so it is a subgroup of $(\mathbb{Z}/D)^\times$, and then it follows (by the first isomorphism theorem) that $(\mathbb{Z}/D)^\times / H \cong \{ \pm 1 \}$, so that this quotient group admits only one non-trivial character, and from this we can conclude that $\chi = \chi_D$. I'm I right? – Adrián Barquero Oct 15 '12 at 3:34
• @AdriánBarquero: Dear Adrián, That's right! Best wishes, – Matt E Oct 15 '12 at 12:16