Legendre symbol

DEFINITION 6.1   Let $p$ be an odd prime. Let $a$ be an integer which is not divisible by $p$. Then we define the Legendre symbol $\left(\frac{a}{p}\right)$ by the following formula.

$${\left(\frac{a}{p}\right)}= \begin{cases}1 & \text{if }(X^2... ...ucible over }\mathbb{F}_p\\ -1 & \text{otherwise} \end{cases}$$

We further define

$${\left(\frac{a}{p}\right)}= 0$$$$\text { if } a \in p\mathbb{Z}.$$

LEMMA 6.2   Let $p$ be an odd prime. Then:

1. ${\left(\frac{a}{p}\right)}= a^{(p-1)/2} \mod p$
2. ${\left(\frac{ab}{p}\right)}= {\left(\frac{a}{p}\right)} {\left(\frac{b}{p}\right)}$

We note in particular that ${\left(\frac{-1}{p}\right)}=(-1)^{(p-1)/2}$.

DEFINITION 6.3   Let $p,\ell$ be distinct odd primes. Let $\lambda$ be a primitive $\ell$-th root of unity in an extension field of ${\mathbb{F}}_p$. Then for any integer $a$, we define a Gauss sum $\tau_a$ as follows.

$$\tau_a=\sum_{t=1}^{\ell-1}{\left(\frac{t}{\ell}\right)}\lambda^{at}$$

$\tau_1$ is simply denoted as $\tau$.

LEMMA 6.4  

1. $\tau_a={\left(\frac{a}{\ell}\right)}\tau$.
2. $\sum_{a=0}^{l-1} \tau_a \tau_{-a}=\ell(\ell-1)$.
3. $\tau^2=(-1)^{(\ell-1)/2}\ell$ ( $=\ell^*$ (say)).
4. $\tau^{p-1}=(\ell^*)^{(p-1)/2}$.
5. $\tau^p=\tau_{p}$.

THEOREM 6.5   ${\left(\frac{p}{\ell}\right)}={\left(\frac{\ell^*}{p}\right)}$ ( where $\ell^*=(-1)^{(\ell-1)/2}\ell$ ) ${\left(\frac{-1}{\ell}\right)}=(-1)^{(\ell-1)/2}$ ${\left(\frac{2}{\ell}\right)}=(-1)^{(\ell^2-1)/8}$

$p$-dependence of zeta functions is important topic. We are not going to talk about that in too much detail but let us explain a little bit.

Let us define the zeta function of a category $\mathcal{C}$ [1].

$$\zeta(s,\mathcal{C})=\prod_{P\in P(\mathcal{C})} (1-N(P)^{-s})^{-1}$$

where $P$ runs over all finite simple objects.

• $P$: finite $\overset{\text{def}}{\iff}$ $N(P)\overset{\rm {def}}{=}\char93 \operatorname{End}(P)<\infty$.
• $P$: simple $\overset{\text{def}}{\iff}$ $\operatorname{Hom}(P,Y)\setminus\{0\}$ consists of mono morphisms.

For any commutative ring $A$, an $A$-module $M$ is simple if and only if $M\cong A/\mathfrak{m}$ for some maximal idea $\mathfrak{m}$ of $A$. We have thus:

$$\zeta(s,(A\operatorname{-modules})) = \prod_{\substack{\mathfrak{m} \in \operatorname{Spm}(A)\\ \char93 (A/\mathfrak{m})<\infty }} (1-\char93 (A/\mathfrak{m})^{-s})^{-1}$$

$$= \prod_{p:\text{prime}} \prod_ {\substack{\mathfrak{m} \in \operat... ...[A/\mathfrak{m}: \mathbb{F}_p]<\infty }} (1-\char93 (A/\mathfrak{m})^{-s})^{-1}$$

$$= \prod_p \prod_ {\substack{\mathfrak{m} \in \operatorname{Spm}(A/p... ...mathfrak{m}: \mathbb{F}_p]<\infty }} (1-\char93 ((A/p)/\mathfrak{m})^{-s})^{-1}$$

$$= \prod_p \zeta(s,(A/p)\operatorname{-modules}).$$

Let us take a look at the last line. It sais that the zeta is a product of zeta's on $A/p$. Let us fix a prime number $p$, put $\bar A=A/p$, and concentrate on $\bar A$ to go on further.

$$\zeta(s,(A/p)\operatorname{-modules}) = \prod_ {\substack{\mathfr... ...athfrak{m}: \mathbb{F}_p]<\infty }} (1-\char93 (\bar A/\mathfrak{m})^{-s})^{-1}$$

$$Z(\operatorname{Spec}(\bar A)/\mathbb{F}_p,T) = \exp(\sum_{r=1}^\infty(\operatorname{Spec}(\bar A)(\mathbb{F}_{p^r}),T))$$

$$= \prod_{\mathfrak{m} \in \operatorname{Spm}(A)} \exp(\sum_{r=1}^\infty(\operatorname{Spec}(\bar A/\mathfrak{m})(\mathbb{F}_{p^r}),T))$$

$$Z(\mathbb{F}_{q^e}/\mathbb{F}_q,T) =\exp(\sum_{e\vert r}\frac{e}{r} T^r) = (1-T^e)^{-1}$$

$$\zeta(s,\mathbb{F}_{p^e}\operatorname{-modules})=Z(\operatorname{Spec}(\mathbb{F}_{p^e})/\mathbb{F}_p,p^s)$$

We conclude:

PROPOSITION 6.6   Let $A$ be a commutative ring. Then:

1. We have a product formula.
   $$\zeta(s,(A\operatorname{-modules}))=\prod_p \zeta(s,(A/p)\operatorname{-modules})$$

2. $\zeta$ is obtained by substituting in the congruent zeta function by .