Weil Conjecture for Curves

Proof. First any image of geometric point $\mathop{\mathrm{Spec}}(\mathbb{F}_{q^{m}})\to X$ is closed since the extension is algebraic. We act $\mathrm{Gal}(\mathbb{F}_{q^{m}}/\mathbb{F_{q}})$ on $\mathop{\mathrm{Hom}}(k(x),\mathbb{F}_{q^{m}})$ by composition on the left. Clearly the action is transitive and has stablizer $\mathrm{Gal}(k(x)/\mathbb{F}_{q})$. So $\left| \mathop{\mathrm{Hom}}(k(x),\mathbb{F}_{q^{m}})\right|=[k(x):\mathbb{F}_{q}]$. ◻

Proof. Note that $\Gamma_{f}$ is obtained by pulling back the diagonal:

$\Delta$ is closed immersion so $\Gamma_{f}$ is a closed immersion, hence $\Gamma_{f}$ is an isomorphism to the schematic image $\overline{\Gamma_{f}(X)}$. Since $X$ is reduced, we may factor $\Gamma_{f}$ through $\Gamma_{f}(X)$, which induces an morphism from $\overline{\Gamma_{f}(X)}$ to $\Gamma_{f}(X)$. From the definition of reduced subscheme, $\overline{\Gamma_{f}(X)}\cong \Gamma_{f}(X)$. ◻

Proof. The fixed point of $X$ under Galois group action $\mathrm{Gal}(\bar{\mathbb{F}_{q}} /\mathbb{F}_{q^{m}})$ is exactly the $\mathbb{F}_{q^{m}}$-points $X(\mathbb{F}_{q^{m}})$. So $x$ is $\mathbb{F}_{q^{m}}$-point if and only if it’s in the intersection of $\Gamma_{\varphi}\cdot \Delta$. It suffices to show each point has multiplicity 1. From the lemma, $\Gamma_{\varphi}$ and $\Delta$ are all isomorphic to $X$, so they are nonsingular. For every point $(\varphi(x),x)\in \Gamma_{\varphi}\cap \Delta$, the tangent space is the product of corresponding tangent space $X$. Note $(d\varphi)_{x}=0$, $\Gamma_{\varphi}$ and $\Delta$ intersect transversally, so each point has single multiplicity. ◻

Proof. Since $\mathrm{rk}(\mathcal{L}_{1})=\frac{\alpha_{2}(\mathcal{L}_{1})}{\alpha_{2}(\mathcal{O}_{X})}$ and $\mathrm{rk}(\mathcal{L}_{2})=\frac{\alpha_{2}(\mathcal{L}_{2})}{\alpha_{2}(\mathcal{O}_{X})}$, we have $\alpha_{2}(\mathcal{L}_{1})=\alpha_{2}(\mathcal{L}_{2})$. The Hilbert polynomial $P(\mathcal{L}_{1})(n)<P(\mathcal{L}_{2})(n)$. Suppose there is a morphism $\psi:\mathcal{L}_{1}\to \mathcal{L}_{2}$. Let $\mathcal{K}$, $\mathcal{F}$ be the kernel and the image of $\psi$, respectively.

We first show the support of $\mathcal{F}$ is $X$. Assume there is a point $x\in X$ that is not in the support of $\mathcal{F}$, since $\mathrm{Supp}\mathcal{F}$ is closed, theres’s an open neighborhood $U$ that has empty intersection with $\mathrm{Supp}\mathcal{F}$. We can choose an affine open set $W=\mathrm{Spec}A$ that has nonempty intersection with $X-U$ and $U$. Pick an nonzero global section $s$ supported on $X-U$ and $f$ is an element in ideal which define $W\cap \mathrm{Supp}s$. Then $s$ is zero in the localization of $D(f)\subset W$, so $sf^{n}=0$ for some $n$. However, $s$ is also in $\mathcal{L}_{1}(W)$, which is torsion-free, contradict to the above argument.

The Hilbert polynomial of $\mathcal{F}$ is also a quadratic function. Since $X$ is integral, the rank of $\mathcal{F}$ is integer, and it is 1. Thus $\mathrm{rk}(\mathcal{K})=0$, which means $\alpha_{2}(\mathcal{K})=0$, and $\dim \mathrm{Supp}\mathcal{K}\leq 1$. Using a similar argument, we can prove that $\mathcal{K}=0$. Therefore, $P(\mathcal{L}_{1})(n)=P(\mathcal{F})(n)>P(\mathcal{L}_{2})(n)$. This contradicts the fact that $\mathcal{F}$ is a subsheaf of $\mathcal{L}_{2}$. ◻

Proof. Assume $\deg D_{i}< m$. We first reduce to the case $\deg D_{i}$ is a negative fixed number. For any divisor $D$ in the family and the corresponding line sheaf $\mathcal{O}_{X}(D)=\mathcal{L}$, consider the exact sequence

$$0\to \mathcal{L}(-m)\xrightarrow{\cdot s} \mathcal{L}\to \mathcal{L}|_{C}\to 0$$

where $s$ is a regular section of $\mathcal{L}$ and $C$ is the support of $s$, by Bertini theorem, such $s$ exists. Then $h^{0}(\mathcal{L})\leq h^{0}(\mathcal{L}(-m))+h^{0}(\mathcal{L}|_{C})$. Note that $\deg_{C}(\mathcal{L}|_{C})=\deg_{C}(D\cap C)=\deg_{X\times X}(D)$. The genus $g_{C}$ is independent on the choice of $C$, and is given by the adjunction formula $g_{C}=1+\frac{1}{2}(K\cdot E_{m} +E_{m}\cdot E_{m})$ where $E_{m}$ is the divisor corresponding to $\mathcal{O}(m)$ and therefore bounded.

If $\deg_{C}(\mathcal{L}|_{C})>2g-2$, then $h^{0}(\mathcal{L}|_{C})=\deg_{C}(\mathcal{L}|_{C})+1-g\leq m+1-g$.

If $\deg_{C}(\mathcal{L}|_{C})\leq 2g-2$, then $h^{0}(\mathcal{L}|_{C})\leq h^{0}(\mathcal{L}|_{C}(2g-1-m))=g$.

Then we have $h^{0}(\mathcal{L}|_{C})$ bounded and we reduce to the case $\deg(D_{i})$ is negative constant by twisting a small number. Since a global section in $h^{0}(\mathcal{O}_{X}(D))$ will induce a morphism from $\mathcal{O}_{X}$ to $\mathcal{O}_{X}(D)$, by above lemma, $\mathcal{O}_{X}(D_{i})$ has no global sections. This finishes the proof. ◻

Proof. By above lemma, $\{h^{0}(nD)\}$ and $\{h^{0}(K-nD)\}$ are all bounded. Using Riemann-Roch theorem on surface,

$$h^{0}(nD)-h^{1}(nD)+h^{0}(K-nD)=\frac{1}{2}nD(nD-K)+1+p_{a}$$

For large $n$, left hand side of the equation is negative, so $n^{2}D\cdot D<0$. ◻

Proof. Let $V$ be the vector space generated by basis $P\times X$, $X\times P'$ and $D$. The intersection product defines a quadratic form

$$M=\begin{pmatrix} 0&1&d_{1}(D)\\ 1&0&d_{2}(D)\\ d_{1}(D)&d_{2}(D)&D\cdot D\end{pmatrix}$$

$\det(M)=2d_{1}(D)d_{2}(D)-D\cdot D$. If $\det (M)<0$, since $M$ is indefinite, we may choose $A_{1},A_{2},A_{3}$ such that $A_{i}\cdot A_{j}=0$, and $A_{i}^{T}MA_{i}=a_{i}$ such that $a_{1}, a_{2}>0$, $a_{3}<0$. Then appropriate linear combination can give a degree 0 divisor with positive self intersection. Contradict to Hodge index theorem. ◻

Proof of Theorem 9. By the above theorem, if we define $D*E=d_{1}(E)d_{2}(D)+d_{1}(D)d_{2}(E)-D\cdot E$, then $D*D\geq 0$. Apply above argument to $X_{\bar{\mathbb{F}_{q}}}\times X_{\bar{\mathbb{F}_{q}}}$ and Frobenius automorphism $\varphi$. Using the Cauchy inequality for bilinear forms, we have $\left| \Gamma_{\varphi}*\Delta \right|\leq \sqrt{(\Delta*\Delta)(\Gamma_{\varphi}*\Gamma_{\varphi})}$, that is

$$\left| \Gamma_{\varphi}\cdot \Delta-1-\deg(\varphi) \right|\leq 2g\sqrt{\deg (\varphi)}$$

◻

Proof. Set $n_{d}$ be the number of closed points with degree $d$ in $X$. By Proposition 4,

$$\begin{aligned} \log(Z_{X}(t))&=\sum_{m\geq 1}\frac{N_{m}}{m}t^{m}=\sum_{m\geq 1}\sum_{d\mid m}\frac{n_{d}\cdot d}{m}t^{m}\\ &= \sum_{d\geq 1}n_{d}\sum_{e} \frac{t^{de}}{e}\\ &=\sum_{d\geq 1}-n_{d}\log(1-t^{d}).\end{aligned}$$

We therefore have the desired formula. ◻

Proof. The number is exactly the size of $\left| \mathcal{O}(D) \right|$. See Ha² Theorem II 7.7. ◻

Proof of Theorem 16 part 1. As above, assume the image of $\deg:\mathop{\mathrm{Pic}}(X)\to \mathbb{Z}$ is $e\mathbb{Z}$. Let $d_{0}$ be the smallest number such that $d_{0}e\geq 2g-1$. Let $h=\left| \mathop{\mathrm{Pic}}^{0}(X) \right|$. Then

$$Z_{X}(t)=\sum_{\deg(D)\leq 2g-2}t^{\deg(D)}+\sum_{d\geq d_{0}}h\frac{q^{de-g+1}-1}{q-1}t^{de}.\qquad (1)$$

$\sum\limits_{\deg(D)\leq 2g-2}t^{\deg(D)}$ is a polynomial in $t^{e}$ of degree less than or equal to $\frac{2g-2}{e}$.

$$\sum_{d\geq d_{0}}h\frac{q^{de-g+1}-1}{q-1}t^{de}=\frac{h}{q-1}(q^{1-g}\frac{(qt)^{d_{0}e}}{1-(qt)^{e}}-\frac{t^{d_{0}e}}{1-t^{e}})$$

is also a rational function in $t^{e}$. So we may write $Z_{X}(t)=\frac{f(t^{e})}{(1-t^{e})(1-q^{e}t^{e})}$.

$$\begin{aligned} \lim\limits_{t\to 1}(t-1)Z_{X}(t)&=\lim\limits_{t\to 1}\frac{h(t-1)}{q-1}(q^{1-g}\frac{(qt)^{d_{0}e}}{1-(qt)^{e}}-\frac{t^{d_{0}e}}{1-t^{e}})\\&=\lim_{t\to 1}-\frac{ht^{d_{0}e}(t-1)}{(1-t^{e})(q-1)}=\frac{h}{e(q-1)}. \end{aligned}$$

$Z_{X}(t)$ has single pole at $t=1$. Consider Zeta function for $X_{\mathbb{F}_{q^{e}}}$. By the universal property of base change $\left| X_{\mathbb{F}_({q}^{e})^{m}} \right| =\left| X(\mathbb{F}_{q^{em}}) \right|$, thus

$$\log Z_{X_{\mathbb{F}_{q^{e}}}}(t^{e})=\sum_{m\geq 1}\frac{\left| X_{\mathbb{F}_{q^{e}}}(\mathbb{F}_{q^{em}}) \right| }{m}t^{em}=\sum_{m\geq 1}\frac{\left| X(\mathbb{F}_{q^{em}}) \right| }{m}t^{em}$$

Let $\xi$ be the primitive root of order $e$. We have $\sum\limits_{i=1}^{e}\xi^{il}=\left\{\begin{aligned} 0& \text{ if } e\nmid l\\ e& \text{ if } e\mid l \end{aligned}\right.$. So

$$\log Z_{X_{\mathbb{F}_{q^{e}}}}(t^{e})=\sum_{i=1}^{e}\sum_{l\geq 1}\frac{\left| X(\mathbb{F}_{q^{l}}) \right| }{l}\xi^{il}t^{l}.$$

$$Z_{X_{\mathbb{F}_{q^{e}}}}(t^{e})=\prod_{i=1}^{e}Z_{X}(\xi^{i}t)=\prod_{i=1}^{e}\frac{f(\xi^{ie}t^{e})}{(1-\xi^{ie}t^{e})(1-q^{e}\xi^{ie}t^{e})}=Z_{X}(t)^{e}.$$

By a similar argument, $Z_{X_{\mathbb{F}_{q^{e}}}}$ also has single pole at $t=1$, so $e=1$. So $Z_{X}(t)=\frac{f(t)}{(1-t)(1-qt)}$ where $f$ is an integer coefficient polynomial. Back to equation (1), we have $\deg f\leq 2g$. ◻

Proof of Theorem 16 part 2. If $g=0$, the only curve is $\mathbb{P}^{1}$. So $N_{m}=\frac{q^{2m}-1}{q^{m}-1}$ and $Z_{\mathbb{P}^{1}}(t)=\exp(\sum\limits_{m\geq 1}\frac{q^{m}+1}{m}t^{m})=\frac{1}{(1-t)(1-qt)}$. $Z_{\mathbb{P}^{1}}(t)$ obviously satisfies the functional equation.

If $g\geq 1$. Rewrite equation (1)

$$\begin{aligned} Z_{X}(t)&=\sum_{d=0}^{2g-2}(\sum_{\mathcal{L}\in \mathop{\mathrm{Pic}}^{d}(X)}\frac{q^{h^{0}(\mathcal{L})}-1}{q-1})t^{d}+\sum_{d\geq 2g-1}h\frac{q^{d-g+1}-1}{q-1}t^{d}\\&=\sum_{d=0}^{2g-2}(\sum_{\mathcal{L}\in \mathop{\mathrm{Pic}}^{d}(X)}\frac{q^{h^{0}(\mathcal{L})}-1}{q-1})t^{d}+\sum_{d\geq 2g-1}\frac{hq^{d-g+1}}{q-1}t^{d}-\sum_{d=0}\frac{ht^{d}}{q-1}\\&=\sum_{d=0}^{2g-2}(\sum_{\mathcal{L}\in \mathop{\mathrm{Pic}}^{d}(X)}\frac{q^{h^{0}(\mathcal{L})}-1}{q-1})t^{d}+\frac{hq^{g}t^{2g-1}}{(1-qt)(q-1)}-\frac{h}{(1-t)(q-1)}. \end{aligned}$$

Let $S_{1}(t)=\sum\limits_{d=0}^{2g-2}(\sum_{\mathcal{L}\in \mathop{\mathrm{Pic}}^{d}(X)}\frac{q^{h^{0}(\mathcal{L})}-1}{q-1})t^{d}$ and $S_{2}(t)=\frac{hq^{g}t^{2g-1}}{(1-qt)(q-1)}-\frac{h}{(1-t)(q-1)}$.

$$\begin{aligned} S_{2}(\frac{1}{qt})&=\frac{hq^{g}(\frac{1}{qt})^{2g-1}}{(1-\frac{1}{t})(q-1)}-\frac{h}{(1-\frac{1}{qt})(q-1)}\\&=q^{1-g}t^{2-2g}(\frac{hq^{g}t^{2g-1}}{(1-qt)(q-1)}-\frac{h}{(1-t)(q-1)})=q^{1-g}t^{2-2g}S_{2}(t).\end{aligned}$$

Note that $\mathcal{L}\mapsto \omega_{X}\otimes \mathcal{L}^{-1}$ gives a bijection between line bundles with degree lies in the interval $[0,2g-2]$. By Serre duality, $h^{0}(\omega_{X}\otimes \mathcal{L}^{-1})=h^{1}(\mathcal{L})=h^{0}(\mathcal{L})+g-1-\deg(\mathcal{L})$.

$$\begin{aligned} S_{1}(\frac{1}{qt})&=\sum_{d=0}^{2g-2}(\sum_{\mathcal{L}\in \mathop{\mathrm{Pic}}^{d}(X)}\frac{q^{h^{0}(\omega_{X}\otimes \mathcal{L}^{-1})}}{q-1})(\frac{1}{qt})^{2g-2-d}\\&=\sum_{d=0}^{2g-2}(\sum_{\mathcal{L}\in \mathop{\mathrm{Pic}}^{d}(X)}\frac{q^{h^{0}(\mathcal{L})+g-1-d}}{q-1})(\frac{1}{qt})^{2g-2-d}\\&=q^{1-g}t^{2-2g}S_{1}(t). \end{aligned}$$

Sum together, we have the functional equation $Z_{X}(\frac{1}{qt})=q^{1-g}t^{2-2g}Z_{X}(t)$. ◻

Proof of Theorem 16 part 3. Write $f(t)=\prod\limits_{i=1}^{2g}(1-\omega_{i}t)$ ($\omega_{i}$ can be 0).

$$Z_{X}(\frac{1}{qt})=\frac{\prod\limits_{i=1}^{2g}(1-\omega_{i}t)}{(1-\frac{1}{qt})(1-\frac{1}{t})}=\frac{q^{1-2g}t^{2-2g}\prod\limits_{i=1}^{2g}(tq-\omega_{i})}{(1-t)(1-qt)}.$$

Therefore

$$q^{g}\prod\limits_{i=1}^{2g}(t-\frac{\omega_{i}}{q})=\prod\limits_{i=1}^{2g}(1-\omega_{i}t)\qquad (2)$$

Observing the degree on each side, an immediate consequence is $\omega_{i}\neq 0$ for all $i$. Therefore $\deg(f)=2g$ and $B_{0}-B_{1}+B_{2}=2-2g$ ◻

Proof of Theorem 16 part 4. By above analysis,

$$Z_{X}(t)=\frac{\prod\limits_{i=1}^{2g}(1-\omega_{i}t)}{(1-t)(1-qt)}=\exp(\sum_{m\geq 1}\frac{N_{m}}{m}t^{m}).$$

$$\begin{aligned} \sum_{m\geq 1}\frac{N_{m}}{m}t^{m}&=\sum_{i=1}^{2g}\log(1-\omega_{i}t)-\log(1-t)-\log(1-qt)\\&=\sum_{m}\frac{1}{m\geq 1}(1+q^{m}-\sum_{i=1}^{2g}\omega_{i}^{m})t^{m}. \end{aligned}$$

So $N_{m}=1+q^{m}-\sum\limits_{i=1}^{2g}\omega_{i}^{m}$ Use Theorem 9, we have $\sum\limits_{i=1}^{2g}\omega_{i}^{m}\leq 2g\sqrt{q^{m}}$. Assume $\omega_{1}$ has the largest absolute value, then $\lim\limits_{m\to \infty}\left| \frac{\sum\limits_{i=1}^{2g}\omega_{i}^{m}}{\omega_{1}^{m}} \right|$ converges to a positive value. So we have $\left| \sum\limits_{i=1}^{2g}\omega_{i}^{m} \right| \geq k\left| \omega_{1}^{m} \right|$ for $m\gg 0$. Thus $\left| \omega_{1} \right|\leq (\frac{2g}{k})^{1 /m} \sqrt{q}$. Take $m\to \infty$, we have $\left| \omega_{1} \right|\leq \sqrt{q}$. By above remark, we know that $\left| \frac{q}{\omega_{i}} \right| =\left| \omega_{j} \right| \leq \sqrt{q}$. Therefore all $\left| \omega_{i} \right| =\sqrt{q}$. ◻