Mixing
Bombieri's Theorem (Weil Conjectures for Curves)
$begingroup$
I am studying Weil's conjectures for curves. Bombieri's theorem says:
Let $C$ be a smooth projective curve of genus g defined over $mathbb{F}_q$. Assume $q> (1+g)^4$ and $q=p^a$ where $a$ is even, and p is a prime number. Then
$$
C(mathbb{F}_q) leq 1+q+(2g+1)q^{frac{1}{2}},
$$
where
$C(mathbb{F}_q)$ is the $mathbb{F}_q$-rational points of $C$.
We have proven that the Riemann hypothesis is equivalent to the existence of a constant $A$ such that
$$
|C(mathbb{F}_{q}) - (1+q)| leq Aq^{frac{1}{2}},
$$
for some $q>>0$.
My question is:
Why is not Bombieri's theorem enough to prove Riemann hypothesis for curves?
If the curve is defined over $mathbb{F}_q$ for some $q$, then it is defined over $mathbb{F}_{q^m}$ for every $m geq 1$. Then we can find $q$ which satisfies the conditions of Bombieri's theorem, and $(2g+1)$ will be the required constant $A$.
Can you provide me with a counterexample? and what is wrong with my argument?
Edit 1: The answer is trivial, I did not notice the absolute value for the equivalence of the RH, so while Bombieri's theorem gives an upper bound for $C(mathbb{F}_q)$, we still need a lower bound to prove RH.
I am still asking about the condition about $q$ in Bombieri's theorem, does my argument work for proving that we can always assume it is fulfilled for any curve or we really need Galois covering?
Edit 2:
Now, I feel stupid! Galois coverings are used to obtain the lower bound, I misread the inequality of another proposition again! My question about the condition about $q$ is still valid though.
algebraic-geometry algebraic-curves projective-varieties
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
$endgroup$
|
show 1 more comment
$begingroup$
I am studying Weil's conjectures for curves. Bombieri's theorem says:
Let $C$ be a smooth projective curve of genus g defined over $mathbb{F}_q$. Assume $q> (1+g)^4$ and $q=p^a$ where $a$ is even, and p is a prime number. Then
$$
C(mathbb{F}_q) leq 1+q+(2g+1)q^{frac{1}{2}},
$$
where
$C(mathbb{F}_q)$ is the $mathbb{F}_q$-rational points of $C$.
We have proven that the Riemann hypothesis is equivalent to the existence of a constant $A$ such that
$$
|C(mathbb{F}_{q}) - (1+q)| leq Aq^{frac{1}{2}},
$$
for some $q>>0$.
My question is:
Why is not Bombieri's theorem enough to prove Riemann hypothesis for curves?
If the curve is defined over $mathbb{F}_q$ for some $q$, then it is defined over $mathbb{F}_{q^m}$ for every $m geq 1$. Then we can find $q$ which satisfies the conditions of Bombieri's theorem, and $(2g+1)$ will be the required constant $A$.
Can you provide me with a counterexample? and what is wrong with my argument?
Edit 1: The answer is trivial, I did not notice the absolute value for the equivalence of the RH, so while Bombieri's theorem gives an upper bound for $C(mathbb{F}_q)$, we still need a lower bound to prove RH.
I am still asking about the condition about $q$ in Bombieri's theorem, does my argument work for proving that we can always assume it is fulfilled for any curve or we really need Galois covering?
Edit 2:
Now, I feel stupid! Galois coverings are used to obtain the lower bound, I misread the inequality of another proposition again! My question about the condition about $q$ is still valid though.
algebraic-geometry algebraic-curves projective-varieties
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
$endgroup$
$begingroup$
The main point is that Riemann-Roch implies $(1-qT)(1-T)zeta_{C/mathbb{F}_q}(T)=exp(sum_{n=1}^infty frac{#C(mathbb{F}_{q^n})-q^n-1}{n} T^n) $ is a polynomial $prod_{j=1}^{2g} (1-a_j T)$ so $#C(mathbb{F}_{q^n}) -q^n-1= -sum_{j=1}^{2g} a_j^n$ and there exists $A_r$ such that $forall n, |#C(mathbb{F}_{q^n}) -q^n-1|le A_r q^{nr}$ iff $forall j,|a_j|le q^{r}$ (so $A_r = 2g$). The Riemann hypothesis is $r=1/2$, equivalently $r -1/2$ arbitrary small, which implies by the functional equation $|a_j|= q^{1/2}$.
$endgroup$
– reuns
Jan 15 at 4:58
$begingroup$
I am sorry, I do not understand how this answers my questions.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:01
$begingroup$
So the RH is equivalent to $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$ for every $n$, not only one. This is of course what Bombieri did, since in his proof $q$ was an arbitrary power of the cardinality of the base field
$endgroup$
– reuns
Jan 15 at 5:07
$begingroup$
Yes, this is true. But from properties of the Riemann zeta function for curves. We can prove that it is sufficient to prove the Riemann hypothesis for only one $q$, (cf. Hansen, Rational points on Curves over Finite Fields, Section 3.1). I do not know f there is another form for Bombieri's theorem, but still my question about the quoted statement.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:20
$begingroup$
No you misunderstood. It is sufficient to prove $|#C(mathbb{F}_{q}) - q-1| le (2g+1)q^{1/2}$ for every curve $C$ defined over $mathbb{F}_{q}$, ie. that for a fixed curve $C$ defined over $mathbb{F}_{q}$, since it is defined over every $mathbb{F}_{q^n}$ and the genus of $C/mathbb{F}_{q}$ and $C/mathbb{F}_{q^n}$ are the same, that for every $n$, $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$
$endgroup$
– reuns
Jan 15 at 5:32
|
show 1 more comment
$begingroup$
I am studying Weil's conjectures for curves. Bombieri's theorem says:
Let $C$ be a smooth projective curve of genus g defined over $mathbb{F}_q$. Assume $q> (1+g)^4$ and $q=p^a$ where $a$ is even, and p is a prime number. Then
$$
C(mathbb{F}_q) leq 1+q+(2g+1)q^{frac{1}{2}},
$$
where
$C(mathbb{F}_q)$ is the $mathbb{F}_q$-rational points of $C$.
We have proven that the Riemann hypothesis is equivalent to the existence of a constant $A$ such that
$$
|C(mathbb{F}_{q}) - (1+q)| leq Aq^{frac{1}{2}},
$$
for some $q>>0$.
My question is:
Why is not Bombieri's theorem enough to prove Riemann hypothesis for curves?
If the curve is defined over $mathbb{F}_q$ for some $q$, then it is defined over $mathbb{F}_{q^m}$ for every $m geq 1$. Then we can find $q$ which satisfies the conditions of Bombieri's theorem, and $(2g+1)$ will be the required constant $A$.
Can you provide me with a counterexample? and what is wrong with my argument?
Edit 1: The answer is trivial, I did not notice the absolute value for the equivalence of the RH, so while Bombieri's theorem gives an upper bound for $C(mathbb{F}_q)$, we still need a lower bound to prove RH.
I am still asking about the condition about $q$ in Bombieri's theorem, does my argument work for proving that we can always assume it is fulfilled for any curve or we really need Galois covering?
Edit 2:
Now, I feel stupid! Galois coverings are used to obtain the lower bound, I misread the inequality of another proposition again! My question about the condition about $q$ is still valid though.
algebraic-geometry algebraic-curves projective-varieties
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
$endgroup$
I am studying Weil's conjectures for curves. Bombieri's theorem says:
Let $C$ be a smooth projective curve of genus g defined over $mathbb{F}_q$. Assume $q> (1+g)^4$ and $q=p^a$ where $a$ is even, and p is a prime number. Then
$$
C(mathbb{F}_q) leq 1+q+(2g+1)q^{frac{1}{2}},
$$
where
$C(mathbb{F}_q)$ is the $mathbb{F}_q$-rational points of $C$.
We have proven that the Riemann hypothesis is equivalent to the existence of a constant $A$ such that
$$
|C(mathbb{F}_{q}) - (1+q)| leq Aq^{frac{1}{2}},
$$
for some $q>>0$.
My question is:
Why is not Bombieri's theorem enough to prove Riemann hypothesis for curves?
If the curve is defined over $mathbb{F}_q$ for some $q$, then it is defined over $mathbb{F}_{q^m}$ for every $m geq 1$. Then we can find $q$ which satisfies the conditions of Bombieri's theorem, and $(2g+1)$ will be the required constant $A$.
Can you provide me with a counterexample? and what is wrong with my argument?
Edit 1: The answer is trivial, I did not notice the absolute value for the equivalence of the RH, so while Bombieri's theorem gives an upper bound for $C(mathbb{F}_q)$, we still need a lower bound to prove RH.
I am still asking about the condition about $q$ in Bombieri's theorem, does my argument work for proving that we can always assume it is fulfilled for any curve or we really need Galois covering?
Edit 2:
Now, I feel stupid! Galois coverings are used to obtain the lower bound, I misread the inequality of another proposition again! My question about the condition about $q$ is still valid though.
algebraic-geometry algebraic-curves projective-varieties
algebraic-geometry algebraic-curves projective-varieties
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
edited Jan 15 at 7:27
Ahmed Elashry
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
asked Jan 15 at 4:23
Ahmed ElashryAhmed Elashry
388
388
$begingroup$
The main point is that Riemann-Roch implies $(1-qT)(1-T)zeta_{C/mathbb{F}_q}(T)=exp(sum_{n=1}^infty frac{#C(mathbb{F}_{q^n})-q^n-1}{n} T^n) $ is a polynomial $prod_{j=1}^{2g} (1-a_j T)$ so $#C(mathbb{F}_{q^n}) -q^n-1= -sum_{j=1}^{2g} a_j^n$ and there exists $A_r$ such that $forall n, |#C(mathbb{F}_{q^n}) -q^n-1|le A_r q^{nr}$ iff $forall j,|a_j|le q^{r}$ (so $A_r = 2g$). The Riemann hypothesis is $r=1/2$, equivalently $r -1/2$ arbitrary small, which implies by the functional equation $|a_j|= q^{1/2}$.
$endgroup$
– reuns
Jan 15 at 4:58
$begingroup$
I am sorry, I do not understand how this answers my questions.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:01
$begingroup$
So the RH is equivalent to $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$ for every $n$, not only one. This is of course what Bombieri did, since in his proof $q$ was an arbitrary power of the cardinality of the base field
$endgroup$
– reuns
Jan 15 at 5:07
$begingroup$
Yes, this is true. But from properties of the Riemann zeta function for curves. We can prove that it is sufficient to prove the Riemann hypothesis for only one $q$, (cf. Hansen, Rational points on Curves over Finite Fields, Section 3.1). I do not know f there is another form for Bombieri's theorem, but still my question about the quoted statement.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:20
$begingroup$
No you misunderstood. It is sufficient to prove $|#C(mathbb{F}_{q}) - q-1| le (2g+1)q^{1/2}$ for every curve $C$ defined over $mathbb{F}_{q}$, ie. that for a fixed curve $C$ defined over $mathbb{F}_{q}$, since it is defined over every $mathbb{F}_{q^n}$ and the genus of $C/mathbb{F}_{q}$ and $C/mathbb{F}_{q^n}$ are the same, that for every $n$, $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$
$endgroup$
– reuns
Jan 15 at 5:32
|
show 1 more comment
$begingroup$
The main point is that Riemann-Roch implies $(1-qT)(1-T)zeta_{C/mathbb{F}_q}(T)=exp(sum_{n=1}^infty frac{#C(mathbb{F}_{q^n})-q^n-1}{n} T^n) $ is a polynomial $prod_{j=1}^{2g} (1-a_j T)$ so $#C(mathbb{F}_{q^n}) -q^n-1= -sum_{j=1}^{2g} a_j^n$ and there exists $A_r$ such that $forall n, |#C(mathbb{F}_{q^n}) -q^n-1|le A_r q^{nr}$ iff $forall j,|a_j|le q^{r}$ (so $A_r = 2g$). The Riemann hypothesis is $r=1/2$, equivalently $r -1/2$ arbitrary small, which implies by the functional equation $|a_j|= q^{1/2}$.
$endgroup$
– reuns
Jan 15 at 4:58
$begingroup$
I am sorry, I do not understand how this answers my questions.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:01
$begingroup$
So the RH is equivalent to $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$ for every $n$, not only one. This is of course what Bombieri did, since in his proof $q$ was an arbitrary power of the cardinality of the base field
$endgroup$
– reuns
Jan 15 at 5:07
$begingroup$
Yes, this is true. But from properties of the Riemann zeta function for curves. We can prove that it is sufficient to prove the Riemann hypothesis for only one $q$, (cf. Hansen, Rational points on Curves over Finite Fields, Section 3.1). I do not know f there is another form for Bombieri's theorem, but still my question about the quoted statement.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:20
$begingroup$
No you misunderstood. It is sufficient to prove $|#C(mathbb{F}_{q}) - q-1| le (2g+1)q^{1/2}$ for every curve $C$ defined over $mathbb{F}_{q}$, ie. that for a fixed curve $C$ defined over $mathbb{F}_{q}$, since it is defined over every $mathbb{F}_{q^n}$ and the genus of $C/mathbb{F}_{q}$ and $C/mathbb{F}_{q^n}$ are the same, that for every $n$, $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$
$endgroup$
– reuns
Jan 15 at 5:32
$begingroup$
The main point is that Riemann-Roch implies $(1-qT)(1-T)zeta_{C/mathbb{F}_q}(T)=exp(sum_{n=1}^infty frac{#C(mathbb{F}_{q^n})-q^n-1}{n} T^n) $ is a polynomial $prod_{j=1}^{2g} (1-a_j T)$ so $#C(mathbb{F}_{q^n}) -q^n-1= -sum_{j=1}^{2g} a_j^n$ and there exists $A_r$ such that $forall n, |#C(mathbb{F}_{q^n}) -q^n-1|le A_r q^{nr}$ iff $forall j,|a_j|le q^{r}$ (so $A_r = 2g$). The Riemann hypothesis is $r=1/2$, equivalently $r -1/2$ arbitrary small, which implies by the functional equation $|a_j|= q^{1/2}$.
$endgroup$
– reuns
Jan 15 at 4:58
$begingroup$
The main point is that Riemann-Roch implies $(1-qT)(1-T)zeta_{C/mathbb{F}_q}(T)=exp(sum_{n=1}^infty frac{#C(mathbb{F}_{q^n})-q^n-1}{n} T^n) $ is a polynomial $prod_{j=1}^{2g} (1-a_j T)$ so $#C(mathbb{F}_{q^n}) -q^n-1= -sum_{j=1}^{2g} a_j^n$ and there exists $A_r$ such that $forall n, |#C(mathbb{F}_{q^n}) -q^n-1|le A_r q^{nr}$ iff $forall j,|a_j|le q^{r}$ (so $A_r = 2g$). The Riemann hypothesis is $r=1/2$, equivalently $r -1/2$ arbitrary small, which implies by the functional equation $|a_j|= q^{1/2}$.
$endgroup$
– reuns
Jan 15 at 4:58
$begingroup$
I am sorry, I do not understand how this answers my questions.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:01
$begingroup$
I am sorry, I do not understand how this answers my questions.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:01
$begingroup$
So the RH is equivalent to $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$ for every $n$, not only one. This is of course what Bombieri did, since in his proof $q$ was an arbitrary power of the cardinality of the base field
$endgroup$
– reuns
Jan 15 at 5:07
$begingroup$
So the RH is equivalent to $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$ for every $n$, not only one. This is of course what Bombieri did, since in his proof $q$ was an arbitrary power of the cardinality of the base field
$endgroup$
– reuns
Jan 15 at 5:07
$begingroup$
Yes, this is true. But from properties of the Riemann zeta function for curves. We can prove that it is sufficient to prove the Riemann hypothesis for only one $q$, (cf. Hansen, Rational points on Curves over Finite Fields, Section 3.1). I do not know f there is another form for Bombieri's theorem, but still my question about the quoted statement.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:20
$begingroup$
Yes, this is true. But from properties of the Riemann zeta function for curves. We can prove that it is sufficient to prove the Riemann hypothesis for only one $q$, (cf. Hansen, Rational points on Curves over Finite Fields, Section 3.1). I do not know f there is another form for Bombieri's theorem, but still my question about the quoted statement.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:20
$begingroup$
No you misunderstood. It is sufficient to prove $|#C(mathbb{F}_{q}) - q-1| le (2g+1)q^{1/2}$ for every curve $C$ defined over $mathbb{F}_{q}$, ie. that for a fixed curve $C$ defined over $mathbb{F}_{q}$, since it is defined over every $mathbb{F}_{q^n}$ and the genus of $C/mathbb{F}_{q}$ and $C/mathbb{F}_{q^n}$ are the same, that for every $n$, $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$
$endgroup$
– reuns
Jan 15 at 5:32
$begingroup$
No you misunderstood. It is sufficient to prove $|#C(mathbb{F}_{q}) - q-1| le (2g+1)q^{1/2}$ for every curve $C$ defined over $mathbb{F}_{q}$, ie. that for a fixed curve $C$ defined over $mathbb{F}_{q}$, since it is defined over every $mathbb{F}_{q^n}$ and the genus of $C/mathbb{F}_{q}$ and $C/mathbb{F}_{q^n}$ are the same, that for every $n$, $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$
$endgroup$
– reuns
Jan 15 at 5:32
|
show 1 more comment
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$begingroup$
The main point is that Riemann-Roch implies $(1-qT)(1-T)zeta_{C/mathbb{F}_q}(T)=exp(sum_{n=1}^infty frac{#C(mathbb{F}_{q^n})-q^n-1}{n} T^n) $ is a polynomial $prod_{j=1}^{2g} (1-a_j T)$ so $#C(mathbb{F}_{q^n}) -q^n-1= -sum_{j=1}^{2g} a_j^n$ and there exists $A_r$ such that $forall n, |#C(mathbb{F}_{q^n}) -q^n-1|le A_r q^{nr}$ iff $forall j,|a_j|le q^{r}$ (so $A_r = 2g$). The Riemann hypothesis is $r=1/2$, equivalently $r -1/2$ arbitrary small, which implies by the functional equation $|a_j|= q^{1/2}$.
$endgroup$
– reuns
Jan 15 at 4:58
$begingroup$
I am sorry, I do not understand how this answers my questions.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:01
$begingroup$
So the RH is equivalent to $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$ for every $n$, not only one. This is of course what Bombieri did, since in his proof $q$ was an arbitrary power of the cardinality of the base field
$endgroup$
– reuns
Jan 15 at 5:07
$begingroup$
Yes, this is true. But from properties of the Riemann zeta function for curves. We can prove that it is sufficient to prove the Riemann hypothesis for only one $q$, (cf. Hansen, Rational points on Curves over Finite Fields, Section 3.1). I do not know f there is another form for Bombieri's theorem, but still my question about the quoted statement.
$endgroup$
– Ahmed Elashry
Jan 15 at 5:20
$begingroup$
No you misunderstood. It is sufficient to prove $|#C(mathbb{F}_{q}) - q-1| le (2g+1)q^{1/2}$ for every curve $C$ defined over $mathbb{F}_{q}$, ie. that for a fixed curve $C$ defined over $mathbb{F}_{q}$, since it is defined over every $mathbb{F}_{q^n}$ and the genus of $C/mathbb{F}_{q}$ and $C/mathbb{F}_{q^n}$ are the same, that for every $n$, $|#C(mathbb{F}_{q^n}) - q^n-1| le (2g+1)q^{n/2}$
$endgroup$
– reuns
Jan 15 at 5:32