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Background

In the course of answering another question (httphttps://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating) I found myself with a curve, that if it had a positive genus, then I could prove something about an interesting number field. Specifically, I was trying to show that, in the terminology of the above question, all biquadratic fields over the rationals have $r=2$.

The curve was defined by a single affine equation with large degrees in $x$ and in $y$ (23 and 25). Giving sage, or to be more precise Singular, the task of computing the genus, it broke down after an hour. The degree is just too much.

Question

But really, all I want is to bound the genus from below. What effective ways are there for doing so?

Possible solution

Compute the number of points in a random small finite field, hoping for no bad reduction, and checking this is much easier than finding all algebraic singular points. If it is not true that $q-2g\sqrt{q}+1 \le |C(\mathbb{F}_q)| \le q+2g\sqrt{q}+1$, then the genus is greater than $g$. Is this correct?

Background

In the course of answering another question (http://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating) I found myself with a curve, that if it had a positive genus, then I could prove something about an interesting number field. Specifically, I was trying to show that, in the terminology of the above question, all biquadratic fields over the rationals have $r=2$.

The curve was defined by a single affine equation with large degrees in $x$ and in $y$ (23 and 25). Giving sage, or to be more precise Singular, the task of computing the genus, it broke down after an hour. The degree is just too much.

Question

But really, all I want is to bound the genus from below. What effective ways are there for doing so?

Possible solution

Compute the number of points in a random small finite field, hoping for no bad reduction, and checking this is much easier than finding all algebraic singular points. If it is not true that $q-2g\sqrt{q}+1 \le |C(\mathbb{F}_q)| \le q+2g\sqrt{q}+1$, then the genus is greater than $g$. Is this correct?

Background

In the course of answering another question (https://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating) I found myself with a curve, that if it had a positive genus, then I could prove something about an interesting number field. Specifically, I was trying to show that, in the terminology of the above question, all biquadratic fields over the rationals have $r=2$.

The curve was defined by a single affine equation with large degrees in $x$ and in $y$ (23 and 25). Giving sage, or to be more precise Singular, the task of computing the genus, it broke down after an hour. The degree is just too much.

Question

But really, all I want is to bound the genus from below. What effective ways are there for doing so?

Possible solution

Compute the number of points in a random small finite field, hoping for no bad reduction, and checking this is much easier than finding all algebraic singular points. If it is not true that $q-2g\sqrt{q}+1 \le |C(\mathbb{F}_q)| \le q+2g\sqrt{q}+1$, then the genus is greater than $g$. Is this correct?
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Dror Speiser
Dror Speiser
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Computationally bounding a curve's genus from below?

Background

In the course of answering another question (http://mathoverflow.net/questions/19638/infinite-collection-of-elements-of-a-number-field-with-very-similar-annihilating) I found myself with a curve, that if it had a positive genus, then I could prove something about an interesting number field. Specifically, I was trying to show that, in the terminology of the above question, all biquadratic fields over the rationals have $r=2$.

The curve was defined by a single affine equation with large degrees in $x$ and in $y$ (23 and 25). Giving sage, or to be more precise Singular, the task of computing the genus, it broke down after an hour. The degree is just too much.

Question

But really, all I want is to bound the genus from below. What effective ways are there for doing so?

Possible solution

Compute the number of points in a random small finite field, hoping for no bad reduction, and checking this is much easier than finding all algebraic singular points. If it is not true that $q-2g\sqrt{q}+1 \le |C(\mathbb{F}_q)| \le q+2g\sqrt{q}+1$, then the genus is greater than $g$. Is this correct?