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Finding Q$Q(\sqrt{-2})$-rational points on X_0$X_0(33)$

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Let $K = Q(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(33)$?

Recall that $X_0(33)$ is of genus $3$ and has the following affine model,

$$y^2 +(-x^4-x^2-1)y = 2x^6-2x^5+11x^4-10x^3+20x^2-11x+8.$$

My attempt at finding $K$-rational points on $X_0(33)$ is as follows: First I find a rational map $f$ from $X_0(33)$ to a quotient curve $E$ of $X_0(33)$ with $E$ an elliptic curve. Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank $0$, $E(K)$ is finite. Then I can use a Grobner basis to determine $f^{-1}(x)$ for every $x \in E(K)$. However in my case $E(K)$ is of rank $1$ and as a result it is computationally infeasible to determine a Grobner basis for every $f^{-1}(x)$ with $x \in E(K)$. I am wondering if there is a work-around this issue?

Moreover I am puzzled that by Falting's theorem the number of $K$-rational points on $X_0(33)$ is finite. Therefore since $E(K)$ is of rank $1$ it means that infinitely many points $x$ of $E(K)$ are such that $f^{-1}(x)$ lies in the same finite set. I don't understand what mechanism makes this possible.

Any help in finding $K$-rational points on $X_0(33)$ would be appreciated.

EDIT: Removed a question after a clarification by Christian Wuthrich.

Let $K = Q(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(33)$?

Recall that $X_0(33)$ is of genus $3$ and has the following affine model,

$$y^2 +(-x^4-x^2-1)y = 2x^6-2x^5+11x^4-10x^3+20x^2-11x+8.$$

My attempt at finding $K$-rational points on $X_0(33)$ is as follows: First I find a rational map $f$ from $X_0(33)$ to a quotient curve $E$ of $X_0(33)$ with $E$ an elliptic curve. Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank $0$, $E(K)$ is finite. Then I can use a Grobner basis to determine $f^{-1}(x)$ for every $x \in E(K)$. However in my case $E(K)$ is of rank $1$ and as a result it is computationally infeasible to determine a Grobner basis for every $f^{-1}(x)$ with $x \in E(K)$. I am wondering if there is a work-around this issue?

Moreover I am puzzled that by Falting's theorem the number of $K$-rational points on $X_0(33)$ is finite. Therefore since $E(K)$ is of rank $1$ it means that infinitely many points $x$ of $E(K)$ are such that $f^{-1}(x)$ lies in the same finite set. I don't understand what mechanism makes this possible.

Any help in finding $K$-rational points on $X_0(33)$ would be appreciated.

Let $K = Q(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(33)$?

Recall that $X_0(33)$ is of genus $3$ and has the following affine model,

$$y^2 +(-x^4-x^2-1)y = 2x^6-2x^5+11x^4-10x^3+20x^2-11x+8.$$

My attempt at finding $K$-rational points on $X_0(33)$ is as follows: First I find a rational map $f$ from $X_0(33)$ to a quotient curve $E$ of $X_0(33)$ with $E$ an elliptic curve. Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank $0$, $E(K)$ is finite. Then I can use a Grobner basis to determine $f^{-1}(x)$ for every $x \in E(K)$. However in my case $E(K)$ is of rank $1$ and as a result it is computationally infeasible to determine a Grobner basis for every $f^{-1}(x)$ with $x \in E(K)$. I am wondering if there is a work-around this issue?

Any help in finding $K$-rational points on $X_0(33)$ would be appreciated.

EDIT: Removed a question after a clarification by Christian Wuthrich.

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Guest
  • 71
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Finding Q(\sqrt{-2})-rational points on X_0(33)

Let $K = Q(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(33)$?

Recall that $X_0(33)$ is of genus $3$ and has the following affine model,

$$y^2 +(-x^4-x^2-1)y = 2x^6-2x^5+11x^4-10x^3+20x^2-11x+8.$$

My attempt at finding $K$-rational points on $X_0(33)$ is as follows: First I find a rational map $f$ from $X_0(33)$ to a quotient curve $E$ of $X_0(33)$ with $E$ an elliptic curve. Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank $0$, $E(K)$ is finite. Then I can use a Grobner basis to determine $f^{-1}(x)$ for every $x \in E(K)$. However in my case $E(K)$ is of rank $1$ and as a result it is computationally infeasible to determine a Grobner basis for every $f^{-1}(x)$ with $x \in E(K)$. I am wondering if there is a work-around this issue?

Moreover I am puzzled that by Falting's theorem the number of $K$-rational points on $X_0(33)$ is finite. Therefore since $E(K)$ is of rank $1$ it means that infinitely many points $x$ of $E(K)$ are such that $f^{-1}(x)$ lies in the same finite set. I don't understand what mechanism makes this possible.

Any help in finding $K$-rational points on $X_0(33)$ would be appreciated.