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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?

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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    $\begingroup$ Not in the same finite set, the points in the preimage of a rational point $x\in E(K)$ will be defined over a larger field than $K$ for all but finitely many $x$. $\endgroup$
    – Chris Wuthrich
    Commented Mar 29, 2020 at 23:15
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    $\begingroup$ Thanks, that's a good point. I edited the question accordingly. $\endgroup$
    – Guest
    Commented Mar 29, 2020 at 23:18

1 Answer 1

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P. Bruin and F. Najman have determined the exceptional quadratic points on $X_0(33)$.

See Table 8 of https://arxiv.org/pdf/1406.0655.pdf

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  • $\begingroup$ This classifies only exceptional points, or am I missing something? $\endgroup$
    – Guest
    Commented Mar 29, 2020 at 23:58
  • $\begingroup$ Yes you are right that they only classify the exceptional points. I will edit the answer accordingly. I will just note that the elliptic curve you found is $X_0(33)^+$ (the quotient of $X_0(33)$ under the Atkin--Lehner involution $\omega_{33}$). This can be compute in Magma via ModularCurveQuotient(33,[33]). This interpretation may help you determine the remaining $\mathbb{Q}(\sqrt{-2})$ points. $\endgroup$
    – Jackson Morrow
    Commented Mar 30, 2020 at 0:47
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    $\begingroup$ It appears that in this case we're lucky: non-exceptional points would have $x$ rational and $-2y_1^2 = P(x)$ where $y_1 = y - \frac12(x^4+x^2+1)$ and $P(y) = x^8 + 10x^6 - 8x^5 + 47x^4 - 40x^3 + 82x^2 - 44x + 33$ is the discriminant of the quadratic in $y$; but $P(x)$ happens to have no real roots, and is thus positive for all $x$, so cannot equal $-2y_1^2$ for any rational $y_1$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 30, 2020 at 2:40
  • $\begingroup$ The name of the linked paper: Bruin and Najman - Hyperelliptic modular curves $X_0(n)$ and isogenies of elliptic curves over quadratic fields. $\endgroup$
    – LSpice
    Commented Aug 1, 2020 at 23:33

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