Timeline for Finding $Q(\sqrt{-2})$-rational points on $X_0(33)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2020 at 23:33 | comment | added | LSpice | The name of the linked paper: Bruin and Najman - Hyperelliptic modular curves $X_0(n)$ and isogenies of elliptic curves over quadratic fields. | |
| Mar 30, 2020 at 2:40 | comment | added | Noam D. Elkies | 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 $y_1$. | |
| Mar 30, 2020 at 0:47 | history | edited | Jackson Morrow | CC BY-SA 4.0 |
corrected an error
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| Mar 30, 2020 at 0:47 | comment | added | Jackson Morrow | 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. | |
| Mar 29, 2020 at 23:58 | comment | added | Guest | This classifies only exceptional points, or am I missing something? | |
| Mar 29, 2020 at 23:46 | history | answered | Jackson Morrow | CC BY-SA 4.0 |