Timeline for What are the $j$-invariants of the genus 1 modular curves?
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13 events
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| Oct 13, 2016 at 6:51 | comment | added | znt | @Jeremy Rouse: you're quite right of course, but let me flag the following spanner which sometimes people throw in to add to the confusion. If $\Gamma$ is $\Gamma(N)$ ($N>2$) then using a full level $N$ structure in the $GL(2)$ theory of canonical models of course spits out a connected but geometrically disconnected curve, defined over $Q$ but whose geom irred components are defined over $Q(\zeta_N)$. However some people use $Y(N)$ to mean the moduli space of ell curve + symplectic full level $N$ structure, which is geom connected and does give a model for the quotient over $Q$. Urk. | |
| Oct 13, 2016 at 2:03 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 13, 2016 at 1:49 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 13, 2016 at 1:47 | answer | added | Jeremy Rouse | timeline score: 6 | |
| Oct 13, 2016 at 1:47 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 13, 2016 at 1:46 | comment | added | stupid_question_bot | @JeremyRouse That's a good point. Here I mean the moduli-theoretic model. I've edited the question accordingly. | |
| Oct 13, 2016 at 1:44 | comment | added | Noam D. Elkies | Not all integral ($X_0(27)$ is the Fermat cubic, with $j=0$), nor all non-integral (if $p\|N$ and $X_0(N)$ has genus $1$ then it has multiplicative reduction mod $p$ so there is a factor of $p$ in the denominator; example: $N=p=11$). | |
| Oct 13, 2016 at 1:40 | history | edited | Will Chen | CC BY-SA 3.0 |
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| Oct 13, 2016 at 1:08 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 12, 2016 at 23:26 | comment | added | Jeremy Rouse | Note that specifying $\Gamma \leq SL_{2}(\mathbb{Z})$ gives a curve $\mathcal{H}/\Gamma$ up to isomorphism over $\mathbb{C}$. This does not specify a $\mathbb{Q}$-structure on the curve. To do so, you need to choose an subgroup $\Gamma \leq GL_{2}(\mathbb{Z}/n\mathbb{Z})$ so that $\det : \Gamma \to (\mathbb{Z}/n\mathbb{Z})^{\times}$ is surjective. For this reason, it is not immediately obvious to me that all of the $j$-invariants are rational, as (I'm pretty sure) that not all of the $\Gamma \leq SL_{2}(\mathbb{Z})$ have models over $\mathbb{Q}$. | |
| Oct 12, 2016 at 20:59 | history | edited | stupid_question_bot | CC BY-SA 3.0 |
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| Oct 12, 2016 at 20:54 | comment | added | znt | For $\Gamma=\Gamma_0(N)$ you can read everything off from (the first few lines of) Cremona's tables. There will be other possibilities for $\Gamma$ though, I guess. | |
| Oct 12, 2016 at 20:00 | history | asked | stupid_question_bot | CC BY-SA 3.0 |