Timeline for Are there nonisotrivial elliptic curves over $\mathbb{G}_m$?
Current License: CC BY-SA 3.0
19 events
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| Jan 5 at 13:52 | comment | added | user267839 | Why is this scheme of triv's of the $\ell$- torsion etale over $\mathbb{G}_m$? Is this a feature of torsion of elliptic curves, or does it hold in greater generality? Eg somethng like: Let $G/S$ finite etale group scheme over base $S$ such that every fiber $G_s$ is isomorphic to a finite constnt scheme $C/ \kappa(s)$, then $\text{Isom}_{S}(G, C_S)$ is etale over $S$ ( where $C_S$ sameconst finite gp scheme over $S$)? Or does it even hold more generally if we instead $C_S$ consider another finite etale $S$-gp scheme $H/S$ with $G_s \cong H_s$ that $\text{Isom}_{S}(G, H)$ is etale over $S$? | |
| Jan 4 at 20:26 | comment | added | Daniel Litt | @user267839: The scheme $\text{Isom}_{\mathbb{G}_m}(E[\ell], (\mathbb{Z}/\ell\mathbb{Z})^2)$. | |
| Jan 3 at 12:33 | comment | added | user267839 | What do you mean by the "scheme of trivializations" of the $\ell$-torsion precisely? | |
| Jan 2 at 22:22 | comment | added | Daniel Litt | @user267839: The key fact here is that any tame cover of $\mathbb{G}_m$ is in fact isomorphic to a disjoint union of $[n_i]$ for some $n_i$ (this follows e.g.~Grothendieck's theorem that specialization maps are surjective for tame fundamental groups). Now the scheme of trivializations of the $\ell$-torsion is étale over $\mathbb{G}_m$, so if you check it is tame you win. | |
| Jan 1 at 21:43 | comment | added | user267839 | #UPDATE: The two points I asked before are clear so far, so my last three comments can be ignored. But could you maybe loose few words on why $[n]$ becomes a trivializing cover for $\ell$ torsion in the sense you explained, when $E[\ell] \to \mathbb{G}_m$ is tamely ramified in "places" $0,\infty$ in above sense? | |
| Dec 31, 2023 at 3:06 | comment | added | user267839 | projective models, ie in case of $\mathbb{G}_m$ the projective line $\mathbb{P}^1$. So, when you say above that " the map $E[\ell]\to \mathbb{G}_m$ has tame ramification in $\infty, 0$", do you tacitly refer to the "extended" map between assoc smooth projective models & their points? Or do I missing your concern? | |
| Dec 31, 2023 at 2:55 | comment | added | user267839 | about the second point I'm not pretty sure if I understand exactly your argument in this context about reasoning about "ramification behavior" at $0, \infty$ of the map $E[\ell] \to \mathbb{G}_m$. You gave the argument that this comes from certain inertia groups of the fundamental group which "detect" them as a kind of - let me say "missing" points. Do I understand the picture correctly that one should think of this as "reasoning about ramification behavior in naive sense of associated maps between the associated unique smooth | |
| Dec 31, 2023 at 2:30 | comment | added | user267839 | alright, so regarding my first question the claim is that one can find a $n$ such the pullback $E'$ of $E/ \mathbb{G}_m$ along covering $\mathbb{G}_m\overset{[n]}\longrightarrow \mathbb{G}_m$ has "trivial $\ell$- torstion" in the sense that there exist an iso between $E'[\ell]$ and constant grp scheme $((\mathbb{Z}/ \ell)^2)_{\mathbb{G}_m}$? That's the issue with trivialization, right? ... and based on this we obtain the map from $\mathbb{G}_m$ to the modular curve. | |
| Dec 29, 2023 at 20:57 | comment | added | Daniel Litt | @user267839 Regarding your first question, this uses the moduli description of modular curves--a map from $X$ to a modular curve is the same as a family of elliptic curves over $X$ with a suitable trivialization of their torsion, essentially by definition. For your second question, the fundamental group of $\mathbb{G}_m$ has inertia subgroups corresponding to the punctures $0, \infty$. | |
| Dec 29, 2023 at 1:38 | comment | added | user267839 | Also, I'm not sure what you mean by that in order to reach the above condition, we must show that for infinitely many $\ell$ the map $E[\ell] \to \mathbb{G}_m$ has tame ramification in $0, \infty$. But there two points not live in the base $\mathbb{G}_m$. Do you ,maybe somehow consider there an extension to base $\mathbb{P}^1$? | |
| Dec 29, 2023 at 1:30 | comment | added | user267839 | @DanielLitt: I'm confused a bit about some steps in your "algebraic" construction. Firstly, why if we manage to find such "big" enough $\ell$ with $(\ell,p)=1$ such that the pullback of $E/ \mathbb{G}_m$ along the multiplication map $[n]$ on the base has only trivial $\ell$ -torsion, then it's "trivialization" induces a map from base $\mathbb{G}_m$ to certain modular curve? Sidenote: what do you mean in this context by "trivialization"? | |
| Dec 13, 2014 at 22:31 | history | edited | Daniel Litt | CC BY-SA 3.0 |
added 1268 characters in body
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| Dec 12, 2014 at 22:56 | vote | accept | Lisa S. | ||
| Dec 12, 2014 at 22:56 | comment | added | Lisa S. | @DanielLitt: Thank you. Of the four given proofs, I like the one that you give in your comment most, so I am going to accept this answer. | |
| Dec 12, 2014 at 7:17 | comment | added | Daniel Litt | You're right of course; forgot about those other pesky covers. | |
| Dec 12, 2014 at 6:46 | comment | added | Noam D. Elkies | Indeed $y^2 = x^3 + x^2 - t$ has discriminant $t$ and $j$-invariant $1/t$ in characteristic $3$. Likewise $y^2 + xy = x^3 + t$ in characteristic $2$. | |
| Dec 12, 2014 at 6:40 | comment | added | user74230 | This algebraic argument might have problems in characteristics 2 and 3. The covering map $\ell:E \rightarrow E$ over $\mathbf{G}_m$ is generally not a Galois covering, just finite etale, corresponding to a representation of $\pi_1(\mathbf{G}_m)$ into ${\rm{GL}}_2(\mathbf{F}_{\ell})$ that might be wildly ramified at $0$ and $\infty$ in those low positive characteristics, so the corresponding cover (which depends on $\ell$) might not be a genus-0 curve. | |
| Dec 12, 2014 at 5:59 | comment | added | Daniel Litt | One can make an algebraic version of this argument as follows, which works in arbitrary characteristic. Some finite etale cover of $\mathbb{G}_m$ trivializes the $\ell$-torsion of $E$, and so after picking a trivialization induces a map from $\mathbb{G}_m$ to a modular curve. But the genus of modular curves tends to infinity, so for large $\ell$, such a map must be constant. | |
| Dec 12, 2014 at 5:54 | history | answered | Daniel Litt | CC BY-SA 3.0 |