Timeline for Discriminant locus of elliptic K3 surfaces
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2020 at 2:18 | comment | added | Noam D. Elkies | The curve is $y^2 = x^3 + (-c_4/48) x + (-c_6/864)$, so $c_4$ is the $x$-coefficient and $c_6$ is the constant coefficient. For us $c_4=0$, which has valuation $+\infty$, so $v(c_4) \geq 3$ holds. The quadruple root of $c_6$ gives $v(c_6) = 4$. | |
| Sep 3, 2020 at 1:07 | comment | added | AmorFati | @NoamD.Elkies I agree that $v(c_6)=4$, but I do not see how $v(c_4)$ is not zero: $v(c_4)$ measures the order of vanishing of the coefficient in the $x$-term, $y^2 = x^3 + (t^2-t)^4$ has no $x$-term... | |
| Aug 18, 2020 at 22:52 | comment | added | Noam D. Elkies | @AmorFati In general the singular fibers are the values of $t$ where the discriminant $\Delta$ vanishes. Here $\Delta$ is a multiple of $(t^2-t)^8$ [really $(t_1 t_0 (t_1-t_0))^8$ where $(t_1:t_0)$ are projective coordinates, so the $t_0=0$ point "$\infty$" is also a zero of $\Delta$; clearly the only others are $t_1=0$ and $t_1=t_0$, i.e. $t=0$ and $t=1$]. The reduction type can be determined by Tate's algorithm. Since we are not in characteristic 2 or 3 here, we can recognize Type IV* from the valuations $v(c_4) \geq 3$, $v(c_6) = 4$. | |
| Aug 18, 2020 at 2:44 | comment | added | AmorFati | @NoamD.Elkies Hi Noam, how do you verify that the singular fibers of $y^2 = x^3 + (t^2-t)^4$ are IV* at $t=0,1, \infty$ and that this fibration has no other singular fibers? | |
| Oct 10, 2018 at 8:47 | comment | added | Ariyan Javanpeykar | Yes, that's right. The argument actually goes back to Shafarevich (as you certainly know), and should appear somewhere in Barry Mazur's paper "Arithmetic on Curves" (projecteuclid.org/euclid.bams/1183553167). | |
| Oct 9, 2018 at 21:53 | comment | added | Noam D. Elkies | Yes, that kind of argument was given in both Daniel Litt's answer to MO 190530 and in mine. (It's related to your "hyperbolic" idea: ${\bf G}_m$ is not hyperbolic, so neither are its unramified covers.) | |
| Oct 9, 2018 at 21:48 | comment | added | Ariyan Javanpeykar | Another "easy" way of showing that $s>2$ is to use that the modular curve $Y(n)$ is of genus at least two when $n>>0$. Indeed, suppose that $s\leq 2$ for $X\to \mathbb{P}^1$. Now, choose $n >0$ such that $X(n)$ has genus at least two. Note that adding level $n$ structure induces a new family $X'\to \mathbb{P}^1$ (because $s<3$), and thus a morphism $\mathbb{P}^1\to X(n)$. | |
| Oct 9, 2018 at 20:59 | history | edited | Noam D. Elkies | CC BY-SA 4.0 |
Added postscript with link to MO 190530 and related discussion. Also: *small* positive characteristic.
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| Oct 9, 2018 at 6:26 | vote | accept | Davide Cesare Veniani | ||
| Oct 8, 2018 at 14:24 | history | answered | Noam D. Elkies | CC BY-SA 4.0 |