The order of the discriminant of a good-reduction elliptic curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:08:33Z http://mathoverflow.net/feeds/question/16743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16743/the-order-of-the-discriminant-of-a-good-reduction-elliptic-curve The order of the discriminant of a good-reduction elliptic curve Chandan Singh Dalawat 2010-03-01T07:01:55Z 2011-01-30T05:02:26Z <p><strong>Notation.</strong> Let $p$ be a prime number, $K$ a finite extension of $\mathbb{Q}_p$ and $E|K$ an elliptic curve which has <em>good reduction.</em> The discriminant $d_{E|K}$ of $E|K$ is an element of the multiplicative group $\mathfrak{o}^\times/\mathfrak{o}^{\times12}$, where $\mathfrak{o}$ is the ring of integers of $K$.</p> <p><strong>Question.</strong> Does the order of $d_{E|K}$ as an element of $\mathfrak{o}^\times/\mathfrak{o}^{\times12}$ show up somewhere ? Is it related to some other invariant of $E|K$ ?</p> <p><strong>Background.</strong> $E$ can be defined over $K$ by a minimal cubic </p> <p>$f=y^2+a_1xy+a_3y-x^3-a_2x^2-a_4x-a_6=0,\ \ (a_i\in\mathfrak{o})$;</p> <p>its discriminant $d_f$ is in $\mathfrak{o}^\times$ (because $E$ has good reduction). If we replace $f$ by another minimal cubic $g$ defining $E$, then $d_f$ gets replaced by $d_g=u^{12}d_f$ for some $u\in\mathfrak{o}^\times$. So the class $d_{E|K}$ of $d_f$ in $\mathfrak{o}^\times/\mathfrak{o}^{\times12}$ depends only on $E|K$, not on the choice of a minimal cubic defining $E$. It can be shown that every class in the finite group $\mathfrak{o}^\times/\mathfrak{o}^{\times12}$ is the discriminant of some good-reduction elliptic curve.</p> <p><strong>Addendum.</strong> As Qing Liu remarks, one may ask, given an elliptic curve $E$ over a finite extension <code>$k|\mathbb{F}_p$</code>, whether the order of the discriminant $d_{E|k}\in k^\times/k^{\times12}$ shows up somewhere. When $p\neq2,3$, the two questions are equivalent.</p> http://mathoverflow.net/questions/16743/the-order-of-the-discriminant-of-a-good-reduction-elliptic-curve/16763#16763 Answer by BCnrd for The order of the discriminant of a good-reduction elliptic curve BCnrd 2010-03-01T17:01:39Z 2010-03-01T20:16:02Z <p>I will give an intrinsic characterization below for what this unit class modulo 12th powers means, which may be viewed as an answer of sorts: it expresses the obstruction to extracting the 12th root of a certain canonical isomorphism between 12th powers of line bundles (and so one could shift the answer to: where does the need to extract such a 12th root come up?)</p> <p>For any ring $R$, the group $R^{\times}/(R^{\times})^{12}$ naturally maps into the degree-1 fppf cohomology of $\mu_{12}$ over ${\rm{Spec}}(R)$, so it classifies isomorphism classes of certain $\mu_{12}$-torsors for the fppf topology over this base. (Namely, those $\mu_{12}$-torsors whose pushout to a $\mathbf{G} _m$-torsor is trivial.)</p> <p>It is the same to use the etale topology when $12$ is a unit in $R$ (as then $\mu_{12}$ is etale over $R$). So the issue is to associate to any elliptic curve $f:E \rightarrow {\rm{Spec}}(R)$ over a ring a canonical $\mu_{12}$-torsor (with the extra property that its pushout to a $\mathbf{G} _m$-torsor is trivial). </p> <p>In the theory of Weierstrass planar models for elliptic curves $E$ over a base scheme $S$ (this includes the condition "good reduction") there is an obstruction to the existence of such a model, namely whether or not the line bundle $\omega_{E/S} = f_{\ast}(\Omega^1_{E/S})$ on $S$ admits a global trivialization. The necessity of such triviality is due to the fact that a Weierstrass model produces a trivialization (the ${\rm{dx}}/(2y+\dots)$ thing), and the sufficiency is explained in Chapter 2 of Katz-Mazur (where they use a choice of trivializing section to distinguish some formal parameters along the origin and pass from this to a Weierstrass model via the relationship between global 1-forms, the relative cotangent space ${\rm{Cot}}_e(E)$ along the identity section $e$, and $\mathcal{O}(ne)/\mathcal{O}((n-1)e) \simeq {\rm{Cot}}_e(E)^{ \otimes -n}$ for $n = 2, 3$). </p> <p>That being said, regardless of whether or not the line bundle $\omega_{E/S}$ is trivial (though it always is when $S$ is local), the line bundle $\omega_{E/S}^{\otimes 12}$ is canonically trivial (in a manner that is compatible with base change and functorial in isomorphisms of elliptic curves): that is the meaning of the classical fact that the product of $\Delta$ with the 12th power of the section ${\rm{d}}x/(2y+\dots)$ is invariant under choice of Weierstrass model. This also underlies Mumford's calculation (recently revisited by Fulton-Olsson) of the Picard group of the moduli stack of elliptic curves as $\mathbf{Z}/12\mathbf{Z}$, which one could regard as providing a distinguished role to that trivialization. Working with the compactified moduli stack over $\mathbf{Z}$ (so allowing generalized elliptic curves with geometrically irreducible but possibly non-smooth fibers, and hence working with relative dualizing sheaf to generalize $\omega_{E/S}$ when allowing non-smooth fibers), the trivialization (which we could generously attribute to Ramanujan) is unique up to a sign, which in turn is nailed down by the Tate curve over $\mathbf{Z}[[q]]$ and the isomorphism of its formal group with $\widehat{\mathbf{G}}_m$. So this trivialization is really a canonical thing, independent of any theory of Weierstrass models. </p> <p>Letting $\theta_{E/S}$ denote this intrinsic trivializing section of $\omega _{E/S}^{\otimes 12}$ as just defined, it is natural to ask if $\theta _{E/S}$ is the 12th power of a trivializing section of $\omega _{E/S}$. Note that this is a nontrivial condition even when $\omega _{E/S}$ is trivial (such as when $S$ is local). Anyway, the functor of such 12th roots is a $\mu _{12}$-torsor over $S$ for the fppf topology (and etale if 12 is a unit on the base), and as such it corresponds to the inverse of the class of $\Delta$ in the question (for which the base was local). So that is an answer of sorts: it describes the obstruction to extracting a 12th root of the canonical trivialization of $\omega^{\otimes 12}$ obtained by pullback from the trivialization over the moduli space of elliptic curves (up to an issue of signs in the exponent). Now does one ever care to extract such a 12th root? That's another matter...</p> http://mathoverflow.net/questions/16743/the-order-of-the-discriminant-of-a-good-reduction-elliptic-curve/53713#53713 Answer by Joe Silverman for The order of the discriminant of a good-reduction elliptic curve Joe Silverman 2011-01-29T13:37:58Z 2011-01-29T13:37:58Z <p>Brian's answer gives a good explanation. Here is a somewhat more prosaic description of how 12 appears when the discriminants are put together globally. Let $K$ be a number field, and for each prime ideal $\mathfrak{p}$ of $K$, let <code>$d_{\mathfrak{p}}$</code> be a minimal discriminant of $E$ at $\mathfrak{p}$. That is, take a Weierstrass equation with $\mathfrak{p}$-integral coefficients such that the discriminant of the equation has minimal valuation. If $E$ has good reduction at $\mathfrak{p}$, then <code>$d_{\mathfrak{p}}$</code> is a unit, but in general it is not. However, <code>$d_{\mathfrak{p}}$</code> is well-defined up to an element of <code>$\mathfrak{o}_{\mathfrak{p}}^{\times12}$</code>. Now we can form the global minimal discriminant, <code>$$ \Delta_{E/K} = \prod_{\mathfrak{p}} \mathfrak{p}^{ord_{\mathfrak{p}}d_{\mathfrak{p}}}. $$</code> If we look at the ideal <em>class</em> of <code>$\Delta_{E/K}$</code> in the ideal class group of $K$, then one can show that there exists an ideal class <code>$\overline{\alpha}_{E/K}$</code> such that <code>$$ \overline{\Delta}_{E/K} = \overline{\alpha}_{E/K}^{12} \quad\mbox{in the ideal class group of } K. $$</code> A useful property of <code>$\overline{\alpha}_{E/K}$</code> is the following theorem: The curve $E/K$ has a global minimal Weierstrass model if and only if <code>$\overline{\alpha}_{E/K}=1$</code>.</p> <p>For details, see Section VIII.8 of <em>The Arithmetic of Elliptic Curves</em>, Springer, GTM 106, 2009.</p> http://mathoverflow.net/questions/16743/the-order-of-the-discriminant-of-a-good-reduction-elliptic-curve/53760#53760 Answer by Chandan Singh Dalawat for The order of the discriminant of a good-reduction elliptic curve Chandan Singh Dalawat 2011-01-30T04:56:46Z 2011-01-30T05:02:26Z <p>This is <em>not</em> an answer but a clarification of my question.</p> <p>In fact it is thinking about Section VIII.8 of your book and especially your 1984 Mathematika <a href="http://ams.org/mathscinet/search/publdoc.html?extend=1&amp;extend=1&amp;pg1=IID&amp;s1=162205&amp;vfpref=html&amp;r=114&amp;mx-pid=804199" rel="nofollow">paper</a> which led to the question.</p> <p>As in your paper, there is an analogy with discriminants of number fields. Keeping to the purely local situation, let $K$ be a finite extension of the $p$-adics, with ring of integers $\mathfrak{o}$, and let $L$ be a finite extension of $K$. Then the discriminant $\delta_{L|K}$ of $L|K$ can be thought of, following <a href="http://ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;review_format=html&amp;s4=frohlich&amp;s5=algebraic&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq&amp;r=8&amp;mx-pid=113876" rel="nofollow">Fröhlich</a>, as an element of the group $K^\times/\mathfrak{o}^{\times 2}$. </p> <p>When $L|K$ is unramified, $\delta_{L|K}$ is an element of $\mathfrak{o}^\times/\mathfrak{o}^{\times 2}$, and its order as an element of this group --- the only possibilities are $1$ and $2$ --- gives us the parity of $[L:K]$. More precisely, $\delta_{L|K}$ has order $1$ if $[L:K]$ is odd, order $2$ if $[L:K]$ is even.</p> <p>Let's now return to our good-reduction elliptic curve $E$ over $K$, whose discriminant $\delta_{E|K}$ is an element of $\mathfrak{o}^\times/\mathfrak{o}^{\times 12}$. The question is, what does the order of the element $\delta_{E|K}$ in the said group --- the possibilities for the order being $1,2,3,4,6,12$ --- tell us about the curve $E$ ? </p> <p>For example, for which curves $E$ is $\delta_{E|K}$ trivial in $\mathfrak{o}^\times/\mathfrak{o}^{\times 12}$ ?</p>