Notation. Let $p$ be a prime number, $K$ a finite extension of $\mathbb{Q}_p$ and $E|K$ an elliptic curve which has good reduction. 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 $K$.

Question. 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 $E|K$ ?

Background. $E$ can be defined over $K$ by a minimal cubic

$f=y^2+a_1xy+a_3y-x^3-a_2x^2-a_4x-a_6=0,\ \ (a_i\in\mathfrak{o})$;

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.

Addendum. As Qing Liu remarks, one may ask, given an elliptic curve $E$ over a finite extension $k|\mathbb{F}_p$, 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.

  • $\begingroup$ Notice that this quantity depends only on the reduction of $E$. So an equivalent form of the question is: if $E$ is an elliptic curve a finite field $k$, what of $E$ is encoded in its discriminant $\in k^*/{k^*}^{12}$ ? $\endgroup$
    – Qing Liu
    Mar 1, 2010 at 9:56
  • $\begingroup$ I'm a bit worried about the primes 2 and 3. For $K=\mathbb{Q}_2$, the group $\mathbb{F}_2^\times/\mathbb{F}_2^{\times12}$ is trivial, whereas $\mathbb{Z}_2^\times/\mathbb{Z}_2^{\times12}$ is not trivial. $\endgroup$ Mar 1, 2010 at 11:24
  • $\begingroup$ Sorry, you are completely right. This bijection with $k^*/{k*}^{12}$ is true only for $p\ne 2,3$. But, if you have an answer over $k$, you have an answer over $K$, and the converse is true if $p\ne 2,3$. $\endgroup$
    – Qing Liu
    Mar 1, 2010 at 13:31
  • $\begingroup$ You are right : for $p\neq2,3$, you may equivalently think of the question as being about discriminants of elliptic curves over the residue field. $\endgroup$ Mar 1, 2010 at 13:51

3 Answers 3


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?)

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.)

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).

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 $n = 2, 3$).

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.

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...

  • 1
    $\begingroup$ All these 12s are very mystifying! $\endgroup$ Mar 1, 2010 at 20:18
  • $\begingroup$ @Mariano. I have some idea of the 12 in the question. That 12 comes out because an elliptic curve made into an affine equation, roughly, (quadratic in y) = (cubic in x), and the possible transformations that preserve this form. Precise details are in Silverman's book on elliptic curves. The 12's in the answer should arise out of the same reasons underlying the 12 in the question, I would imagine. $\endgroup$
    – Regenbogen
    Mar 1, 2010 at 20:38
  • 1
    $\begingroup$ @Brian: Thank you very much for your wonderful answer, placing this little question in a wider context. $\endgroup$ Mar 2, 2010 at 3:06

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 $d_{\mathfrak{p}}$ 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 $d_{\mathfrak{p}}$ is a unit, but in general it is not. However, $d_{\mathfrak{p}}$ is well-defined up to an element of $\mathfrak{o}_{\mathfrak{p}}^{\times12}$. Now we can form the global minimal discriminant, $$ \Delta_{E/K} = \prod_{\mathfrak{p}} \mathfrak{p}^{ord_{\mathfrak{p}}d_{\mathfrak{p}}}. $$ If we look at the ideal class of $\Delta_{E/K}$ in the ideal class group of $K$, then one can show that there exists an ideal class $\overline{\alpha}_{E/K}$ such that $$ \overline{\Delta}_{E/K} = \overline{\alpha}_{E/K}^{12} \quad\mbox{in the ideal class group of } K. $$ A useful property of $\overline{\alpha}_{E/K}$ is the following theorem: The curve $E/K$ has a global minimal Weierstrass model if and only if $\overline{\alpha}_{E/K}=1$.

For details, see Section VIII.8 of The Arithmetic of Elliptic Curves, Springer, GTM 106, 2009.


This is not an answer but a clarification of my question.

In fact it is thinking about Section VIII.8 of your book and especially your 1984 Mathematika paper which led to the question.

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 Fröhlich, as an element of the group $K^\times/\mathfrak{o}^{\times 2}$.

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.

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 $E$ ?

For example, for which curves $E$ is $\delta_{E|K}$ trivial in $\mathfrak{o}^\times/\mathfrak{o}^{\times 12}$ ?


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