The Stacks project

105.6 A smooth cover

The last thing we have to do is find a smooth cover of $\mathcal{M}_{1, 1}$. In fact, in some sense the existence of a smooth cover implies1 the key fact! In the case of elliptic curves we use the Weierstrass equation to construct one.


\[ W = \mathop{\mathrm{Spec}}(\mathbf{Z}[a_1, a_2, a_3, a_4, a_6, 1/\Delta ]) \]

where $\Delta \in \mathbf{Z}[a_1, a_2, a_3, a_4, a_6]$ is a certain polynomial (see below). Set

\[ \mathbf{P}_ W^2 \supset E_ W : zy^2 + a_1 xyz + a_3 yz^2 = x^3 + a_2x^2z + a_4xz^2 + a_6z^3. \]

Denote $f_ W : E_ W \to W$ the projection. Finally, denote $0_ W : W \to E_ W$ the section of $f_ W$ given by $(0 : 1 : 0)$. It turns out that there is a degree $12$ homogeneous polynomial $\Delta $ in $a_ i$ where $\deg (a_ i) = i$ such that $E_ W \to W$ is smooth. You can find it explicitly by computing partials of the Weierstrass equation – of course you can also look it up. You can also use pari/gp to compute it for you. Here it is

\begin{align*} \Delta & = -a_6a_1^6 + a_4a_3a_1^5 + ((-a_3^2 - 12a_6)a_2 + a_4^2)a_1^4 + \\ & (8a_4a_3a_2 + (a_3^3 + 36a_6a_3))a_1^3 + \\ & ((-8a_3^2 - 48a_6)a_2^2 + 8a_4^2a_2 + (-30a_4a_3^2 + 72a_6a_4))a_1^2 + \\ & (16a_4a_3a_2^2 + (36a_3^3 + 144a_6a_3)a_2 - 96a_4^2a_3)a_1 + \\ & (-16a_3^2 - 64a_6)a_2^3 + 16a_4^2a_2^2 + (72a_4a_3^2 + 288a_6a_4)a_2 + \\ & -27a_3^4 - 216a_6a_3^2 -64a_4^3 - 432a_6^2 \end{align*}

You may recognize the last two terms from the case $y^2 = x^3 + Ax + B$ having discriminant $-64A^3 - 432B^2 = -16(4A^3 + 27B^2)$.

Lemma 105.6.1. The morphism $W \xrightarrow {(E_ W, f_ W, 0_ W)} \mathcal{M}_{1, 1}$ is smooth and surjective.

Proof. Surjectivity follows from the fact that every elliptic curve over a field has a Weierstrass equation. We give a rough sketch of one way to prove smoothness. Consider the sub group scheme

\[ H = \left\{ \left( \begin{matrix} u^2 & s & 0 \\ 0 & u^3 & 0 \\ r & t & 1 \end{matrix} \right) \middle | \begin{matrix} u\text{ unit} \\ s, r, t\text{ arbitrary} \end{matrix} \right\} \subset \text{GL}_{3, \mathbf{Z}} \]

There is an action $H \times W \to W$ of $H$ on the Weierstrass scheme $W$. To find the equations for this action write out what a coordinate change given by a matrix in $H$ does to the general Weierstrass equation. Then it turns out the following statements hold

  1. any elliptic curve $(E, f, 0)/S$ has Zariski locally on $S$ a Weierstrass equation,

  2. any two Weierstrass equations for $(E, f, 0)$ differ (Zariski locally) by an element of $H$.

Considering the fibre product $S \times _{\mathcal{M}_{1, 1}} W = \mathit{Isom}_{S \times W}(E \times W, S \times E_ W)$ we conclude that this means that the morphism $W \to \mathcal{M}_{1, 1}$ is an $H$-torsor. Since $H \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is smooth, and since torsors over smooth group schemes are smooth we win. $\square$

Remark 105.6.2. The argument sketched above actually shows that $\mathcal{M}_{1, 1} = [W/H]$ is a global quotient stack. It is true about 50% of the time that an argument proving a moduli stack is algebraic will show that it is a global quotient stack.

[1] This is a bit of a cheat because in checking the smoothness you have to prove something close to the key fact – after all smoothness is defined in terms of fibre products. The advantage is that you only have to prove the existence of these fibre products in the case that on one side you have the morphism that you are trying to show provides the smooth cover.

Comments (2)

Comment #3761 by Rohrbach on

There is a typo in the Weierstrass equation for , the term should be .

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 072S. Beware of the difference between the letter 'O' and the digit '0'.