Let $\mathcal{M}_{1,1}$ be the moduli stack of elliptic curves (over the complex numbers). Define $$\begin{eqnarray*} X &:=& \Bbb{A}^1_{\lambda} - \{0,1\}\\ X' &=& \Bbb{A}^1_{\lambda'} - \{0,1\}.\end{eqnarray*} $$ There are morphisms $X \to \mathcal{M}_{1,1}$ and $X' \to \mathcal{M}_{1,1}$ given by the families of curves $$\begin{eqnarray*} E_\lambda := V(y^2 - x(x-1)(x-\lambda)) \\ E_{\lambda' }:= V(y^2 - x(x-1)(x-\lambda')). \end{eqnarray*}$$

By results of Grothendieck, we know that the fiber product $\text{Isom}(E_{\lambda}, E_{\lambda'}) := X \times_{\mathcal{M}_{1,1}} X'$ is a scheme. Its $T$-points are given by $$\text{Isom}(E_{\lambda}, E_{\lambda'})(T) = (T \to X ,T \to X', {E_{\lambda}}_T \stackrel{\simeq}{\to} {E_{\lambda'}}_T ) .$$ The isomorphism above between ${E_{\lambda}}_T$ and ${E_{\lambda'}}_T$ is a $T$-isomorphism.

My goal is to try and understand why $X \to \mathcal{M}_{1,1}$ is not étale. To do this, it is enough to show that $\text{Isom}(E_{\lambda},E_{\lambda'}) \to X$ is not \'{e}tale.

Since the automorphisms of any elliptic curve in Legendre form are given by $y \mapsto cy$ and $x \mapsto ax +b$, I can see that the scheme $\text{Isom}(E_{\lambda},E_{\lambda'})$ is given by the following conditions in $\Bbb{A}^5:$

$$\text{Isom}(E_{\lambda},E_{\lambda'}) = \operatorname{Spec} \frac{\Bbb{C}[\lambda, \frac{1}{\lambda}, \frac{1}{1-\lambda},\lambda', \frac{1}{\lambda'}, \frac{1}{1-\lambda'},a,\frac{1}{a},b,c]}{(j(\lambda) - j(\lambda'), f_1,f_2,f_3,f_4)}. $$

The polynomials $f_1,f_2,f_3,f_4$ are obtained from equating the coefficients of the relation $$ x(x-1)(x-\lambda') = \frac{(ax+b)(ax+b-1)(ax+b-\lambda)}{c^2}.$$

Explicitly, they are given by:

$$\begin{eqnarray*} f_1 &=& a^3 - c^2 \\ f_2 &=& 3a^2b - a^2 \lambda - a^2 + a^3(\lambda' + 1) \\ f_3 &=& 3ab^2 - 2ab\lambda - 2ab + a\lambda -a^3\lambda'\\ f_4 &=& b^3 - b^2\lambda - b^2 + b\lambda. \end{eqnarray*} $$

Now if I compute the fiber of the map $\text{Isom}(E_{\lambda}, E_{\lambda'}) \to X$ over the $\lambda = -1$ ($j = 1728$), I get the non-reduced scheme

$$\text{Isom}(E_{\lambda}, E_{\lambda'})_{-1} = \operatorname{Spec} \frac{\Bbb{C}[\lambda', \frac{1}{\lambda'}, \frac{1}{1-\lambda'},a,\frac{1}{a}, b,c]}{ \left((2 \lambda'-1)^2 (\lambda'+1)^2 (\lambda'-2)^2,f_1,f_2',f_3',f_4'\right)}$$ where

$$\begin{eqnarray*} f_1 &=& a^3 - c^2 \\ f_2' &=& 3b +a (\lambda'+1) \\ f_3' &=& 3b^2 - a^2\lambda' - 1\\ f_4' &=& b^3 - b. \end{eqnarray*} $$ Hence $X \to \mathcal{M}_{1,1}$ is ramified.

However:On the other hand, I have computed the cardinality of the fiber $\text{Isom}(E_{\lambda},E_{\lambda'}) \to X$ to always be 12.

Indeed consider the $\Bbb{C}$-point of $X$ corresponding to $\lambda = -1$. There are three possibilities for $\lambda'$, namely $-1,2,1/2$. An elliptic curve with $j$-invariant 1728 has automorphism group of order 4, and so the fiber over $-1$ has cardinality $3\times 4 = 12$. The story is the same for the other values of $\lambda$.

My question is:Why am I always getting 12? I am not taking into account some non-reduced issue here? I am also confused because in my head, the fiber cardinality should jump for a ramified morphism.

EditI was wrong previously. The fiber over $\lambda = -1$ is reduced, as Macaulay2 tells me (using the command isNormal) that the same ring with coefficients in $\Bbb{Q}$ is normal, hence reduced. Tensoring with $\Bbb{C}$ over $\Bbb{Q}$ still preserves reducedness (since $\Bbb{Q}$ is perfect). The key point is that the element $(2\lambda'-1)(\lambda'+1)(\lambda'-2)$ is already in the ideal $(f_1,f_2',f_3',f_4')$ (also confirmed by Macaulay2).

doesacquire "Legendre form" over an etale cover. Do you know the size of the automorphism group of a "Legendre structure"? $\endgroup$inside$E_\lambda$? Is that the same as $E_\lambda$ not being isotrivial? $\endgroup$as deformations) and formally smooth of relative dimension 1 then a formal deformation over a complete dvr that is nontrivial to first order must be a universal deformation. This is a basic exercise. I don't claim any relation to isotriviality, which cannot be detected at the infinitesimal level. (I'm not sure what you mean by "inside $E_{\lambda}$.) $\endgroup$9more comments