What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$?

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.

Edit I 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).

• Forget about the messy equations. Etaleness of a map to the DM moduli stack says exactly that formal fibers of the family are the universal deformations. In this case it is formally smooth of relative dimension 1, so it is equivalent to check if the first-order deformation of each geometric fiber is nontrivial. Bringing in $X'$ is a red herring; forget about it. But why do you think it is not etale? After all, every elliptic curve over a $\mathbf{Z}[1/2]$-scheme does acquire "Legendre form" over an etale cover. Do you know the size of the automorphism group of a "Legendre structure"? – nfdc23 Nov 8 '15 at 11:00
• @nfdc23 I'm very much a beginner in deformation theory - Why would having nontrivial first order deformations imply that the formal fibers are universal deformations? Here, do you mean "nontrivial first order deformations" inside $E_\lambda$? Is that the same as $E_\lambda$ not being isotrivial? – Will Chen Nov 8 '15 at 13:39
• @oxeimon: I meant it as I wrote it: if the deformation theory is "rigid" (no nontrivial automorphisms 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 $E_{\lambda}$.) – nfdc23 Nov 8 '15 at 16:10
• @BenLim How do you know it is non-reduced? I think your first relation (involving $\lambda'$) is actually in the ideal generated by $f_1',\dots,f_4'$, in which case it actually is reduced. – Charles Rezk Nov 9 '15 at 21:43
• @CharlesRezk I think you're right. Macaulay2 says that $(2\lambda'-1)(\lambda'+1)(\lambda'-2)$ is in that ideal. – Ben Lim Nov 9 '15 at 22:23

Let $Leg: \mathbb P^1-\{0,1,\infty\}\to \mathcal M_{1,1}$ be the Legendre map. (This associates to $\lambda$ the elliptic curve given by $y^2 = x(x-1)(x-\lambda)$.)

1. The coarse moduli space map $j:\mathcal M_{1,1}\to \mathbb A^1$ is of degree $1/2$.

2. The morphism $\mathbb P^1-\{0,1,\infty\}\to \mathbb A^1$ is of degree $6$. (Indeed, given a $j$-invariant, there are precisely 6 possibilities for the $\lambda$-invariant of that curve: $\lambda$, $1/\lambda$, $1-\lambda$, $1/(1-\lambda)$, $\lambda/(1-\lambda)$ and $(1-\lambda)/\lambda$.) In other words, the degree of $j\circ L$ is $6$.

It follows that the degree of $L$ is the degree of $j\circ L$ divided by the degree of $j$. This gives $6\times 2 = 12$.

• @BenLim Concerning degrees of maps of stacks: consider $G$ a finite group. What would you say the degree of $\{pt\} \to BG$ is? And what about $BG\to \{pt\}$? – Ariyan Javanpeykar Nov 8 '15 at 20:24
• @BenLim Concerning your confusion: The Legendre map is etale. An elliptic curve in Legendre form has rational $2$-torsion, and any isomorphism of an elliptic curve respecting all of the $2$-torsion points has to be $\pm 1$. A proof of this can be found in Katz-Mazur. (Compare this to the fact that an automorphism of an elliptic curve respecting all of the $n$-torsion points ($n>2$) has to be trivial.) – Ariyan Javanpeykar Nov 8 '15 at 20:27
• Ok. However, if it indeed is \'{e}tale, then why am I getting that the fiber of $\text{Isom}(E_{\lambda}, E_{\lambda'}) \to X$ over $\lambda = -1$ is non-reduced? – Ben Lim Nov 8 '15 at 20:28
• @BenLim (Here's me trying to explaining degrees.) The map $\{pt\} \to [\{pt\}/G]$ is of degree $\# G$. The composition with the coarse map $[\{pt\}/G] \to \{pt\}$ is the identity. So this means that the "degree" of $BG \to \{pt\}$ is $1/\# G$. You could think of the "point" of $BG$ as a $1/\# G$-th point. There are some other answers on Mathoverflow explaining this. Look up "groupoid cardinality" for instance. – Ariyan Javanpeykar Nov 8 '15 at 20:31
• @BenLim I can't see where your mistake is in your computation at the moment. In any case, there has to be a mistake because $Isom_S(E,E^\prime)\to S$ is a finite unramified morphism of schemes, whenever $S$ is a scheme and $E$ and $E^\prime$ are elliptic curves over $S$. (Note that $Isom_S(E,E^\prime)$ is the sheaf of isomorphisms respecting the zero section.) Are you sure you are computing the scheme-theoretic fibre correctly? – Ariyan Javanpeykar Nov 8 '15 at 20:43