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).
Best Answer
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 $y^2 = x(x-1)(x-\lambda)$.)
The coarse moduli space map $j:\mathcal M_{1,1}\to \mathbb A^1$ is of degree $1/2$.
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 $6\times 2 = 12$.