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I am trying to understand in a very down to earth way how the definition of stack (really, prestack, i.e. Homs form a sheaf) allows for the presence of twists; specifically in the example of the moduli stack $\mathcal{M}_{1,1}$ of elliptic curves, thought of a pseudofunctor $\mathcal{M}_{1,1}:(Sch_{et}) \to (Groupoids)$ from schemes to groupoids; (with some choice of cleavage).

Concretely, let $E_a$ and $E_b$ be the elliptic curves (over $\mathbb{Q}$) with Weierstrass equations $x^3-x =2y^2$ and $x^3-x=y^2$ respectively. The curves $E_a$ and $E_b$ are not isomorphic over $\mathbb{Q}$, but become isomorphic over $K:=\mathbb{Q}[\sqrt{2}]$, and let $$i:E_a' \to E_b'$$ (where $E_j'$ denotes a pullback of $E_j$ along $\mathrm{Spec}(K) \to \mathrm{Spec} \mathbb{Q}$) be the isomorphism $(x, y) \mapsto (x, \sqrt{2}y)$. So $i$ is a morphism in the groupoid $\mathcal{M}_{1,1}(K)$ and I'm trying to understand why it doesn't satisfy the prestack condition to descend to an isomorphism in $\mathcal{M}_{1,1}(\mathbb{Q})$. Let $pr_1, pr_2:\mathrm{Spec}( K \otimes_{\mathbb{Q}} K) \to \mathrm{Spec}(K)$ be the two projections. Let $(E_a')_1$ denote $pr_1^*E_a'$ and so on. Pulling back $i$ via these two projections we have $$i_1=pr_1^*i:(E_a')_1 \to (E_b')_1$$ $$i_2=pr_2^*i:(E_a')_2 \to (E_b')_2$$ We have objects with descent data $$(E_a', \alpha_{12}:(E_a')_1 \to (E_a')_2)$$ $$(E_b', \beta_{12}:(E_b')_1 \to (E_b')_2)$$ where $\alpha, \beta$ are the canonical isomorphisms between pullbacks. The map $i$ defines a morphism between these two objects with descent data if the diagram

$$ \require{AMScd} \begin{CD} (E_a')_1 @>{\alpha_{12}}>> (E_a')_2\\ @V{i_1}VV @VV{i_2}V \\ (E_b')_1 @>{\beta_{12}}>> (E_b)'_2 \end{CD}$$

commutes, but if I understand correctly, we don't want it to commute because then the fact that $\mathcal{M}_{1,1}$ is a stack would imply that $i$ descends to an isomorphism $E_a \to E_b$ (over $\mathbb{Q}$). So my question is, how to see that the diagram does not commute? Or maybe it does and my logic is mistaken?

(I am following the excellent exposition here by Thomas Barnett-Lamb; on p.31 he asks the above question, but I want to see his answer of "As usual for ´etale coverings which are Galois, the compatibility criterion on an isomorphism φ boils down to a Galois condition" more concretely, specifically as I translated above. I know $K \otimes_{\mathbb{Q}} K \simeq K \times K$ via $x\otimes y \mapsto (xy, x\sigma(y))$ where $\sigma$ is the automorphism $(a + b\sqrt{2} \mapsto a - b\sqrt{2})$ and think this may have something to do with the answer.)

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    $\begingroup$ Your origin is $(x,y)=(0,0)$, right? If so, then already the derivative at the origin of $i$, considered as an element of $\text{Hom}_{\mathbb{Q}}(\mathfrak{m}_{E_b,0}/\mathfrak{m}^2_{E_b,0}, \mathfrak{m}_{E_a,0}/\mathfrak{m}^2_{E_a,0})\otimes_{\mathbb{Q}}K$, is not the base change of any element of $\text{Hom}_{\mathbb{Q}}(\mathfrak{m}_{E_b,0}/\mathfrak{m}^2_{E_b,0}, \mathfrak{m}_{E_a,0}/\mathfrak{m}^2_{E_a,0})$. $\endgroup$ – Jason Starr Nov 26 '16 at 19:07
  • $\begingroup$ @JasonStarr I'm convinced $i$ doesn't come from anything over $\mathbb{Q}$. I'm more interested in understanding in this simple case why the morphism $i$ doesn't induce a morphism of descent data. Does your remark help with that? $\endgroup$ – usr0192 Nov 26 '16 at 20:03
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    $\begingroup$ The rule $(E,0)\mapsto \mathfrak{m}_{E,0}/\mathfrak{m}^2_{E,0}$ is a contravariant functor from descent data of elliptic curves to descent data of one-dimensional vector spaces. Since $i^*_0\in \text{Hom}_{\mathbb{Q}}(\mathfrak{m}_{E_b,0}/\mathfrak{m}^2_{E_b,0}, \mathfrak{m}_{E_a,0}/\mathfrak{m}^2_{E_a,0})\otimes_{\mathbb{Q}}K$ is not in the image of $\text{Hom}_{\mathbb{Q}}(\mathfrak{m}_{E_b,0}/\mathfrak{m}^2_{E_b,0}, \mathfrak{m}_{E_a,0}/\mathfrak{m}^2_{E_a,0})$, it is not part of a descent datum. $\endgroup$ – Jason Starr Nov 26 '16 at 20:56

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