## 104.5 The definition

We'll formulate it as a definition and not as a result since we expect the reader to try out other cases (not just the stack $\mathcal{M}_{1, 1}$ and not just $\mathit{Sch}$ the category of all schemes).

Definition 104.5.1. We say $\mathcal{M}_{1, 1}$ is an algebraic stack if and only if

1. We have descent for objects for the étale topology on $\mathit{Sch}$.

2. The key fact holds.

3. there exists a surjective and smooth morphism $S \to \mathcal{M}_{1, 1}$.

The first condition is a “sheaf property”. We're going to spell it out since there is a technical point we should make. Suppose given a scheme $S$ and an étale covering $\{ S_ i \to S\}$ and morphisms $e_ i : S_ i \to \mathcal{M}_{1, 1}$ such that the diagrams

$\xymatrix{ S_ i \times _ S S_ j \ar[rd]_{e_ i \circ \text{pr}_1} \ar[rr]_{\text{id}} & & S_ i \times _ S S_ j \ar[ld]^{e_ j \circ \text{pr}_2} \\ & \mathcal{M}_{1, 1} }$

commute. The sheaf condition does not guarantee the existence of a morphism $e : S \to \mathcal{M}_{1, 1}$ in this situation. Namely, we need to pick witnesses $\alpha _{ij}$ for the diagrams above and require that

$\text{pr}_{02}^*\alpha _{ik} = \text{pr}_{12}^*\alpha _{jk} \circ \text{pr}_{01}^*\alpha _{ij}$

as witnesses over $S_ i \times _ S S_ j \times _ S S_ k$. I think it is clear what this means... If not, then I'm afraid you'll have to read some of the material on categories fibred in groupoids, etc. In any case, the displayed equation is often called the cocycle condition. A more precise statement of the “sheaf property” is: given $\{ S_ i \to S\}$, $e_ i : S_ i \to \mathcal{M}_{1, 1}$ and witnesses $\alpha _{ij}$ satisfying the cocycle condition, there exists a unique (up to unique isomorphism) $e : S \to \mathcal{M}_{1, 1}$ with $e_ i \cong e|_{S_ i}$ recovering the $\alpha _{ij}$.

As you can see even formulating a precise statement takes a bit of work. The proof of this “sheaf property” relies on a fundamental technique in algebraic geometry, namely descent theory. My suggestion is to initially simply accept the “sheaf property” holds, and see what it implies in practice. In fact, a certain amount of mental agility is required to boil the “sheaf property” down to a manageable statement that you can fit on a napkin. Perhaps the simplest variant which is already a bit interesting is the following: Suppose we have a finite Galois extension $L/K$ of fields with Galois group $G = \text{Gal}(L/K)$. Set $T = \mathop{\mathrm{Spec}}(L)$ and $S = \mathop{\mathrm{Spec}}(K)$. Then $\{ T \to S\}$ is an étale covering. Let $(E, f, 0)$ be an elliptic curve over $L$. (Yes, this just means that $E \subset \mathbf{P}^2_ L$ is given by a Weierstrass equation and $0$ is the usual point at infinity.) Denote $E_\sigma = E \times _{T, \mathop{\mathrm{Spec}}(\sigma )} T$ the base change. (Yes, this corresponds to applying $\sigma$ to the coefficients of the Weierstrass equation, or is it $\sigma ^{-1}$?) Now, suppose moreover that for every $\sigma \in G$ we are given an isomorphism

$\alpha _\sigma : E \longrightarrow E_\sigma$

over $T$. The cocycle condition above means in this situation that

$(\alpha _\tau )^\sigma \circ \alpha _\sigma = \alpha _{\tau \sigma }$

for $\sigma , \tau \in G$. If you've ever done any group cohomology then this should be familiar. Anyway, the “glueing” condition on $\mathcal{M}_{1, 1}$ says that if you have a solution to this set of equations, then there exists an elliptic curve $E'$ over $S$ such that $E \cong E' \times _ S T$ (it says a little bit more because it also tells you how to recover the $\alpha _\sigma$).

Challenge: Can you prove this entirely using only elliptic curves defined in terms of Weierstrass equations?

Comment #1679 by Joseph Gunther on

Minor typo: in the second-to-last sentence, the last $E$ should be an $E'$.

Comment #6608 by Jonas Ehrhard on

The cocycle condition for the $\alpha_{ij}$ misses a ${}^*$ I think.

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