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$\DeclareMathOperator\Spec{Spec}$Let $L/K$ be a field extension, and let $\mathcal{M}$ be some moduli stack (for example, the stack of genus $g$ curves).

Let $X$, $X'$ be two objects of $\mathcal{M}$ over $K$, giving us two morphisms $X,X':\Spec K\rightarrow\mathcal{M}$. Suppose their pullbacks $X_L$, $X'_L$ are isomorphic, which is to say that the two composed morphisms $$\Spec L\rightarrow\Spec K\rightrightarrows \mathcal{M}$$ are 2-isomorphic. Now, I sort of want to say that because $\Spec L\rightarrow\Spec K$ is an epimorphism, that $X$, $X'$ must have determined 2-isomorphic morphisms, and hence were already isomorphic (over $K$) in the first place. …But this is obviously wrong (for example, take $\mathcal{M}$ to be the moduli stack of elliptic curves, and $X$, $X'$ to be two nonisomorphic (over $K$) elliptic curves with the same $j$-invariant).

Where exactly is the problem?

For example,

  1. Is $p : \Spec L\rightarrow\Spec K$ not an epimorphism in the category of algebraic stacks?
  2. Perhaps the right question is — Is $p$ a 2-epimorphism in the 2-category of algebraic stacks? What is a down-to-earth definition of a 2-epimorphism anyway? (nlab was not especially helpful in this regard.)
  3. Does the problem that arises in this situation disappear if we assume that $L$, $K$ are both algebraically closed?
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  • $\begingroup$ You need extra conditions on your 2-arrow, namely coherence conditions, for it to descend. $\endgroup$
    – David Roberts
    Aug 10, 2016 at 22:32
  • $\begingroup$ @DavidRoberts Can you elaborate on what you mean by coherence conditions? $\endgroup$ Aug 10, 2016 at 23:48
  • $\begingroup$ Hmm, maybe I was too quick. I think you need Spec(L) --> Spec(K) to be a regular epimorphism, hence K --> L to be an effective monomorphism. I'm not sure about such things in algebra... $\endgroup$
    – David Roberts
    Aug 11, 2016 at 5:41
  • $\begingroup$ @DavidRoberts So, according to nlab, in a category with pullbacks, (eg, the category of schemes), regular epimorphisms are the same as effective epimorphisms, and iirc finite etale morphisms are effective epi's, but the statement is definitely false for finite etale (ie, separable) extensions of fields. $\endgroup$ Aug 11, 2016 at 17:17
  • $\begingroup$ Sorry, then I'm out of ideas. $\endgroup$
    – David Roberts
    Aug 11, 2016 at 22:26

2 Answers 2

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Let more generally $f : T \to S$ be a fpqc covering and $X,X' \in \mathcal{M}(S)$ with $f^* X \cong f^* X'$ in $\mathcal{M}(T)$. By definition(*) of a prestack in the fpqc-topology, a necessary and sufficient condition for $X \cong X'$ (inducing the isomorphism $f^* X \cong f^* X'$) is that the following diagram commutes, where $p_1,p_2 : T \times_S T \rightrightarrows T$ are the two projections:

$$\begin{array}{ccc} p_1^* f^* X & \rightarrow & p_1^* f^* X' \\ \downarrow && \downarrow \\ p_2^* f^* X & \rightarrow & p_2^* f^* X' \end{array}$$

Vertically, we have the isomorphisms induced by $f p_1 = f p_2$. Horizontally, we have the isomorphisms induced by $f^* X \cong f^* X'$.

(*) A prestack (resp. stack) $\mathcal{M}$ is defined to be a fibered category such that for every covering $f:T \to S$ the functor $\mathcal{M}(S) \to \mathcal{M}_{\mathrm{descent}}(f)$ into the category of descent data is fully faithful (resp. an equivalence of categories).

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    $\begingroup$ I appreciate your answer, though I'm not sure how it addresses my question... $\endgroup$ Sep 17, 2016 at 22:16
  • $\begingroup$ You ask "what's the problem", the answer is "we have to work in the category of descent data, not just in $\mathcal{M}(L)$". Also, I've replaced by $\mathrm{Spec}(L) \to \mathrm{Spec}(K)$ by $f : T \to S$ since fields don't play any essential role here. $\endgroup$
    – HeinrichD
    Sep 17, 2016 at 22:23
  • $\begingroup$ I mean I agree with everything you said in your answer, but can you explain why we "have to work in the category of descent data?" For example, by Yoneda, there's a natural equivalence of categories between $Hom(S,\mathcal{M})$ and $\mathcal{M}(S)$. What you wrote is that $\mathcal{M}(S)\cong\mathcal{M}_{descent}(T\rightarrow S)$, which I understand. So my question effectively becomes: What exactly about my argument involving epimorphisms and universal properties breaks down when we translate everything into (e.g.) descent data? $\endgroup$ Sep 17, 2016 at 22:47
  • $\begingroup$ I wouldn't say that "epimorphism" (either in $1$- or $2$-categorical setting) explains what is going on here. In fact, there is no property at all on $f:T \to S$ which guarantees $$f^* X \cong f^* X' \text{ in } \mathcal{M}(T) \Longrightarrow X \cong X' \text{ in } \mathcal{M}(S),$$ expect for something trivial such as "$f$ is a split epimorphism". This is because $\mathcal{M}(T)$ is the "wrong" category. $\endgroup$
    – HeinrichD
    Sep 18, 2016 at 7:27
  • $\begingroup$ Thanks for your comment. I've asked an updated question here: mathoverflow.net/questions/250182/… $\endgroup$ Sep 18, 2016 at 16:17
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$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\Aut{Aut}$I think the issue here is that points of algebraic stacks have nontrivial automorphisms. In your example, the issue is that the 2 isomorphism is defined over $L$ but not over $K$, i.e., the automorphism group of that point change its structure upon a field extension. Certainly, for schemes, the map $\Spec L \to \Spec K$ is an epimorphism, since the automorphism group is trivial so it's defined over the integer. Now, if I want to remove that issue, I may want to assume that the automorphism group bears a trivial action of $\Aut_K(L)$. For nice algebraic stacks, the stabilizer groups are algebraic groups, so if $K$ is algebraically closed, then its structure is already stable under any field extension, i.e., its structure will not change anymore upon field extensions, this reduce the problem to the case where $L/K$ is algebraic.

Also, note that in the general case, where the source is no longer the spectrum of a field but only an algebraic stacks, then we can test whether two morphisms are the same on smooth presentations of the source algebraic stacks, and this reduces the problem to the case where the source is an affine scheme. Then we need to study the Galois action on both the set of 1-morphisms and also 2-morphisms. I guess that is the rough idea.

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  • $\begingroup$ TeX note: \DeclareMathOperator is specifically meant for, well, what its name suggests; so prefer \DeclareMathOperator\Spec{Spec}, later followed by $\DeclareMathOperator\Spec{Spec}\Spec L$ \Spec L (which automatically spaces as desired!), to $Spec \text{ } L$ Spec \text{ } L. For a one-off, you can use something like \operatorname{Spec} L instead. I have edited accordingly. $\endgroup$
    – LSpice
    Oct 7, 2023 at 0:05

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