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 $X,X':\text{Spec }K\rightarrow\mathcal{M}$. Suppose their pullbacks $X_L,X'_L$ are isomorphic, which is to say that the two composed morphisms $$\text{Spec }L\rightarrow\text{Spec K}\rightrightarrows \mathcal{M}$$ are 2-isomorphic. Now, I sort of want to say that because $\text{Spec }L\rightarrow\text{Spec }K$ 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 $p : \text{Spec }L\rightarrow\text{Spec }K$ 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?
  • $\begingroup$ You need extra conditions on your 2-arrow, namely coherence conditions, for it to descend. $\endgroup$ – David Roberts Aug 10 '16 at 22:32
  • $\begingroup$ @DavidRoberts Can you elaborate on what you mean by coherence conditions? $\endgroup$ – stupid_question_bot Aug 10 '16 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 '16 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. $\endgroup$ – stupid_question_bot Aug 11 '16 at 17:17
  • $\begingroup$ Sorry, then I'm out of ideas. $\endgroup$ – David Roberts Aug 11 '16 at 22:26

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).

  • 1
    $\begingroup$ I appreciate your answer, though I'm not sure how it addresses my question... $\endgroup$ – stupid_question_bot Sep 17 '16 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 '16 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? $\endgroup$ – stupid_question_bot Sep 17 '16 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 '16 at 7:27
  • $\begingroup$ Thanks for your comment. I've asked an updated question here: mathoverflow.net/questions/250182/… $\endgroup$ – stupid_question_bot Sep 18 '16 at 16:17

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