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$\DeclareMathOperator{\Spec}{Spec}$ Let $X$ be an algebraic stack.

Is there is a well-defined notion of the residue field of a point $x \in |X|$?

Attempts:

  1. Recall that a point on a stack is an equivalence class of morphisms $\Spec k \to X$ from fields $k$. The issue is that it is not clear that there is a minimal choice of $k$ to warrant being called the residue field.
  2. There is also the notion of a residual gerbe on a stack; but again it is not clear whether this comes with some kind of canonical field of definition which is compatible with 1.
  3. If $X$ has a coarse moduli space $X^c$, then one could define the residue field of $x$ to be the residue field of the image of $x$ in $X^c$. This is well-defined, but seems to lose some of the subtle properties of stacks and again it's not clear whether it is compatible with 1. and 2.

I'm happy to assume my stack is sufficiently nice (e.g. smooth, DM,..)

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  • $\begingroup$ Given a point, one can take the category of pairs of a field $k$ and $k$-point of $X$ equivalent to your point, with morphisms respecting the $k$-point, and then take "global sections" of the functor sending a pair to $k$, i.e. take natural transformations from the constant functor 1 to this functor. But I guess this would be equivalent to option 3. $\endgroup$
    – Will Sawin
    Aug 3 at 13:25
  • $\begingroup$ @DanielLoughran - why do you ask? Do you want to define the "Hasse-Weil L-function" $L([X], s)$ of an algebraic stack $[X]$? Do you want to "count the number of elements" in the "residue field" $\kappa(x)$? $\endgroup$
    – user122276
    Aug 3 at 13:28
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    $\begingroup$ @hm2020: The residue field is such a fundamental concept in scheme theory I want to see whether there is a stacky analogue. $\endgroup$ Aug 3 at 13:59
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By definition, a residue field is an equivalence class of morphisms $\operatorname{Spec} k \to X$, i.e. of pairs of a field $k$ and an object in $X(k)$

We can upgrade that equivalence class into a category: Given fields $k$, $L$ and objects $a \in X(k) , b\in X(L)$, a morphism is a map $s \colon k \to L $ together with an isomorphism $s^* a \to b$.

The key property that the residue field $F$ should have is that for every $k$-point of $X$ we obtain a map $F \to k$.

I claim we should define the residue field as the universal object with this property.

In other words, an element of the residue field is an assignment to each pair $k, a \in X(k)$ an element $\alpha_a \in k$, compatible in the obvious way with morphisms: For $s \colon (k, a) \to (L,b)$ a morphisms, we have $s(\alpha_a) = \alpha_b$.

Elements of the residue field form a ring as there is an obvious notion of addition and multiplication. To check that they form a field, we need to check that every element is either invertible on every $k$-point or zero on every $k$-point, but this follows from the fact that we are working with a single equivalence class.

So we indeed have a universal notion of the residue field.

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  • $\begingroup$ Let me clarify: where do you use the fact that $X$ is an algebraic stack? Or you show that the residue field exists for any presheaf of groupoids? $\endgroup$
    – Z. M
    Aug 3 at 15:09
  • $\begingroup$ @Z.M This notion of the residue field exists for any presheaf of groupoids. $\endgroup$
    – Will Sawin
    Aug 3 at 15:41
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    $\begingroup$ I guess this is equivalent to the global sections of the structure sheaf on the residual gerbe. $\endgroup$
    – Will Sawin
    Aug 3 at 15:44
  • $\begingroup$ Very slick construction, thanks. Just so I completely understand: your definition of the residue field $\kappa(x)$ does not necessarily come equipped with a morphism whose image is $x$? $\endgroup$ Aug 3 at 18:33
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    $\begingroup$ @DanielLoughran Correct. I think an example where this will not exist, because there is no minimal field as in 1, is a gerbe over $\mathbb Q$ or another nice field associated to a nontrivial class in $H^2(\mathbb Q, \mu_2)$. There are many extensions of degree $2$ over which this gerbe trivializes and hence the stack has a point, but no extension of lesser degree, and thus no minimal one. $\endgroup$
    – Will Sawin
    Aug 3 at 19:32

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