TITLE: Residue field of point on an algebraic stack
QUESTION [10 upvotes]: $\DeclareMathOperator{\Spec}{Spec}$ Let $X$ be an algebraic stack.
Is there is a well-defined notion of the residue field of a point $x \in |X|$?
Attempts:

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.
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.
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,..)

REPLY [5 votes]: 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.