The Stacks project

Lemma 39.23.6. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(A)$ and $R = \mathop{\mathrm{Spec}}(B)$ are affine and $s, t : R \to U$ finite locally free. Let $C \subset A$ be as in (39.23.0.1). Then $U \to M = \mathop{\mathrm{Spec}}(C)$ has the following properties:

  1. the map on points $|U| \to |M|$ is surjective and $u_0, u_1 \in |U|$ map to the same point if and only if there exists a $r \in |R|$ with $t(r) = u_0$ and $s(r) = u_1$, in a formula

    \[ |M| = |U|/|R| \]
  2. for any algebraically closed field $k$ we have

    \[ M(k) = U(k)/R(k) \]

Proof. Since $C \to A$ is integral (Lemma 39.23.4) and injective we see that $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(C)$ is surjective, see Algebra, Lemma 10.36.17. Thus $|U| \to |M|$ is surjective.

Let $k$ be an algebraically closed field and let $C \to k$ be a ring map. Since surjective morphisms are preserved under base change (Morphisms, Lemma 29.9.4) we see that $A \otimes _ C k$ is not zero. Now $k \subset A \otimes _ C k$ is a nonzero integral extension. Hence any residue field of $A \otimes _ C k$ is an algebraic extension of $k$, hence equal to $k$. Thus we see that $U(k) \to M(k)$ is surjective.

Let $a_0, a_1 : A \to k$ be two ring maps. If there exists a ring map $b : B \to k$ such that $a_0 = b \circ t^\sharp $ and $a_1 = b \circ s^\sharp $ then we see that $a_0|_ C = a_1|_ C$ by definition. Thus the map $U(k) \to M(k)$ equalizes the two maps $R(k) \to U(k)$. Conversely, suppose that $a_0|_ C = a_1|_ C$. Let us name this algebra map $c : C \to k$. Consider the diagram

\[ \xymatrix{ & & B \ar@{-->}[lld] \\ k & & A \ar@<0.5ex>[ll]^{a_0} \ar@<-0.5ex>[ll]_{a_1} \ar@<1ex>[u] \ar@<-1ex>[u] \\ & & C \ar[u] \ar[llu]^ c } \]

If we can construct a dotted arrow making the diagram commute, then the proof of part (2) of the lemma is complete. Since $s : A \to B$ is finite there exist finitely many ring maps $b_1, \ldots , b_ n : B \to k$ such that $b_ i \circ s^\sharp = a_1$. If the dotted arrow does not exist, then we see that none of the $a'_ i = b_ i \circ t^\sharp $, $i = 1, \ldots , n$ is equal to $a_0$. Hence the maximal ideals

\[ \mathfrak m'_ i = \mathop{\mathrm{Ker}}(a_ i' \otimes 1 : A \otimes _ C k \to k) \]

of $A \otimes _ C k$ are distinct from $\mathfrak m = \mathop{\mathrm{Ker}}(a_0 \otimes 1 : A \otimes _ C k \to k)$. By Algebra, Lemma 10.15.2 we would get an element $f \in A \otimes _ C k$ with $f \in \mathfrak m$, but $f \not\in \mathfrak m_ i'$ for $i = 1, \ldots , n$. Consider the norm

\[ g = \text{Norm}_{s^\sharp \otimes 1}(t^\sharp \otimes 1(f)) \in A \otimes _ C k \]

By Lemma 39.23.2 this lies in the invariants $C^1 \subset A \otimes _ C k$ of the base change groupoid (base change via the map $c : C \to k$). On the one hand, $a_1(g) \in k^*$ since the value of $t^\sharp (f)$ at all the points (which correspond to $b_1, \ldots , b_ n$) lying over $a_1$ is invertible (insert future reference on property determinant here). On the other hand, since $f \in \mathfrak m$, we see that $f$ is not a unit, hence $t^\sharp (f)$ is not a unit (as $t^\sharp \otimes 1$ is faithfully flat), hence its norm is not a unit (insert future reference on property determinant here). We conclude that $C^1$ contains an element which is not nilpotent and not a unit. We will now show that this leads to a contradiction. Namely, apply Lemma 39.23.5 to the map $c : C \to C' = k$, then we see that the map of $k$ into the invariants $C^1$ is injective and moreover, that for any element $x \in C^1$ there exists an integer $n > 0$ such that $x^ n \in k$. Hence every element of $C^1$ is either a unit or nilpotent.

We still have to finish the proof of (1). We already know that $|U| \to |M|$ is surjective. It is clear that $|U| \to |M|$ is $|R|$-invariant. Finally, suppose $u_0, u_1 \in U$ maps to the same point $m \in M$. Then the induced field extensions $\kappa (u_0)/\kappa (m)$ and $\kappa (u_1)/\kappa (m)$ are algebraic (as $A$ is integral over $C$ as used above). Hence if $k$ is an algebraic closure of $\kappa (m)$ then we can find $\kappa (m)$-embeddings $\overline{u}_0 : \kappa (u_0) \to k$ and $\overline{u}_1 : \kappa (u_1) \to k$. These determine $k$-valued points $\overline{u}_0, \overline{u}_1 \in U(k)$ mapping to the same point of $M(k)$. By part (2) we see that there exists a point $\overline{r} \in R(k)$ with $s(\overline{r}) = \overline{u}_0$ and $t(\overline{r}) = \overline{u}_1$. The image $r \in R$ of $\overline{r}$ is a point with $s(r) = u_0$ and $t(r) = u_1$ as desired. $\square$


Comments (4)

Comment #1543 by jojo on

At the beginning of the proof it should say that is surjective not the other way around, I think.

Comment #6499 by Taeyeoup Kang on

Sorry, is it clear that (2) implies (1)? I think I can prove this as follows, but is there a simpler argument to show this one?

Proof of (2) (1) : Since both are locally of finite type, so is . Then (2) gives that is surjective(See 81.5.20 and 81.5.19 for details). Also we have a surjection by 64.4.3. Then the composition gives that is surjective, which means that is injective. The surjectivity of is clear from the surjectivity of .

Comment #6561 by on

OK, I gave a direct argument as we can't use the references you mentioned since they come later in the Stacks project. Thanks! See changes here.

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  • 6 comment(s) on Section 39.23: Finite flat groupoids, affine case

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