Lemma 39.23.5. 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). Let $C \to C'$ be a ring map, and set $U' = \mathop{\mathrm{Spec}}(A \otimes _ C C')$, $R' = \mathop{\mathrm{Spec}}(B \otimes _ C C')$. Then
The maps $s, t, c$ induce maps $s', t', c'$ such that $(U', R', s', t', c')$ is a groupoid scheme. Let $C^1 \subset A'$ be the $R'$-invariant functions on $U'$.
The canonical map $\varphi : C' \to C^1$ satisfies
for every $f \in C^1$ there exists an $n > 0$ and a polynomial $P \in C'[x]$ whose image in $C^1[x]$ is $(x - f)^ n$, and
for every $f \in \mathop{\mathrm{Ker}}(\varphi )$ there exists an $n > 0$ such that $f^ n = 0$.
If $C \to C'$ is flat then $\varphi $ is an isomorphism.
Proof.
The proof of part (1) is omitted. Let us denote $A' = A \otimes _ C C'$ and $B' = B \otimes _ C C'$. Then we have
\[ C^1 = \{ a \in A' \mid (t')^\sharp (a) = (s')^\sharp (a) \} = \{ a \in A \otimes _ C C' \mid t^\sharp \otimes 1(a) = s^\sharp \otimes 1(a) \} . \]
In other words, $C^1$ is the kernel of the difference map $(t^\sharp - s^\sharp ) \otimes 1$ which is just the base change of the $C$-linear map $t^\sharp - s^\sharp : A \to B$ by $C \to C'$. Hence (3) follows.
Proof of part (2)(b). 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 also $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(C')$ is surjective as a base change of a surjective morphism (Morphisms, Lemma 29.9.4). Hence $\mathop{\mathrm{Spec}}(C^1) \to \mathop{\mathrm{Spec}}(C')$ is surjective also. This implies (2)(b) holds for example by Algebra, Lemma 10.30.6.
Proof of part (2)(a). By Lemma 39.23.3 our groupoid scheme $(U, R, s, t, c)$ decomposes as a finite disjoint union of groupoid schemes $(U_ r, R_ r, s, t, c)$ such that $s, t : R_ r \to U_ r$ are finite locally free of rank $r$. Pulling back by $U' = \mathop{\mathrm{Spec}}(C') \to U$ we obtain a similar decomposition of $U'$ and $U^1 = \mathop{\mathrm{Spec}}(C^1)$. We will show in the next paragraph that (2)(a) holds for the corresponding system of rings $A_ r, B_ r, C_ r, C'_ r, C^1_ r$ with $n = r$. Then given $f \in C^1$ let $P_ r \in C_ r[x]$ be the polynomial whose image in $C^1_ r[x]$ is the image of $(x - f)^ r$. Choosing a sufficiently divisible integer $n$ we see that there is a polynomial $P \in C'[x]$ whose image in $C^1[x]$ is $(x - f)^ n$; namely, we take $P$ to be the unique element of $C'[x]$ whose image in $C'_ r[x]$ is $P_ r^{n/r}$.
In this paragraph we prove (2)(a) in case the ring maps $s^\sharp , t^\sharp : A \to B$ are finite locally free of a fixed rank $r$. Let $f \in C^1 \subset A' = A \otimes _ C C'$. Choose a flat $C$-algebra $D$ and a surjection $D \to C'$. Choose a lift $g \in A \otimes _ C D$ of $f$. Consider the polynomial
\[ P = \text{Norm}_{s^\sharp \otimes 1}(t^\sharp \otimes 1(x - g)) \]
in $(A \otimes _ C D)[x]$. By Lemma 39.23.2 and part (3) of the current lemma the coefficients of $P$ are in $D$ (compare with the proof of Lemma 39.23.4). On the other hand, the image of $P$ in $(A \otimes _ C C')[x]$ is $(x - f)^ r$ because $t^\sharp \otimes 1(x - f) = s^\sharp (x - f)$ and $s^\sharp $ is finite locally free of rank $r$. This proves what we want with $P$ as in the statement (2)(a) given by the image of our $P$ under the map $D[x] \to C'[x]$.
$\square$
Comments (2)
Comment #4121 by Ko Aoki on
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