The Stacks project

Proof. Consider the commutative diagram

of Lemma 39.13.4. Think of $f \in \Gamma (U, \mathcal{O}_ U)$. The commutativity of the top part of the diagram shows that $\text{pr}_0^\sharp (t^\sharp (f)) = c^\sharp (t^\sharp (f))$ as elements of $\Gamma (R \times _{S, U, t} R, \mathcal{O})$. Looking at the right lower cartesian square the compatibility of the norm construction with base change shows that $s^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f))) = \text{Norm}_{\text{pr}_1^\sharp }(c^\sharp (t^\sharp (f)))$. Similarly we get $t^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f))) = \text{Norm}_{\text{pr}_1^\sharp }(\text{pr}_0^\sharp (t^\sharp (f)))$. Hence by the first equality of this proof we see that $s^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f))) = t^\sharp (\text{Norm}_{s^\sharp }(t^\sharp (f)))$ as desired. $\square$

Proof. By Morphisms, Lemma 29.48.5 there exists a decomposition $U = \coprod \nolimits _{r \geq 0} U_ r$ such that $s : s^{-1}(U_ r) \to U_ r$ is finite locally free of rank $r$. As $s$ is surjective we see that $U_0 = \emptyset $. Note that $u \in U_ r \Leftrightarrow $ if and only if the scheme theoretic fibre $s^{-1}(u)$ has degree $r$ over $\kappa (u)$. Now, if $z \in R$ with $s(z) = u$ and $t(z) = u'$ then using notation as in Lemma 39.13.4

is the base change of both $s^{-1}(u) \to \mathop{\mathrm{Spec}}(\kappa (u))$ and $s^{-1}(u') \to \mathop{\mathrm{Spec}}(\kappa (u'))$ by the lemma cited. Hence $u \in U_ r \Leftrightarrow u' \in U_ r$, in other words, the open subsets $U_ r$ are $R$-invariant. In particular the restriction of $R$ to $U_ r$ is just $s^{-1}(U_ r)$ and $s : R_ r \to U_ r$ is finite locally free of rank $r$. As $t : R_ r \to U_ r$ is isomorphic to $s$ by the inverse of $R_ r$ we see that it has also rank $r$. $\square$

Proof. First, by Lemma 39.23.3 we know that $(U, R, s, t, c)$ is a disjoint union of groupoid schemes $(U_ r, R_ r, s, t, c)$ such that each $s, t : R_ r \to U_ r$ has constant rank $r$. As $U$ is quasi-compact, we have $U_ r = \emptyset $ for almost all $r$. It suffices to prove the lemma for each $(U_ r, R_ r, s, t, c)$ and hence we may assume that $s, t$ are finite locally free of rank $r$.

Assume that $s, t$ are finite locally free of rank $r$. Let $f \in A$. Consider the element $x - f \in A[x]$, where we think of $x$ as the coordinate on $\mathbf{A}^1$. Since

is also a groupoid scheme with finite source and target, we may apply Lemma 39.23.2 to it and we see that $P(x) = \text{Norm}_{s^\sharp }(t^\sharp (x - f))$ is an element of $C[x]$. Because $s^\sharp : A \to B$ is finite locally free of rank $r$ we see that $P$ is monic of degree $r$. Moreover $P(f) = 0$ by Cayley-Hamilton (Algebra, Lemma 10.16.1). $\square$

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

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

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 $D[x] \to C'[x]$. $\square$

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

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

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

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. $\square$

Proof. Set $M = \mathop{\mathrm{Spec}}(C)$ and $M' = \mathop{\mathrm{Spec}}(C')$. Write $U = \mathop{\mathrm{Spec}}(A)$, $U' = \mathop{\mathrm{Spec}}(A')$, $R = \mathop{\mathrm{Spec}}(B)$, and $R' = \mathop{\mathrm{Spec}}(B')$. We will use the results of Lemmas 39.23.4, 39.23.5, and 39.23.6 without further mention.

Assume $C$ is a strictly henselian local ring. Let $p \in M$ be the closed point and let $p' \in M'$ map to $p$. Claim: in this case there is a disjoint union decomposition $(U', R', s', t', c') = (U, R, s, t, c) \amalg (U'', R'', s'', t'', c'')$ over $(U, R, s, t, c)$ such that for the corresponding disjoint union decomposition $M' = M \amalg M''$ over $M$ the point $p'$ corresponds to $p \in M$.

The claim implies the lemma. Suppose that $M_1 \to M$ is a flat morphism of affine schemes. Then we can base change everything to $M_1$ without affecting the hypotheses (1) – (4). From Lemma 39.23.5 we see $M_1$, resp. $M_1'$ is the spectrum of the $R_1$-invariant functions on $U_1$, resp. the $R'_1$-invariant functions on $U'_1$. Suppose that $p' \in M'$ maps to $p \in M$. Let $M_1$ be the spectrum of the strict henselization of $\mathcal{O}_{M, p}$ with closed point $p_1 \in M_1$. Choose a point $p'_1 \in M'_1$ mapping to $p_1$ and $p'$. From the claim we get

and correspondingly $M'_1 = M_1 \amalg M''_1$ as a scheme over $M_1$. Write $M_1 = \mathop{\mathrm{Spec}}(C_1)$ and write $C_1 = \mathop{\mathrm{colim}}\nolimits C_ i$ as a filtered colimit of étale $C$-algebras. Set $M_ i = \mathop{\mathrm{Spec}}(C_ i)$. The $M_1 = \mathop{\mathrm{lim}}\nolimits M_ i$ and similarly for the other schemes. By Limits, Lemmas 32.4.11 and 32.8.11 we can find an $i$ such that

We conclude that $M'_ i = M_ i \amalg M''_ i$. In particular $M' \to M$ becomes étale at a point over $p'$ after an étale base change. This implies that $M' \to M$ is étale at $p'$ (for example by Morphisms, Lemma 29.36.17). We will prove $U' \cong M' \times _ M U$ after we prove the claim.

Proof of the claim. Observe that $U_ p$ and $U'_{p'}$ have finitely many points. For $u \in U_ p$ we have $\kappa (u)/\kappa (p)$ is algebraic, hence $\kappa (u)$ is separably closed. As $U' \to U$ is étale, we conclude the morphism $U'_{p'} \to U_ p$ induces isomorphisms on residue field extensions. Let $u' \in U'_{p'}$ with image $u \in U_ p$. By assumption (3) the morphism of scheme theoretic fibres $(s')^{-1}(u') \to s^{-1}(u)$, $(t')^{-1}(u') \to t^{-1}(u)$, and $G'_{u'} \to G_ u$ are isomorphisms. Observing that $U_ p = t(s^{-1}(u))$ (set theoretically) we conclude that the points of $U'_{p'}$ surject onto the points of $U_ p$. Suppose that $u'_1$ and $u'_2$ are points of $U'_{p'}$ mapping to the same point $u$ of $U_ p$. Then there exists a point $r' \in R'_{p'}$ with $s'(r') = u'_1$ and $t'(r') = u'_2$. Consider the two towers of fields

whose “ends” are the same as the two “ends” of the two towers

These two induce the same maps $\kappa (p') \to \kappa (r')$ as $(U'_{p'}, R'_{p'}, s', t', c')$ is a groupoid over $p'$. Since $\kappa (u)/\kappa (p)$ is purely inseparable, we conclude that the two induced maps $\kappa (u) \to \kappa (r')$ are the same. Therefore $r'$ maps to a point of the fibre $G_ u$. By assumption (3) we conclude that $r' \in (G')_{u'_1}$. Namely, we may think of $G$ as a closed subscheme of $R$ viewed as a scheme over $U$ via $s$ and use that the base change to $U'$ gives $G' \subset R'$. In particular we have $u'_1 = u'_2$. We conclude that $U'_{p'} \to U_ p$ is a bijective map on points inducing isomorphisms on residue fields. It follows that $U'_{p'}$ is a finite set of closed points (Algebra, Lemma 10.35.9) and hence $U'_{p'}$ is closed in $U'$. Let $J' \subset A'$ be the radical ideal cutting out $U'_{p'}$ set theoretically.

Second part proof of the claim. Let $\mathfrak m \subset C$ be the maximal ideal. Observe that $(A, \mathfrak m A)$ is a henselian pair by More on Algebra, Lemma 15.11.8. Let $J = \sqrt{\mathfrak m A}$. Then $(A, J)$ is a henselian pair (More on Algebra, Lemma 15.11.7) and the étale ring map $A \to A'$ induces an isomorphism $A/J \to A'/J'$ by our deliberations above. We conclude that $A' = A \times A''$ by More on Algebra, Lemma 15.11.6. Consider the corresponding disjoint union decomposition $U' = U \amalg U''$. The open $(s')^{-1}(U)$ is the set of points of $R'$ specializing to a point of $R'_{p'}$. Similarly for $(t')^{-1}(U)$. Similarly we have $(s')^{-1}(U'') = (t')^{-1}(U'')$ as this is the set of points which do not specialize to $R'_{p'}$. Hence we obtain a disjoint union decomposition

This immediately gives $M' = M \amalg M''$ and the proof of the claim is complete.

We still have to prove that the canonical map $U' \to M' \times _ M U$ is an isomorphism. It is an étale morphism (Morphisms, Lemma 29.36.18). On the other hand, by base changing to strictly henselian local rings (as in the third paragraph of the proof) and using the bijectivity $U'_{p'} \to U_ p$ established in the course of the proof of the claim, we see that $U' \to M' \times _ M U$ is universally bijective (some details omitted). However, a universally bijective étale morphism is an isomorphism (Descent, Lemma 35.25.2) and the proof is complete. $\square$

Proof. During this proof we will write $s, t : A \to B$ instead of $s^\sharp , t^\sharp $, and similarly $c : B \to B \otimes _{s, A, t} B$. We write $p_0 : B \to B \otimes _{s, A, t} B$, $b \mapsto b \otimes 1$ and $p_1 : B \to B \otimes _{s, A, t} B$, $b \mapsto 1 \otimes b$. By Lemma 39.13.5 and the definition of $C$ we have the following commutative diagram

Moreover the tow left squares are cocartesian in the category of rings, and the top row is isomorphic to the diagram

which is an equalizer diagram according to Descent, Lemma 35.3.6 because condition (2) implies in particular that $s$ (and hence also then isomorphic arrow $t$) is faithfully flat. The lower row is an equalizer diagram by definition of $C$. We can use the $x_ i$ and get a commutative diagram

where in the right vertical arrow we map $x_ i$ to $x_ i$, in the middle vertical arrow we map $x_ i$ to $t(x_ i)$ and in the left vertical arrow we map $x_ i$ to $c(t(x_ i)) = t(x_ i) \otimes 1 = p_0(t(x_ i))$ (equality by the commutativity of the top part of the diagram in Lemma 39.13.4). Then the diagram commutes. Moreover the middle vertical arrow is an isomorphism by assumption. Since the left two squares are cocartesian we conclude that also the left vertical arrow is an isomorphism. On the other hand, the horizontal rows are exact (i.e., they are equalizers). Hence we conclude that also the right vertical arrow is an isomorphism. $\square$

Proof. During this proof we use the notation $s, t : A \to B$ instead of the notation $s^\sharp , t^\sharp $. By Lemma 39.20.3 it suffices to show that $C \to A$ is finite locally free and that the map

is an isomorphism. First, note that $j$ is a monomorphism, and also finite (since already $s$ and $t$ are finite). Hence we see that $j$ is a closed immersion by Morphisms, Lemma 29.44.15. Hence $A \otimes _ C A \to B$ is surjective.

We will perform base change by flat ring maps $C \to C'$ as in Lemma 39.23.5, and we will use that formation of invariants commutes with flat base change, see part (3) of the lemma cited. We will show below that for every prime $\mathfrak p \subset C$, there exists a local flat ring map $C_{\mathfrak p} \to C_{\mathfrak p}'$ such that the result holds after a base change to $C_{\mathfrak p}'$. This implies immediately that $A \otimes _ C A \to B$ is injective (use Algebra, Lemma 10.23.1). It also implies that $C \to A$ is flat, by combining Algebra, Lemmas 10.39.17, 10.39.18, and 10.39.8. Then since $U \to \mathop{\mathrm{Spec}}(C)$ is surjective also (Lemma 39.23.6) we conclude that $C \to A$ is faithfully flat. Then the isomorphism $B \cong A \otimes _ C A$ implies that $A$ is a finitely presented $C$-module, see Algebra, Lemma 10.83.2. Hence $A$ is finite locally free over $C$, see Algebra, Lemma 10.78.2.

By Lemma 39.23.3 we know that $A$ is a finite product of rings $A_ r$ and $B$ is a finite product of rings $B_ r$ such that the groupoid scheme decomposes accordingly (see the proof of Lemma 39.23.4). Then also $C$ is a product of rings $C_ r$ and correspondingly $C'$ decomposes as a product. Hence we may and do assume that the ring maps $s, t : A \to B$ are finite locally free of a fixed rank $r$.

The local ring maps $C_{\mathfrak p} \to C_{\mathfrak p}'$ we are going to use are any local flat ring maps such that the residue field of $C_{\mathfrak p}'$ is infinite. By Algebra, Lemma 10.159.1 such local ring maps exist.

Assume $C$ is a local ring with maximal ideal $\mathfrak m$ and infinite residue field, and assume that $s, t : A \to B$ is finite locally free of constant rank $r > 0$. Since $C \subset A$ is integral (Lemma 39.23.4) all primes lying over $\mathfrak m$ are maximal, and all maximal ideals of $A$ lie over $\mathfrak m$. Similarly for $C \subset B$. Pick a maximal ideal $\mathfrak m'$ of $A$ lying over $\mathfrak m$ (exists by Lemma 39.23.6). Since $t : A \to B$ is finite locally free there exist at most finitely many maximal ideals of $B$ lying over $\mathfrak m'$. Hence we conclude (by Lemma 39.23.6 again) that $A$ has finitely many maximal ideals, i.e., $A$ is semi-local. This in turn implies that $B$ is semi-local as well. OK, and now, because $t \otimes s : A \otimes _ C A \to B$ is surjective, we can apply Algebra, Lemma 10.78.8 to the ring map $C \to A$, the $A$-module $M = B$ (seen as an $A$-module via $t$) and the $C$-submodule $s(A) \subset B$. This lemma implies that there exist $x_1, \ldots , x_ r \in A$ such that $M$ is free over $A$ on the basis $s(x_1), \ldots , s(x_ r)$. Hence we conclude that $C \to A$ is finite free and $B \cong A \otimes _ C A$ by applying Lemma 39.23.8. $\square$