Geometric arcs and fundamental groups of rigid spaces

Definition A.1

A rigid 𝐾-space is curve-connected if, for every two classical points x 0 , x 1 of 𝑋, there exists a sequence of morphisms C i → X ( i = 0 , … , n ) over 𝐾 where each C i is a connected affinoid rigid 𝐾-curve and we have x 0 ∈ im ⁡ ( C 0 → X ) , x 1 ∈ im ⁡ ( C n → X ) , and im ⁡ ( C i → X ) ∩ im ⁡ ( C i − 1 → X ) ≠ ∅ for i = 1 , … , n .

Proposition A.2

Let 𝑋 be a connected rigid 𝐾-space. Then 𝑋 is curve-connected.

Remark A.3

(a) The analogous result was proven by de Jong in [16, Theorem 6.1.1] for quasi-compact rigid 𝐾-spaces in the case when 𝐾 is discretely valued. Our proof follows his, but requires some nontrivial alterations due to the fact that O K is non-Noetherian. It is also worth pointing out that Berkovich obtained similar results for Berkovich spaces (see [7, Theorem 4.1.1]) which, in the language of adic spaces, are partially proper over Spa ⁡ ( K ) .

(b) The referee has suggested to us a shorter, simpler proof using more advanced machinery. See Remark A.7.

Proof

Idea of the proof. If 𝑋 is a connected affine variety, one can use the Bertini theorem to find a connected hypersurface passing through two given points. In the situation at hand, if X = Spa ⁡ ( A ) is affinoid, we can construct a suitable connected hyperplane section of its reduction X ~ = Spec ⁡ ( A ~ ) and lift it to the formal model Spf ⁡ ( A ∘ ) . The main difficulty, resolved carefully below, is to ensure that the resulting hypersurface remains connected on the generic fiber.

Step 1 [We may assume that 𝑋 is affinoid]. Let 𝑈 be the union of all connected quasi-compact opens containing x 0 . Then 𝑈 is (open and) closed: if y ∉ U and 𝑉 is a connected affinoid neighborhood of 𝑦, then V ∩ U = ∅ ; otherwise, V ∪ U is quasi-compact and connected. Since 𝑋 is connected, x 1 ∈ U , i.e. x 0 and x 1 admit a connected quasi-compact neighborhood 𝑊. Write W = ⋃ i = 1 n X i with X i affinoid, which we may assume are connected and X i ∩ X i − 1 ≠ ∅ (in particular, the intersection has classical points), and it is enough to show the result for the X i .

Step 2 [We may assume that 𝑋 is geometrically connected and geometrically normal]. In the proof, we can freely pass to a finite extension 𝐿 of 𝐾 because every connected component of X L surjects onto 𝑋. By [13, § 3.2], we may assume that 𝑋 is geometrically connected. By [13], affinoid 𝐾-algebras are excellent and we have the normalization map X n → X , a finite surjective morphism. Therefore, by [13, Theorem 3.3.6], after replacing 𝐾 with K 1 / p n (if char ⁡ K = p > 0 ), we may assume that 𝑋 is geometrically normal.

Note that, a priori, the replacement of 𝐾 with K 1 / p n is not permissible, as the curves we obtain may not be rigid 𝐾-curves if [ K 1 / p n : K ] is infinite. That said, let X ′ = X K 1 / p . The absolute Frobenius F : X → X factors through the relative Frobenius F X / K : X → X ′ and the base change map π : X ′ → X . Each of those maps is a (universal) homeomorphism, so in particular, X ′ is connected. Suppose that C ′ → X ′ is a map from a connected curve connecting the unique points x i ′ ∈ X ′ over x i ∈ X . Let C → X be its base change along F X / K ; then C → C ′ is a homeomorphism, so 𝐶 is connected. This argument is inspired by [16, § 6.2].

Step 3 [Reduced Fiber Theorem]. Write X = Spa ⁡ ( A ) . By the Reduced Fiber Theorem [26, 11], passing to a finite extension of 𝐾, we may assume that A ∘ is a topologically finitely presented O K -algebra such that A ∘ ⊗ k is geometrically reduced and the formation of A ∘ is compatible with further finite extensions of 𝐾.

Write A ~ = A ∘ ⊗ k and X ~ = Spec ⁡ ( A ~ ) , and set d = dim X . We denote the irreducible components of X ~ by Z 1 , … , Z r . Note that we have d = dim Z i for all 𝑖. To see this, let U ~ i ⊆ X ~ be an affine open subset contained in Z i so that dim U ~ i = dim Z i and let U i ⊆ Spf ⁡ ( A ∘ ) be the corresponding open formal subscheme. Then the generic fiber U i , η is an affinoid subdomain of 𝑋 and hence has dimension 𝑑. We conclude by [14, Theorem A.2.1].

Step 4 [Noether normalization]. Since X ~ is geometrically reduced, Lemma A.4 below yields a finite and generically étale morphism f ~ : X ~ → A k d (if 𝑘 is finite, we might need to pass to a finite extension of 𝐾). Choosing lifts of f ~ * ⁢ ( x i ) to A ∘ , we lift f ~ to a map

which we claim is finite, and étale at the points of X ~ where f ~ is étale. Then the morphism f η : X → D K d induced by K ⁢ ⟨ x 1 , … , x d ⟩ → A is finite as well.

To prove this claim, set B ∘ = O ⁢ ⟨ x 1 , … , x d ⟩ and let f 0 : B ∘ / ϖ ⁢ B ∘ → A ∘ / ϖ ⁢ A ∘ be the induced map. Since f 0 is a thickening of f ~ in the sense of [49, Tag 04EX] and A ∘ / ϖ ⁢ A ∘ is of finite presentation over O K / ϖ ⁢ O K , by [49, Tag 0BPG], the finiteness of f ~ implies the finiteness of f 0 . By [24, Chapter I, Proposition 4.2.3], this implies that 𝑓 is finite and therefore also finitely presented [9, Theorem 7.3/4].

Let now x ∈ X ~ ⊆ Spec ⁡ ( A ∘ ) be a point where f ~ is étale; we will show that 𝑓 is étale at 𝑥. Set y = f ⁢ ( x ) . As 𝑓 is finitely presented and Spec ( A ∘ ) y → Spec ( k ( y ) ) is étale at 𝑥 (as it agrees with f ~ ), it is enough by [49, Tag 01V9] to show that O Spec ⁡ ( B ∘ ) , y → O Spec ⁡ ( A ∘ ) , x is flat. Note that

and the right-hand side is flat by Gabber’s lemma [9, Lemma 8.2/2], but the map is also local by [9, Remark 7.2/1]. Thus, O Spec ⁡ ( A ∘ ) , x → O Spf ⁡ ( A ∘ ) , x is faithfully flat, and similarly, the map O Spec ⁡ ( B ∘ ) , y → O Spf ⁡ ( B ∘ ) , y is faithfully flat. Finally, the map O Spf ⁡ ( B ∘ ) , y → O Spf ⁡ ( A ∘ ) , x is flat by [24, Proposition I, 5.3.11] and [49, Tag 0CF4]. Note that the compositions

are equal. By the discussion above, the second composition is flat, and thus the first. As

is faithfully flat, we see O Spec ⁡ ( B ∘ ) , y → O Spec ⁡ ( A ∘ ) , x is flat as desired.

Step 5. We claim that it is enough to show that

if d ≥ 2 , there is a hyperplane H ⊆ D K d such that f η − 1 ⁢ ( H ) ⊆ X is connected.

Here, by a hyperplane, we mean the zero set H = V ⁢ ( h ) of a linear form h = ∑ i = 1 d a i ⁢ x i − a with a i , a ∈ O K with a i not all zero in 𝑘.

The argument for this is exactly as in [16]: we prove the entire theorem by induction on d = dim X ≥ 0 . Since the assertion is clear for d ≤ 1 , we assume d ≥ 2 . For the induction step, we find 𝐻 as in the claim, and since f η − 1 ⁢ ( H ) is curve-connected by the induction assumption, it suffices to connect a given classical point x ∈ X to a point y ∈ f η − 1 ⁢ ( H ) by a curve in 𝑋. Let p : D K d → H be a linear projection (so p | H = id H ); then C = p − 1 ⁢ ( p ⁢ ( f ⁢ ( x ) ) ) is a connected curve (in fact, isomorphic to D L 1 for a finite extension L / K ) connecting f ⁢ ( x ) with 𝐻. Let C ′ be the connected component of f − 1 ⁢ ( C ) containing 𝑥. We claim that dim C ′ = 1 ; otherwise, 𝑥 is an isolated point of f − 1 ⁢ ( C ) , which is impossible because the map

is open by [49, Tag 0F32]^[13]. The map C ′ → C is finite and hence surjective, and therefore C ′ connects 𝑥 with a point in f − 1 ⁢ ( H ) .

In the following steps, we shall construct a hyperplane H ~ ⊆ A k d and show that every hyperplane H ⊆ D K d lifting H ~ has connected preimage in 𝑋.

Step 6 [Construction of 𝐻]. For a decomposition I ∪ J = { 1 , … , r } with 𝐼, 𝐽 non-empty and disjoint, we set

treated as a reduced subscheme of X ~ . We claim as in [16, § 6.4] that dim Z I , J = d − 1 for every such I , J . To this end, note that X ~ ∖ Z I , J is disconnected, and hence so is U η ⊆ X , where U ⊆ Spf ⁡ ( A ∘ ) is the open formal subscheme supported on X ~ ∖ Z I , J . But if dim Z I , J < d − 1 , then by [39, Satz 2], we have O ⁢ ( U η ) = A , which does not have nontrivial idempotents.

We denote by T ~ ⊆ X ~ the closed subscheme where f ~ is not étale, and by S ~ ⊆ X ~ the closed subscheme where f ~ is not flat. By construction, we have dim T ~ < d . By Miracle Flatness [49, Tag 00R4], S ~ is the non-Cohen–Macaulay locus of X ~ , see [49, Tag 00RE]. Since X ~ is reduced, it is S 1 , see [49, Tag 031R], and therefore we have dim S ~ < d − 1 .

By the Bertini theorem [32, Théorème 6.3], a generic hyperplane (which exists after replacing 𝑘 by a finite extension if 𝑘 is finite) H ~ = V ⁢ ( h ~ ) ⊆ A k d , h ~ = ∑ i = 1 d a ~ i ⁢ x i − a ~ ( a ~ i , a ~ ∈ k ) satisfies the following properties.

1. The intersections f ~ − 1 ⁢ ( H ~ ) ∩ Z i ( i = 1 , … , r ) are irreducible of dimension d − 1 and generically étale over H ~ .

2. For every decomposition I ∪ J = { 1 , … , r } with 𝐼 and 𝐽 non-empty and disjoint, the intersection f ~ − 1 ⁢ ( H ~ ) ∩ Z I , J has a component of dimension d − 2 , and all such components are generically flat over H ~ .

Indeed, [32, Théorème 6.3 (1b), (4)] shows that an open set of hyperplanes will have the property that f ~ − 1 ⁢ ( H ~ ) ∩ Z i are irreducible of dimension d − 1 . To ensure étaleness in (1), it suffices to choose H ~ not contained in f ~ ⁢ ( T ~ ) . If V ⊆ Z I , J is an irreducible component of dimension d − 1 , then by [32, Théorème 6.3 (1b)] for a generic H ~ , the preimage f ~ − 1 ⁢ ( H ~ ) ∩ V is of dimension d − 2 . To ensure flatness in (2), we choose a hyperplane whose intersection with f ~ ⁢ ( V ) is not contained in f ~ ⁢ ( S ~ ) .

We let h = ∑ i = 1 d a i ⁢ x i − a ( a i , a ∈ O K ) be any lifting of ℎ, and set H = V ⁢ ( h ) ⊆ D K d . In the remaining two steps, we shall prove that f η − 1 ⁢ ( H ) ⊆ X is connected. By the claim in Step 5, this will finish the proof. To this end, if f η − 1 ⁢ ( H ) is disconnected, then we have Spec ⁡ ( A ∘ / h ⁢ A ∘ ) = T 1 ∪ T 2 for two non-empty closed subsets T 1 , T 2 such that T 1 ∩ T 2 is non-empty and set-theoretically contained in X ~ . In the final two steps, we shall derive a contradiction.

Step 7 [ T 1 ∩ T 2 does not contain any f ~ − 1 ⁢ ( H ~ ) ∩ Z i ]. Let ξ i be the generic point of f ~ − 1 ⁢ ( H ~ ) ∩ Z i ; we will show that ξ i ∉ T 1 ∩ T 2 . It is enough to show that A ξ i ∘ / h ⁢ A ξ i ∘ is a domain. Indeed, since dim ( T 1 ) = dim ( T 2 ) = d if ξ i ∈ T 1 ∩ T 2 , then Spec ⁡ ( A ξ i ∘ / h ⁢ A ξ i ∘ ) would contain multiple components of Spec ⁡ ( A ∘ / h ⁢ A ∘ ) , but this would contradict that A ξ i ∘ / h ⁢ A ξ i ∘ is a domain. To see that A ξ i ∘ / h ⁢ A ξ i ∘ is a domain, consider the localization 𝑅 of O K ⁢ ⟨ x 1 , … , x d ⟩ at the generic point of H ~ and the map R → A ξ i ∘ induced by 𝑓. Since 𝑓 is étale in a neighborhood of ξ i , the induced map R / h ⁢ R → A ξ i ∘ / h ⁢ A ξ i ∘ is the composition of an étale morphism and a localization; in particular, it is weakly étale. Since O K ⁢ ⟨ x 1 , … , x d ⟩ / h ≃ O K ⁢ ⟨ y 1 , … , y d − 1 ⟩ is normal, so is R / h ⁢ R . Thus, by [49, Tag 0950], we conclude that A ξ i ∘ / h ⁢ A ξ i ∘ is normal as well; in particular, it is a domain.

Step 8 [The contradiction]. It follows from the previous step that if we set

then these form a nontrivial partition of { 1 , … , r } . Therefore, T 1 ∩ T 2 ∩ X ~ is contained in Z I , J . Note though that, since T 1 and T 2 both have codimension one in Spec ⁡ ( A ∘ ) , the intersection T 1 ∩ T 2 ∩ X ~ necessarily contains a component 𝑉 of f ~ − 1 ⁢ ( H ~ ) ∩ Z I , J . This component can be taken to have codimension two in X ~ , and by our choice of hyperplane, we know that this component is generically Cohen–Macaulay and so not contained in S ~ .

Let 𝜁 be the generic point of 𝑉. By the previous paragraph, we know that 𝜁 is not contained in S ~ , and therefore A ~ ζ is a two-dimensional, Cohen–Macaulay, local ring. Moreover, h ~ is a non-zero divisor of A ~ ζ contained in its maximal ideal. Note that, since A ~ ζ / h ~ ⁢ A ~ ζ is not zero-dimensional, its maximal ideal contains some non-zero divisor. Let g ~ in A ~ ζ be the lift of such an element. Then ( g ~ , h ~ ) is a regular sequence in A ~ ζ . Let 𝜛 be the pseudouniformizer from above and choose a lift 𝑔 of g ~ in A ζ ∘ . Note then that ( ϖ , g , h ) is a regular sequence in A ζ ∘ . Indeed, this is clear since A ζ ∘ / ϖ ⁢ A ζ ∘ is a local ring and ( g , h ) have images in this ring that, in the further quotient ring, A ~ ζ have images forming a regular sequence. Therefore, ( ϖ , g ) is a regular sequence in the two-dimensional A ζ ∘ / h ⁢ A ζ ∘ . The fact that we can permute a regular sequence follows from the argument given in [49, Tag 00LJ]. Indeed, while A ζ ∘ / h ⁢ A ζ ∘ is not Noetherian, it is coherent (e. g. by [24, Chapter 0, Corollary 9.2.8]), and so the annihilator of any element of A ζ ∘ / h ⁢ A ζ ∘ is finitely generated, which is all the referenced argument requires.

By Lemma A.5 below, we know that W = Spec ⁡ ( A ζ ∘ / h ⁢ A ζ ∘ ) ∖ V ⁢ ( ϖ , g ) is connected. Note though that, since T 1 ∪ T 2 = Spec ⁡ ( A ∘ / h ⁢ A ∘ ) by construction, we have T 1 ′ ∪ T 2 ′ = W , where T i ′ = T i ∩ W . Each T i ′ is closed in 𝑊, and we have that T 1 ′ ∩ T 2 ′ ⊆ V ⁢ ( ϖ ) ∩ W . However, we have V ( ϖ ) ∩ W ) = ∅ : note that V ⁢ ( ϖ ) ∩ W = Spec ⁡ ( A ~ ζ / h ~ ⁢ A ~ ζ ) ∖ V ⁢ ( g ~ ) and this set is contained in the union of ξ i ’s that are in the intersection T 1 ∩ T 2 . As shown above, the set of those ξ i ’s is empty. Thus, T 1 ′ and T 2 ′ form a disconnection of 𝑊, a contradiction. ∎

Lemma A.5

Let 𝑅 be a local ring, let x , y ∈ R be such that ( x , y ) and ( y , x ) are regular sequences, and set W = Spec ⁡ ( R ) ∖ V ⁢ ( x , y ) . Then Γ ⁢ ( W , O W ) = R ; in particular, 𝑊 is connected.

Proof

Since W = D ⁢ ( x ) ∪ D ⁢ ( y ) , we have Γ ⁢ ( W , O W ) = ker ⁡ ( R x × R y → R x ⁢ y ) . As 𝑥 is a non-zero divisor, R → R x is injective, and hence so is R → Γ ⁢ ( W , O W ) . To show it is surjective, take an element ( a / x n , b / y m ) ∈ R x × R y in the kernel, i.e. ( x ⁢ y ) N ⁢ ( a ⁢ y m − b ⁢ x n ) = 0 . Since x ⁢ y is a non-zero divisor, we have a ⁢ y m = b ⁢ x n . We may assume that n = 0 or that a ∉ x ⁢ R . If n > 0 , then we have a ⁢ y m = 0 in R / x ⁢ R , and hence a ∈ x ⁢ R since 𝑦 is a non-zero divisor in R / x ⁢ R . Therefore, n = 0 , analogously m = 0 , and hence a = b . ∎

Corollary A.6

Let 𝑋 be a connected rigid 𝐾-space. Fix maximal points 𝑥 and 𝑦 in 𝑋. Then there exists a complete extension 𝐿 of 𝐾 and smooth connected affinoid 𝐿-curves C i with maps C i → X such that, for all 𝑖, we have that im ⁢ ( C i → X ) ∩ im ⁢ ( C i + 1 → X ) is non-empty, x ∈ im ⁢ ( C 1 → X ) , and y ∈ im ⁢ ( C m → X ) .

Proof

As in the second step of the proof of Proposition A.2, we may assume that 𝑋 is geometrically connected. Let us then note that, by Gruson’s theorem [27, Théorème 1], one has that the completed tensor product k ⁢ ( x ) ⊗ ^ K k ⁢ ( y ) is non-zero. Thus, by taking a point of Spa ⁡ ( k ⁢ ( x ) ⊗ ^ K k ⁢ ( y ) ) (which exists by [4, Theorem 1.2.1]) and looking at its residue field, we get a valued extension 𝐿 of 𝐾 containing both k ⁢ ( x ) and k ⁢ ( y ) . This gives two maps Spa ⁡ ( L ) ⇉ X . They give rise to two 𝐿-points x ′ and y ′ of X L mapping to 𝑥 and 𝑦, respectively. We then note that, by Proposition A.2, there exists smooth, connected, affinoid curves C 1 , … , C n over 𝐿 and morphisms C i → X L satisfying x ′ ∈ im ⁢ ( C 1 → X L ) , y ′ ∈ im ⁡ ( C n → X L ) , and

is non-empty. Clearly, then the compositions C i → X L → X satisfy the desired properties. ∎

Remark A.7

The referee pointed out to us a different (and in many ways simpler) proof of Proposition A.2, avoiding formal models and hence non-Noetherian difficulties. Instead, it relies on the foundational work of Ducros in [19, 21] surrounding the study of flatness in Berkovich analytic geometry. Because we think that both approaches contain valuable techniques, we have elected to include them both.

The key idea is to locally produce a smooth morphism with connected fibers to a curve, and then use the existence of multisections to deduce the curve-connectedness of the space from that of the fibers, so that we conclude by induction on dimension. This is enabled by two results of Ducros, or rather their following corollaries for rigid 𝐾-spaces.

For visual aid,

We shall now deduce these propositions from their Berkovich geometry versions, which requires some preparations that the reader may safely skip. We shall use the equivalence of categories

from [31, Remark 8.3.2], defined on affinoids by Spa ( A ) Berk = M ( A ) . In particular, 𝑋 is affinoid if and only if X Berk is affinoid, and a map 𝜑 of rigid 𝐾-spaces is a closed immersion if and only if φ Berk is.

By [31, Lemma 8.1.7], the composition of ( − ) Berk with the forgetful functor to topological spaces agrees with universal separated quotient functor [ − ] from § 2.1. In particular, as observed in [31], we have that 𝑋 is, respectively, connected and quasi-compact if and only if X Berk is (see Proposition 2.2.3). Finally, one sees from [31, Proposition 8.3.7] that 𝑋 is good if and only if X Berk is good.

Moreover, as the source category and target category are closed under fiber products with the category of all rigid 𝐾-spaces and strictly 𝐾-analytic spaces by [31, Lemma 5.1.4] and the fact that the natural map | Y × X X ′ | → | Y | × | X | | X ′ | is compact in the sense of [5, § 1.4], respectively, the functor ( − ) Berk commutes with fiber products in the naive sense. Continuing further, one sees that, for a morphism 𝜑 in the source category, one has that Δ φ Berk = Δ φ Berk and thus, from our previous observations concerning Zariski closed embeddings, that 𝜑 is separated if and only if φ Berk is.

The following lemma lists additional properties of the functor ( − ) Berk which will be useful in the proof. For part (1), recall that, by definition, a morphism of rigid 𝐾-spaces φ : Y → X is flat if and only if, for every y ∈ Y , the stalk O Y , y is flat over O X , φ ⁢ ( y ) . In Berkovich geometry, such a naive notion is problematic (see [19, § 4.4]), and the actual definition of flat maps of 𝐾-analytic spaces is more involved; in particular, it imposes stability under base change (see [19, Definition 4.1.8]).

We now return to the second proof of Proposition A.2. In the process of the proof, we shall use the following reduction steps.

1. If X = ⋃ X i is an open cover with X i connected and the conclusion of Proposition A.2 holds for the X i , then it holds for 𝑋 (see Step 1 of the previous proof).

2. If X ′ → X is a surjective map with X ′ connected, and the conclusion of Proposition A.2 holds for X ′ , then it holds for 𝑋.

3. If 𝑋 is irreducible and U ⊆ X is a non-empty Zariski open subspace, and the conclusion of Proposition A.2 holds for 𝑈,^[18] then it holds for 𝑋.

To see (C), it is enough to show that, for any classical point x ∈ X ∖ U , there exists a classical point 𝑦 of 𝑈 which is connected to 𝑥 by curves. By (A), we may assume that 𝑋 is affinoid. Let f : X → D K n be a finite map, which exists by Noether normalization (cf. [10, Corollary 6.1.2/2]). As 𝑋 is irreducible, we know that X − U has strictly smaller dimension than 𝑋, and thus f ⁢ ( X − U ) is a proper Zariski closed subset of D K n (cf. [10, Proposition 9.6.3/3]). Take a point 𝑝 in D K n − f ⁢ ( X − U ) and a line ℓ in D L n connecting f ⁢ ( x ) and 𝑝, where 𝐿 is a finite extension of 𝐾 containing the residue fields of f ⁢ ( x ) and 𝑝. Consider then any connected component 𝐶 of X × D K n ℓ . Observe that, as C → ℓ is finite, it is clear that 𝐶 is an affinoid curve, and this map is surjective since the image is a Zariski closed subset (by [10, Proposition 9.6.3/3]) of dimension 1. In particular, the image of C → X must contain a point 𝑞 of f − 1 ⁢ ( p ) which cannot be in X − U and so must be in 𝑈.