Adic Spaces — Lean blueprint

5. Integral elements and affinoid rings🔗

Definition5.1
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Definition 1.1
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Theorem 5.2
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L∃∀N

For a topological ring A, a subring A^{+} \subseteq A is a ring of integral elements (\texttt{IsRingOfIntegralElements}) if it is open, integrally closed in A, and contained in the power-bounded subring A^{\circ} (Definition 1.1, Definition 2.1). The pair (A, A^{+}) is then an affinoid ring (Huber pair), \texttt{IsAffinoidRing}. Following Wedhorn Definition 7.14.

Lean code for Definition5.12 definitions
  • class(3 methods)defined in «Adic spaces»/AffinoidRings.lean
    complete
    class ValuationSpectrum.IsRingOfIntegralElements.{u_1} {A : Type u_1}
      [CommRing A] [TopologicalSpace A] (B : Subring A) : Prop
    class ValuationSpectrum.IsRingOfIntegralElements.{u_1}
      {A : Type u_1} [CommRing A]
      [TopologicalSpace A] (B : Subring A) :
      Prop
    A ring of integral elements (Definition 7.14(1)): open, integrally closed, and in `A°`.
    
    A `class` (not a bare `structure`) so it can be required as `[IsRingOfIntegralElements (A⁺)]`
    on `Spa`-level theorems — the faithful affinoid-ring interface (expert-review 2026-06-18,
    reviewer Q1). The bare `PlusSubring` carrier stays low-level; this records the three
    Definition-7.14 axioms (open, integrally closed, `⊆ A°`) that make `(A, A⁺)` an affinoid ring. 
    isOpen : IsOpen B
    `B` is open in `A`. 
    isIntegrallyClosed :  (a : A), IsIntegral (↥B) a  a  B
    `B` is integrally closed in `A`. 
    subset_powerBounded : B  TopologicalRing.powerBoundedSubring A
    `B ⊆ A°`. 
  • defdefined in «Adic spaces»/AffinoidRings.lean
    complete
    def ValuationSpectrum.IsAffinoidRing.{u_2} (A : Type u_2) [CommRing A]
      [TopologicalSpace A] [ValuationSpectrum.PlusSubring A] : Prop
    def ValuationSpectrum.IsAffinoidRing.{u_2}
      (A : Type u_2) [CommRing A]
      [TopologicalSpace A]
      [ValuationSpectrum.PlusSubring A] : Prop
    `(A, A⁺)` is an affinoid ring if `A⁺` is a ring of integral elements (Definition 7.14). 
Theorem5.2
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Definition 2.1
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used by 0L∃∀N

Every ring of integral elements A^{+} is contained in A^{\circ} and contains all topologically nilpotent elements: A^{\circ\circ} \subseteq A^{+} \subseteq A^{\circ} (Definition 5.1, Definition 2.1). Following Wedhorn Remark 7.15.

Lean code for Theorem5.22 theorems
  • theoremdefined in «Adic spaces»/AffinoidRings.lean
    complete
    theorem ValuationSpectrum.IsRingOfIntegralElements.le_powerBoundedSubring.{u_1}
      {A : Type u_1} [CommRing A] [TopologicalSpace A] {B : Subring A}
      (hB : ValuationSpectrum.IsRingOfIntegralElements B) :
      B  TopologicalRing.powerBoundedSubring A
    theorem ValuationSpectrum.IsRingOfIntegralElements.le_powerBoundedSubring.{u_1}
      {A : Type u_1} [CommRing A]
      [TopologicalSpace A] {B : Subring A}
      (hB :
        ValuationSpectrum.IsRingOfIntegralElements
          B) :
      B 
        TopologicalRing.powerBoundedSubring A
    Any ring of integral elements is contained in `A°` (Remark 7.15(1)). 
  • theoremdefined in «Adic spaces»/AffinoidRings.lean
    complete
    theorem ValuationSpectrum.IsRingOfIntegralElements.topologicallyNilpotentElements_subset.{u_1}
      {A : Type u_1} [CommRing A] [TopologicalSpace A] {B : Subring A}
      (hB : ValuationSpectrum.IsRingOfIntegralElements B) :
      TopologicalRing.topologicallyNilpotentElements A  B
    theorem ValuationSpectrum.IsRingOfIntegralElements.topologicallyNilpotentElements_subset.{u_1}
      {A : Type u_1} [CommRing A]
      [TopologicalSpace A] {B : Subring A}
      (hB :
        ValuationSpectrum.IsRingOfIntegralElements
          B) :
      TopologicalRing.topologicallyNilpotentElements
          A 
        B
    A ring of integral elements contains all topologically nilpotent elements. 
Proof for Theorem 5.2
uses 0

By definition A^{+} \subseteq A^{\circ}. Conversely a topologically nilpotent element t satisfies t^n \to 0, so 1 - t is a unit and t is integral over A^{+} after clearing by an open neighbourhood; openness and integral-closedness of A^{+} then force t \in A^{+}. Hence A^{\circ\circ} \subseteq A^{+}. Following Wedhorn Remark 7.15.

Definition5.3
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Definition 2.1
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used by 1L∃∀N

A Huber ring A is uniform if A^{\circ} is bounded (\texttt{IsUniform}); an affinoid ring (A, A^{+}) is stably uniform if every rational localisation stays uniform (\texttt{IsStablyUniform}) (Definition 5.1, Definition 2.1). Stable uniformity is the standard hypothesis guaranteeing the structure presheaf is a sheaf. Following Wedhorn Definitions 7.36 and 7.37.

Lean code for Definition5.32 definitions
  • class(1 method)defined in «Adic spaces»/Uniform.lean
    complete
    class TopologicalRing.IsUniform.{u} (A : Type u) [CommRing A]
      [TopologicalSpace A] : Prop
    class TopologicalRing.IsUniform.{u} (A : Type u)
      [CommRing A] [TopologicalSpace A] : Prop
    A topological ring `A` is **uniform** if the set `A°` of power-bounded elements is
    bounded (Definition 7.36 of Wedhorn). 
    isBounded_powerBounded : TopologicalRing.IsBounded (TopologicalRing.powerBoundedSubring A)
    The set `A°` of power-bounded elements is bounded. 
  • class(1 method)defined in «Adic spaces»/Uniform.lean
    complete
    class TopologicalRing.IsStablyUniform.{u} (A : Type u) [CommRing A]
      [TopologicalSpace A] [IsTopologicalRing A]
      [ValuationSpectrum.PlusSubring A] [IsHuberRing A] : Prop
    class TopologicalRing.IsStablyUniform.{u}
      (A : Type u) [CommRing A]
      [TopologicalSpace A]
      [IsTopologicalRing A]
      [ValuationSpectrum.PlusSubring A]
      [IsHuberRing A] : Prop
    A Huber pair `(A, A⁺)` is **stably uniform** if for every rational localization
    `(A, A⁺) → (B, B⁺)`, the localization ring `B` (with the localization topology) is
    uniform (Definition 7.37 of Wedhorn). 
    presheafValue_isUniform :  (D : ValuationSpectrum.RationalLocData A),
      TopologicalRing.IsBounded (TopologicalRing.powerBoundedSubring (ValuationSpectrum.presheafValue D))
    The presheaf value (= completion of the localization) is uniform for every
    rational localization datum `D`.