Algebraic Zariski's Main Theorem #
The statement of Zariski's main theorem is the following:
Given a finite type R-algebra S, and p a prime of S such that S is quasi-finite at R,
then there exists a f ∉ p such that S[1/f] is isomorphic to R'[1/f] where R' is the integral
closure of R in S.
We follow https://stacks.math.columbia.edu/tag/00PI and proceed in the following steps
Algebra.ZariskisMainProperty.of_adjoin_eq_top: The case whereS = R[X]/I. The key isPolynomial.not_ker_le_map_C_of_surjective_of_quasiFiniteAtwhich shows that there exists someg ∈ Isuch that some coefficientgᵢ ∉ p. Then one basically takesf = gᵢandgbecomes monic inR[1/gᵢ][X]up to some minor technical issues, and thenS[1/gᵢ]is basically integral overR[1/gᵢ].Algebra.ZariskisMainProperty.of_algHom_polynomial: The case whereSis finite overR⟨x⟩for somex : S. The following key results are first esablished:
isStronglyTranscendental_mk_radical_conductor: Let𝔣be the conductor ofx(i.e. the largestS-ideal inR⟨x⟩).xas an element ofS/√𝔣is strongly transcendental overR.Algebra.not_quasiFiniteAt_of_stronglyTranscendental: IfSis reduced, thenx : Sis not strongly transcendental overR. One first reduces to whenR ⊆ Sare domains, and then to whenRis integrally closed. A going down theorem is now available, which could be applied toPolynomial.map_under_lt_comap_of_quasiFiniteAt:(p ∩ R)[X] < p ∩ R<x>to get a contradiction. The second result applied toS/√𝔣together with the first result implies thatpdoes not contain𝔣. The claim then follows fromLocalization.localRingHom_bijective_of_not_conductor_le.
Algebra.ZariskisMainProperty.of_algHom_mvPolynomial: The case whereSis finite overR⟨x₁,...,xₙ⟩. This is proved using induction onn.
Main definiton and results #
Algebra.ZariskisMainProperty: We say that anRalgebraSsatisfies the Zariski's main property at a primepofSif there existsr ∉ pin the integral closureS'ofRinS, such thatS'[1/r] = S[1/r].Algebra.ZariskisMainProperty.of_finiteType: IfSis finite type overRand quasi-finite atp, thenZariskisMainPropertyholds.Algebra.QuasiFiniteAt.exists_fg_and_exists_notMem_and_awayMap_bijective: IfSis finite type overRand quasi-finite atp, then there exists a subalgebraS'ofRthat is finitely generated as anR-module, and somer ∈ S'such thatr ∉ pandS'[1/r] = S[1/r].
We say that an R algebra S satisfies the Zariski's main property at a prime p of S
if there exists r ∉ p in the integral closure S' of R in S, such that S'[1/r] = S[1/r].
Equations
- Algebra.ZariskisMainProperty R p = ∃ (r : ↥(integralClosure R S)), ↑r ∉ p ∧ Function.Bijective ⇑(Localization.awayMap (integralClosure R S).val.toRingHom r)
Instances For
Given a map φ : R[X] →ₐ[R] S. Suppose t = φ r / φ p is integral over R[X] where
p is monic with deg p > deg r, then t is also integral over R.
The algebraic version of Zariski's Main Theorem:
Given a finite type R-algebra S that is quasi-finite at a prime p,
there exists a f ∉ p such that S[1/f] is isomorphic to R'[1/f] where R' is the integral
closure of R in S.