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By Antoine Chambert-Loir

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For a functor F∶ C → D to be an equivalence of categories, it is necessary and sufficient that it be fully faithful and essentially surjective. Proof. — Let G∶ D → C be a functor such that F and G are quasi-inverse. For every object P of D, F ○ G(P) is isomorphic to P, hence F is essentially surjective. Moreover, for every objects M, N of C , the functor G ○ F, being isomorphic to idC , induces a bijection from C (M, N) to itself. This bijection is the composition of the map ΦF ∶ C (M, N) → D(M, N) induced by F and of the map ΦG ∶ D(M, N) → C (M, N) induced by G.

One thus has σ(b) ∈ B. This shows that σ(B) ⊂ B. Similarly, one has σ −1 (B) ⊂ B, hence B ⊂ σ(B). 13. 4), the map → Spec(A) is surjective, so that the fiber (aφ)−1 (x) is non-empty. Let y, y′ be two elements of this fiber; let q, q′ be the corresponding prime ideals of B. Let b ∈ q′ . The product a = ∏σ∈G σ(b) is an element of F which is fixed by G. By Galois theory, it is radicial over K: there exists an integer q ⩾ 1 such that a q ∈ K. ) Since b is integral over A, each σ(b) is integral over A, and a is integral over A, as well as a q .

If n = 0, then A is a field, hence dim(A) = 0. Let us assume that n ⩾ 1. The chain p0 ⊂ p1 of prime ideals is maximal among those ending at p1 . Since every maximal chain of prime ideals ending at p1 begins at (0) = p0 , one has ht(p1 ) = 1. Moreover, the quotient ring A/p1 is an integral domain and a finitely generated K-algebra. In this ring, the increasing sequence p1 /p1 ⊂ ⋅ ⋅ ⋅ ⊂ pn /p1 is a maximal chain of prime ideals. By induction, one has dim(A/p1 ) = n − 1. Consequently, n = 1 + dim(A/p1 ) = 1 + dim(A) − ht(p1 ) = dim(A), as was to be shown.

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Algebraic Geometry of Schemes [Lecture notes] by Antoine Chambert-Loir


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