By Antoine Chambert-Loir
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Extra resources for Algebraic Geometry of Schemes [Lecture notes]
Consequently, there exists a′ ∈ A p such that aa′ ∈ q. Observe that a′ ∈/ q, for otherwise, one would have a′ ∈ q ∩ A = p, which does not hold. Since q is a prime ideal, a ∈ q. 5). — Let B be a ring, let A be a subring of B. If B is integral over A, then dim(A) = dim(B). Proof. — Let q0 ⊊ ⋅ ⋅ ⋅ ⊊ qn be a chain of prime ideals of B. Let us intersect these ideals with A; this gives an increasing family (q0 ∩ A) ⊂ ⋅ ⋅ ⋅ ⊂ (qn ∩ A) of prime ideals of A. 4, this is even a chain of prime ideals, so that dim(A) ⩾ dim(B).
It is faithful, because a group morphism is determined by the map between the underlying sets. It is however not full because there are maps between two (non-trivial) groups which are not morphism of groups. 4). — The construction of the spectrum of a ring defines a contravariant functor from the category Ring of rings to the category Top of topological spaces. In the other direction, set O(X) to be the ring of continuous complex-valued functions on a topological space X. If f ∶ X → Y is a continuous map of topological spaces, let f ∗ ∶ O(Y) → O(X) be the morphism of rings given by f ∗ (u) = u ○ f .
Moreover, the family (Jm )m⩾0 is increasing. Since A is noetherian, there exists an integer n such that Jm = Jn for every integer m such that m ⩾ n. Let Q be the finite set Q = Q0 ∪ ⋅ ⋅ ⋅ ∪ Qn and let I′ be the ideal of A[X] it generates. One has I′ ⊂ I; it suffices to prove that I = I′ . Let thus P ∈ I and let us prove by induction on deg(P) that P ∈ I′ . Let m = deg(P), and let a ∈ Jm be the leading coefficient of P. Let p = min(m, n); one has a ∈ J p . By definition of Q p , there exists a polynomial P′ of degree p which is a linear combination of polynomials in Q p (hence an element of I′ ) whose leading coefficient is equal to a.