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

By Antoine Chambert-Loir

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The height (resp. the coheight) of x is the supremum of the length of chains ending (resp. starting) at x. They are denoted ht(x) and coht(x) respectively. 2). — Let X be a topological space. The Krull dimension of X, denoted dim(X), is the dimension of the set C(X) of all irreducible closed subsets of X, ordered by inclusion. Let Z be a closed irreducible subset of X. The codimension of Z in X, denoted codim(Z), is the coheight of Z in the partially ordered set C. 11. DIMENSION 33 d) If Y and Z are irreducible closed subsets of X such that Y ⊂ Z, then dim(Y) ⩽ dim(Z) and codim(Z) ⩽ codim(Y).

The element a is not a unit (for, otherwise, b = 0, a contradiction); consequently, V(a) ≠ ∅. Similarly, V(b) is neither empty, nor equal to Spec(A). This implies that Spec(A) is not irreducible. 3). — Let X be an irreducible topological space and let U be a non-empty open subset of X. a) The open subset U is dense in X, and is irreducible; b) The map Z ↦ Z ∩ U defines a bijection between the set of irreducible closed subsets of X which meet U and the set of irreducible closed subsets of U. Its inverse bijection is given by Z ↦ Z.

Then, Y ⊂ Z ∩ U ⊂ Z′ ∩ U = Y, hence Y = Z ∩ U. 10. IRREDUCIBLE COMPONENTS 29 Since Y is irreducible, it is non-empty, hence the set Z is not empty. Let Z1 and Z2 be closed subsets of X such that Z ⊂ Z1 ∪ Z2 . Then Y = U ∩ Z ⊂ (U ∩ Z1 ) ∪ (U ∩ Z2 ). Since Y is irreducible, one has Y ⊂ U ∩ Z1 or Y ⊂ U ∩ Z2 . In the first case, Z1 is a closed subset of X containing Y, hence Z ⊂ Z1 ; in the other case, Z ⊂ Z2 . This shows that Z is irreducible. We may now conclude the proof of the proposition. By what precedes, setting α(Y) = Y defines a map from the set of irreducible closed subsets of U to the set of irreducible closed subsets of X.