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Lecture 20 October, 23 1 Point Set Topology Let X be a topological space, then Z is locally closed if it is an open set in the induced topology on Z. In other words, Z = U ∩ C where U is open in X and C ≡ Z is closed in X. Assume that X has dcc on closed sets (Noetherian on ideals) then a subset of X, Z is constructible if it is a finite union of locally closed sets. Proposition 1. • Z contains an open subset of Z. • Z is a finite union of disjoint locally closed sets. Proposition 2. The set of constructible sets of X is closed under operations finite union ∪, finite intersection ∩, and complement X − Z. Proof of Proposition 1. Let Z be locally closed, the Z = S C1 ∪ · · · ∪ Cn is a union of a finite number of irreducible sets. It follows that Z = (Z ∩ Ci ), where Z ∩ Ci is an open subset in Ci . So Z is a union of disjoint locally closed sets. Assume that Z = Z1 ∪ · · · ∪ Zn is the union of locally closed sets, then Zi = Ci are irreducible. If C1 ⊂ C2 ∪ · · · ∪ Cr , then D = C2 ∪ · · · ∪ Cr is a closed set and U = Z − (Z ∩ D) is an open set in Z, as desired. So we may assume that C1 6⊂ C2 ∪ · · · ∪ Cr . Let Y1 = C1 −(D∩C1 ) be an open set of C1 . Then A1 = C1 −Y1 and B1 = C1 −Z1 are closed sets in C1 . But since C1 is closed in Z, it follows that A1 and B1 are closed in Z. Then A1 ∪ B1 is a proper closed set in C1 , otherwise it would not be irreducible. Hence Z − (A1 ∪ B1 ) is an open non-empty set of Z and we are done. Definition. A quasi-projective variety X is a locally closed subset of Pn . Examples of them are: projective variety, affine variety. If we consider A2 − {0}, then it is a quasiprojective variety that is neither affine nor projective. 1 2 Main Theorem Theorem. Let φ : A → B be a homomorphism of finitely generated algebras and let f : Y → X be a morphism of their varieties. Then the image of f (Y ) is a constructible set of X. Proof. We will use Noetherian (acc) induction on B or dcc on varieties of Y , this will be the same as induction on dimension. We may assume that Y is an irreducible affine variety. It follows from the fact that a finite union of constructible sets is a constructible set and every variety is a union of finite number of irreducible varieties. Thus we may assume that B is a domain. The base of induction is clear, the image of an irreducible set of dimension zero is the image of a point, which must be a point in X. Let A = A/ker(φ), then the corresponding variety C is a closed set of X, and Y via f factors through C. If f (Y ) is constructible in C then it is constructible in X. So we can replace A by A, that is start with φ : A → B being injective. Suppose s ∈ B, s 6= 0. Then the variety of Bs = B[s−1 ] = Y − V (s), since we localized B. Note that the variety of B = B/sB is V (s). Thus Y can be written as the union of varieties of Bs and B/sB. We would like to show that f (Y ) is constructible. But the image of V (s) is constructible by Noetherian induction. Thus we are allowed to localize B at s ∈ B in order to simplify the problem. If s ∈ A, s 6= 0, consider a map φs : As → Bs , then the corresponding variety Xs is an open set because Xs = X − (set of zeros ofs in X) and the set of zeros of s in X is a closed set. So if the image is constructible in Xs , then it is constructible in X. This follows from the observation that if a set is open in Xs , then it is open in X. If a set is constructible in Xs , say Z is an intersection of an open set and a closed set Us ∩ Cs . Then Us and Xs − Cs are open in X and Z is constructible in X, since Z = Us ∩ (X − (Xs − Cs )) that is an intersection of an open and a closed set. Thus we are allowed to localize A in order to simplify the problem. So far these were preliminary observations. Our plan is to choose s ∈ A appropriately and localize such that Ys → Xs will be surjective. Let K be a fraction field of A. We extend the map φ : A → B between domains to a map from K → BK , where BK is a fraction ring consisting of {u−1 b | b ∈ B, u ∈ A, u 6= 0}. We simplified notation s−1 φ(b) and for the rest of the proof we think of the image of A as an inclusion of A in B. 2 Note that B is a finite type algebra, so BK will be also a finite type algebra. Using Noether Normalization Theorem: there exist independent y1 , . . . , yd ∈ BK such that K[y1 , . . . , yd ] ⊂ BK and B K is a finite module over K[Y ]. So any element P in BK can written as a finite sum Pi (Y )vi , where Pi (Y ) ∈ K[Y ] and {vi ∈ BK } is the generating set. Since BK is a finite module over K[Y ], it follows that BK is integral over K[Y ]. Let zi be one of generators of B, then there exists a monic polynomial with coefficients in K[Y ], X n + P1 (Y )X n−1 + · · · + Pn (Y ) = 0 such that zi is the solution of it. There is a finite number of polynomials Pi (Y ) and all they have finite number of coefficients in K. We localize at all u ∈ A that appear in all denominators of all coefficients. This yields that zi will be integral over A[Y ] after localizing A. Applying same procedure for all of generators of B, we get that B after localization is integral over A[Y ] (A is also considered after localization). Hence B is a finitely generated module over A[Y ], so f is a surjective map. Since f is surjective, Y entirely maps to X and we conclude the proof. 3