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(AS) 2 Lecture 2. Sheaves: a Potpourri of Algebra, Analysis and Topology (UW) 11 Lecture 3

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(AS) 2 Lecture 2. Sheaves: a Potpourri of Algebra, Analysis and Topology (UW) 11 Lecture 3 Contents Lecture 1. Basic notions (AS) 2 Lecture 2. Sheaves: a potpourri of algebra, analysis and topology (UW) 11 Lecture 3. Resolutions and derived functors (GL) 25 Lecture 4. Direct Limits (UW) 32 Lecture 5. Dimension theory, Gr¨obner bases (AL) 49 Lecture6. Complexesfromasequenceofringelements(GL) 57 Lecture7. Localcohomology-thebasics(SI) 63 Lecture 8. Hilbert Syzygy Theorem and Auslander-Buchsbaum Theorem (GL) 71 Lecture 9. Depth and cohomological dimension (SI) 77 Lecture 10. Cohen-Macaulay rings (AS) 84 Lecture 11. Gorenstein Rings (CM) 93 Lecture12. Connectionswithsheafcohomology(GL) 103 Lecture 13. Graded modules and sheaves on the projective space(AL) 114 Lecture 14. The Hartshorne-Lichtenbaum Vanishing Theorem (CM) 119 Lecture15. Connectednessofalgebraicvarieties(SI) 122 Lecture 16. Polyhedral applications (EM) 125 Lecture 17. Computational D-module theory (AL) 134 Lecture 18. Local duality revisited, and global duality (SI) 139 Lecture 19. De Rhamcohomologyand localcohomology(UW) 148 Lecture20. Localcohomologyoversemigrouprings(EM) 164 Lecture 21. The Frobenius endomorphism (CM) 175 Lecture 22. Some curious examples (AS) 183 Lecture 23. Computing localizations and local cohomology using D-modules (AL) 191 Lecture 24. Holonomic ranks in families of hypergeometric systems 197 Appendix A. Injective Modules and Matlis Duality 205 Index 215 References 221 1 2 Lecture 1. Basic notions (AS) Definition 1.1. Let R = K[x1,...,xn] be a polynomial ring in n variables over a field K, and consider polynomials f ,...,f R. Their zero set 1 m ∈ V = (α ,...,α ) Kn f (α ,...,α )=0,...,f (α ,...,α )=0 { 1 n ∈ | 1 1 n m 1 n } is an algebraic set in Kn. These are our basic objects of study, and include many familiar examples such as those listed below. Example 1.2. If f ,...,f K[x ,...,x ] are homogeneous linear polynomials, 1 m ∈ 1 n their zero set is a vector subspace of Kn. If V, W are vector subspaces of Kn, then we have the following inequality of vector space dimensions: rankK(V W ) > rankK V + rankK W n. ∩ − An easy way to see this inequality is via the exact sequence β 0 V W α V W V + W 0, −−−−→ ∩ −−−−→ ⊕ −−−−→ −−−−→ where α(u) = (u,u) and β(v, w)= v w. Then − rankK(V W ) = rankK V W rankK(V + W ) ∩ ⊕ − > rankK V + rankK W n. − Example 1.3. A hypersurface is a zero set of one equation. The circle of unit radius in R2 is a hypersurface—it is the zero set of the polynomial x2 + y2 1. − Example 1.4. If f K[x ,...,x ] is a homogeneous polynomial of degree d, then ∈ 1 n f(α1,...,αn) = 0 implies that d f(cα1,...,cαn)= c f(α1,...,αn)=0 for all c K. Hence if an algebraic set V Kn is the zero set of homogeneous ∈ ⊂ polynomials, then, for all (α1,...,αn) V and c K, we have (cα1,...,cαn) V . In this case, the algebraic set V is said∈ to be a cone∈ . ∈ Example 1.5. For integers m,n > 2, consider the set V Kmn of all m n ⊂ × matrices over K which have rank less than a fixed integer t. A matrix has rank less than t if and only if its size t minors (i.e., the determinants of t t submatrices) all equal zero. Take × R = K[x 1 6 i 6 m, 1 6 j 6 n], ij | which is a polynomial ring in mn variables arranged as an m n matrix. Then V m n × is the solution set of the t t polynomials which arise as the size t minors of the matrix (x ). Hence V is an algebraic set (in fact, a cone) in Kmn. ij Exercise 1.6 ([94]). Let K be a finite field. (1) For every point p Kn, construct a polynomial f K[x ,...,x ] such that ∈ ∈ 1 n f(p) = 1 and f(q) = 0 for all points q Kn p . ∈ \{ } (2) Given a function g : Kn K, show that there is a polynomial f −→ ∈ K[x ,...,x ] with f(p)= g(p) for all p Kn. 1 n ∈ (3) Prove that any subset of Kn is the zero set of a single polynomial. Remark 1.7. One may ask: is the zero set of an infinite family of polynomials also the zero set of a finite family? To answer this, recall that a ring is Noetherian if all its ideals are finitely generated, and that the polynomial ring R = K[x1,...,xn] is Noetherian by the Hilbert basis theorem, [4, Theorem 7.5]. Let a R be the ideal ⊂ 3 generated by a possibly infinite family of polynomials gλ . The zero set of gλ is the same as zero set of all polynomials in the ideal a.{ But} a is finitely generated,{ } say a = (f ,...,f ), so the zero set of g is precisely the zero set of the finitely 1 m { λ} many polynomials f1,...,fm. Given a set of polynomials f1,...,fm generating an ideal a R, we denote n ⊂ k their zero set in K by Var(f1,...,fm) or by Var(a). Note that Var(f) = Var(f ) for any integer k > 1, hence if a and b are ideals with the same radical, then Var(a) = Var(b). A theorem of Hilbert states that over an algebraically closed field, the converse is true as well: Theorem 1.8 (Hilbert’s Nullstellensatz). Let R = K[x1,...,xn] be a polynomial ring over an algebraically closed field K. If Var(a) = Var(b) for ideals a, b R, then rad a = rad b. Consequently the map a Var(a) is a containment reversing⊂ 7→ n bijection between the set of radical ideals of K[x1,...,xn] and algebraic sets in K . For a proof, solve [4, Problem 7.14]. In particular, the theorem above tells us when polynomial equations have a common solution, and the following corollary is also (and perhaps more appropriately) referred to as the Nullstellensatz: Corollary 1.9. Let R = K[x1,...,xn] be a polynomial ring over an algebraically closed field K. Then polynomials f1,...,fm R have a common zero if and only if (f ,...,f ) = R. ∈ 1 m 6 Proof. Var(R) = ∅, so Var(a) = ∅ for an ideal a if and only if rad a = R, which occurs if and only a = R. Corollary 1.10. Let R = K[x1,...,xn] be a polynomial ring over an algebraically closed field K. Then the maximal ideals of R are precisely the ideals (x α ,...,x α ) where α K. 1 − 1 n − n i ∈ Proof. Let m be a maximal ideal of R. Then m = R so there exists a point (α ,...,α ) Var(m). But then 6 1 n ∈ Var(x α ,...,x α ) Var(m), 1 − 1 n − n ⊆ so m (x1 α1,...,xn αn). Since each is a maximal ideal of R, they must be equal.⊆ − − Exercise 1.11 ([94]). Let K be a field which is not algebraically closed. Prove that any algebraic set in Kn is the zero set of a single polynomial f K[x ,...,x ]. ∈ 1 n Krull dimension of a ring We would like a notion of dimension for algebraic sets which agrees with vec- tor space dimension if the algebraic set is a vector space, and gives a suitable generalization of the inequality in Example 1.2. The situation is certainly more complicated than with vector spaces; for example, not all points of an algebraic set are “similar”—the algebraic set defined by xy = 0 and xz = 0 is the union of a line and a plane. To obtain a good theory of dimension, we recall some notions from commutative algebra. 4 Definition 1.12. Let R be a ring. The spectrum of R, denoted Spec R, is the set of prime ideals of R with the Zariski topology, which is the topology where the closed sets are V (a)= p Spec R a p for ideals a R. { ∈ | ⊆ } ⊆ It is easily verified that this is indeed a topology: the empty set is both open and closed, an intersection of closed sets is closed since λ V (aλ)= V ( λ aλ), and the union of two closed sets is closed since T S V (a) V (b)= V (a b)= V (ab). ∪ ∩ The height of a prime ideal p, denoted height p, is the supremum of integers t such that there exists a chain of prime ideals p = p ) p ) p ) p , where p Spec R. 0 1 2 · · · t i ∈ The height of an arbitrary ideal a R is ⊂ height a = inf height p p Spec R, a p . { | ∈ ⊆ } The Krull dimension of R is dim R = sup height p p Spec R . { | ∈ } Note that for every prime ideal p of R, we have dim Rp = height p. Example 1.13. The prime ideals of Z are (0) and (p) for prime integers p. Conse- quently the longest chains of prime ideals in Spec Z are those of the form (0) ( (p), and so dim Z = 1. More generally, if R is a principal ideal domain which is not a field, then dim R = 1. Theorem 1.14 (Krull). Let R be a Noetherian ring. If an ideal a ( R is generated by n elements, then each minimal prime p of a has height p 6 n. In particular, every ideal a ( R has finite height. The above theorem implies that every proper principal ideal of a Noetherian ring has height at most one, which is Krull’s principal ideal theorem. For a proof of Theorem 1.14, see [4, Corollary 11.16]. While it is true that every prime ideal in a Noetherian ring has finite height, the Krull dimension is the supremum of the heights of prime ideals, and this supremum may be infinite, see [4, Problem 11.4] for an example due to Nagata.
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