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Algebraic Geometry Notes Algebraic Geometry Notes Sean Sather-Wagstaff Department of Mathematics, 412 Minard Hall, North Dakota State University, Fargo, North Dakota 58105-5075, USA E-mail address: Sean.Sather-Wagstaff@ndsu.edu URL: www.ndsu.edu/pubweb/~ssatherw 1 March 26, 2012 Contents Contents i Preface 1 1 Affine Space 3 1.1 Algebraic Subsets . 3 1.2 Zariski Topology . 6 1.3 Geometric Ideals . 8 1.4 Hilbert’s Nullstellensatz . 10 1.5 Irreducible Closed Subsets . 11 1.6 Finding Irreducible Components . 16 2 Projective Space 23 2.1 Motivation . 23 n 2.2 Projective Space Pk ................................... 24 2.3 Homogeneous Polynomials . 25 n 2.4 The Zariski Topology on Pk ............................... 28 n 2.5 Geometric Ideals in Pk .................................. 32 2.6 Projective Nullstellensatz . 33 2.7 Irreducible Closed Subsets . 35 2.8 Regular Functions . 37 2.9 Finding Irreducible Components . 38 3 Sheaves 41 3.1 Presheaves . 41 n 3.2 Regular Functions on Ak ................................. 42 Index 45 i Introduction What is Algebraic Geometry? The study of geometric objects determined by algebraic “data”, i.e. polynomials. Some examples are lines in R2, conics in R2, planes in R3, spheres, ellipsoids, etc. in R3. Geometric objects of interest include: solution sets to systems of polynomial equations, study them using algebraic techniques. For example: is the solution set finite or infinite? Example 0.0.1 Let f; g 2 C[x; y; z] and let V = f(x; y; z) 2 C3 j f(x; y; z) = 0 = g(x; y; z)g. Assume that V 6= ; (Hilbert’s Nullstellensatz says that this is equivalent to (f; g) 6= C[x; y; z]). Then V is infinite. This is an application of two algebraic results: Hilbert’s Nullstellensatz and Noether’s Normalization Lemma. 1 Chapter 1 Affine Space 1.1 Algebraic Subsets Notation: Throughout these notes, k will represent a field. n n Definition 1.1.1 Given an integer n > 1, the set Ak = k = f(a1; : : : ; an) j a1; : : : ; an 2 kg is a affine n-space over k. Example 1.1.2 n = n (as a set), and 1 = k (as a set). AR R Ak n Fact 1.1.3 Given f 2 k[x1; : : : ; xn] and a = (a1; : : : ; an) 2 Ak , the element f(a) 2 k is well- n defined, i.e. f : Ak ! k is a well-defined function. These are called “regular functions.” k[x1; : : : ; xn] n is the ring of regular functions on Ak . Note: Different polynomials can describe the same function. Example 1.1.4 Let k = Z=pZ for a prime p, f(x) = x, and g(x) = xp. Fermat’s Little Theorem implies that xp = x for all x 2 k. n Definition 1.1.5 For each S ⊆ k[x1; : : : ; xn] set V (S) = fa 2 Ak j f(a) = 0 8 f 2 Sg. V is for “variety” or “vanishing.” V (S) is the solution set to the system of polynomial equations ff = 0 j f 2 Sg and is called the vanishing locus for S. Notation: If S = ff1; : : : ; fmg, we write V (f1; : : : ; fm) instead of V (ff1; : : : ; fmg). n Example 1.1.6 V (0) = Ak = V (;) and V (1) = ; = V (k[x1; : : : ; xn]). In 2 : V (ax + by + c) = line, V (x2 + y2 − 1) = circle, and similarly for other conics. AR In 3 : V (ax + by + cz + d) = plane, V (ax + by + cz + d; αx + βy + γz + δ) is a line as AR long as the two planes are distinct and non-parallel, V (a2x2 + b2y2 + c2z2 − d2) = ellipsoid where a; b; c; d 6= 0. 3 1. Affine Space 1 1 Example 1.1.7 In Ak: either V (S) = ;, V (S) = Ak or V (S) is finite, and for every finite set V there exists S such that V = V (S). n Definition 1.1.8 A subset V ⊆ Ak is an algebraic subset if there exists S ⊆ k[x1; : : : ; xn] such that V = V (S). 0 0 Lemma 1.1.9 Let S ⊆ S ⊆ k[x1; : : : ; xn]. Then V (S) ⊇ V (S ). Proof: Exercise Proposition 1.1.10 Let S ⊆ k[x1; : : : ; xn] and I = (S) ⊆ k[x1; : : : ; xn]. Then V (S) = V (I). Proof: Since I = (S) ⊇ S, the previous lemma implies V (I) ⊆ V (S). For the other containment, let a 2 V (S). Then for all f 2 S, f(a) = 0. Therefore for all Pm f1; : : : ; fm 2 S, for all g1; : : : ; gm 2 k[x1; : : : ; xn], and an arbitrary element h = i=1 gifi 2 I, we have m X h(a) = gi(a)fi(a) = 0: i=1 Thus a 2 V (I) and V (S) = V (I). Notation 1.1.11 We will denote the ring R := k[x1; x2; : : : ; xm]. Proposition 1.1.12 (a) For each Si ⊂ R and Ii = (Si)R V (S1) [ V (S2) [···[ V (Sm) = V (I1) [ V (I2) [···[ V (Im) = V (I1 \ I2 \···\ Im) = V (I1I2 ··· Im): (b) For all λ 2 Λ let Sλ ⊆ R and Iλ = (Sλ)R: T T λ2Λ V (Sλ) = λ2Λ V (Iλ) P = V λ2Λ Iλ S = V λ2Λ Sλ where Λ is an index set that is not necessarily finite. n (c) The set of algebraic subsets of Ak is closed under finite unions and arbitrary intersections. Proof: (a) We first note that V (S1) [ V (S2) [···[ V (Sm) = V (I1) [ V (I2) [···[ V (Im) since V (St) = V (It) by the previous proposition. Next we notice that Ij ⊇ I1 \ I2 \···\ Im for all j. Hence by Lemma 1.1.9 we have V (Ij) ⊆ V (I1 \ I2 \···\ Im) S for all j. Therefore j V (Ij) ⊆ V (I1 \ I2 \···\ Im). 4 1.1. Algebraic Subsets Now since I1I2 ··· Im ⊆ I1 \ I2 \···\ Im, by Lemma 1.1.9 we know V (I1I2 ··· Im) ⊇ V (I1 \ I2 \···\ Im): n S Finally let a 2 Ak n j V (Ij). Then we have a 62 V (Ij) for all j. Therefore for all j there exists fj 2 Ij such that fj(a) 6= 0. So let f = f1f2 ··· fm 2 I1I2 ··· Im. Then f(a) = f1(a)f2(a) ··· fm(a) 6= 0: Therefore a 62 V (I1 · I2 ··· Im). Hence what we have shown is V (I1) [ V (I2) [···[ V (Im) ⊆ V (I1 \ I2 \···\ Im) ⊆ V (I1I2 ··· Im) ⊆ V (I1) [ V (I2) [···[ V (Im) giving us equality at each stage. T T (b) Again by Lemma 1.1.9, λ2Λ V (Sλ) = λ2Λ V (Iλ) since V (Sλ) = V (Iλ) for all λ 2 Λ. T P Next we will show λ2Λ V (Iλ) = V ( λ2Λ Iλ). We will first show (⊇). Here we note X Iµ ⊆ Iλ for all µ 2 Λ: λ2Λ P T P Hence V (Iµ) ⊇ V ( λ2Λ Iλ) for all µ (again by Lemma 1.1.9). Therefore µ V (Iµ) ⊇ V ( λ2Λ Iλ): T (⊆) Let a 2 V (Iλ). Then a 2 V (Iλ) for all λ. Therefore for all fλ 2 Iλ, fλ(a) = 0. So P λ2Λ let f 2 λ2Λ Iλ. Then finite X f = fλ λ2Λ where fλ 2 Iλ for all λ. Therefore X X f(a) = fλ = 0 = 0: λ2Λ λ2Λ P Hence a 2 V ( λ2Λ Iλ) as desired. Now for the last equality we have S S P V λ2Λ Sλ = V λ2Λ Sλ R = V λ2Λ Iλ S P since λ2Λ Sλ R = λ2Λ Iλ. n Definition 1.1.13 A hypersurface in Ak is a subset of the form V (f) for a single f. Corollary 1.1.14 Let V be an algebraic subset. Then V is a finite intersection of hypersurfaces. Proof: Since V is algebraic and, V = V (S) = V (I) where I = (S)R. The Hilbert Basis Theorem allows us to write I = (f1; f2; : : : ; fn)R. Now apply V (−) to get: V (I) = V (f1; f2; : : : ; fm) = V (f1R + f2R + ··· + fmR) = V (f1R) \ V (f2R) \···\ V (fmR) = V (f1) \ V (f2) \···\ V (fm): 5 1. Affine Space Hence V is a finite intersection of hypersufaces. n Notation: Given a 2 Ak , let ma := (x1 − a1; x2 − a2; : : : ; xj − aj; : : : ; xn − an)R. Fact 1.1.15 ma ⊆ R is maximal because φa : R ! k : f 7! f(a) is a ring epimorphism such that ma = Ker(φa). Proposition 1.1.16 fag = V (ma) = V (x1 − a2; x2 − a2; : : : ; xn − an). That is every singleton is algebraic. Proof: First notice that by the Proposition 1.1.10 we have the second equality. So we will just show fag = V (ma). (⊆) Since a satisfies xi − ai for all i, we have a 2 V (x1 − a1; x2 − a2; : : : ; xn − an) and fag ⊆ V (x1 − a1; x2 − a2; : : : ; xn − an). (⊇) If b 2 V (x1 − a1; x2 − a2; : : : ; xn − an), then b satisfies xi − ai for all i. Therefore bi − ai = 0 ) bi = ai for all i ) b = a: Hence V (x1 − a1; x2 − a2; : : : ; xn − an) ⊆ fag. n Corollary 1.1.17 Every finite subset of Ak is algebraic. Proof: falgebraic subsetsg contains all singletons and is closed under finite unions. 1.2 Zariski Topology n Definition 1.2.1 The Zariski Toplogy on Ak . n • A subset V ⊆ Ak is closed if it is algebraic. n n • A subset U ⊆ Ak is open if Ak n U is closed, i.e. algebraic. A set is open if and only if its complement is an algebraic set. n n Notation: For all f 2 R, set Uf := fa 2 Ak jf(a) 6= 0g = Ak n V (f). Therefore Uf is open in the Zariski Topology.
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