Introduction to the Basics of Algebraic Geometry, Computational Tools and Geometric Applications Janko BÄohm University of Saarland 17.05.2004 Abstract This is the manuscript for a talk given in a seminar on computer aided geometric design at the University of Saarland. The aim of the talk was to introduce the basic concepts of algebraic geometry, the computational tools, i.e. resultants and Groebner bases, and their geometric applications. Contents 1 Overview 2 2 Basics of algebra and geometry 2 2.1 A±ne varieties . 2 2.2 Varieties de¯ned by ideals . 6 2.3 Hilbert Basis Theorem . 7 2.4 The ideal of a variety . 8 2.5 The Nullstellensatz . 11 2.6 Projection and elimination . 12 2.7 Some remarks on projective geometry . 13 2.7.1 Projective space and projective varieties . 13 2.7.2 The Theorem of Bezout . 14 3 Computational tools 15 3.1 Resultants . 15 3.2 Groebner bases . 18 1 4 Geometric Applications 22 4.1 Implicitation and birational geometry computations . 22 1 Overview The main topics: ² Basics of Algebra and Geometry A±ne varieties Varieties de¯ned by ideals Hilbert Basis Theorem The ideal of a variety Nullstellensatz Projection and elimination Some remarks on projective geometry ² Computational Tools Resultants Groebner bases ² Geometric Applications Implicitation and birational geometry computations Intersection computations The genus of a curve Parametrization of curves and surfaces 2 Basics of algebra and geometry 2.1 A±ne varieties Let k be a ¯eld. De¯nition 1 An a±ne variety is the common zero locus of polynomials f1; :::; fr 2 k [x1; :::; xn] V (f1; :::; fr) = ff1 = 0; :::; fr = 0g 2 Example 2 ² V (1) = ; ² V (0) = kn ² Linear Algebra: For linear fi this is the solution space of an inhomo- geneous system of linear equations. Here we know, how to decide if V (f1; :::; fr) is empty and describe V (f1; :::; fr) by a parametrization applying Gauss algorithm. x3¡1 ² The graph of a function: For example the graph of y (x) = x : is V (xy ¡ x3 + 1). ² Plane curve: Consider V (f) for one equation in the plane, e.g. f = y2 ¡ x3 ¡ x2 + 2x ¡ 1: ² Surfaces in k3: Consider V (g) for one equation in 3-space ( an algebraic 3 set with 1 equation is called a hypersurface), e.g. g = f ¡ z + z2. ² Twisted cubic: Consider the curve C = V (y ¡ x2; z ¡ x3) ½ k3 given by 2 equations in 3-space. ² There are also strange examples, e.g. V (xz; yz) = V (z) [ V (x; y) decomposes into the x ¡ y-plane V (z) and the z-axis V (x; y) 4 Varieties, which do not admit a further decomposition are called irre- ducible, otherwise reducible. Example 3 Consider the following sets given by parametrizations: ² Bezier spline: The curve C ½ k2 parametrized by 3 2 2 3 X (t) = x0(1 ¡ t) + 3x1t(1 ¡ t) + 3x2t (1 ¡ t) + x3t 3 2 2 3 Y (t) = y0(1 ¡ t) + 3y1t(1 ¡ t) + 3y2t (1 ¡ t) + y3t 2 with t 2 k goes through the points (x0; y0) ; (x3; y3) 2 k and the tangent lines at these points go through (x1; y1) resp. (x2; y2) See for example the standard vector graphics programs. ² Whitney umbrella: The surface S ½ k3 given by the parametrization X (s; t) = s ¢ t Y (s; t) = s Z (s; t) = t2 with (s; t) 2 k2. 5 C and S are also varieties. To see this, we have to describe them by implicit equations, i.e. as V (f1; :::; fr). Techniques to do this will be one of the topics. The implicit description would allow us for example to check easily, if a given point lies on the variety. 2.2 Varieties de¯ned by ideals First recall, that an ideal I ½ R inside a ring R is an additive subgroup of R s.t. R-multiples of elements in I again lie inside of I. An important observation is, that the common zeroset of polynomials f1; :::; fr only depends on the ideal I = hf1; :::; fri ½ k [x1; :::; xn] generated by f1; :::; fr. The reason is, that if f1 (p) = 0; :::; fr (p) = 0 any polynomial linear combination also vanishes in p: =0 Xr z }| { si (p) fi (p) = 0 i=1 for all si 2 k [x1; :::; xn]. Hence any di®erent set of generators of I gives the same zeroset and we should de¯ne: De¯nition 4 For an ideal I ½ k [x1; :::; xn] we de¯ne V (I) = fx 2 kn j f (x) = 08f 2 Ig Example 5 Consider I = hf1; f2i with 2 2 f1 = 2x ¡ 3y + 10 2 2 f2 = 3x ¡ y + 1 Gaussian elimination applied to the linear system of equations 2X ¡ 3Y + 10 = 0 3X ¡ Y + 1 = 0 shows that ­ ® I = x2 ¡ 1; y2 ¡ 4 6 hence V (I) = f(1; 2) ; (¡1; 2) ; (1; ¡2) ; (¡1; ¡2)g consists of 4 points. 3 2 1 y0 ±1 ±2 ±3 ±3 ±2 ±1 0 1 2 3 x The de¯nition of V (I) naturally raises the question, whether any ideal in k [x1; :::; xn] has a ¯nite set of generators, hence any V (I) = V (f1; :::; fr) with some f1; :::; fr 2 k [x1; :::; xn]. This is answered by the Hilbert Basis Theorem: 2.3 Hilbert Basis Theorem In k [x] (principal ideal domain) every ideal has a single generator, which one can ¯nd by successively using the Euclidian algorithm to compute the gcd. What about polynomials in several variables? De¯nition 6 A ring R is called Noetherian, if every ideal is ¯nitely generated or equivalently if R contains no in¯nitely properly ascending chain of ideals I1 $ I2 $ I3 $ ::: (Exercise: recall the proof of the equivalence from your algebra lecture). You can imagine the analogous de¯nition for modules. Theorem 7 (Hilbert Basis Theorem) If R is Noetherian then so is R [x]. We skip the Proof. Let I ½ R [x] an ideal. The set of lead coe±cients of I generate an ideal L ½ R which is ¯nitely generated by some g1; :::; gr 2 R since R is Noetherian. By de¯nition for each gi there is an mi 2 N0 s.t. mi I 3 fi = gix + lower order terms 7 and let J = hf1; :::; fri ½ R [x] and m = max fm1; :::; mrg. Modulo the generators of J we can reduce the degree of elements of I to degree < m. The R-module M generated by 1; x; :::; xm¡1 is ¯nitely generated hence also M \I is ¯nitely generated by some h1; :::; hl 2 R [x] and I = hf1; :::; fr; h1; :::; hli. 2.4 The ideal of a variety Given an ideal I, we formed the set of common zeros V (I) of the elements of I. Conversely given some set S ½ kn, we can consider the set of polynomials I (S) vanishing on S, which is indeed an ideal (exercise: prove this). n De¯nition 8 For S ½ k let I (S) = ff 2 k [x1; :::; xn] j f (x) = 08x 2 Sg. Obviously it holds, that Remark 9 J1 ½ J2 ) V (J2) ½ V (J1) S1 ½ S2 ) I (S2) ½ I (S1) So one can start to study a given variety by studying the hypersurfaces, in which it is contained V (J) ½ V (f) e.g. the twisted cubic lies inside V (y ¡ x2) and V (z ¡ x3). By the above correspondence this naturally leads to the ideal membership problem: Given an ideal J ½ R check, if a given f 2 R lies inside J. For k [x] this is solved by the division with remainder with respect to the single generator of J (found by the Euclidian algorithm). How do the two processes forming I (S) and V (J) relate to each other? One could expect, that J = I (V (J)) for any ideal J, but this is not true: Consider for example the ideal J = hx2i ½ k [x], where V (J) = f0g ½ k, so I (V (J)) = hxi. At least we can note, that if f 2 J and p 2 V (J), then f (p) = 0 so f 2 I (V (J)), i.e. Remark 10 For every ideal J J ½ I (V (J)) What about the relation the other way around? Since every polynomial in I (S) is vanishing on S, we note: Remark 11 For any set S ½ kn S ½ V (I (S)) If S = V (J) we can get the other inclusion by applying V to J ½ I (V (J)), hence 8 Remark 12 If S is a variety i.e. S = V (J) for some ideal J, then S = V (I (S)). By playing with V and I you can easily check: Remark 13 V (I (S)) is the smallest variety containing S. (If S is any set and V (J) some variety with S ½ V (J) ½ V (I (S)) then I (V (J)) ½ I (S) so V (I (S)) ½ V (I (V (J))) = V (J) ½ V (I (S)) so V (I (S)) = V (J)). Example 14 Take S = f(x; 0) 2 R2 j 0 < x < 1g. Then I (S) = hyi and V (I (S)) is the whole x-axis. For a small part S of a circle V (I (S)) gives the whole circle The blue + cyan part is V (I (S)) for the blue Bezier spline S: If you know about topology, then note, that we can consider V (I (S)) as the closure of S in a suitable topology, the Zariski topology. You can imagine, that as the closed sets we should take the a±ne varieties.