Introductory Geometry: Algebraic Geometry

1 Introductory GeometryCourse No. 100 351 Fall 2005 Second Part: Algebraic GeometryMichael StollContents1. What Is Algebraic Geometry ?22. Affine Spaces and Algebraic Sets33. Projective Spaces and Algebraic Sets64. Projective Closure and Affine Patches95. Morphisms and Rational Maps116. Curves Local Properties147. B ezout s Is Algebraic Geometry ?Linear Algebracan be seen (in parts at least) as the study of systems of linearequations. In geometric terms, this can be interpreted as the study of linear (oraffine) subspaces ofCn(say). Algebraic Geometrygeneralizes this in a natural way be looking at systems ofpolynomialequations. Their geometric realizations (their solution sets inCn, say)are calledalgebraic questions one can study in various parts of mathematics lead in a naturalway to (systems of) polynomial equations, to which the methods of AlgebraicGeometry can be Geometry provides a translation betweenalgebra(solutions of equations)andgeometry(points on Algebraic varieties).

2 The methods are mostly Algebraic ,but the Geometry provides the to Differential Geometry , in Algebraic Geometry we consider a ratherrestricted class of manifolds those given by polynomial equations (we canallow singularities , however). For example,y= cosxdefines a perfectly nicedifferentiable curve in the plane, but not an Algebraic return, we can get stronger results, for example a criterion for the existence ofsolutions (in the complex numbers), or statements on the number of solutions (forexample when intersecting two curves), or classification some cases, there are close links between both worlds. For example, a compactRiemann Surface ( , a one-dimensional complex manifold) is the same as a(smooth projective) Algebraic curve we do not have much time in this course, we will mostly look at the simplestnontrivial (but already very interesting case), which is to consideroneequation intwovariables. Such an equation describes aplane Algebraic will usexandyas the simplest examples are provided by the equationsy= 0 andx= 0; theydescribe thex-axis andy-axis, respectively.

3 More generally, alineis given by anequationax+by=cwithaandbnot both equationx2+y2= 1 describes the unit circle. Note that the set of its realpoints (x,y) R2is compact, but its set of complex points is not there are two branches extending to infinity, withx/ytending toiand to irespectively. Itturns out that we can compactify the set of complex points by throwing in twoadditional points at infinity corresponding to these two formally, we introduce theprojective planeP2as the set of points (x:y:z)with (x,y,z) C3\{0}, where we identify (x:y:z) and ( x: y: z) for C\{0}. We find the usualaffine planeA2=C2withinP2as the subset ofpoints (x:y: 1); the points (x:y: 0) form the line at infinity , and there is onepoint for each direction in the affine plane. The unit circle acquires the two newpoints (1 :i: 0) and (1 : i: 0).Aprojectiveplane curve is now given by ahomogeneouspolynomial in thethreevariablesx,y,z. To obtain it from the original affine equation, replacexandybyx/zandy/z, respectively and multiply by a suitable power ofzto cancel the3denominators.

4 For the unit circle we obtainx2+y2=z2; a general line inP2isgiven byax+by+cz= 0 witha,b,cnot all zero. (The line at infinity has equationz= 0, for example.)One of the great advantages ofP2overA2is that inP2any pair of distinct lineshas exactly one common point there is no need to separate the case of parallellines; every pair of lines stands on the same fact that two lines always intersect in exactly one point has a far-reachinggeneralization, known asB ezout s Theorem. It says that two projective planecurvers of degreesmandnintersect in exactlymnpoints (counting multiplicitiescorrectly).The first question towards a classification of Algebraic curves one could ask is toorder them in some way according to their complexity. Roughly, one would expectthat the curve is more complicated when the degree of its defining polynomial islarge. However, this is not true in general, for example, a curvey=f(x) can betransformed to the liney= 0 by a simple substitution, no matter how large thedegree offis.

5 But it is certainly true that a curve given by an equation of lowdegree cannot be very turns out that there is a unique discrete invariant of an Algebraic curve: itsgenusg. The genus is a nonnegative integer, and for a plane curve of degreed,we haveg (d 1)(d 2)/2. So lines (d= 1) and conic sections (d= 2) are ofgenus zero, whereas a general cubic curve (d= 3) will have genus one. Some cubiccurves will have genus zero, however; it turns out that these are the curves havingasingular point, where the curve is not smooth (not a manifold in the DifferentialGeometry sense). In general, there is a formula relating the degreedof aprojectiveplane curve, its genusgand contributions Passociated to its singular pointsP:g=(d 1)(d 2)2 P [Iterationz7 z2+c; to be added] Spaces and Algebraic SetsIn the following, we will do everything over the fieldCof complex reason for this choice is thatCisalgebraically closed, , it satisfies the Fundamental Theorem of Algebra C[x]be a non-constant polynomial.

6 Thenfhas a induction, it follows that every non-constant polynomialf C[x] splits intolinear factors:f(x) =cn j=1(x j)wheren= degfis the degree,c C and the j everything we do would work as well over any other algebraically closedfield (of characteristic zero).4 The first thing we have to do is to provide the stage for our objects. They willbe the solution sets of systems of polynomial equations, so we need the space ofpoints that are potential ,An, is the setCnof alln-tuples ofcomplex numbers. Note thatA0is just one point (the empty tuple).A1is alsocalled theaffine line,A2theaffine (1)LetS C[x1,..,xn] be a subset. The(affine) Algebraic setdefined bySisV(S) ={( 1,.., n) An:f( 1,.., n) = 0 for allf S}.IfI= S is the ideal generated byS, thenV(S) =V(I). Note thatV( ) =V(0) =AnandV({1}) =V(C[x1,..,xn]) = .An non-empty Algebraic set is called analgebraic varietyif it is not theunion of two proper Algebraic subsets.(2)LetV Anbe a subset.

7 TheidealofVis the setI(V) ={f C[x1,..,xn] :f( 1,.., n) = 0 for all ( 1,.., n) V}.It is clear thatI(V) is indeed an ideal ofC[x1,..,xn]. that the finite union and arbitrary intersection of algebraicsets is again an Algebraic set we have j JV(Sj) =V( j JSj)V(S1) V(S2) =V(S1S2) whereS1S2={fg:f S1,g S2}.Since the fullAnand the empty set are also Algebraic sets, one can define a topologyonAnin which the Algebraic sets are exactly the closed sets. This is called theZariski Algebraic sets are closed in the usual topology (the solutionset off= 0 is closed as a polynomialfdefines a continuous function), this newtopology is coarser than the usual obviously haveS1 S2= V(S1) V(S2) andV1 V2= I(V1) I(V2).By definition, we haveS I(V(S)) andV V(I(V)).Together, these implyV(I(V(S))) =V(S) andI(V(I(V))) =I(V).This means that we get an inclusion-reversing bijection between Algebraic sets andthose ideals that are of the formI(V).Hilbert s Nullstellensatztells us what theseideals an ideal,V=V(I).

8 Iff I(V), thenfn Ifor somen can deduce thatI(V(I)) = rad(I) ={f C[x1,..,xn] :fn Ifor somen 1}5is theradicalofI. Note that rad(I) is an ideal (Exercise). HenceI=I(V(I)) ifand only ifIis aradical ideal, which means thatI= rad(I); equivalently,fn Ifor somen 1 impliesf I. Note that rad(rad(I)) = rad(I) (Exercise).So we see thatI7 V(I),V7 I(V) provide an inclusion-reversing bijectionbetween Algebraic sets and radical ideals ofC[x1,..,xn]. Restricting this to alge-braic varieties, we obtain a bijection between Algebraic varieties andprime idealsofC[x1,..,xn] ( , idealsIsuch thatfg Iimpliesf Iorg I).Note thatC[x1,..,xn] is anoetherianring; therefore every ideal is finitely gener-ated. In particular, takingS to be a finite generating set of the ideal S , we seethatV(S) =V(S ) every Algebraic set is defined by afiniteset of us consider the Algebraic sets and varieties in the affine Algebraic set is given by an ideal ofC[x]. NowC[x] is a principal ideal domain,hence every idealIis generated by one element:I= f.

9 Iff= 0, then thealgebraic set is all ofA1. So we assume nowf6= 0. Then the ideal is radical ifand only iffhas no multiple roots, and the Algebraic set defined by it is just thefinite set of points corresponding to the roots off; these arenpoints, wherenis the degree off. (This set is empty whenn= 0, ,fis constant.) So thealgebraic sets inA1are exactly the finite subsets and the whole line. It is thenclear that the Algebraic varieties inA1are the whole line and single points (andindeed, the prime ideals ofC[x] are the zero ideal and the ideals generated by alinear polynomialx ). consider the affine planeA2. The planeA2itself is an al-gebraic set A2=V( ). Any single point ofA2is an Algebraic set (even analgebraic variety) {( , )}=V(x ,y ). Therefore all finite subsets ofA2are Algebraic sets. Is there something in between finite sets and the whole plane?Yes: we can consider something likeV(x) orV(x2+y2 1). We get an algebraicset that is intuitively one-dimensional.

10 Here we look at ideals F generatedby a single non-constant polynomialF C[x,y]. Such an ideal is radical iffFhas no repeated factors in its prime factorization (recall thatC[x,y] is a uniquefactorization domain), and it is prime iffFis irreducible. We callV(F) anaffineplane Algebraic curve;the curve is calledirreducibleifFis examples of affine plane Algebraic curves are thelinesV(ax+by c) (with(a,b)6= (0,0)) or the unit circle V(x2+y2 1), which is a special case of aquadricorconic section a curveV(F), whereFhas (total) degree 2. Note thatthe real picture inR2of the unit circle is misleading: it does not show the twobranches tending to infiniy with asymptotes of slopeiand i!One can show that a general proper Algebraic subset ofA2is a finite union ofirreducible curves and , we need to introduce two more notions that deal with the functions wewant to consider on our Algebraic sets. As this is algebra, the only functions wehave at our disposal are polynomials and quotients of polynomials.