Notes on basic algebraic geometry - Purdue …

1 Notes on basic algebraic geometryJune 16, 2008 These are my Notes for an introductory course in algebraic geometry . I havetrodden lightly through the theory and concentrated more on examples. Someexamples are handled on the computer using Macaulay2, although I use this asonly a tool and won t really dwell on the computational course, any serious student of the subject should go on to learn aboutschemes and cohomology, and (at least from my point of view) some of theanalytic theory as well. Hartshorne [Ht] has become the canonical introductionto the first topic, and Griffiths-Harris [GH] the Affine algebraic sets.. Weak Nullstellensatz.. Zariski topology.. The Cayley-Hamilton theorem.

2 Affine Varieties.. Hilbert s Nullstellensatz.. Nilpotent matrices..122 Projective Projective space.. Projective varieties.. Projective closure.. Miscellaneous examples.. Grassmanians.. Elimination theory.. Simultaneous eigenvectors..233 The category of Rational functions.. Quasi-projective varieties.. Graphs..284 Dimension Dimension.. Dimension of fibres.. Simultaneous eigenvectors (continued)..325 Differential Tangent spaces.. Singular points.. Singularities of nilpotent matrices.. Bertini-Sard theorem..382 Chapter 1 Affine algebraic setsLetkbe a field. We writeAnk=kn, and call thisndimensional affine spaceoverk. Letk[x1.]

3 Xn] ={ .. xinn| k}be the polynomial ring. Givena= (ai) An, we can substitutexibyai kinfto obtain an element denoted byf(a) oreva(f), depending on our mood. Apolynomialfgives a functionev(f) :Ank kdefined bya7 eva(f).Givenf k[x1, .. xn], define it zero set byV(f) ={a Ank|f(a) = 0}At this point, we are going to need to assume something about our field. Thefollowing is easy to prove by induction on the number of variables. We leavethis as an algebraically closed andfnonconstant, thenV(f) k[x1, .. xn], then letV(S) be the set of common zerosV(S) = f SV(f)A set of this form is calledalgebraic. I want to convince you that algebraic setsabound in Fermat curve of degreedisV(xd1+xd2 1) A2.

4 Moregenerally, a Fermat hypersurface is given byV(xd1+xd2+.. xdn 1).Example us identifyAn2kwith the setM atn n(k)ofn set of singular matrices is algebraic since it is defined by the vanishing ofthe determinantdetwhich is a the setSLn(k) An2of matrices with determinant1is algebraic since it s justV(det 1).The set of nonsingular matricesGLn(k) is not an algebraic subset ofM atn n(k).However, there is useful trick for identifying it with an algebraic subset ofAn2+1=An2 image ofGLn(k)under the mapA7 (A,1/det(A))identifies it with the algebraic set{(A, a) An2+1|det(A)a= 1}Example the set ofm nmatricesM atm n(k).Then the set of matrices of rank ris algebraic . This is because it is definedby the vanishing of the(r+ 1) (r+ 1)minors, and these are polynomials inthe entries.

5 Notice that the set of matrices with rank equalris not set of pairs(A, v) M atn n(k) knsuch thatvis aneigenvector ofAis algebraic , since the condition is equivalent torank(A, v) An2kbe the set of matrices which are nilpotent oforderi, matricesAsuch thatAi= 0. These are doing the next example, let me remind you about resultants. Giventwo polynomialsf=anxn+.. a0andg=bmxm+.. b0 Suppose, we wanted to test whether they had a common zero, say . Thenmultiplyingf( ) =g( ) = 0 by powers of yieldsan n+an 1 n 1+.. a0= 0..an n+m+an 1 n+m 1+..= 0bm m+..b0= 0..bm n+m+..= 0We can treat this as a matrix equation, with unknown vector ( n+m, n+m 1, .. ,1) the a solution to exist, we would need the determinant of the coefficient ma-trix, called the resultant offandg, to be zero.

6 The converse, is also true (fork= k) and can be found in most standard algebra texts. Thus:Example the set of pairs(f, g)withA(n+1)+(m+1). The set ofpairs with common zeros is can use this to test whether a monic polynomial ( a polynomial withleading coefficient 1)fhas repeated root, by computing the resultant offandits derivativef . This called the discriminant off. Alternatively, if we writef(x) = (x ri), the discriminantdisc(f) = i<j(ri rj)2. This can bewritten as a polynomial in the coefficients offby the theorem on elementarysymmetric set of monic polynomials with repeated roots is a analgebraic subset call a mapF:An Amamorphismif it is given byF(a) = (f1(a).)

7 Fm(a))for polynomialsfi k[x1, .. xn]. Clearly the preimage under a regular map ofan algebraic set is algebraic . Let us identifyAn2with the set ofn nmatricesonce again. To every matrix,Awe can associate its characteristic polynomialdet(tI A). We thus get a morphismch:An2 Angiven by taking thecoefficients of this polynomial other than the leading coefficient which is justone. ThereforeExample set of matrices inAn2with repeated eigenvalues is analgebraic set. More explicitly it is the zero set of the discriminant of the char-acteristic (A2)3with the set of triples of points in the plane. Whichof the following is algebraic :a)The set of triples of distinct )The set of triples(p1, p2, p3)withp3=p1+ )The set of triples of colinear thatV(S) =V( S ), where S ={ hifi|hi k[x1.]}

8 Xn], fi S}is the ideal generated byS. Therefore by the Hilbert basis theorem, whichsays thatk[x1, .. xn]is Noetherian, we find that any algebraic set is de-fined by a finite number of Weak NullstellensatzRecall that the a (commutative)k-algebra is a commutative ringRwith a ringhomomorphismk R. For example,k[x1, .. xn] is ak-algebra. A homomor-5phism ofk-algebras is a ring homomorphismR Ssuch thatk ????????R//Scommutes. Let s make a simple observation at this point:Lemma mapf7 ev(f)is a homomorphism ofk-algebras fromk[x1, .. xn]to the algebra ofk-valued functions that this homomorphism is injective ifkis infinite, butnot in general.(In view of this, we will eventually stop distinguishing betweenfandev(f)whenkis infinite.)

9 Let s suppose thatSis explicity given to us as a finite set of can now ask is there an algorithm to decide whenV(S) this nonempty?Here are some closed fields: Yes by Hilbert s Nullstellensatz (see below). fields: Yes, since there are only a finite number of points to : Yes, by : Unknown! (However, Matjisevich proved that there is no algorithmforZ, or equivalently Hilbert s 10th problem has a negative solution. Soit s reasonable to expect that it would be no forQas well.)Theorem (Weak Hilbert Nullstellensatz).Ifkis algebraically closed,thenV(S) = iff there existsf1.. fN Sandg1.. gN k[x1, .. xn]suchthat figi= 1 The German word nullstellensatz could be translated as zero set theorem.

10 The Weak Nullstellensatz can be rephrased asV(S) = iff S = (1). Sincethis result is central to much of what follows,we will assume thatkis alge-braically closed from now onunless stated otherwise. To get an algorithmas claimed above, we need an effective form of the nullstellensatz:Theorem (Hermann).If(f1, .. fN) = (1), then there existsgi, withdegree bounded by a computable constant depending onmax{degfi}, such that figi= the ringR=k[x1, .. xn]/ S Lemma (kany field.)eva:k[x1, .. xn] kfactors throught the canon-ical mapk[x1, .. xn] Riffa V(S). throughRiffeva( S ) = 0 ifff(a) = 0, f S iffa V(S).In view of the lemma, we can viewevaas a homomorphism ofR kwhena V(S).