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1 CLASSICAL ALGEBRAIC GEOMETRY Daniel Plaumann Universität Konstan Summer HILBERT POLYNOMIALS T An ane variety V A n with vanishing ideal I(V) K[,, n ] is completely determined by its coordinate ring A(V) = K[x,,x n ]I(V), which is the ring of regular functions on V By Hilbert s Nullstellensat, the points of V correspond to the maximal ideals of A(V) Two ane varieties V and W are isomorphic if and only if their coordinate rings are isomorphic K-algebras If X P n is a projective variety with vanishing ideal I(X) K[Z,,Z n ],the corresponding quotient is the ring S(X) = K[Z,,Z n ]I(X), the homogeneous coordinate ring of X However, unlike the coordinate ring of an ane variety, its elements are not functions on X Furthermore, it is not invariant under isomorphisms of projective varieties For example, the homogeneous coordinate ring of P is K[X, Y], while that of the twisted cubic C P is S(C) =K[Z, Z, Z, Z ](F, F, F ), where F = Z Z Z, F = Z Z Z Z, F = Z Z Z (It is not hard to check that (F, F, F ) is indeed a radical ideal) e twisted cubic is isomorphic to P,butS(C) is not isomorphic to K[X, Y] at is because the ane cone Ĉ dened by F, F, F in A, whose ane coordinate ring is S(C),isnot isomorphic to A, the ane cone of P ; one way of showing this is to examine the tangent space of Ĉ at the origin, which shows that Ĉ is singular (Exercise) So the homogeneous coordinate ring is not a ring of functions and not an invariant of a projective variety up to isomorphism Rather, it encodes information about the embedding of the variety into projective space T H F Let X P n be a projective variety with vanishing ideal I(X) K[Z,,Z n ] For d, we denote by I(X) d the dth homogeneous part I(X) K[Z,,Z n ] d of I(X) Since I(X) is a homogeneous ideal, the homogeneous coordinate ring S(X)

2 HILBERT POLYNOMIALS is a graded ring with decomposition S(X) = S(X)d d where S(X)d = K[Z,, Z n ]d I(X)d -dimensional KEach homogeneous part I(X)d is a linear subspace of the n+d n vector space K[Z,, Z n ]d e dimension of I(X)d is the number of independent hypersurfaces of degree d containing X e Hilbert Function h X of X counts these hypersurfaces by returning the co-dimension of I(X)d, ie hx N N m dims(x)m Examples () Suppose X consists of three points in P en the value h X () tells us exactly whether or not those three points are collinear Namely, if the three points are collinear if they are not h X () = On the other hand, h X () = no matter what To see this, let p, p, p be the three points and consider the map φ K[Z, Z, Z ] K F (F(p ), F(p ), F(p )) Now for any choice of two of the points p, p, p there is some quadratic form that vanishes at these two points but not at the other one us φ is surjective, and its kernel is three-dimensional e same argument shows h X (m) = for all m () If X P consists of four points, there are again two possible Hilbert functions, namely for m = h X (m) = for m = for m if the points are collinear and h X (m) = for m = for m if they are not () We will discuss the following general statement in the exercises: Proposition Let X Pn be a set of d points If m d, then h X (m) = d ere is an abuse of notation here: Since the p i are points in the projective plane, a polynomial F has no well-dened value However, we can pick vectors v i K with p = [v i ] and take F(p i ) to mean F(v i ) Of course, the value depends on the choice of v i, but as long as we only care about the dimension of kernels and images, this does not matter Since this occurs rather frequently, it is too cumbersome to always choose representatives of points

3 SATURATION OF HOMOGENEOUS IDEALS () Next, let X P be a curve given by some irreducible homogeneous polynomial F of degree d e mth homogeneous part I(X)m then consists of all polynomials of degree m divisible by F So we can identify I(X)m with K[Z, Z, Z ]m d for m d, so that m d + dim(i(x)m ) =, hence m+ m d + d(d ) h X (m) = =d m So for m d, h X is a polynomial function of degree It turns out that the last observation comes from a general fact eorem Let X Pn be a projective variety with Hilbert function h X en there exists a polynomial p X in one variable with rational coecients such that h X (m) = p X (m) for all suciently large m N e degree of p X is the dimension of X e polynomial p X is called the Hilbert polynomial of X While we will not need the result in this generality, its proof as outlined in [Ha] uses several important ideas, some of which we have already seen, and is very interesting in its own right We will discuss it aer some preparation Examples () Let X Pn be a nite set of d points By Prop, we have h X (m) = d for all m d So p X is the constant polynomial d () If X P is a curve given by a homogeneous polynomial F of degree d as above, we have just seen that the Hilbert function satises h X (m) = d m d(d ) for all m d us d(d ) p X (t) = d t () Let us determine the Hilbert polynomial of the rational normal curve C in d P Under the Veronese map vd P Pd given by [X, X ] [Xd, Xd X,, X Xd, Xd ] the restriction of a homogeneous polynomial F K[Z,, Z n ] of degree m to C = vd (P ) is a homogeneous polynomial of degree d m in X, X It is not hard to check that this is bijection, in other words S(C)m is isomorphic to K[X, X ]d m us h X (m) = p X (m) = d m + S If X Pn is a projective variety, its vanishing ideal I(X) is a homogeneous ideal Of course, any homogeneous ideal I with I = I(X) also denes X A particular such ideal arises by deleting lower homogeneous parts of X We denote by (Z,, Z n ) the ideal generated by the variables, called the irrelevant ideal (because its ero set in Pn is empty) For k, let J = I(X) (Z,, Z n )k, so that J = I(X)ℓ, ℓk

4 HILBERT POLYNOMIALS ie the rst k homogeneous parts have been deleted It is clear that J = I(X), since F k I for all F I(X), so in particular, J still denes X However, it has a stronger property Given F I(X) and any form G K[Z]k of degree k, we have FG J is has the following consequence: If we dehomogenie and pass from Pn to one of the standard ane-open subsets U i = {[Z] Pn Z i } An, we in fact still obtain the radical ideal: Take i = for simplicity, then we pass from homogeneous polynomials in K[Z] = K[Z,, Z n ] to all polynomials in K[] = K[,, n ], where i = Z i Z Now given f K[] with f X U =, we have F(Z) = Zd f (Z Z,, Z n Z ) I(X), hence Zk F J But the dehomogeniation of Zk F is f erefore, the dehomogenisations of I(X) and J are the same Example For a simple example, take the ideal J generated by Z and Z Z in K[Z, Z ] en V(I) = {P} with P = [, ], but I(P) = (Z, Z ) J In fact, J = I(P) (Z, Z ) If we dehomogenie with respect to Z, we send Z to = Z Z and Z to, so that J becomes the ideal generated by and in K[], which is the vanishing ideal () of P U A in K[] To formalie this, we introduce the following notion: Let J K[Z,, Z n ] be an ideal en Sat(J) = F K[Z,, Z n ] there exists k such that (Z,, Z n )k F J is an ideal, called the saturation of J If Sat(J) = J, then J is called saturated Lemma For any two homogeneous ideals I, J K[Z,, Z n ], the following are equivalent: () I and J have the same saturation () ere exists d such that I m = J m holds for all m d () I and J generate the same ideal locally, ie they are the same in every localiation K[Z Z i,, Z n Z i ], i =,, n Proof Exercise Examples Let C Pd be the rational normal curve Its vanishing ideal I(C) = K[Z,, Z n ] is generated by the d polynomials Fi, j = Z i Z j Z i Z j+, i j d We have shown that for d =, all three polynomials F,, F,, F, are needed to cut out the curve But in general, one can show that the d polynomials Fi,i for i =,, d and Fi,i+ for i =,, d generate an ideal with saturation I(C), but fail to generate I(C) if d (cf [Ha], Exercise /) We will need a strengthening of Bertini s theorem Consider rst the following Example Let X be the plane curve in P dened by F = Z Z Z e line Wα dened by L α = Z αz for α K will meet X in the two points [α, ± α, ] For α, these are two distinct points and the vanishing ideal I(X Wα ) is generated by F and L α (is means that (Z αz, Z αz ) (Z αz, Z + αz ) = (Z Z Z, Z αz ); check!) However, the line W is tangent to X, which is reected

5 PROOF OF THE MAIN THEOREM ON HILBERT POLYNOMIALS in the fact that (F, L ) = (Z Z Z, Z ) is not the vanishing ideal (Z, Z ) of the point {[,, ]} = X W, since (F, L ) does not contain Z eorem Let X Pn be an irreducible projective variety of dimension k with vanishing ideal I(X) e general linear subspace W Pm of dimension n k satises Sat(I(X) + I(W)) = I(X W) Sketch of proof We know from m that the general linear subspace W of dimension n k intersects X in nitely many points X W = {p,, pd } By Bertini s theorem, X W will also be smooth for general W With our denition of smoothness, this is an empty statement, since nitely many points are always a smooth variety However, what Bertini s theorem really says in this case is that the intersection of X W is transversal for general W, which means that p,, pd are smooth points of X and the projective tangent space T p i (X) meets W only at p i, for i =,, d Using the local description of tangent space from Prop, one can show that this implies that I(X) and I(W) generate I(X W) locally P H To show that the Hilbert function eventually agrees with a polynomial, we need one more elementary lemma Lemma Let f Z Z be a function and assume that there exist m Z and a polynomial p() Q() such that f (m + ) f (m) = p(m) for all m m en there exists a polynomial q Q[] of degree deg(p) + such that f (m) = q(m) for all m m Proof For every k N, let Fk () = = ( )( k + ) k k! Since Fk () has degree exactly k, it is clear that = F, F,, Fk form a basis of Q[]k So write p = ki= c k i Fi with c i Q Using the notation ( s)() = s( + ) s() for s Q[], we nd us p = q, where Fk = + = = Fk k k k q = c + + c k k+ It follows that ( f q )(m) = for all m m is means that ( f q )(m) is a constant c k+ for all m m, hence f (m) = q (m) + c k+ for all m m, so that q = q + c k+ satises the claim We are now ready for the proof of m

6 HILBERT POLYNOMIALS eorem Let X Pn be a projective variety with Hilbert function h X en there exists a polynomial p X in one variable with rational coecients such that h X (m) = p X (m) for all suciently large m N e degree of p X is the dimension of X Proof Let k = dim(x) and let W be a linear subspace of dimension n k that meets X in nitely many points and satises Sat(I(X), I(L)) = I(X W) Such W exists by m (Note that k is the highest dimension of any irreducible component of X) Let L,, L k be linear forms that generate I(W) Consider the chain of ideals I(X) = I () I () I (k) K[Z,, Z n ] given by I ( j) = (I(X), L,, L j ) Let S ( j) = K[Z,, Z n ]I ( j) and h ( j) (m) = dim(s ( j) )m, so that h() (m) = h X (m) Since the saturation of I (k) is I(X W), the function h (k) agrees with the Hilbert function h X W for all suciently large arguments Also, since X W consists of nitely many points, we know that h X W (m) = d for large values of m, where d = (X W) (cf Exercise ) Now since the dimension of X W is, it follows that the variety V(I ( j) ) must have dimension exactly k j (Reason: e intersection of a variety of dimension ℓ with a hyperplane has dimension at least ℓ So if the dimension between V(I () ) = X and V(I (k) ) = X W drops from k to, it must drop by one in each step) In particular, this means that the image of L j+ in S ( j) is not a ero-divisor, since otherwise L j+ would generate a ero-dimensional ideal in S ( j) us multiplication by L j+ denes an inclusion (S ( j) )m (S ( j) )m for all m e quotient (S ( j) )m (S ( j) )m with respect to this inclusion is exactly (S ( j+) )m is tells us that h( j+) (m) = h ( j) (m) h( j) (m ) Since h(k) is constant for large arguments, we can apply Lemma inductively and conclude that h () for large arguments agrees with a polynomial of degree k Remark While the Hilbert function of a projective variety eventually agrees with a polynomial, it is in general very hard to predict at what degree the equality occurs When trying to analye the above proof to extract that degree, the problem lies in the fact that we have to take the saturation of the radical ideal of X with the added linear forms It is a non-trivial fact that given any polynomial p Q[], there exists a constant m such that h X (m) = p(m) holds for all m m and all varieties X with p X = p Very little is known in general about the sie of this m, even though this is a highly relevant question for modern constructions like the so-called Hilbert-scheme

7 THE DEGREE OF A PROJECTIVE VARIETY T Together with the dimension, the degree is an important invariant of projective varieties Unlike the dimension, however, it is not invariant under isomorphism but rather depends on the embedding into projective space A hypersurface in Pn is dened by a square-free homogeneous polynomial F K[Z,, Z n ], which is unique up to scaling e degree of the hypersurface is then the degree of X We would like to extend this denition to general subvarieties of Pn We say that X Pn has pure dimension k if every irreducible component of X has dimension k Proposition Let X Pn be a projective variety of pure dimension k en there exists a number d such that the general subspace of dimension n k intersects X in exactly d points at number is equal to k! times the leading coecient of the Hilbert polynomial of X e number d is called the degree of X Proof We know that the general subspace of dimension n k meets X in nitely many points We need to show that this number does not depend on the choice of subspace, ie there is some open subset U of G(n, n k) such that the number of points in X W is the same for all W U We could argue using m Alternatively, the proof of m shows that if W is any subspace meetings X transversally (ie satisfying the conclusion of m ), then the number of points in X W is equal to the kth dierence function of the Hilbert function is is exactly k! times the leading coecient of the Hilbert polynomial It is not hard to see that this denition of degree agress with the old one for hypersurfaces: If X is a hypersurface dened by a square-free homogeneous polynomial F, then for a general line W Pn, the restriction FW will have deg(f)-many distinct roots (We can argue with the discriminant as we did in the case of curves in Chapter ) us deg(f) is the degree of X Examples () If X Pn is set of d points, the degree is d by denition Indeed, we determined in Example () that the Hilbert polynomial of X is the constant d () If X is a plane curve dened by a square-free polynomial F of degree d, we determined in Example () that the Hilbert polynomial is p X (t) = d t d(d ) is conrms that the degree of X is d (It would not be hard to do a similar computation in the case of hypersurfaces in Pn ) () Likewise, it follows from () that the degree of the rational normal curve in Pd is d Again, it is not hard to verify directly that the general hyperplane in Pd meets the rational normal curve in d distinct points () In Exercise, we showed that the Hilbert polynomial of the Segre variety Σ m,n = σ(pm + Pn ) is a polynomial of degree m + n with leading coecient m!n! us the degree of Σ m,n is (m + n)!m!n! = m+n m is means that F has distinct irreducible factors, so that the ideal (F) is radical

8 HILBERT POLYNOMIALS B e simplest form of Béout s theorem asserts that two plane projective curves X and Y of degree d and e that intersect transversally have exactly d e intersection points In the case of curves, transversal intersection just means that I(X Y) = I(X) + I(Y) Here is the precise statement: eorem (Béout) Let X and Y be curves in P without common components en X Y consists of at most deg(x) deg(y) points, with equality if and only if X and Y intersect transversely Proof Let d = deg(x), e = deg(y), I(X) = (F), I(Y) = (G) with F, G K[Z] = K[Z, Z, Z ] and put I = (F, G) K[Z] We wish to compute the Hilbert polynomial of X Y = V(I) To do this, let m d + e and consider the linear map α K[Z]m d K[Z]m e K[Z]m (A, B) AF + BG e image of α is exactly I m We have α(a, B) = if and only if AF = BG Since F and G are coprime by hypothesis, this implies GA and FB Putting A = A G and B = B F, we nd A = B So the kernel of α consists exactly of all elements of the form (RG, RF) for R K[Z]m d e erefore, we have dim(i m ) = dim(k[z]m d K[Z]m e ) dim K[Z]m d e m d + m e+ m d e+ m+ + = de = dim K[Z]m de = us codim(i m ) = de for all m d + e Now if X and Y intersect transversely, then I is the vanishing ideal of X Y We then have codim(i m ) = h X Y (m), hence p X Y = de is shows that X Y is -dimensional of degree de, as claimed If X and Y do not intersect transvesely, then I(X Y) = (F, G) and strictly includes (F, G) erefore, h X Y (m) must be strictly less than de for large m e statement of Béout s theorem can be made more precise and can also be generalied to higher dimensions We indicate some such statements () A more precise version for curves takes multiplicities into account is generalies the fact that a polynomial of degree d in one variable has exactly d eros, counted with mulitplicities If X and Y are two curves without common components and p X Y, the intersection multiplicity of X and Y at p is denoted m p (X, Y) eorem Let X and Y be curves in P without common components en deg(x) deg(y) = m p (X, Y) p X Y e intersection multiplicity m p (X, Y) is dened locally: If p U i A and f resp g are polynomials dening X U i resp Y U i, then m p (X, Y) is dened as the dimension of the K-vector space O p ( f, g), where O p is the local ring k[, ]m p With a little bit of local algebra, this denition is quite simple to handle X and Y intersect transversely if and only if every point P X Y is a smooth point of X and Y and the tangent lines T p X and T p Y are distinct Alternatively,

9 BÉZOUT THEOREMS () e most straightforward generaliation of m to higher dimensions is the following eorem Let X,, X n be hypersurfaces in Pn and assume that Z = X X n is nite with I(Z) = ni= I(X i ) en Z consists of deg(x ) deg(x n ) points is becomes harder to prove than in the case of curves because if I(X i ) = (Fi ) and we look at the map (A,, A n ) ni= A i Fi as in the proof of m, it is more dicult to compute the dimension of the kernel in degree m e appropriate technical tool is the Kosul complex, sketched at the end of Chapter in [Ha] () If the intersection is allowed to be higher-dimensional, things become more complicated For example, we know that if X and Y are two distinct quadric surfaces in P, then X Y consists of a twisted cubic C and a line L So we nd deg(x) deg(y) = = deg(c) + deg(l) = deg(x Y), as expected On the other hand, the fact that the intersection with another quadric Z containing C denes C, which is of degree, while deg(x) deg(y) deg(z) =, is less easy to explain in this way () In general, the condition that I(X) and I(Y) generate I(X Y) needs to be replaced by the following: Let X, Y Pn and let Z,, Z k be the irreducible components of X Y We say that X and Y intersect generically transversely if for each i the general point p on Z i is a smooth point of X and Y and the tangent spaces T p (X) and T p (Y) span T p (Pn ) eorem Let X and Y be subvarieties of pure dimensions k and ℓ in Pn with k + ℓ n and suppose they intersect generically transversely en deg(x Y) = deg(x) deg(y) In particular, if k + ℓ = n, this implies that X Y consists of deg(x) deg(y) points () Again, if we are content with an inequality, we can get away with less We say that X and Y intersect properly if dim(x) + dim(y) n and every irreducible component of X Y has the expected dimension, ie dim(x) + dim(y) n eorem Let X, Y be subvarieties of Pn of pure dimension intersecting properly, then deg(x Y) deg(x) deg(y) () ere is a corresponding version with intersection multiplicities that looks analogous to the case of curves eorem Let X, Y be subvarieties of Pn of pure dimension intersecting properly, then deg(x) deg(y) = m Z (X, Y) deg(z) Z X Y where the sum is taken over all irreducible components Z of X Y and m Z (X, Y) is the intersection multiplicity of X and Y along Z However, it becomes much more complicated in general to correctly dene the intersection multiplicity, and, naturally, also to prove the statement

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