Lecture 12-13 Dimension Introduction We use throughout the Noetherian condition as a finiteness assumption for rings and modules: it is convenient to use, and in practice it covers most of the cases we need. I treat dimension theory (originally due to Krull) following [Matsumura, Chap 5]. Intuitively, there are 3 different basic points of view, and we set these up in rigorous algebra, and show that they coincide in appropriate situations. (1) Dimension n means that a ball of radius r has volume ~ r^n, or the number of lattice points in a simplex of size is (r + bit choose n) ~ (const.)*r^n/n! + l.o.t. (2) A n-dimensional body is something you can cut by a hyperplane (setting a single equation equal to zero) to get down to n-1 dimensions, and so on to dimension 0, that is, a point or a finite set of points. (3) Krull dimension n is the longest chain of prime ideals P0 in P1 in .. Pn or the longest chain of irreducible subvarieties pt in curve in surface in .. n-dimensional subvariety in a variety. ==== Graded ring I work mostly with NN-graded rings and modules NN = {0,1,2,..} (gradings by more general semigroups are also useful in different contexts) A graded ring is R = directsum_{n>=0} R_n where R is a ring, and on the right, the R_n are additive subgroups with ring multiplication doing R_n1 x R_n2 -> R_{n1+n2}. It follows of course that R0 is a subring, with 1_R in R0, and each Rn is an R0-module. Also, J = directsum_{n>=0} R_n is an ideal such that R/J = R0. Elements of Rn are homogeneous of degree n. Any f in R is uniquely expressible as a finite sum f = sum_n fn with fn homogeneous of degree n (the homogeneous _component_ or homogeneous _piece_ of f). In dimension theory, we often restrict to the case that R0 is an Artinian ring, or even just R0 = k is a field. (There will eventually be a way around this.) ==== Lemma [Standard elementary fact] A graded ring R is Noetherian if and only if R_0 is a Noetherian subring and R is generated as an R_0 algebra by finitely many homogeneous elements xi in R_di both implications are straightforward (please think them through). ==== Why graded rings? grading ~ representation of CC^x ~ equivalence relation defining PP^n The equivalence relation defining PP^n is an action of the algebraic group Gm(k) = GL(1,k), vulgarly known as CC^x. Projective space PP^n has homogeneous coordinates (x0,..n), with a point P in PP^n the ratio (x0 : x1 :.. xn). The xi are not meaningful, but the ratio xi : xj or the rational function xi/xj is well defined if xj(P) <> 0. If f in k[x0,..xn] is homogeneous of deg d, changing all xi |-> la*xi then changes f(x0..n) |-> la^d*f, so the condition f(P) = 0 is well defined. Implicit in this is the key point that the algebraic group Gm is _reductive_. Every representation of Gm splits into 1-dimensional representations, and every 1-dimensional representation of Gm is given by a homomorphism Gm -> Gm, which is an nth power map z |-> z^n for some n in ZZ. Making a vector space V into a representation of Gm is exactly the same as putting a ZZ-grading on V: V = sum_{n in ZZ} Vn, where Gm acts on Vn by v |-> la^n*v. ==== Graded modules and graded ideals Given R as above, a graded module M over R is an R-module with a grading M = directsum_n M_n, with the ring operation doing R_a x M_n -> M_{a+n}. For some purposes, we might want to allow the grading of M to have negative pieces M_n (win n<0), but a finite module M only has finitely many, and the textbooks assume n >= 0 for ease of notation when setting up the theory. A homogeneous ideal or graded ideal I in R is a particular case of graded submodule. It is an ideal I in R that is of the form I = sum In where In = I intersect Rn. Another way of saying this: for every element f in I, write f = sum fn in the graded structure of R; then the homogeneous pieces fn of f are also in I. In the I <-> V correspondences between subvarieties V in PP^n_k and ideals I in k[x0.. xn] (the homogeneous coordinate ring of PP^n), note that the I taking part are the homogeneous ideals with nilrad(I) not contained in (x0,.. xn), meaning that they define a nonempty subvariety Cone(V) in k^{n+1} \ 0 that is invariant under the Gm action, the _affine cone_ over V. ==== Hilbert series: For the present, consider an NN-graded ring R = sum Rn with the additional condition that R0 is an Artinian ring (e.g., simply a field). Write l(P) = length_R0(P) for the length of a finite R0 module; if R0 = k is a field, l(P) is just the dimension of P as a k-vector space. Suppose also that R is Noetherian, so generated over R0 by finitely many homogeneous elements {xi in R_di}, that is: R = R0[x1,.. xr]/IR (where IR is the ideal of relations) (1) When there are different di, we use _weight_ (or weighted degree) as another word for degree, to avoid the ambiguity between homogeneous and weighted homogeneous monomial: a monomial prod xi^ai has weight sum ai*di. For given weight n, there are only finitely many monomials prod xi^ai of degree sum ai*di = n, so each homogeneous piece Rn of R is finite over R0, and of finite length. This gives us something to count. The same applies to a finite R-module M: each Mn is a finite R0 module. The Hilbert series of M is defined by setting Pn(M) = length(Mn) for n >= 0, and P(M,t) = sum Pn(M)*t^n. This is a formal power series in t, a generating series for the infinitely many coefficients. Its big selling point is that it takes a closed form, giving lots of useful information for little effort. ==== Theorem Let R be as in (1). Then P(M,t) is a rational function of t with denominator product_{i=1}^r (1-t^di), where di = weight xi. That is P(M,t) = H(M,t) / prod_{i=1}^r (1-t^di). Here H(M,t) is a polynomial with integer coefficients (called in different contexts the Hilbert numerator). ==== The "straight" homogeneous case. The case all di = 1 is important. It is known (and argued below) that 1/(1-t)^{m+1} = sum_{n >= 0} (n+m choose n)*t^n. (*) To get the hang of it, write it out for m = 0, 1, 2. Corollary. When all di = 1, suppose that the numerator H(M,t) has degree D. Then P_n(M) is a polynomial in n for n >= D. Proof of corollary. In fact if the numerator H(M,t) = a0 + a1*t + .. aD*t^D, then P(M,t) = H(M,t)/(1-t)^{m+1} = H(M,t) * RHS of (*), so that when n >= D the coefficient of t^n is a0*(n+m choose n) + a1*(n+m-1 choose n) + .. + aD*(n_m-D choose n), which is a polynomial. Proof of Theorem. The method of proof is induction over the number r of generators of R. If r = 0 then M is a finite graded module over the Artinian ring R0, so is a sum of finite length R0-modules Mn in finitely many degrees n, and P(M,t) is just a polynomial. ==== Inductive step. The main step is the following induction: consider multiplication by xr mu_{xr}: M -> M. This increases degrees by dr. Now the kernel K = ker mu_{xr} in M is the submodule annihilated by xr, so is a graded module over Rbar = R/(xr). Also, by construction, N = coker mu_{xr} = M/(xr*M) is a graded R-module annihilated by xr. Write the resulting exact sequence 0 -> K -> M -xr-> M -> N -> 0. (2) Note that M is a graded module over R, whereas by what I just said, K and N are graded modules over the quotient ring Rbar = R/(xr), which is a graded R0-algebra of the same shape as (1), but now with only r-1 generators. Thus by induction on r, I can assume that the result is already known for both K and N: the Hilbert series for K and N have the form P(K,t) = H(K,t) / prod_{i=1}^{r-1} (1-t^di), and (3) P(N,t) = H(N,t) / prod_{i=1}^{r-1} (1-t^di). where the numerators are polynomials with integer coefficients. Since mu_{xr} increases degrees by dr, the exact sequence (2) breaks up according to weighted degrees as exact sequences 0 -> K_{n-dr} -> M_{n-dr} -> M_n -> N_n -> 0 (4) of R0-modules of finite length, one for each n. This gives the equality l(M_n) - l(M_{n-dr}) = N_n - K_{n-dr} for each n. (5) Now sum (3) * t^n gives (1-t^dr) * P(M,t) = P(N,t) - t^{-dr} * P(K,t). (6) This is the result we wanted. In fact both terms on the r-h.s. are of the form polynomial (or possibly Laurent polynomial) divided by prod^{r-1} (1-t^di). Therefore P(M,t) equals polynomial / prod^r (1-t^{di}). QED ==== Explicit cases The classic case is R0 = k and all the di = 1. This means discussing rings of the form k[x0,.. xn] with the ordinary homogeneous grading, or its quotients k[x0,.. xn]/I with I a homogeneous ideal. k[x0,.. xn] is the homogeneous coordinate ring of projective space PP^n_k over k. For a prime ideal I (over algebraically closed k for simplicity), k[x0,.. xn]/I is the homogeneous coordinate ring of the projective variety X = V(I) in PP^n. To avoid complications, I only treat the case X = V(f) where f is an irreducible homogeneous polynomial of degree m, so that X in PP^n is a hypersurface of degree m. ==== Proposition (1) Set R = k[x0,.. xn]. Its rth homogeneous component is the vector space S^r(x0,.. xn) = {x0^r, x0^{r-1}*x1,.. xn^r} with basis the (n+r choose n) monomials of degree d in R. Therefore R has Hilbert series P(R,t) = sum_{r >= 0} (n+r choose n)*t^r = 1/(1-t)^{n+1}, and dim R = n+1. Notice that the dimension Pr = (n+r choose n) = r^n/n! + l.o.t. (Here the denominator n! is the volume of the n-dimensional simplex as a fraction of the n-dimensional unit cube.) (2) Set Rf = R/(f) with f as above. Then P(Rf,t) = (1-t^m) / (1-t)^{n+1} = (1+t+.. t^{m-1}) / (1-t)^n. Its order of growth has the leading term m * r^{n-1} / (n-1)!. ==== Proof Start from the formal expansion for a single variable: 1/(1-x) = 1 + x + x^2 + .. x^n + .. If there are 2 variables, 1/((1-x)*(1-y)) is the formal product of the two power series = (1 + x + x^2 + .. x^n + ..)*(1 + y + y^2 + .. y^n + ..) = sum_{i,j} x^i*y^j, or in other words, the sum of all monomials in k[x,y] each with coefficient 1. In the same way 1/(prod (1-xi)) = sum m m in k[x0..xn] a monomial Now substitute all xi = t. The left-hand side is 1/(1+t)^{n+1}. The r-h.s is a sum of t^r taken over all monomials prod xi^ai with sum ai = t^r. In other words, it is a power series in t, with the coefficient of t^r equal to the number of monomials of degree r. That is statement (1). The number of monomials is (n+r choose n): think of the rule for Pascal's triangle: on multiplying the series for 1/(1-t)^{n-1} by the series for 1/(1-t), the coefficient of t^r is the cumulative sum of the coefficients of all t^i in 1/(1-t)^{n-1} from i = 0..r. Thus the coefficient of t^r equals the coefficient of t^{r-1} in the product plus the coefficient of t^r in the first factor. Or in formulas (n+r-1 choose n) + (n+r-1 choose n-1) = (n+r choose n) LHS is (n+r-1)! / n!.(r-1)! + (n+r-1)! / (n-1)! r! In the first term mult top and bottom by r; in the second term mult top and bottom by n. This gives (r+n) * (n+r-1)! / n! * r! = RHS. For (2), Rf = R/R*f (divided by the principal ideal (f)). The exact sequence 0 -> R*f -> R -> Rf -> 0 splits into homogeneous terms 0 -> R_(r-m) -> R_r -> Rf_r -> 0. Or the monomials of degree r in R map to a generating set of Rf_r, with multiples of f giving linear dependant relations. So dim Rf_r = (n+r choose r) - (n-m choose r) or = (n+r choose r) if n < m. This gives P(Rf,t) = (1-t^m)*P(R,t) = (1-t^m)/(1-t)^(n+1) = (1+t+..+t^(m-1))/(1-t)^n. The conclusion is dimension one less, but degree *m. ==== Weighted homomogeneous case. When the di are not all equal to 1, you don't expect Pn(M) to be a polynomial in n. [A&M] and [Ma] say nothing about the general case, but move on to the next topic. I mention the following: Theorem [Zariski] R = R0[x1,..xk]/I and M as above, with deg xi = di. Separate the n into the different congruence classes i mod N = prod di. Then there are polynomials p_i such that Pn(M) = p_i(n) for n == i mod N and n >> 0. In other words, P_n is not given by a polynomial, but by a "periodic choice of polynomial": there exists N, and polynomials p0,p1,.. p_{N-1} s.t. for n >> 0, Pn(M) = p_i for n == i mod n.