Now that we know that minimal free resolutions exist and are unique for all finitely generated gradedR-modules, we can talk about the minimal graded free resolution of a graded module.
Graded free resolutions of graded R-modules take the form
· · · Fi · · · F1 F0 M M j∈Z R(−j)βi,j M j∈Z R(−j)β1,j M j∈Z R(−j)β0,j ∼ = =∼ =∼
The exponents βi,j of the minimal free resolution, called Betti numbers,
completely determine the objects in the resolution up to isomorphism, and they are an important invariant of a module. The Betti number βi,j tells
us that the ith object in the minimal free resolution has βi,j generators of
3.5. BETTI NUMBERS 33
We often display the Betti numbers in a Betti table:
0 1 . . . i . . . 0 β0,0 β1,1 . . . βi,i . . . 1 β0,1 β1,2 . . . βi,i+1 . . . .. . ... ... ... j β0,j β1,j+1 . . . βi,i+j . . . .. . ... ... ...
Note that the entry in theith column,jth row isβi,i+j, notβi,j. One reason
for this is the following property:
Proposition 3.13. If βi,j = 0 for all j < d and some fixedi, thenβi+1,j+1 =
0 for all j < d.
Remark. In terms of the Betti table, this means that if one column is entirely 0 above some row, then the next column to the right (and, by induction, all columns to the right) are also entirely 0 above that same row.
Proof. The minimality condition of the free resolution means that a sum- mandR(−j) ofFi must be mapped into mFi−1. The map preserves degrees,
and m is precisely the subset of R generated by polynomials with positive degree. Therefore the image of a generator of R(−j) must have degree at most −j −1, so since the maps of a minimal resolution take generators to generators bijectively, no βi+1,j+1 can be non-zero for j < d.
The ideas in the proof of Proposition 3.10 give us a way of computing the Betti numbers from an arbitrary free resolution:
Proposition 3.14. βi,j = dimkTorRi (k, M)j
Proof. If F is the minimal free resolution of M, then TorRi (k, M) is the ith homology of F ⊗k. Since F was minimal, the maps in F ⊗k are all zero by the definition of a minimal resolution, so the ith homology of F ⊗k is simply Fi ⊗k. The dimension of the degree j component of this is exactly
Chapter 4
Apolarity
The polar pairing is a bilinear operation that lets us treat polynomials as “inverses”. We can associate a polynomial f with the set of inverse polyno- mials that annihilate it in this system, which we call theapolar ideal — this set has some important properties. In particular, there is a bijection between homogeneous polynomials (up to scaling) and these sets, which implies that we lose no information about the polynomial by looking at its apolar ideal instead.
4.1
The polar pairing
Definition. Let R = k[x1, . . . , xn] and S = k[x−11, . . . , x
−1
n ] be subrings of
the fraction field K =k[x1, x1−1, . . . , xn, x−n1] of R.
Suppose f ∈ R and g ∈ S are monomials. Define the binary operation
∗:S×R →R in the following way:
g∗f =
(
gf if this is an element of R, and
0 otherwise.
Extend linearly to all polynomials. This givesRthe structure of anS-module — in particular, we have the useful fact that f ∗(g ∗h) = (f g)∗h for all
f, g∈S and h∈R.
We will often write capital letters instead of inverses: so x−i1 = Xi, and S =k[X1, . . . , Xn].
4.1. THE POLAR PAIRING 35 Example. IfR =k[x1, x2, x3] and S =k[X1, X2, X3], X1∗x31 =x −1 1 x31 =x21 X2∗x1x2 =x1 X1∗x2x3 = 0 (X1+X2)∗(x1x23+x 2 2x3) =x23+x2x3 (X2+X3)∗(x2−x3) = 1−1 = 0
Remark. Observe that Xi acts like the partial differential operator ∂x∂
i on
polynomials, except that it doesn’t change coefficients. That is,
Xi∗xni =x n−1 i , whereas ∂ ∂xi xni =nxn−i 1.
We will sometimes abuse notation and write Xi as ∂x∂
i, treating this as “dif-
ferentiation without coefficients”, where this won’t cause confusion. In par- ticular, if f is a polynomial where no variable has exponent greater than 1, differentiation without coefficients agrees with the usual differentiation.1 Remark. GiveS the analogous grading toR, so that a monomialX1d1· · ·Xdn
n
has degreed=d1+· · ·+dn. Iff ∈Randg ∈Sare homogeneous polynomials
with degrees d and d0 respectively, then g ∗f is clearly homogeneous with degree d−d0, since g∗f is made up of monomials that either have degree
d−d0, or are zero.
An immediate consequence of this observation is that if d < d0, we must have g∗f = 0, since there are no non-zero polynomials in R with negative degree. Also, if d = d0, then g∗f has degree 0, so it is a constant, that is, an element of k. This suggests the following definition:
Definition. The polar pairing is the map h·,·i:Sd×Rd →k wherehg, fi= g∗f.
1The lack of coefficients is not an irrelevant detail — the systems with and without
coefficients are truly different algebras. For example, with coefficients,
∂2 ∂x2− ∂ ∂y ∂ ∂z (x2+yz) = 2−1 = 1,
but without coefficients,
36 CHAPTER 4. APOLARITY
Recall that we defined a bilinear form in Section 1.3.
Proposition 4.1. The pairing h·,·i is a non-degenerate, symmetric bilinear form, under the identification of S with R that associates Xi with xi.
Proof. The bilinearity follows immediately from the definition.
Since this is a bilinear form, by Proposition 1.10 we can compute the matrix of the form, with respect to the standard basis forR of monomials.
Observe that if f and g are both monomials of degree d with coefficient 1, if f 6= g, by the pigeonhole principle there must be a variable that has a higher exponent in g than in f, so g ∗f = 0; otherwise, if f = g, then
g ∗f = 1. Therefore the matrix of the form with respect to the basis of monomials is the identity matrix.
The identity matrix is definitely symmetric and invertible, so by Propo- sition 1.11, the form is symmetric and non-degenerate.
Remark. This pairing gives an injection Rd → Sd∨, where S ∨
d is the vector
space dualSd∨ = Homk(Sd, k), given byf 7→ h·, fi, which is the mapg 7→g∗f.
There is an injection in the other direction, so it follows that this map is an isomorphism betweenRd and Sd∨.