*Proof.* Let vanish at with multiplicity one. Then, vanishes at with multiplicity .

- the set can be defined by polynomials in ;
- the set is invariant under the -action.

*Proof.* Suppose vanishes on . Let , and set . If is defined by polynomials in then and so is Galois-invariant.

Conversely, if is Galois-invariant, then if vanishes on then does also. Denoting to be the elementary symmetric polynomials of for , each and they vanish on . We claim that these polynomials define . For, if every vanished at a point , then every is zero for they are the roots of the polynomial . So the define .

Let be a field and be a standard graded polynomial ring, i.e.,

We will be working with *homogeneous* ideals and *graded* modules and rings. Geometrically, we are working with *projective* varieties in *projective* space .

Let be a finitely generated graded -module. Graded means the module can be decomposed as follows:

Since we haven’t really discussed this yet, just think of as a quotient ring by a homogeneous ideal, or the homogeneous ideal itself.

Recall the following definition:

**Definition.** The (projective) *Hilbert function of * is .

This is non-trivial to compute, and so we’d like a better way to compute this. The motivation to do so stems from the fact that the Hilbert function is an invariant on modules, and also because it (asymptotically) contains other geometric invariants like dimension, degree, and genus.

**Example** (Free modules)**.** Recall that a module is *free* if it is the direct sum of copies of . Thus, its Hilbert function is just its rank times the Hilbert function of .

Consider when the rank is one. Then, we claim that . We just have to count the ways to get degree monomials from variables by using the “stars and bars” argument. We have “stars” that we want to separate into “bins” which we separate by “bars.” To count how many ways to do this, we count the number of ways we can put stars into bins that are non-empty, and then subtract one star from each bin to allow for empty ones. But this is the same as considering the way you can put bars in gaps to make the bins. Thus, we have as desired.

Hilbert’s idea was to compute by comparing with free modules, using a *free resolution*, since as we have seen the free case is very easy to understand. To adapt the corresponding notion for affine/ungraded objects to graded modules in graded polynomial rings, we need some new terminology.

For any graded module , denote the module shifted by : , i.e., the degree elements in become degree (); think of this as shifting every summand in the decomposition into graded parts to the left spots.

**Example** (a principal ideal)**.** The free -module of rank generated by an element of degree is . The easy example is that the ideal is isomorphic as a graded -module to .

**Definition.** A *graded free resolution* of is an exact sequence of degree-0 maps between graded free modules such that : .

**How to compute the (graded) free resolution:**

- Given homogeneous elements of degree that generate as an -module, we can define a map from the graded free module onto by sending the th generator to . Note we need the grade shifts to make sure our maps are degree-preserving.
- Let be defined as ; this is also finitely generated by the Hilbert basis theorem. The elements of are called the
*syzygies*of . This can be done with Buchberger’s algorithm. - Choosing finitely many homogeneous syzygies that generate , we can define map of graded -modules with image . Continuing in this way we construct a graded free resolution of . Programs like Macaulay2 use an improvement of Buchberger’s algorithm for free resolutions due to Schreyer [Sch91, App.].
- Note that this process stops because of the Hilbert syzygy theorem [Eis05, Thm. 1.1].

A free resolution is an example of a *complex* of graded modules, i.e., a chain of graded modules with (grade-preserving) maps between them such that the composition of two adjacent maps is always zero.

**Example** (twisted cubic, [Eis05, Exc. 2.8])**.** Recall that the (projective) twisted cubic is defined by the ideal [Har95, Ex. 1.10]. You can also see this by taking the projective closure of the affine twisted cubic. This is a Gröbner basis (check e.g. with Macaulay2) with the GrRevLex order in the usual ordering .

This gives our first map in the free resolution:

This completes to a free resolution as follows:

To show this is exact, since our maps are grade-preserving, it suffices to show that the maps between the degree pieces of each free module are exact as -linear maps. Exactness at since the matrix

is of full rank. Exactness at follows since the matrix

has rank , and by the rank-nullity theorem.

We can check this with Macaulay2 (modulo change of bases):

i1 : R = QQ[x_0..x_3,MonomialOrder=>GLex] o1 = R o1 : PolynomialRing i2 : M = coker matrix{{x_1*x_3-x_2^2,-x_0*x_3+x_1*x_2,x_0*x_2-x_1^2}} o2 = cokernel | x_1x_3-x_2^2 -x_0x_3+x_1x_2 x_0x_2-x_1^2 | 1 o2 : R-module, quotient of R i3 : d = res M 1 3 2 o3 = R <-- R <-- R <-- 0 0 1 2 3 o3 : ChainComplex i4 : d.dd 1 3 o4 = 0 : R <----------------------------------------------- R : 1 | x_0x_2-x_1^2 x_0x_3-x_1x_2 x_1x_3-x_2^2 | 3 2 1 : R <--------------------- R : 2 {2} | x_2 -x_3 | {2} | -x_1 x_2 | {2} | x_0 -x_1 | 2 2 : R <----- 0 : 3 0 o4 : ChainComplexMap

Note that in general, a graded complex of graded modules (e.g. a free resolution) is exact if and only if the maps restricted to each degree piece of the modules is exact. We used above the fact that a short exact sequence of vector spaces is exact if and only if (the rank-nullity theorem). By splitting up a complex into short exact sequences like the one above, we have that a complex of vector spaces is exact if and only if , nothing that we require that our complex is finite.

This suggests that we can write the Hilbert function as , and this sum makes sense since our resolution is finite by the Hilbert syzygy theorem.

**Theorem.**If the graded -module has finite free resolution , with each a finitely generated free module , then .

*Proof.*, so it suffices to show , but decomposing as a direct sum and shifting back gives us *exactly* the calculation we did as before.

**Corollary.** There is a polynomial (called the *Hilbert polynomial of *) such that, if has free resolution as above, then for .

*Proof.* The binomial coefficient is a polynomial when .

**Example** (the twisted cubic, continued)**.** The Hilbert function for the twisted cubic is given by , which for is the polynomial . The in the first term corresponds to the degree, the largest exponent corresponds to the dimension, and the genus is .

We can check this with Macaulay2:

i5 : hilbertPolynomial(M) o5 = - 2*P + 3*P 0 1 o5 : ProjectiveHilbertPolynomial i6 : hilbertPolynomial(M, Projective => false) o6 = 3i + 1 o6 : QQ[i]

**Example** (Koszul complexes, [Eis05, p. 4])**.** When our polynomials are sufficiently different (precisely, if they form a *regular sequence*), then computing the free resolution becomes *much* easier, by using something called *Koszul complexes*. For example, let , and consider the quotient ring . We then have the resolution

**Example** (Monomial ideals, [BPS98, Ex. 3.4])**.**For a sufficiently general monomial ideals, we can can calculate the free resolution using simplicial methods. Let . Then, we can construct what is called the Scarf complex:

The basic idea is to connect vertexes that represent generators when they share variables. The triangle is labeled by , the edges of the triangle by , and the other edge by . Then the free resolution becomes

Note that this is the simplicial homology complex. This ends up having a Hilbert polynomial of .

[AM69] | M. F. Atiyah and I. G. Macdonald. Introduction to commutative algebra. Reading, Mass.: Addison-Wesley Publishing Co., 1969. |

[BPS98] | D. Bayer, I. Peeva, and B. Sturmfels. “Monomial resolutions.” Math. Res. Lett. 5.1-2 (1998), pp. 31–46. |

[Eis05] | D. Eisenbud. The geometry of syzygies: A second course in commutative algebra and algebraic geometry. Graduate Texts in Mathematics 229. New York: Springer-Verlag, 2005. |

[Eis95] | D. Eisenbud. Commutative algebra: With a view toward algebraic geometry. Graduate Texts in Mathematics 150. New York: Springer-Verlag, 1995. |

[Har95] | J. Harris. Algebraic geometry: A first course. Graduate Texts in Mathematics 133. Corrected reprint of the 1992 original. New York: Springer-Verlag, 1995. |

[Sch91] | F.-O. Schreyer. “A standard basis approach to syzygies of canonical curves.” J. Reine Angew. Math. 421 (1991), pp. 83–123. |

*Proof.* Let have coordinates , and let where . Then, . We claim . But any has coordinates , and clearly any vanishes on , and so . Conversely, note that any point satisfies , and if , say, then setting gives maps to .

*Proof.* The Veronese embedding is defined by , . If is defined by , then , and so defines a hypersurface . Thus .

- Let be the homomorphism defined by sending to , and let be the kernel of . Then is a homogeneous prime ideal, and so is a projective variety in .
- Show that the image of is exactly . (One inclusion is easy. The other will require some calculation).
- Now show that is a homeomorphism of onto the projective variety .
- Show that the twisted cubic curve in (Ex. 2.9) is equal to the 3-uple embedding of in , for suitable choice of coordinates.

*Proof of a.* is homogeneous since it is generated by homogeneous elements in , for each is sent to an element of degree . is prime since maps onto a subring of , which is an integral domain, so is prime by the first isomorphism theorem.

*Proof of b.* , so . Conversely, . If , then it is identically zero for all , so , i.e. .

*Proof of c.* By *b* it suffices to show that the map is continuous with continuous inverse.

*Proof of d.* maps , which is the twisted cubic curve by Ex. 2.9*b*.

- Show that the following two conditions are equivalent for a variety in :
- can be generated by linear polynomials.
- can be written as an intersection of hyperplanes.

In this case we say that is a

*linear variety*in . - If is a linear variety of dimension in , show that is minimally generated by linear polynomials.
- Let be linear varieties in , with . If , then . Furthermore, if , then is a linear variety of dimension .

*Proof of a.* for linear implies for hyperplanes. If , then , where we note each is a linear polynomial. Note we can drop the radical since our generators are linear.

*Proof of b.* Suppose can be generated by , i.e., $\dim S(Y) > r+1$. By Problem this implies $\dim Y > r$, a contradiction.

*Proof of c.* Consider their respective affine cones ; they have dimension respectively by Problem 2.10. Then, their intersection in has dimension by the hypothesis . Since , we finally have that by Problem 2.10. is linear by *a*.

- Show that is an algebraic set in , whose ideal is equal to , considered as an ordinary ideal in .
- is irreducible if and only if is.

Sometimes we consider the projective closure of in . This is called the *projective cone* over .

*Proof of a.* since consists of all scalar multiples of points in , i.e., all points in that are zeros of the generators of .

*Proof of b.* is irreducible if and only is prime if and only if is prime (by *a*) if and only if is irreducible.

*Proof of c.* by Problem 2.6.

- Show that is the ideal generated by , using the notation of the proof of (2.2).
- Let be the twisted cubic of (Ex. 1.2). Its projective closure is called the
*twisted cubic curve*in . Find generators for and , and use this example to show that if generate , then do*not*necessarily generate .

*Proof of a.* Suppose . Then vanishes on , so . Since , we have that . Similarly, if , then it is the sum of polynomials of the form , each one of which is in after homogenization, and so .

*Proof of b.* Recall . Now , and so , but this is not equal to .

*Proof.* by Problem 2.6. By Proposition 1.13, this is true if and only if the affine cone is defined by as the zero set of a nonconstant irreducible polynomial , which is true if and only if is defined by the homogenization of .