1 Monomials

For all these talks, F will be a finitely generated free module over the base ring under discussion. F will always have a specified basis, usually {e_{i}}_{i∈I} for some finite index set I. Frequently the ring will be S = k[x_{1}, …, x_{n}] where k is a field. Sometimes k is required to be closed, but that will be announced.

Definition 1.1 Let F be free over S = k[x_{1}, …, x_{n}]. A monomial in F is an element of F of the form m · e_{i} where m is a monomial in S. A monomial submodule is a submodule of F generated by monomials. Let m, n be monomials in S, and u, v ∈ k. We say u · m · e_{i} divides v · n · e_{j} iff i = j and n is a multiple of m. We write \frac{vne_{j}}{ume_{i}} = vu^{−1}(n/m) ∈ S. We will write x = (x_{1}, …, x_{n}). For α ∈ ℕ^{n}, x^{α} = x_{1}^{α_{1}} · . . . · x_{n}^{α_{n}}.

Note 1.2 If M ⊆ F is monomial,

f ∈ M ⟺ all monomials of f are in M

and for a monomial m we have

m · e ∈ M ⟺ M = ∑_{j∈I} m_{j}e_{j} and some m_{j}e_{j} divides me.

Definition 1.3 Let M be monomial. A set of minimal generators for M is a set of monomials which satisfies either of the two equivalent properties

i) it generates M and is minimal as such,

ii) it is the set of monomials in M which are minimal in M with respect to divisibility.

Proof. if i) is true, and our set is {m_{i}e_{i}}, no m_{i}e_{i} may divide m_{j}e_{j} for i ≠ j. So they are at least a subset of the set from ii). But if one were missing, they could not generate M.

If ii) is true, we have a generating set by the preceeding Note. This will be minimal as generating set, as otherwise by the same Note we should have m_{j}e_{j}|m_{i}e_{i} for some i ≠ j. □

Definition 1.4 An admissible (monomial, multiplicative) order on S is a total order on the monomials in S satisfying

• x^{A} > 1 for all A ≠ 0, • x^{A} > x^{B} implies x^{A+C} > x^{B+C} for all exponent vectors A, B, C ≥ 0.

(A, B, C ∈ ℕ^{n}.)

The typical examples, and the useful ones too, are graded lexicographic order (lex) and reverse graded lexicographic order (revlex), where

• x^{A} >_{lex} x^{B}, if either deg x^{A} > deg x^{B} (we say that >_{lex} refines the partial order given by degree), or if the degrees are equal and A−B has a positive first nonzero component;

• and x^{A} >_{revlex} x^{B}, if either deg x^{A} > deg x^{B} (again >_{revlex} refines the partial order given by degree), or if the degrees are equal and A − B has a negative last nonzero component.

Definition 1.5 The initial term of f ∈ S is the largest monomial in f, written in(f). Given an ideal I in S, we denote with in(I) the ideal generated by the elements in(x) with x running through I.

Now we start doing something. Note that for a monomial ideal J the monomials not in J constitute a vspace basis for S/J. Something like it works for the general case too, but it is not quite as nice. Let us assume, that the ideal I is fixed and given and that the ideal J contains at least one monomial from each polynomial in I. Let B be the set of monomials not in J. Then I claim that B constitutes a subset of a vspace basis for S/I.

Proof. If x + ∑ b_{i}α_{i} = 0 in S for x ∈ I, b_{i} ∈ B, then no b_{i} is in J. As their sum is a polynomial in I, at least one of them has to be in J by the hypothesis. By contradiction x = 0 and all α_{i} = 0. □

Now let us assume that B is indeed a basis for S/I. Then J has to be minimal in the above sense. That is, we may not drop one of the monomials in J.

Proof. Suppose we could. Then we may add a monomial to B. That is, there is a generator j for J such that J = (j, j_{1}, …, j_{r}) and (j_{1}, …, j_{r}) contains a monomial of each element of I and j ∉ (j_{1}, …, j_{r}). It follows that j + I is not in the vspace span of B. (Otherwise:

j + x + ∑ α_{i}b_{i} = 0 ∈ S

for x ∈ I and by hypothesis there is at least one of the monomials, say j′, of x that will have to belong to (j_{1}, …, j_{r}). But then j′ is also one of the b_{i} or equals j. Both is impossible. Hence x does not have monomials. But j is certainly different from all the b_{i} so ht we have a contradiction here.)

As therefore the span of B in S/I is not the whole ring, B was not a basis.□

Example 1.6 Let I = (m_{1} + m_{2}). Any J in the above sense will have to contain either m_{1} or m_{2}. In the case where m_{1}|m_{2}, assume we have J = (m_{1}). The corresponding B to this J is NOT a basis, as B will contain neither m_{1} nor m_{2} and so the span of B, modulo I, willnot contain m_{1} + I. On the other hand, if we take J to be (m_{2}), then B is indeed a basis, as then m_{1} ∈ B.

Example 1.7 Let m_{1,2,3} be distinct monomials, of degree d. Set I = (m_{1} + m_{2}, m_{2} + m_{3}, and all monomials of degree > d). Then for each of m_{1} +m_{2} and m_{2} +m_{3} we have to pick one of the monomials. Assume we take m_{1} and m_{2}. Then, as m_{1} − m_{3} ∈ I, we have to take either m_{1} or m_{3} as well. It would be convenient not to take m_{3} as we strive for minimality. So any receipe that produces minimal J should detect that it has to take m_{1} here.

So if m_{1} is preferable to m_{2} and m_{2} preferable to m_{3}, then we better have m_{1} preferable to m_{3}.

Maybe we want to take a total order on the monomials, and then each time take the largest for each element of I.

If we do that, then the order > should be able to survive multiplication, as e.g. if I = (m_{1} + m_{2}) and m_{1} > m_{2}, we find m_{1} in J. As J is an ideal, for all monomials n, m_{1}n ∈ J. Then we certainly would not want to take m_{2}n for n(m_{1} + m_{2}) ∈ I. Hence we require nm_{1} > nm_{2} for all n.

Definition 1.8 Let F be finite free over S. A monomial order > on F is a total order on the monomials of F, M(F), for which

m_{1} > m_{2} ⇒ nm_{1} > nm_{2}

for all monomials n, m_{1}, m_{2}.

Lemma 1.9 Any monomial order on F satisfies DCC.

Proof. Let X ⊆ M(F). The submodule ∑_{x∈X} S · x is f.g., as F is Noetherian. In particular, there is Y ⊆ X such that ∑_{x∈X} Sx = ∑_{y∈Y} Sy and Y is finite. So the least element y of Y is in X. And it will be the least element in X too, as all monomials in ∑_{x∈Y} S · x = ∑_{x∈X} S · x will at least be equal to y. If X is a chain, this shows that all chains are bounded below in M(F). By Zorn’s lemma, > satisfies DCC. □

Note 1.10 We frequently used the idea of a term rather then a monomial: a term is a monomial in F multiplied with a nonzero element in k. We define for two terms um and vn, u, v ∈ k, m, n ∈ M(F), that um ≥ vn if m ≥ n. Note that this even for total > does not induce a partial order on the terms, as for example um ≥ vm ≥ um for all nonzero u, v ∈ k.

Definition 1.11 As for elements of S, we define the initial term in(f) of f to be the term corresponding to the largest monomial in f. For a submodule M ⊆ F we define in M to be the monomial module generated by all in(f) where f runs through M.

Note that for monomial submodules or ideals, M = in(M) for any monomial order.

Fact 1.12 Let f ∈ F, p ∈ S. Then in(fp) = in(f) times v where v is the term in p for which in f · v is maximal. The statement here is, that in(f) divides in(fg).

Proof. Let v′ be another term in p, and m be a term in f. Then

v · in(f) ≥ v′ in(f)  (1)
≥ v′m  (2)

by the choice of v and the property of monomial orders. □

One of the nice features of monomial orders, and the underlying property for all applications, is reflected in the

Theorem 1.13 (Macaulay) Let F be finite free over S and let M be a submodule of F. Let > be a monomial order on F and B the monomials not in in_{>}(M). Then B is a vspace basis for F/M.

Proof. Remember that in_{>}(M) contains each time the largest term of an element in M. So this is what we would have recommended after example 1.6.

For the proof, we show first the linear independence. Let p = ∑ u_{i}m_{i} where m_{i} ∈ B, p ∈ M, u_{i} not all equal to 0 in k. Then in_{>}(p) ∈ in_{>} M and in(p) is one of the m_{i} times a unit in k. But then in(p) ∈ B. As p ∈ M, in(p) ∉ B by hypothesis. By contradiction the cosets of the elements of B have to be linearly independent.

That they span indeed F/M shows the following. Suppose not. Let f be minimal with respect to

• not in the span of B ∪ M,

• minimal initial term.

(Remember that we have shown that monomial orders satisfy DCC!) Then in(f) ∉ B as otherwise f − in(f) is in F \ span(M ∪ B) which has smaller initial term than f. By construction of B, in(f) ∈ M. But this is clearly impossible as again f − in(f) has smaller initial term than f. □

Fact 1.14 The orders lex and revlex have the following characteristic properties:

• if in_{lex}(f) ∈ k[x_{l}, … x_{n}] for homogeneous f, then f ∈ k[x_{l}, …, x_{n}],

• if in_{revlex}(f) ∈ (x_{l}, …, x_{n}) for homogeneous f, then f ∈ (x_{l}, …, x_{n}).

Proof.

• in_{lex}(f) is the term with “the biggest variable”. As lex means x_{1} > x_{2} > …, in_{lex}(f) ∈ k[x_{l}, …, x_{n}] implies that all other terms in f will have to contain only variables “to the right of x_{l}”by homogeneity.

Assume that m >_{lex} m′ and that they have the same degree. Then m has higher powers in variables with smaller index.

Conversely, compare the coprime monomials m and m′ by looking at in_{>}(m + m′) for some order > satisfying the given characteristic property of lex. Then in_{>}(m + m′) cannot be m′ as otherwise m + m′ would be in k[x_{l}, …, x_{n}] where x_{l} is the smallest variable occuring in m′ so that m ∈ k[x_{l}, …, x_{n}] as well. As we are assuming m, m′ to be coprime, we must then have m <_{lex} m′, a contradiction. This being true for coprime monomials, it has to be true for not-coprime monomials as well, as monomial orders respect multiplication with the same element. Finally, for any monomials m, n ∉ k we have mn > n by the monomial order property. That finishes the proof.

• in_{revlex}(f) is the term in f which has the least number of small variables. (f was homogeneous!) So if in_{revlex}(f) ∈ (x_{l}, …, x_{n}), and the l here is the smallest l for which we can say that, all other monomials in f have to have a divisor in {x_{l}, …, x_{n}}.

Conversely, assume that > satisfies the given property. Let m, m′ be monomials of the same degree with m >_{revlex} m′. That is to say, that m has less of the smaller variables. We show that then m > m′. We assume again first that m, m′ are coprime. If m > m′ is wrong, m′ > m is right as monomial orders are total. As we assume coprimeness, and m >_{revlex} m′, m′ contains the smalles variable x_{l} of all the ones that occur in either m or m′. So m′ ∈ (x_{l}, …, x_{n}). But as m′ = in_{>}(m + m′), m + m′ is in the ideal (x_{l}, …, x_{n}) by hypothesis on >. That is impossible, because it has to be true for m then as well. In other words, m >_{revlex} m′ ⇒ m > m′ in any order satisfying the given property. □

Definition 1.15 Let λ : ℤ^{r} → ℤ be a linear functional. We write x^{α} >_{λ} x^{β} for α, β ∈ ℤ^{r} if λα > λβ in ℤ. The induced partial order on monomials is called a weight order and λ is the weight.

Given a finite number of weight orders, λ_{1}, …, λ_{s}, one can define the lexicographic product of them as follows. Given two monomials x^{α}, x^{β} we set x^{α} >_{λ} x^{β} if

• λ_{1}α > λ_{1}β or

• λ_{1}α = λ_{1}β and λ_{2}α > λ_{2}β or

• ⋮

• λ_{i}α = λ_{i}β ∀ i = 1, …, s − 1 and λ_{n}α > λ_{n}β.

Robbiano proved in 1987 that any monomial order on r variables is the lexicographic product of r weight orders.

2 Gröbner Bases

We keep our notation of a finite free module F on which a monomial order > works.

Definition 2.1 Let M ⊆ F and G = {g_{1}, …, g_{r}} ⊆ M. G is called a Gröbner basis for M, if

(in(g_{1}), …, in(g_{r})) = in(M).

Example 2.2 If n = 0, that is S = k, then F is a finite dimensional vspace. The only existing monomials then are the basis elements e_{i} of F. So a set {g_{i}} is in essence a matrix with entries in k. Now for example any matrix that comes in echelon form, will represent a Gröbner basis, provided e_{1} > e_{2} > … holds. This is because any linear combination of rows will have the leading term of the highest row. That is, the initial module of a module spanned by an echelon matrix is exactly the module generated by the leading terms of the rows. Which is the definition of Gröbner.

Note, that for any M ⊆ F, we can indeed find a finite Gröbner basis, as F is Noetherian, and therefore in(M) ⊆ in(F) = F is finitely generated.

Lemma 2.3 Let N ⊆ M ⊆ F. Then

in_{>}(N) = in_{>}(M) ⇒ N = M.

Proof. Suppose not. So there is f ∈ M \ N. Choose the f which has minimal initial term. (Again, > is Artinian!) Then in_{>}(f) ∈ in_{>}(M) = in_{>}(N). So there is g ∈ N, in_{>}(f) = in_{>}(g). As f ∉ N, neither is f − g. But the difference will have initial term strictly smaller than f. By contradiction, M = N. □

Remark 2.4 The above lemma is very helpful in order to answer the question “Given f ∈ F, M ⊆ F, is f ∈ M?”. It will be so, iff in(M) = in(M + Sf). This is the same test, only we have simply monomials, and then is really easy to do, see 1.2.

Definition 2.5 A reduced Gröbner basis is a Gröbner basis for which in(g_{i})| in(g_{j}) implies i = j for two elements g_{i}, g_{j} in the basis.

Now we come to a tremenously important

Proposition 2.6 Let f, g_{1}, …, g_{t} ∈ F. Then there exist f′ ∈ F, f_{1}, …, f_{t} ∈ S with

f′ = f − ∑ f_{i}g_{i}

where

• no monomial of f′ is in (in(g_{1}), …, in(g_{t})) ⊆ F,

• in(f) ≥ in(f_{i}g_{i}) for all i.

f′ is then called a remainder with respect to {g_{1}, …, g_{t}} and f − ∑ f_{i}g_{i} a standard expression for f′.

Proof. Write f′_{p} = f − ∑_{i=1}^{p} m_{i}g_{s_{i}} which is just f for p = 0. Then let m be the largest term of f′_{p} which is divisible by one of the in(g_{i}), say by in(g_{j}). Then choose s_{p+1} = j, m_{p+1} = \frac{m}{in(g_{i})}. Construct so a sequence of f′_{p}. The process will terminate either if none of the monomials of f′_{p} is divisible by any of the initial terms of the g_{i}, or when f′_{p} = 0. In either case we are happy. So we only have to be sure that the process terminates at all. And it does, since {m_{p}} is a decreasing sequence with respect to >.

As in each step we we choose the cofactor for g_{i} to be a monomial such that the product is not bigger than f′_{p} (see 1.12), we certainly will have f_{i}g_{i} ≤ f for all i. □

Note, that this is a generalization of the Euclidean algorithm, and even the proof works like the proof of that.

Remark 2.7 Except for effectivity (possibly) there is no reason to always take the largest monomial in f′_{p} that is divided by one of the in(g_{i}). We could take them in any order. It turns out that (at least for theoretical purposes) it sometimes is useful to take that with the smallest i or something like that.

The above division algorithm now leads to an algorithm for the production of Gröbner bases. In order to describe that algorithm, we need some notation. Let F as usual be a finite free S-module. Let g_{1}, …, g_{t} be a subset of F, none of them zero. Set M = (g_{1}, …, g_{t}) ⊆ F. Let E = ⨁_{1}^{t} Sε_{i}. Write ϕ : E → F by ϕ(ε_{i}) = g_{i}.

Now, whenever two initial terms in(g_{i}) and in(g_{j}) involve the same basis element e_{k} of F, then we write m_{i,j} = in(g_{i})/ gcd(in(g_{i}), in(g_{j})) Note that m_{i,j} ≠ m_{j,i}. We set σ_{i,j} = m_{j,i}ε_{i} − m_{i,j}ε_{j}. Then in(σ_{i,j} < in(m_{j,i}g_{i}) = in(m_{i,j}g_{j}). Using the preceeding theorem, we may find a standard expression

h_{i,j} = (m_{j,i}g_{i} − m_{j,i}g_{j}) − ∑ f_{k}^{i,j}g_{k}.

If the initial terms of g_{i}, g_{j} involve two different basis elements of F, we set h_{i,j} = 0. We will need the following

Lemma 2.8 Let the notation be as described above. Then the kernel of ϕ is the module generated by the σ_{i,j}.

Proof. It is rather obvious that Σ = ∑_{i,j} S · σ_{i,j} ⊆ ker ϕ.

Let σ = ∑ p_{i}ε_{i} be a kernel element. Let u be a monomial in F which is in one of the p_{j}m_{j}. For all i we let p_{i,u} be the term of p_{i} such that p_{i,u}m_{i} is a scalar times u. (If that does not happen for particular i, set p_{i,u} = 0.) Then we will have that ∑ p_{i,u}m_{i} = 0 as m_{i} are monomials. So it is enough to show that ∑ p_{i,u}ε_{i} ∈ Σ.

Let τ = ∑ p_{i,u}ε_{i}. If there is no nonzero term at all in τ, we are done. Otherwise, we will have, say, p_{i,u} ≠ 0. But then necessarily there is some other p_{j,u} different frm 0. Then we find κ · u = p_{j,u}m_{j} = p_{j,u} · m_{j,i} · gcd(m_{i}, m_{j}) which implies that u = κ^{−1}p_{j,n}m_{j,i} gcd(m_{i}, m_{j}). Then we may use σ_{i,j} to cancel the term ε_{j}p_{j,u} and we note that the only new term in the sum then will be a multiple of ε_{j}. That is, we may continue with the reduction process, until only one of the ε_{j} is left in the sum. Its coefficient then has to be 0, in other words τ ∈ ∑ S · σ_{i,j}. □

Theorem 2.9 (The Buchberger Criterion) Using the notation established before the lemma, {g_{1}, …, g_{t}} = G is a Gröbner basis, if and only if all the h_{i,j} are equal to 0.

Proof. We set (g_{1}, …, g_{t}) = M. By construction we have h_{i,j} ∈ M.

If G is Gröbner, and there is one of the h_{i,j} different from 0, we must have 0 ≠ in(h_{i,j}) ∈ in(M) = (in(g_{1}), …, in(g_{t})) as G Gröbner. But this contradicts the remainder definition, which says that no monomial of the remainder is divisible by any of the leading terms of the basis. So G Gröbner implies h_{i,j} = 0.

Now assume that G is not Gröbner, but all h_{i,j} = 0. Then there is 0 ≠ f = ∑ f_{i}g_{i} ∈ M which has initial term in(f) ∉ (in(g_{1}), …, in(g_{t})). Amongst all these f choose one for that the largest monomial m in {in(f_{i}g_{i})} is minimal. Let then ∑′ f_{ν}g_{ν} be the subsum where in(f_{ν}g_{ν}) is a scalar multiple of m. Then in(f_{ν}g_{ν}) = u_{ν} in(g_{ν}) for some term u_{ν} in f_{ν}.

Suppose, α = ∑′ u_{ν} in(g_{ν}) ≠ 0. Then α = in(f). This means m|α. But α = in(f) and m = u_{ν} · in(g_{ν}) up to a unit, so that in(g_{ν})| in(f). But we chose f not to be in (in(g_{1}), …, in(g_{t})) which contradicts what we found. Hence we must have ∑′ u_{ν} in(g_{ν}) = 0. This is a nontrivial representation of 0, in terms of the in(g_{i}), as all we have here are leading terms of g_{ν} and of f_{ν} and if they all are 0, f = 0. So we are looking at a syzygy on the in(g_{i}) here. By lemma 2.8, we can write

∑^{′} u_{ν}ε_{ν} = ∑ σ_{i,j}a_{i,j}

where all a_{i,j} are in k. This is an equality in E = ⨁_{1}^{t} S · ε_{i}, and so we might use ϕ : ε_{i} → g_{i} instead of ψ = (ε_{i} → in(g_{i})) to conclude that

∑^{′} u_{ν}g_{ν} = ∑(m_{j,i}g_{i} − m_{i,j}g_{j})a_{i,j}  (3)
= ∑ (∑ f_{k}^{i,j}g_{k}) a_{i,j} as all h_{i,j} are supposed to equal 0 (4)
=^{!} ∑ g_{k}h_{k}.  (5)

As f_{k}^{i,j} is a division element for ϕ(σ_{i,j}), in(f_{k}^{i,j}g_{k}) ≤ in(ϕ(σ_{i,j})) by construction. This implies that in(g_{k}h_{k} < m as the a_{i,j} are scalars.

Now we subtract ∑′ u_{ν}g_{ν} − ∑ g_{k}h_{k} from the expression f = ∑ f_{i}g_{i} we started with. That will cancel the leading terms ∑′ u_{ν} in(g_{ν}) in the expression for f (namely all such in which the m occurs). That is, f can be written with smaller largest occuring terms, contradicting the choice of f. So G must be Gröbner. □

Now we are able to actually produce Gröbner bases.

Algorithm 2.10 Let g_{1}, …, g_{t} generate M ⊆ F, F finite free with monomial order >.

• Compute the reduced leading terms m_{i,j} = in(g_{i})/ gcd(in(g_{i}), in(g_{j})) whenever in(g_{i}), in(g_{j}) do involve the same basis element of F;

• for these i, j compute remainders h_{i,j} = (m_{i,j}g_{j} − m_{j,i}g_{i}) − ∑ f_{k}^{i,j}g_{k} according to proposition 2.6;

• if all h_{i,j} = 0, stop;

• otherwise, add all of the h_{i,j} to {g_{1}, …, g_{t}} and repeat the previous steps.

Note that the ideal (in(g_{1}), …, in(g_{t})) is growing each time as we add the h_{i,j} which according to the definition of remainders cannot be in (in(g_{1}), …, in(g_{t})) if we do not stop. As S is Noetherian, this will at some stage terminate. When it does, by the Buchberger criterion, {g_{1}, …, g_{t}} will be a Gröbner basis for (g_{1}, …, g_{t}). And as all h_{i,j} we added are clearly in M, the algorithm indeed produces a Gröbner basis for M.

Gröbner bases enjoy the folowing nice property.

Lemma 2.11 If G = {g_{i}} is a Gröbner basis for M ⊆ F, and f ∈ F, then any two remainders of f with respect to G are equal.

Proof. Let h, h′ be two different remainders to f. Then h−h′ will be in M and different from 0. As G is Gröbner for M, the initial term of h − h′ is divisible by one of the in(g_{i}). But the initial term of h−h′ is either a monomial in h or a monomial in h′ or a monomial in both. So in either case, there is a monomial in h or h′ divisible by one of the in(g_{i}). Which contradicts the definition of a remainder.

Remark 2.12 For lex, there is a sometimes sharp upper bound of the number of elements in a Gröbner basis. Let r be the number of variables, d the maximal degree of the given generators of M, s the degree of the Hilbert polynomial of M. Then the constructed Gröbner basis of 2.10 will have at most [(r + 1)(d + 1) + 1]^{2^{s+1}·(r+1)} elements, which of course is terrible.

3 Syzygies

We begin this section with

Theorem 3.1 (Schreyer) Let g_{1}, …, g_{t} be a Gröbner basis for M. Assume that if i < j and in(g_{i}), in(g_{j}) involve the same basis element of F, the division algorithm 2.6 yields the expression

0 = (m_{j,i}g_{i} − m_{i,j}g_{j}) − ∑ f_{k}^{i,j}g_{k},

with in(f_{k}^{i,j}g_{k}) ≤ in(m_{j,i}g_{i} − m_{i,j}g_{j}) that is, a syzygy of the form

τ_{i,j} = m_{j,i}ε_{i} − m_{i,j}ε_{j} − ∑ f_{k}^{i,j}ε_{k}

on {g_{i}} (which is possible as the g_{i} are Gröbner by hypothesis). Let > be the order on Σ = ⨁ Sε_{i} defined by mε_{i} > m′ε_{j} if in(mg_{i}) > in(m′g_{j}) in F, or if in(mg_{i}) = in(m′g_{j}) up to a scalar and i < j for monomials m, m′ ∈ S. (Note that this order is monomial: it is certainly total, and if we have M_{1} ≥ M_{2} for monomials in Σ, then mM_{1} ≥ mM_{2} as > is monomial in F and by the same token we find mM_{2} ≥ M_{2}.)

Under these conditions, {τ_{i,j}} is a Gröbner basis for the syzygy module on the {g_{i}}. Furthermore, in(τ_{i,j}) = m_{j,i}ε_{i} for all i < j.

Proof. First we prove the last assertion. We have m_{j,i} in(g_{i}) = m_{i,j} in(g_{j}). By construction of the f_{k}^{i,j}, we also have m_{j,i} in(g_{i}) > anything in (f_{k}^{i,j}g_{k}) as this is a consequence of the division algorithm. Hence in(τ_{i,j}) = m_{j,i}ε_{i} or −m_{i,j}ε_{j} by definition of order in Σ. But if i < j, m_{j,i}ε_{i} > m_{i,j}ε_{j}.

Now we show that τ_{i,j} are a Gröbner basis for the syzygies on the g_{i}. For, let τ = ∑ f_{ν}ε_{ν} be a syzygy, f_{ν} ∈ S. We need to show that in(τ) is in the module generated by the in(τ_{i}) which is the set of m_{j,i}ε_{i}.

For all indices ν, set u_{ν}ε_{ν} to be in(f_{ν}ε_{ν}). These terms do not cancel, as they work on different ε_{ν}. Hence for some i, in(τ) = u_{i}ε_{i}. Let σ = ∑′ u_{ν}ε_{ν}, where the sum goes over all ν for which u_{ν}g_{ν} = u_{i}g_{i} up to scalars. In other words, σ is a syzygy on in(g_{ν}) with ν ≥ i. By lemma 2.8 this implies that σ is a linear combination of the σ_{k,l}, k > l where σ_{k,l} = m_{k,l}ε_{l} − m_{l,k}ε_{k}. But the leading term of σ_{k,l} is m_{k,l}ε_{l}, which is also in(τ_{k,l}). So in(τ) is a multiple of one of the in(σ_{k,l}), so one of in(τ_{k,l}) and that finishes the proof by 2.3 □

Corollary 3.2 With the notation as before, assume that the g_{i} have been arranged so that whenever in(g_{i}) and in(g_{j}) involve the same basis element of F, say in(g_{i}) = n_{i}e, in(g_{j}) = n_{j}e, then i < j ⇒ n_{i} > n_{j} in not graded lex. (Note that we do not require the order on F to be lex!)

If the variables x_{1}, …, x_{s} are missing from the initial terms of the g_{i}, then the variables x_{1}, …, x_{s+1} are missing from the in(τ_{i,j}), and furthermore

F/(g_{1}, …, g_{t}) = F/M

has a free resolution of length most n − s.

Proof. By the previous theorem, in(τ_{i,j} = m_{j,i}ε_{i}, m_{j,i} = m_{j}/ gcd(m_{i}, m_{j}) unless τ_{i,j} = 0. Let i < and τ_{i,j} ≠ 0. As m_{i} = in(g_{i}) ≥ in(g_{j}) = m_{j}, the power of x_{s+1} in m_{i} will be at least equal to the power in m_{j} as we assume that m_{i} > m_{j} in ungraded lex and the first s variables are missing. But the leading term of τ_{i,j} is m_{j,i} = m_{j}/ gcd(m_{j}, m_{i}) which implies that x_{s+1} cannot be a factor in m_{j,i}.

Now suppose none of the variables appears in m_{j,i}. That is, s = n. Then we have to show that F/M is free. As all in(g_{i}) are scalars then, in(M) is the free submodule of F generated by the e_{i} which appear in {in(g_{i})}. Let F′ be the free submodule spanned by the other e_{i}. By theorem 1.13, F/M has a vspace basis consisting of the monomials not in M. But that are the monomials in F′. Hence F/M is isomorphic to F′ which is free.

Now for arbitrary s consider

F_{i} ⟶^{ϕ_{i}} F_{i−1} → ⋯ ⟶^{ϕ_{2}} F_{1} = ⨁ Sε_{i} ⟶^{ϕ_{1}} F = F_{0} ⟶^{ϕ_{0}} F/M → 0

which is formed in the following inductive process. Choose a monomial order in F_{i}, find a Gröbner basis for the kernel of ϕ_{i}, which are the syzygies on the generators of F_{i−1}, arrange the basis to satisfy the hypothesis of the corollary. Then F_{i+1} will be the free module over the generators of the syzygies in F_{i}. Now using the first statement of the corollary i times gives that the first s + i variables are missing from the initial terms of the constructed Gröbner basis for ker ϕ_{i}. So the kernel of ϕ_{n−s} will have a Gröbner basis in which no variable occurs. By the case s = n, F′_{n−s} = im ϕ_{n−s} = F_{n−s}/ ker ϕ_{n−s} is free. So we may replace F_{n−s} by F′_{n−s} and are done. □

Let us do an example here. Let F = S = k[x, y]. Choose also g_{1} = x^{2}, g_{2} = xy + y^{2}. Then, if the order is lex with x > y, in(g_{1}) = x^{2}, in(g_{2}) = xy. The gcd of both is x. So, working along the Buchberger algorithm, we calculate in(g_{2})/x · g_{1} − in(g_{1})/x · g_{2} = −xy^{2}. The leading term xy^{2} of this is divisible by in(g_{2}). So we continue with −xy^{2} = y^{3} − yg_{2}. The leading term y^{3} of the remainder y^{3} is not divisible by either of the leading terms we have. By the Buchberger criterion, {g_{1}, g_{2}} was not Gröbner. Set g_{3} = y^{3}. Note that g_{3} = yg_{2} + (yg_{1} − yg_{2}). This produces a syzygy

τ_{1,2} = (yε_{1} − xε_{2}) + (yε_{2}) + (−ε_{3}).

As g_{1}, g_{3} are coprime monomials, the corresponding syzygy τ_{1,3} has the form y^{3}ε_{1} − x^{2}ε_{3}. If we start the Buchberger algorithm with g_{2}, g_{3}, we find g_{2} · in(g_{3})/y−g_{3}·in(g_{2})/y = y^{2}(xy+y^{2})−xy^{3} = y^{4} = y·g_{3} as gcd(in(g_{2}), in(g_{3})) = y. The syzygy τ_{2,3} is therefore (y^{2}ε_{2} − xε_{3}) − yε_{3}.

So if we take the basis G = {g_{1,... ,3}}, all the σ_{i,j} have a division algorithm that leads to 0. The Buchberger criterion asserts, that G is Gröbner. Furthermore, the syzygies τ_{1,2}, τ_{1,3}, τ_{2,3} generate the syzygies on G.

It could be that we are intersted in the syzygies on the originally given g_{1}, g_{2}. Then, as in the Euckidean algorithm, we work back to where we started: τ_{1,2} gives an expression for g_{3}, so ε_{3} −(yε_{1} −xε_{2})+yε_{2} is the key. We substitute into τ_{1,3} to get y^{3}−x^{2}(yε_{1}−xε_{2})−x^{2}yε_{2} = ε_{1}(y^{3}−x^{2}y)+ε_{2}(x^{3}−x^{2}y). If we substitute in τ_{2,3}, we get y^{2}ε_{2} − (x + y)(yε_{1} − xε_{2}) − (x + y)yε_{2} = ε_{1}(−xy − y^{2}) + ε_{2}x^{2}.

These two will be the generators for the syzygies on g_{1}, g_{2}. Note that after the replacement of g_{3}, we have τ_{1,3} = (x−y)τ_{2,3}, which is no surprise considering the fact that k[x, y] is UFD and x^{2} and xy + y^{2} are relatively prime.

4 The command ‘modulo’

Over the ring R, we might want to look at the syzygies of generators of modules. A module inside a free module is a matrix of generators for the module. Another way of describing a module is a matrix that if interpreted as a map between free R-modules has the module in question as cokernel. (For the same module M these two matrices will be completely different.) By abuse of notation we will sometimes say that a given matrix I is a module I, namely the module generated by the columns, and that it represents the module which is the cokernel.

Let I ∈ R^{n,r}, n the rows, r the columns. A syzygy on I is then a ‘vector’g ∈ R^{r} where ∑ g_{i}f_{i} = 0 and f_{i} are the columns of I. So the syzygy module on I is the kernel of R^{r} → R^{n} given by e_{i} → I · e_{i} as matrix product.

Definition 4.1 Let a, b be two matrices over R, a ∈ R^{m,r}, b ∈ R^{m,s}. Then we define mod (a, b) to be any matrix over R the columns of which represent the kernel of the map R^{r} ⟶^{a} R^{m} ⟶^{nat} R^{m}/ im(b) where im(b) is the module spanned by the columns of b in R^{m}.

In what is to follow, we will set R = S/(h_{1}, …, h_{p}), S = k[x_{1}, …, x_{n}] as usual.

Proposition 4.2 mod (a, b) is computable.

Proof. Write H = (h_{1}, …, h_{p}), so R = k[x_{1}, …, x_{n}]/H. Suppose a ∈ R^{m,r}, b ∈ R^{m,s}. Then let A, B be two matrices over S the image of which under the natural projection S → R are a and b respectively. Let C = H^{m} which is a submodule of

S^{m} given by an m × mp matrix of the form ( \begin{matrix}(H) \\ & (H) \\ & & (H) \end{matrix} ). Write D =

(A|B|C). This will live inside S^{m,r+s+mp}. Now compute a Gröbner basis for the syzygies on the columns of D. Write syzygies as (α_{1}, …, α_{r}, β_{1}, …, β_{s}, γ_{1}, …, γ_{mp})^{T}.

Claim: Each such syzygy corresponds to a column vector (α_{1}, …, α_{r})^{T} which generate the kernel of a : R^{r} → R^{m}/ im(b).

This is, because if (α_{1}, …, α_{r}, β_{1}, …, β_{s}, γ_{1}, γ_{mp})^{T} is a syzygy, then

A · (α_{i}) + B · (β_{i}) + C · (γ_{i}) = 0,

that is, A · (α_{i}) ∈ im(B) + im(C) which in turn implies A · (α_{i}) = 0 in S^{m}/(im(B) + im(C)) = R^{m}/ im(B) as R^{m} = S^{m}/H^{m}.

On the other hand, if α ∈ ker(a), then A · α = 0 in S^{m}/(im(B) + im(C)). That is, this α will come from one of the syzygies we computed for the columns of D. In other words, mod (a, b) is given by the vectors which are the first r coordinates of the generators for the syzygy module syz(D). □

We will now list a couple of applications of the command modulo.

• ∩

Let R be an affine ring, I, J, K ideals in R. Let L = I∩J ∩K. (That we use 3 ideals is of no importance, it is only an example. Any finite number is fine.) Then L will be the kernel of the map ϕ : R → R/I⊕R/J ⊕R/K, 1 → (1 + I) ⊕ (1 + J) ⊕ (1 + K). We will calculate this kernel. Let I, J, K be generated by the elements of the row vectors a, b, c. Then consider the matrix

a ⊕ b ⊕ c = ( \begin{matrix}( a ) & 0 & 0 \\ 0 & ( b ) & 0 \\ 0 & 0 & ( c ) \end{matrix} ) .

Then M = mod (( \begin{matrix}1 \\ 1 \\ 1 \end{matrix} ) , a ⊕ b ⊕ c) is a matrix the columns of which

generate the kernel of ϕ. In other words, they generate L.

• :

Let I = (a_{1}, …, a_{r}), J = (b_{1}, …, b_{s}) be two ideals. Then the quotient ideal I : J will be equal to ker(R → R/I ⊕ ⋯ ⊕ R/I) (s copies) where 1 → (b̄_{1}, …, b̄_{s}). This is because the kernel of that map will consist exactly of all the ring element which annihilate (I + J/I). This way of writing I : J makes application of mod possible: we only have to take the columns of the matrix mod (b^{T} , a ⊕ a ⊕ ⋯ ⊕ a) where the direct sum symbol operating on a has the same meaning as in the preceeding item.

• ann

Let M be a module which is represented by the matrix a, that is it is the cokernel of the map induced by a between the free modules F and G, a : G → F. Then ann(M) = 0 :_{R} M. For example, suppose we take

a = ( \begin{matrix}x & y & z \\ y & z & x \end{matrix} ). Then f is in the annihilator of M, if fe_{1} = 0, fe_{2} = 0

where e_{1} = \binom{1}{0} + im(a), e_{2} = \binom{0}{1} + im(a). The funny thing is, that we may express this annihilator as the syzygies on the matrix

( \begin{matrix}1 & x & y & z & 0 & 0 & 0 \\ 0 & y & z & x & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & x & y & z \\ 1 & 0 & 0 & 0 & y & z & x \end{matrix} )

Namely, each syzygy on this matrix (on the columns of course) will in its first coordinate have an element that annihilates e_{1} because of the two first lines and annihilates e_{2} beacuse of the two lowest lines. So annihilates M. On the other hand, each ring element that annihilates M will give rise to a syzygy, as each such element multiplied with e_{1} and e_{2} has to give something in im(a).

We will try to explain this a bit. (Emphasis on try.) The first column of the above matrix is a map that sends R into the module R^{2} ⊗ R^{2}^{∗} where the star means the dual module to R^{2}. In particular, it is some kind of a diagonal map, only that all the places where a 1 occurs correspond to e_{i} ⊗ e_{i}^{∗}. The reason for this is the canonical isomorphism hom_{R}(M, N) ≅ M^{∗} ⊗ N. We have the identity map on R^{2} which will correspond to the element f_{1} ⊗ e_{1} + f_{2} ⊗ e_{2} in R^{2}^{∗} ⊗ R^{2} where f_{1}(r, s) = r, f_{2}(r, s) = s. This, if carried out will of curse give the identity matrix. Now then we look at the map that carries R into R^{2} ⊗ R^{2}^{∗} which maps 1 onto this identity matrix. The map will have to be a 4 × 1 matrix, and exactly the first column of our matrix. We call this diagonal map Δ.

The other stuff in that matrix will be the matrix which represents the map a ⊗ id_{R^{2}}^{∗} which will be of type 2 · 2 × 3 · 2. After all this setup, the annihilator of M (which was the cokernel of a) will consist of all elements of R which if multiplied with any of the generators for R^{2} are in the image

of a. As the generators of R^{2} will show up as (\begin{matrix}( & {1 \atop 0} & ) \\ ( & {0 \atop 1} & ) \end{matrix}) in the first column of

our matrix, ann(M) = mod (Δ, a ⊗ id_{R^{2}}^{∗}).

In general, for a : G → F, Δ : R → F ⊗ F^{∗}, A ⊗ F^{∗}G ⊗ F^{∗} → F × F^{∗} we have ann(coker(a)) = mod (Δ, a ⊗ F^{∗}) where F^{∗} stands for the identity map on F^{∗}.

• Ext

Note first that syz(d) = mod (0, d) for any matrix d. Let M be the cokernel of the matrix d = d_{1} : F_{1} → F_{0}. Then we define inductively matrices d_{i} : F_{i} → F_{i−1} which are maps between finite free modules, each time just using the syzygies on the kernel of the preceeding map. Let us say that d_{i} is a r_{i−1} × r_{i} matrix. So we find a resolution for M of the form

→ F_{j+1} ⟶^{d_{j+1}} F_{j} → … → F_{1} ⟶^{d_{1}} F_{0} → M → 0.

Then by definition, Ext^{j}(M, R) = ker(hom(d_{j+1}, R))/ im(hom(d_{j}, R)). Let us just write ker d_{j+1}^{∗}/ im d_{j}^{∗} for that. d_{i+1}^{∗} will be the adjoint of d_{i+1} and therefore of size r_{i+1} × r_{i}.

Now by definition of a syzygy, d_{j+1}^{∗} ∘ syz(d_{j+1}^{∗}) = 0. Let syz(d_{i+1}) be of size r_{j} × s_{j+1}. Set X = mod (syz(d_{j+1}^{∗}), d_{j}^{∗}). This will be the collection of columns that generate the kernel of

\xymatrix{R^{s_{j+1}} \ar[r]^{syz(d_{j+1}^{∗})} & F_{j}^{∗}/ im(d_{j}^{∗})}

Say X is of type s_{j+1} × t_{j+1}. I claim that X represents ker(d_{j+1}^{∗})/ im(d_{j}^{∗}), that is, R^{s_{j+1}}/X is isomorphic to Ext^{j}(M, R).

The proof goes as follows. Let v ∈ R^{s_{j+1}}. Then syz(d_{j+1}^{∗}) takes v into R^{r_{j}} = F_{j}. As the image of v will be a linear combination of syzygies on d_{j+1}^{∗}, it is in ker(d_{j+1}^{∗}). Let now v be specifically in the module generated by the columns of X. By definition of X and the modulo command, syz(d_{j+1}^{∗}) carries v into im(d_{j}^{∗}). Hence

syz(d_{j+1}^{∗}) : R^{s_{j+1}}/ mod (syz(d_{j+1}^{∗}), d_{j}^{∗}) ⟶ ker(d_{j+1}^{∗})/ im(d_{j}^{∗})

is induced.

By definition of modulo again, all the elements which go into im(d_{j}^{∗}) under syz(d_{j+1}^{∗}) are exactly X. So the above map is injective.

Finally, the map syz(d_{j+1}^{∗}) : R^{s_{j+1}} → ker(d_{j+1}^{∗}) is surjective because any kernel element will be a syzygy on d_{j+1}^{∗} which is exactly the stuff in the span of syz(d_{j+1}^{∗}). As the induced map is from a quotient of R^{s_{j+1}} to a quotient of ker(d_{j+1}^{∗}), it will be surjective too.

• hom

Let a : F_{1} → F_{0}, b : G_{1} → G_{0} be maps between free modules. Set M = coker(a), N = coker(b). Suppose we are interested in hom(M, N). Again we will visualize hom(M, N) as M^{∗} ⊗ N.

Any f ∈ F_{0}^{∗} ⊗ G_{0} is a map from F_{0} into G_{0} and hence one into N. So we have λ : F_{0}^{∗} ⊗ G_{0} → hom(F_{0}, N) working by ∑ f_{i}^{∗} ⊗ g_{i} → (x → ∑ g_{i} · f_{i}^{∗}(x) + im(b)). As F_{0} is projective, this is surjective.

Consider now the subset of F_{0}^{∗} ⊗ G_{0} for which if F_{0} ∋ x = a · y, y ∈ F_{1}, then ∑ f_{i}^{∗}(ay)g_{i} ∈ im(b). As a is a map from F_{1} to F_{0}, the interpretation hom(a, G_{0}) ≅ a^{∗} ⊗ G_{0} turns that into

(a^{∗} ⊗ G_{0}) ∘ (∑ f_{i}^{∗} ⊗ g_{i}) (y) ∈ im(b)

for all y ∈ F_{1}. So we are looking for all those maps which send all F_{1} into im(b) if combined with a. (The things we are trying to determine are the ∑ f_{i}^{∗} ⊗ g_{i}!) So we are talking about mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b). This is because any map that sends F_{1} completely into im(b) will have to factor through b by projectivity of F_{1}.

Any element in mod (a^{∗} ⊗G_{0}, F_{1}^{∗} ⊗b) will induce a map F_{0} → G_{0}/ im(b) under which im(a) is annihilated. We restrict the map λ from above to X = mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b). As im(a) dies under each element of X, we get induced maps between M and N. So we have

λ : mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b) → hom(M, N).

This map is surjective as the following shows. Let f : M → N be given. So we have F_{0} ⟶^{π} F_{0}/ im(a) ⟶^{f} G_{0}/ im(b). This will complete to

\xymatrix{ & F_{0} \ar@{-->}[dl]_{g} \ar[d]^{f∘π} \\ G_{0} \ar[r]_{π′} & G_{0}/ im(b) \ar[r] & 0}

so that the diagram commutes by projectivity of F_{0}.

But π′g = fπ implies π′g(im(a)) = 0 or in other words g(im(a)) ⊆ im(b). By isomorphy of F_{0}^{∗} ⊗ G_{0} and hom(F_{0}, G_{0}) there is corresponding to g a F_{0}^{∗} ⊗ G_{0} ∋ ∑ f_{i}^{∗} ⊗ g_{i} with (a^{∗} ⊗ G_{0})(∑ f_{i}^{∗} ⊗ g_{i}) mapping into im(F_{0}^{∗} ⊗ b) which therefore is in mod (a^{∗} ⊗ G_{0}, F_{0}^{∗} ⊗ b).

Let us look for the kernel of λ_{|_{X}}. ∑ f_{i}^{∗} ⊗ g_{i} ∈ ker λ ⟺ ∑ f_{i}^{∗}(x)g_{i} ∈ im(b)∀x ∈ F_{0}. To say that the image of ∑ f_{i}^{∗} ⊗ g_{i} is contained in the image of b is the same as saying that ∑ f_{i}^{∗} ⊗ g_{i} is in the image of F_{0}^{∗} ⊗ b (as F_{1} is free).

Therefore, ker λ ∩ X is generated by the columns of mod ( mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b), F_{0}^{∗} ⊗ b).

Now we do a little detour. Let X, Y be submodules of free modules and F → X be a surjection where F is free. Then this map is given by the matrix X. Then mod (X, Y) = ker(F ⟶^{X} X → (X + im(Y))/ im(Y)). Then clearly F/ mod (X, Y) will be isomorphic to X modulo the image of mod (X, Y) in X. Note that mod (X, Y) is a map from another free module G into F.

This being true for all modules, we have shown that hom(M, N) is isomorphic to mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b) modulo mod ( mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b), F_{0}^{∗} ⊗ b), which is isomorphic to the cokernel of mod ( mod (a^{∗} ⊗ G_{0}, F_{1}^{∗} ⊗ b), F_{0}^{∗} ⊗ b).

As we can construct modulo for any two finite matrices of correct size, ann, ∩, :, Ext^{j}(−, R) and Hom are all constructible.

5 Miscellanea

The first results in this section will have to do with elimination theory.

Fact 5.1 Let ϕ : A → B be a ring map. Write A = k[x_{1}, …, x_{m}]/I, B = k[y_{1}, …, y_{n}]/J. Let ϕ be given by ϕ(x̄_{i}) = f_{i}(y) + J. Then ker ϕ is generated by

(J + (x_{1} − f_{1}(y), …, x_{m} − f_{m}(y))) ∩ k[x_{1}, …, x_{m}].

(That is clear as the x_{i} are indeterminates.)

Definition 5.2 Let S = k[x_{1}, …, x_{r}], T = S[y_{1}, …, y_{s}]. A monomial order on T satisfies the elimination property with respect to y_{1}, …, y_{s}, if

E: If f ∈ T and in(f) ∈ S then f ∈ S.

The easiest examples for orders satisfying E are

• the lexicographic order with respect to any initial set of variables,

• any order which is the refinement of the partial order corresponding to total degree in x refined by any monomial order will satisfy E with respect to y where x means {x_{i}} and y means {y_{j}}.

Proposition 5.3 Let > be monomial on T and suppose that > has property E with respect to y. If I ⊆ T is an ideal then, with respect to the order on S induced by >, we have

in(I ∩ S) = in(I) ∩ S.

Furthermore, if g_{1}, …, g_{t} is Gröbner for I, and g_{1}, …, g_{u} are those g_{i} which do not involve y, then g_{1}, …, g_{u} is Gröbner for I ∩ S.

Proof. Set J = I ∩ S. Clearly in(J) ⊆ in(I) ∩ S. We will show that the in(g_{i}) for i ≤ u generate in(I) ∩ S. By lemma 2.3 this will prove both statements.

Suppose, m ∈ in(I)∩S. As the g_{i} are Gröbner for I, m is a multiple of in(g_{i}) for some i ≤ t. As m ∈ S, we must have in(g_{i}) ∈ S. By elimination property, g_{i} ∈ S whence i ≤ u as required. □

Remark 5.4 This generalizes to submodules of free modules and their restriction to modules over S.

There are more applications of elimination like things, like projective closure, ideal at infinity and such like. But they seem to be special results rather than general ideas. (To me anyway.)

Now we turn to the Hilbert function and related topics. To get started, we cite a theorem in [1].

Theorem 5.5 Let S = k[x_{1}, …, x_{n}] be a graded rind, M a finitely generated S-module. Then there is a Hilbert series

∑_{−∞}^{∞} dim_{k}(M_{d})t^{d} = H_{M}(t)

which satisfies

• H_{M}(t) = P (t)/(1−t)^{n} where the n is the number of indeterminates above, and P is a polynomial,

• for i ≥ d where d is the multiplicity of 1 in P, dim_{k}(M_{d})i^{d} = h_{i} where again h is a polynomial,

Now we can prove our own things about Hilbert functions.

Theorem 5.6 Let S be as before. Assume further that > is a monomial order on S^{m}, and let I be a graded submodule of S^{m}. Then

dim_{k}(S^{m}/I)_{d} = dim_{k}(S^{m}/ in(I))_{d}.

Proof. We have (S^{m}/I)_{d} ≅ (S^{m})_{d}/I_{d} as k-vspaces, and similarly (S^{m}/ in(I))_{d} ≅ (S^{m})_{d}/(in(I))_{d} as vspaces. The theorem of Macaulay 1.13 asserts that the monomials of (S^{m})_{d} which are not in (in(I))_{d} are a vspace basis for (S^{m})_{d}/I_{d}. But clearly the same is true for (S^{m})_{d}/(in(I))_{d}. So the spaces have the same dimension. □

The above result shows that we should try to find a way to determine the Hilbert function of quotients of S^{m} by monomial modules. But, as h_{i}(M) + h_{i}(S^{m}/M) = h_{i}(S^{m}) = mh_{i}(S), it is enough to do it for monomial submodules. A monomial submodule is generated by monomials m_{j}e_{i}^{j} where e_{i}^{j} = e_{i}. Then if I = ∑ Sm_{j}e_{i}^{j}, I decomposes into the direct sum ⨁ I_{i}e_{i} where I_{i} = ∑ Sm_{i}^{j}. For the Hilbert function this implies that H_{I}(t) = ∑ H_{I_{i}}(t). Therefore it is enough to find Hilbert functions of monomial ideals.

As 0 → J → S → S/J → 0 exact shows that H_{S}(t) − H_{S/J}(t) = H_{J}(t) and H_{S}(t) = \binom{t+n}{n}, it is enough to deal with H_{S/J} for monomial J. Let J be generated by m_{1}, …, m_{γ}. Pick a minimal generator m^{∗} among these, with degree δ (Remember that we have an order > on S!) Suppose m^{∗} = m_{γ}. Set J′ the ideal generated by {m_{1}, …, m_{γ}} \ m^{∗}. Then

\xymatrix{S(−d) \ar[r]^{ϕ} & S/J′ \ar[r]^{ψ} & S/J \ar[r] & 0 \\  & s + J′ \ar[r] & s + J \\ s \ar[r] & s · m^{∗} + J′}

will be exact as

ker ψ = J/J′  (6)
= {s + J′ : s ∈ J}  (7)
= {s + J′ : s = j′ + s′ · m^{∗}}  (8)
= im(ϕ).  (9)

Furthermore,

ker(ϕ) = {m ∈ S : mm^{∗} ∈ J′}  (10)
= ∑_{i=1}^{γ−1}(m_{i}S : m^{∗}S)  (11)

= ∑_{i=1}^{γ−1} (m_{i}/ gcd(m_{i}, m^{∗})) · S =: K  (12)

by 1.2. NOTE, that J′ and K are generated by fewer elements than J. We have constructed exact

\xymatrix{0 \ar[r] & (S/K)(−d) \ar[r] & S/J′ \ar[r] & S/J \ar[r] & 0 \\  & s + K \ar[r] & sm^{∗} + J′ \\  &  & s + J′ \ar[r] & s + J}

Hence H_{S/J}(t) = H_{S/J′}(t) − H_{S/K}(t − d). Reductively we only have to (keep track of the confusion and) be able to find Hilbert functions of principal monomial ideals. But

H_{S/mS}(t) = \{ \begin{matrix}H_{S}(t) & : & t < deg m, \\ H_{S}(t) − H_{S(−d)}(t) = H_{S}(t) − H_{S}(t − d) & : & t ≥ deg m, \end{matrix}

as mS ≅ S(d) as graded modules.

Now we will say some things concerning regular sequences, generic coordinates and flat families.

Lemma 5.7 Let F be a graded finite free S-module ordered by revlex. Let M ⊆ F be a homogeneous submodule (that is, M is generated by products of basis elements of F and homogeneous polynomials in S = k[x_{1}, …, x_{n}]). Then for all 1 ≤ s ≤ n we have

in(M + (x_{s+1}, …, x_{n})F) = in(M) + (x_{s+1}, …, x_{n})F.

Proof. The inclusion ⊇ is clear. Let f ∈ M + (x_{s+1}, …, x_{n})F be a homogeneous element. We prove that in(f) ∈ in(M) + (x_{s+1}, …, x_{n})F. Write f = g + h, g ∈ M, h ∈ (x_{s+1}, …, x_{n})F in such a way that the largest monomial of g that involves x_{s+1}, …, x_{n} is minimal. If in(f) ≠ in(g), then all terms of g are < in(f). As there has to be an initial term in(g), they are < something in (x_{s+1}, …, x_{n})F. By revlex definition and homogeneity of g, any monomial in g will have to contain a factor in {x_{s+1}, …, x_{n}}. But that contradicts the choice of g, unless g = 0. In that case f ∈ (x_{s+1}, …, x_{n})F.

Otherwise we have in(g) = in(f) ∈ in(M) which proves in(M+(x_{s+1}, …, x_{n})F) ⊆ in(M) + (x_{s+1}, …, x_{n})F.

By 2.3, we are done. □

Theorem 5.8 Let F be finite free with a monomial order, possibly different from revlex. Let G = {in(g_{1}), …, in(g_{t})} minimally generate N ⊆ F. Put M = ∑ g_{i} · F.

i) If x_{s}, …, x_{n} are mising in {in(g_{i})}, then x_{s}, …, x_{n} is a regular sequence on F/N and conversely.

ii) If x_{s}, …, x_{n} is a regular sequence on F/N, and G is Gröbner for M, then x_{n}, …, x_{s} (reverse order!) is a regular sequence on F/M.

iii) If G is Gröbner and homogeneous and the order is revlex, then the converse of ii) is true. (Note that a reduced Gröbner basis at the same time is Gröbner for M and minimally generating in(M) with its leading terms.)

Proof.

i) If x_{s}, …, x_{n} are missing in the leading terms of G, then x_{s+i} · ḡ = ¯0 for ḡ = g + (x_{s}, …, x_{s+i−1})F + N means x_{s+i} · g = ∑_{j=s}^{s+i−1} x_{j}f_{j} + n. For any basis element e_{k} of F this means that the component of gx_{s+i} is in (N_{k} +(x_{s}, …, x_{s+i−1})F), N_{k} being the module {he_{k} : ∃n ∈ N, n_{k} = h} = N ∩ Se_{k} by monomiality of N. For the initial term of x_{s+i}g this means that in(∑_{j=s}^{s+i−1} x_{j}f_{j} + n) is divisible by x_{s+i}. As neither of the x_{j} nor any of the in(g_{j}) are, the f_{j} responsible for the initial term of g splits into x_{s+i} · f′. In other words, g − x_{j}x_{s+i}f′ will belong to the same coset as g. So we can choose g to have smaller initial term than we did. Repeat to get a g without initial term (remember 1.9). So the sequence is regular.

On the other hand, if x_{s}, …, x_{n} is a regular sequence on F/N and in(g_{i}) = x_{j} · m for some i larger than s, by definition of regular we must have m ∈ N + (x_{s}, …, x_{j−1})F. Let us pick amongst all such cases a minimal j. Then m cannot have divisors x_{k} for s ≤ k ≤ j − 1. Hence m ∈ N. But then m has to be a multiple of one of the in(g_{k}) where k ≠ j obviously. This contradicts the minimality assumption to the generating set in(G) for N.

ii) Suppose ∑_{i=s}^{n} x_{i}f_{i} ∈ M. Proceeding by induction, it is our goal to show f_{s} ∈ M + (x_{s+1}, …, x_{n})F. G generates M. We replace each f_{i} by a remainder with respect to g_{1}, …, g_{t}. That will not change the above relation. Then no term of any f_{i} is divisible by any of the in(g_{j}). As x_{n}, …, x_{s} is regular on F/N by hypothesis, no term of any x_{j}f_{j} is in N = in(M) as the x’s are missing in in(G) by i). Since ∑_{i=s}^{n} x_{i}f_{i} ∈ M, the leading term will be in in(M). That is only possible if there is no leading term, hence ∑ x_{j}f_{j} = 0. Since x_{n}, …, x_{s} is a regular sequence on F, f_{s} ∈ (x_{s+1} …, x_{n})F which is what we wanted to prove.

iii) If s = n, we have to show x_{n}f ∈ N ⇒ f ∈ N. Assume f is a monomial. x_{n}f = in(g) for some g ∈ M. Then in(g) ∈ x_{n} · F. As g is homogeneous and revlex we find g ∈ x_{n}F, so g = x_{n}h. Since x_{n} is not a zerodivisor on F/M, h ∈ M. Hence f = in(h) ∈ in(M).

For general f, x_{n}f ∈ in(M) implies x_{n} · f̂ ∈ in(M) for any monomial of f by monomiality of in(M). So we are done.

If s < n, we use the preceding lemma for induction. Let j ∈ {n, …, s}. Assume x_{j}f_{j} ∈ N + (x_{n}, …, x_{j+1})F. Since N and (x_{n}, …, x_{j+1})F are monomial, we may assume that f_{j} is a monomial. Then x_{j}f_{j} ∈ in(M) + (x_{n}, …, x_{j+1})F = in(M +(x_{n}, …, x_{j+1})) by the lemma. So x_{j}f_{j} = in(g) with g ∈ M + (x_{n}, …, x_{j+1})F. g can be chosen homogeneous as the module it lives in is homogeneous by hypothesis. By revlex and x_{j}f_{j} = in(g), we conclude that g contains only monomials which are multiples of x_{j}, …, x_{n}. Using (x_{n}, …, x_{j+1})F we may reduce g to a multiple of x_{j}. g = x_{j}f̃, g ∈ M + (x_{n}, …, x_{j+1})F. By hypothesis x_{j} is regular on F/(M + (x_{n}, …, x_{j+1})F), so f̃ ∈ M + (x_{n}…. , x_{j+1})F. Hence f = in(f̃) = in(g)/x_{j} ∈ N + (x_{j+1}, …, x_{n})F so that x_{n}, …, x_{s} is a regular

sequence on F/N by induction. By part i, we are done. □

Note 5.9 In the theorem we just proved, it might happen that x_{s}, …, x_{n} is not regular on F/M though other sequences might. But, if there is any sequence regular of (maximal) length n − s + 1 at all, then after a transformation in Gl_{n}(k), the given one will be of the form in the theorem. Moreover, the set of such sequences is an orbit of an open subset of Gl_{n}(k) in the Zariski topology. Such things are called “generically true”.

Given an ideal I = (f_{1}, …, f_{s}) in S together with the quotient ring S/I = R, it happens that people want to look at the associated graded ring gr_{𝔪}(R) = ⨁_{ℕ} (𝔪^{n}R/𝔪^{n+1}R). It appears that the associated graded ring is related to the ideal in S generated by the pieces of lowest degree in I, which I do not know. Let us call this lowest pieces ideal to I the bottom ideal of I, written I_{bottom} or I′. For an element f ∈ S we write F for the homogenized version of it with respect to the new variable x_{0}. If I = (f_{1}, …, f_{s}) then I^{∗} = (F_{1}, …, F_{s}).

Using this notation we have

Proposition 5.10 Let {G_{1}, …, G_{t}} be Gröbner for (F_{1}, …, F_{s}) ⊆ S[x_{0}] with respect to some order that refines the partial order on monomials defined by deg_{x_{0}}. We set g_{i} = G_{i}(1, x_{1} …, x_{n}). Then I′ = (g_{1bottom}, … g_{tbottom}).

Proof. Let g ∈ I. Then we have to show, that g_{bottom} ∈ (g_{1bottom}, …, g_{tbottom}). Write g = ∑ p_{i}f_{i}. Let G, P_{i}, F_{i} be the homogenized polynomials. So x_{0}^{a}G = ∑ P_{i}F_{i} ∈ (F_{1}, …, F_{s}), a ∈ ℕ. Then G is going to be x_{0}^{b}g_{bottom}+ things of lower degree in x_{0} and higher in other variables. As the G_{i} are Gröbner for I^{∗}, we must have x_{0}^{a}G = ∑ Q_{i}G_{i} and in(Q_{i}G_{i}) ≤ in(x_{0}^{a}G) for all i, Q_{i} ∈ S[x_{0}]. But then, g_{bottom} must be the sum of products of highest degree in x_{0} in Q_{i}G_{i}. (By homogeneity, such are the things with lowest degree otherwise.) Setting x_{0} = 1 in those, we get g_{bottom} as the sum of products of lowest degree in Q(1, x_{1}, …, x_{n})G_{i}(1, x_{0}, …, x_{n}) which we write as q_{i}g_{i}, which are in (g_{1bottom}, …, g_{tbottom}). □

Let us take a look at projective closures. Given I ⊆ S let I_{s}^{∗} be the ideal generated by the s-homogenizations of the elements of I with respect to the first s variables in S with the additional variable x_{0}. (E.g, x^{3} +z → x^{3} +x in I_{1}^{∗} and I_{2}^{∗} but to x^{3} + w^{2}z in I_{3}^{∗}.) The ideal I_{s}^{∗} corresponds to the closure of V (I) in ¶^{s} × A^{n−s}. Let I_{s}^{∞} be the ideal generated by the pieces of maximal s-degree in each element in I. (That is, degree in the first s variables only counts.)

Proposition 5.11 Let T = S[x_{0}]. If > is a monomial order on T refining the partial order given by the s-degree, S inherits the given order on T as subring, and g_{1}, …, g_{t} is Gröbner for I ⊆ S,

i) I_{s}^{∗} is generated by the s-homogenizations of the g_{i};

ii) I_{s}^{∞} is generated by the g_{i⊤} which are the pieces of g_{i} which have maximal degree in x_{1} …, x_{s}.

Proof. Note that in(g_{i}) = in(g_{i}^{∗}).

i) Let ^{∗} denote the homogenization with respect to the s-degree. We have to show that g_{1}^{∗}, …, g_{t}^{∗} generate the initial ideal of I_{s}^{∗}. Let m ∈ in(I_{s}^{∗}). Then m = in(f), f ∈ I_{s}^{∗}. Set f̃ = f(1, x_{1}, …, x_{s}) ∈ I. So f̃ = ∑ h_{i}g_{i}, where we assume that this is a standard expression (with remainder 0 as the g_{i} are Gröbner.) . Then the top degree term in f̃ is the same as in f since f̃ is the homogenization of f. (Or maybe it is an x_{0}-multiple.) In either case, in(f) = in(f̃) · x_{0}^{a} = in(h_{i}) in(g_{i})x_{0}^{a} for one of the i in the sum.

ii) Let f ∈ I, f = ∑ h_{i}g_{i} be a standard expression. Then deg_{s}(f) ≥ deg_{s}(g_{i}h_{i}) for all i. We restrict the sum to all the i for which we have equality above. Then f_{⊤} = ∑′ h_{i⊤}g_{i⊤} which is what we wanted to prove.

□

Gröbner bases may also be used to calculate saturations of modules with respect to given ideals.

Proposition 5.12 Let > be monomial on finite free F over T = S[y]. Assume that for all homogeneous (usual sense) f ∈ T we have that in(f) ∈ (y) ⇒ f ∈ (y) (e.g, revlex with y as smallest). If g_{1}, …, g_{t} is homogeneous Gröbner for M ⊆ F, and g_{i} is divisible by y^{ε_{i}}, 1 ≤ i ≤ t, but no higher power, then {g_{i}/y^{ε_{i}}} is Gröbner for M : y^{∞} = ⋃_{1}^{∞} M : (y^{j}).

Proof. By construction, g_{i}/y^{ε_{i}} is not in yM. Hence by the assumption on the order in F, in(g_{i}/y^{ε_{i}}) ∉ yM. So y^{ε_{i}} is the largest power dividing in(g_{i}).

Certainly g_{i}/y^{ε_{i}} ∈ (M : y^{∞}). Let y^{ε} · f ∈ M. Then in(y^{ε}f) = y^{ε} in(f) and since {g_{i}} are Gröbner, in(y^{ε}f) ∈ in(M) = (in(g_{1}), …, in(g_{t})). The latter module is monomial, so y^{ε} in(f) = m · in(g_{i}) for some i. By the UFD property in T, in(g_{i})/y^{ε_{i}} = in(g_{i}/y^{ε_{i}}) divides in(f) by the first paragraph.

We have shown that {in(g_{i})/y^{ε}} generate in(M : y^{∞}); by lemma 2.3 we are done. □

The last item in the list of possible applications of Gröbner bases, apart from the radical in the next section, is the lifting of give homomorphisms. Let

\xymatrix{ & G \ar[d]_{γ} \ar@{-->}[dl]_{ψ} \\ H \ar[r]^{ϕ} & F}

be given, where all modules are free and im(γ) ⊆ im(ϕ). The problem is to construct a lift G → H. Let g_{1}, …, g_{s} be the images of a chosen basis for H. Make it a Gröbner basis using the Buchberger algorithm. Let us say, h_{1}, …, h_{t} is the result of that. Then define H′ = ⨁_{1}^{t} Se_{j} be the free module over this generating set and define also ϕ′ : H′ → F by ϕ′(e_{j}) = h_{j}.

When we do the Buchberger process in order to make a Gröbner basis out of the g_{i}, we get expressions for the h_{j} in terms of the g_{i}. So there is a map α : H′ → H, ϕ′ = ϕ ∘ α by sending the preimage of h_{i} (namely e_{i}) to the same linear combination, only using elements in H instead of F. That is, h_{j} = ∑ β_{i}^{j}g_{i}, then e_{j} → ∑ β_{i}^{j}χ_{i} where χ_{i} is a preimage for g_{i} under ϕ, the element in our chosen basis for H mapping to g_{i}.

Now for ε_{i} ∈ G, we use the division algorithm to find γ(ε_{i}) = ∑ p_{j}h_{j}, which works because im γ ⊆ im ϕ. Then we finally define ψ′ : G → H′ by ε_{i} → ∑ p_{j}e_{j}. This gives ϕ′ ∘ ψ′ = γ. Then α ∘ ψ′ will be the map we wanted. Picture:

\xymatrix{H′ \ar@{-->}[d]_{α} \ar[dr]^{ϕ′} & G \ar[d]^{γ} \ar@{-->}[l]_{ψ′} \\ H \ar[r]_{ϕ} & F}

6 The Radical of an Equidimensional Ideal

For this section we will assume that k is algebraically closed and of characteristic zero.

We assume that I is equidimensional. That is, all prime divisors of I have the same dimension d. We set R = S/I and write I = (f_{1}, …, f_{r}). To these functions we look at the Jacobian

J(f) = ((∂f_{j}/∂x_{i}))

which will have size n × r. We define I ⊆ J_{d}(I) ⊆ J_{d+1}(I) ⊆ … ⊆ J_{n}(I) ⊆ S by J_{a}(I) = I + Δ_{a}(J(f)), Δ_{a}(J(f)) being all (n − a) × (n − a)-minors of J(f). We set dim(J_{n}(I)) = −1.

Definition 6.1 P ∈ A^{n} is called singular, if f(P) = 0∀f ∈ J_{d}(I). That is, rank(J) < n−d in P. We call Z(J_{d}(I)) the singular locus of I, denoted sing(I), sing(R) or sing(R/I).

Proposition 6.2 Let I = ⋂_{1}^{s} Q_{i} be a irredundant primary decomposition for the equidimensional ideal I. Let X = Z(I), X_{i} = Z(Q_{i}) as ringed spaces. Then sing(X) = ⋃_{1}^{s} sing(X_{i}) ∪ ⋃_{i<j} X_{i} ∩ X_{j}.

Proof. sing(X) can be thought of to be the set of maximal ideals in R for which R_{𝔪} is not regular. So for I ⊆ 𝔪 we have either a unique index i : Q_{i} ⊆ 𝔪 which implies R_{𝔪} ≅ (S/Q_{i})_{𝔪} and so 𝔪 ∈ sing(X) iff 𝔪 ∈ sing(X_{i}), or there are at least two indices i, j : Q_{i} ⊆ m ⊇ Q_{j}, which means that R_{𝔪} cannot be a domain and hence not regular. □

Example 6.3 Let I = (x^{2}, xy, y^{2}) in k[x, y, z]. Then J_{1}(I) = I since it is I plus the 2×2-minors of the matrix

( \begin{matrix}2x & y & 0 \\ 0 & x & 2y \\ 0 & 0 & 0 \end{matrix} )

which are all in I. As dim(I) = 1, all points in Z(I) are singular.

Proposition 6.4 For equidimensional I we have I is radical if and only if dim(J_{d}(I)) < d.

Proof. R is reduced if and only if R_{P} is regular for all minimals primes of R and I has no embedded components.

Suppose first that I = \sqrt{(}I). Then dim(sing(X_{i})) < dim(X_{i}) = d since X_{i} is a variety. Furthermore, P_{i} ⊈ P_{j} for associated primes of I, and so X_{j} ⊈ X_{i}. Therefore dim(X_{i} ∩ X_{j}) < d which means that the singular locus of I has dimension less than d.

Now let dim(sing(I)) < d. I cannot have embedded components as it is equidimensional. Each X_{i} contains a nonsingular point by dimension argument. In such a point we will have regular R_{𝔪} isomorphic to (S/Q_{i})_{𝔪} since 𝔪 cannot contain any other P_{j} by the proof of the previous proposition.

By Serre’s theorem, ((S/Q_{i})_{𝔪})_{P_{i}} ≅ (S/Q_{i})_{P_{i}} is regular, and by the criterion stated above R is reduced. □

Lemma 6.5 I equidimensional, P a minimal prime to I, Q the P -component. Then the rank of J(Q) is equal to r on a generic subset of Z(P) if and only if (I + Δ_{n−r}J(I))(S/I)_{P} = (S/I)_{P} and (I + Δ_{n−r−1})(I))(S/I)_{P} ≠ (S/I)_{P}.

Proof. Since P is minimal, (S/I)_{p} ≅ (S/Q)_{P}. Also, Δ_{n−r}(J(I)) contains products of derivatives of I. In S_{P} we have I · S_{P} = Q · S_{P} and hence J(I)S_{P} = J(Q)S_{P}. Therefore

(I + Δ_{i}(J(I)))(S/I)_{P} = (Q + Δ_{i}(J(I)))(S/Q)_{P}  (13)
= Q + the (n − i) × (n − i) determinants of J(I) in (S/Q)_{P} (14)
= Q + the (n − i) × (n − i) determinants of J(Q) in (S/Q)_{P} (15)
= (Q + Δ_{i}(J(I)))(S/Q)_{P} .  (16)

So we may assume that I is P -primary, I = Q. Then as a set Z(I) is a variety. Such things are regular on a generic set. Let g = 0 describe sing(P). Then in each point T on D_{+}(g) we have (S/P)_{𝔪(T)} is regular of dimension dim P.

Let us now look at the second condition. (Q + Δ_{n−r}(J(Q)))(S/Q)_{P} = (S/Q)_{P} iff there is x ∈ Δ_{n−r}(J(Q)) such that \frac{x+Q}{1+Q} is a unit. This implies that x ∉ P. Therefore x ≠ 0 on Z(P) = Z(Q). So there has to be a r by r minor in the Jacobian for Q which is not in P. So J(Q) contains r rows that are linearly independent on a generic set in Z(P) which happens if and only if rank J(Q) ≥ r on a generic subset of Z(P).

Also, (Q + Δ_{n−r−1}(J(Q)))(S/Q)_{P} ≠ (S/Q)_{P} if and only if there exists y ∉ P with y · Δ_{n−r−1}(J(Q)) ⊆ P since anything is finitely generated and P is prime. In other words, if and only if Δ_{n−r−1}(J(Q)) ⊆ P. Or, if and only if all (r + 1) × (r + 1) minors of J(Q) are zero on Z(P) which happens if and only if the rank of J(Q) is at most r on any generic set in Z(P). □

Proposition 6.6 I equidimensional of dimension d. If dim(J_{a}(I)) = d and dim(J_{a+1}(I)) < d, then \sqrt{I} = \sqrt{I : J_{a}(I)}.

Proof. Since I : J_{a}(I) can only have fewer associated primes than I it will be enough to show that P ∈ ass(I) ⇒ P ∈ ass(I : J_{a}(I)).

I am not able to prove that. See the notes by Stillman (which don’t tell me anything.) □

Fact 6.7 If we use the notation of the preceeding proposition, in particular concerning a, then

• I ⊈ (I : J_{a}(I)) ⊆ \sqrt{I},

• dim(J_{a}(I)) < d ⇒ I : J_{a}(I) = I.

This is both more or less by construction.

Example 6.8 That dim(J_{a}(I)) = d does not imply \sqrt{I : J_{a}(I)} = \sqrt{I} as the following example shows. We really need dim(J_{a+1}(I)) < d. Namely, let in k[x, y, z] I be the zerodimensional ideal 𝔪^{3} = (x, y, z)^{3}. Then Δ_{3}(J(I)) ⊆ 𝔪^{3}, Δ_{2}(J(I)) = 𝔪^{4}, Δ_{1}(J(I)) = 𝔪^{2}, Δ_{0}(J(I)) = 𝔪^{0} = S, as the Jacobian

( \begin{matrix}3x^{2} & 2xy & 2xz & 0 & y^{2} & 0 & 0 & z^{2} & 0 & yz \\ 0 & x^{2} & 0 & 3y^{2} & 2xy & 2yz & 0 & 0 & z^{2} & xz \\ 0 & 0 & x^{2} & 0 & 0 & y^{2} & 3z^{2} & 2xz & 2yz & xy \end{matrix} )

of I shows. So J_{0}(I) = J_{1}(I) = 𝔪^{3}, J_{2}(I) = 𝔪^{2}, J_{3}(I) = S. So I : J_{0}(I) = I : J_{1}(I) = S, I : J_{2}(I) = 𝔪, I : J_{3}(I) = I, so that we don’t want to take d = 0 or 1.

Using what we have (not) proved so far, we may state

Algorithm 6.9 (The Radical of an Equidimensional Ideal) Input: I ⊆ k[x_{1}, …, x_{n}] equidimensional of dimension d. Output: \sqrt{I}. begin: Compute J_{d}(I). If dim J_{d}(I) < d, return I. Otherwise find minimal a with d ≤ a ≤ n − 1, dim J_{a}(I) > dim J_{a+1}(I). return: radical(I : J_{a}(I)).

end.

This will work, provided we can compute dimensions and quotients. Which we can by the sections 4 and 5. As I : J_{a}(I) ⊃ I, the process will terminate.

Note also, that if dim(J_{a}(I)) < d, then setting I_{1} = I : J_{a}(I) we find J_{a+1}(I) ⊆ J_{a+1}(I_{1}) which is to say that dim(J_{a+1}(I_{1})) < d and therefore the ”a” of I_{1} is smaller than the a for I since dim(I_{1}) = dim(I) by 6.6.

Let’s do two examples here.

Example 6.10 Let I = (f^{100}) with f square free. Then J_{n−1}(I) = I+Δ_{1}(J(I)) = (f_{x_{1}}, …, f_{x_{n}})· f^{99} + (f^{100}) and J_{n−m}(I) = I for all m > 1. Hence a = n − 1 and I_{1} = I : J_{n−1}(I) ⊆ (f) by inspection. By 6.6, \sqrt{I} = \sqrt{I : J_{n−1}(I)}. As (f) is radical, we have equality. That is, we needed one step to find the radical. The same is true if f is not square free. Only the radical is not (f) then.

Another example is the following. Suppose R = k[a, b, c, d, e]. Let I = (a + b + c + d + e, ab + bc + cd + de + ea, abc + bcd + cde + dea + eab, abcd + bcde + cdea + deab + eabc). Using Macaulay, we find codim(I) = 4. So dim(I) = 1. That means I is a complete intersection. In particular, it is equidimensional. Working along the above algorithm, we find dim(J_{1}(I)) = 1, dim(J_{2}(I)) = 0, again using Macaulay. So a = 1. It turns out that codim(J_{1}(I : J_{1}(I))) = 5. In other words, I : J_{1}(I) = \sqrt{I}.

7 Ext and removing lowdimensional components

Definition 7.1 Let I = ⋂_{1}^{s} Q_{i} be a minimal primary decomposition for I ⊆ S. Suppose 0 ≤ r ≤ s and i > r if and only if dim Q_{i} < e for some natural e. Then we set ⋂_{1}^{r} Q_{i} =: Top_{e}(I) = ⋂_{dim P_{i}≥e} Q_{i}.

In what follows, we set codim(M) to be the codimension of the annihilator ideal of M, and ass(M) is the set of associated primes to ann(M).

Proposition 7.2 Let I ⊆ S be an ideal.

i) Ext_{S}^{j}(S/I, S) ≠ 0 ⇒ codim Ext_{S}^{j}(S/I, S) ≥ j,

ii) ass_{j} Ext_{S}^{j}(S/I, S) = ass_{j}(I) where ass_{k}(M) is the subset of ass(M) which have codimension k.

Proof. For the proof, we first recall Serre’s theorem:

R is regular local ⇒ any localisation of R at a prime ideal is regular local.

Furthermore, observe that depth R_{P} = 0 if and only if P ∈ ass R. We also will need the Auslander-Buchsbaum-formula pd_{R}(M) + depth(M) = dim R whenever R is regular local and M is finitely generated. Note that for any prime in S, S_{P} is regular.

Then we have codim P < j implies pd_{S_{P}} (S/I)_{P} < j since dim S_{P} is just codim P and is by the above formula an upper bound for the projective dimension. Therefore, Ext_{S_{P}}^{j} ((S/I)_{P} , S_{P}) (which is as S- or S_{P} -module isomorphic

to (Ext_{S}^{j}(S/I, S))_{P} ) has to be zero by a resolution argument. Now suppose

that Ext_{S}^{j}(S/I, S) is nonzero. So it has a nontrivial annihilator which will have associated primes. Let P be one of them. Then (Ext_{S}^{j}(S/I, S))_{P} will by the above be zero whenever codim P < j. But the localisation of a module at one of its associated primes is never 0. So codim P ≥ j.

Assume that codim P = j. Then (,as codim Ext_{S}^{j}(S/I, S) ≥ j,)

P ∈ ass Ext_{S}^{j}(S/I, S) ⇔ Ext_{S}^{j}(S/I, S)_{P} ≠ 0  (17)
⇔ Ext_{S_{P}}^{j} ((S/I)_{P} , S_{P}) ≠ 0  (18)
⇔ pd_{S_{P}} (S/I)_{P} ≥ j  (19)

( ⇒ is clear. For ⇐, use the notation S_{P} = R, (S/I)_{P} = M. R is local Noetherian. dim R = j. Therefore pd M = j as it is f.g. and therefore the formula applies. As R is finitely generated too, the lemma of Auslander and Goldman^{1} implies that Ext_{R}^{j}(M, R) ≠ 0)

(20)

⇔ depth_{S_{P}} (S/I)_{P} = 0  (21)
by Auslander-Buchsbaum again  (22)
⇔ all elements in P S_{P} are zerodivisors on (S/I)_{P} (23)
⇔ P ∈ ass I.  (24)

(25)

□

Therefore we may state the following algorithm.

Algorithm 7.3 (The Topdimensional Part of an Ideal) Input: I ⊆ k[x_{1}, …, x_{n}] = S, e ∈ ℕ. Output: Top_{e} I. begin: For j down from n to n − e + 1 do Repeat Get a free resolution of S/I. Get Ext_{S}^{j}(S/I, S). If codim Ext_{S}^{j}(S/I, S) = j then com- pute J := ann Ext_{S}^{j}(S/I, S) and set I := I : J. Until codim Ext_{S}^{j}(S/I, S) > j.

Return I. end.

This algorithm works because for j = n we eliminate all primary components of I which have codimension n. That follows from the repeated application of the preceding proposition: as long as I has a component in codimension n, we find a module which has codimension j, compute its annihilator and divide. As the annihilator will have Associated primes of dimension n equal to the associated primes of I in codimension n, repeated application will cancel these. As the annihilator has codimension n, we will not change any other primary component. Inductively we get rid of all components of codimension at least n − e + 1 without changing the others. Hence what remains is exactly Top_{e}(I).

1Lemma: Let (A, 𝔪) be a local Noetherian commutative ring and N a finite A-module of finite homological dimension n. If M is a nonzero finitely generated A-module , then Ext_{A}^{n}(N, M) ≠ 0.

If we want to use this algorithm, we have to be able to compute annihilators, ideal quotients, codimension and extension modules. Which we can as shown before.

Now look at ⊤_{e}(I) = ⋂_{dim P_{i}≥e} Q_{i} and ⊤_{e+1}(I) = ⋂_{dim P_{i}≥e+1} Q_{i}. Let us divide the two. For any primary component Q_{i} of I which has dimension e + 1 we have Q_{i} : ⋂_{dim P_{i}≥e+1} Q_{i} = S as the second ideal is contained in the first. For any Q_{i} of dimension e which is embedded, there is going to be another primary component Q_{j} thew radical of which is contained in P_{i}. That is to say dim Q_{j} < dim Q_{i}. Then ⋂_{dim P_{i}≥e+1} Q_{k} ⊆ Q_{j} ⊆ P_{i}. But then a sufficiently high power of ⋂_{dim P_{i}≥e+1} Q_{k} will have to be in Q_{i}. In other words, repetition of division will lead to a quotient of all nonembedded components o dimension e of I. This will be equidimensional and hence we can apply the algorithm of the previous section.

References

[1] H. Matsumura, Kakan kan ron, Cambridge University Press 1986.