Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...
Transcript of Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...
![Page 1: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/1.jpg)
Lecture 1: Lattice ideals and lattice basis ideals
Jurgen HerzogUniversitat Duisburg-Essen
August 17-24Moieciu de Sus, Romania
![Page 2: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/2.jpg)
Outline
Toric ideals
Lattice ideals
Lattice basis ideals
![Page 3: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/3.jpg)
Outline
Toric ideals
Lattice ideals
Lattice basis ideals
![Page 4: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/4.jpg)
Outline
Toric ideals
Lattice ideals
Lattice basis ideals
![Page 5: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/5.jpg)
Toric ideals
Let K be a field. We denote by S = K [x1, . . . , xn] the polynomialring in the variables x1, . . . , xn. A binomial belonging to S is apolynomial of the form u − v , where u and v are monomials in S .
![Page 6: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/6.jpg)
Toric ideals
Let K be a field. We denote by S = K [x1, . . . , xn] the polynomialring in the variables x1, . . . , xn. A binomial belonging to S is apolynomial of the form u − v , where u and v are monomials in S .
A binomial ideal is an ideal of S generated by binomials. Anybinomial ideal is generated by a finite number of binomials.
![Page 7: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/7.jpg)
Toric ideals
Let K be a field. We denote by S = K [x1, . . . , xn] the polynomialring in the variables x1, . . . , xn. A binomial belonging to S is apolynomial of the form u − v , where u and v are monomials in S .
A binomial ideal is an ideal of S generated by binomials. Anybinomial ideal is generated by a finite number of binomials.
An important class of binomial ideals are the so-called toric ideals.
![Page 8: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/8.jpg)
In order to define toric ideals we let A = (aij) 1≤i≤d1≤j≤n
be a
d × n-matrix of integers and let
aj =
a1ja2j...adj
, 1 ≤ j ≤ n
be the column vectors of A.
![Page 9: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/9.jpg)
In order to define toric ideals we let A = (aij) 1≤i≤d1≤j≤n
be a
d × n-matrix of integers and let
aj =
a1ja2j...adj
, 1 ≤ j ≤ n
be the column vectors of A.
We write Zd×n for the set of d × n-matrices A = (aij) 1≤i≤d1≤j≤n
with
each aij ∈ Z.
![Page 10: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/10.jpg)
In order to define toric ideals we let A = (aij) 1≤i≤d1≤j≤n
be a
d × n-matrix of integers and let
aj =
a1ja2j...adj
, 1 ≤ j ≤ n
be the column vectors of A.
We write Zd×n for the set of d × n-matrices A = (aij) 1≤i≤d1≤j≤n
with
each aij ∈ Z.
As usual a · b =∑n
i=1 aibi denotes the inner product of the vectorsa = (a1, . . . , an)
t and b = (b1, . . . , bn)t . Here ct denotes transpose
of a vector c.
![Page 11: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/11.jpg)
A matrix A = (aij) 1≤i≤d1≤j≤n
∈ Zd×n is called a configuration matrix if
there exists c ∈ Qd such that
aj · c = 1, 1 ≤ j ≤ n.
![Page 12: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/12.jpg)
A matrix A = (aij) 1≤i≤d1≤j≤n
∈ Zd×n is called a configuration matrix if
there exists c ∈ Qd such that
aj · c = 1, 1 ≤ j ≤ n.
For example, A =
(
1 3 20 2 1
)
is a configuration matrix, while
(a1, . . . , an) ∈ Z1×n is a configuration matrix if and only ifa1 = a2 = . . . = an 6= 0.
![Page 13: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/13.jpg)
A matrix A = (aij) 1≤i≤d1≤j≤n
∈ Zd×n is called a configuration matrix if
there exists c ∈ Qd such that
aj · c = 1, 1 ≤ j ≤ n.
For example, A =
(
1 3 20 2 1
)
is a configuration matrix, while
(a1, . . . , an) ∈ Z1×n is a configuration matrix if and only ifa1 = a2 = . . . = an 6= 0.
Now let T = K [t±11 , . . . , t±1
d ] be the Laurent polynomial ring overK in the variables t1, . . . , tn, and let A ∈ Zd×n with columnvectors aj .
![Page 14: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/14.jpg)
We define a K -algebra homomorphism
π : S → T with xj 7→ taj .
![Page 15: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/15.jpg)
We define a K -algebra homomorphism
π : S → T with xj 7→ taj .
The image of π is the K -subalgebra K [ta1 , . . . , tan ] of T , denoted
K [A]. We call K [A] the toric ring of A.
![Page 16: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/16.jpg)
We define a K -algebra homomorphism
π : S → T with xj 7→ taj .
The image of π is the K -subalgebra K [ta1 , . . . , tan ] of T , denoted
K [A]. We call K [A] the toric ring of A.
For the configuration matrix A of the above example we haveK [A] = K [t1, t
31 t
22 , t
21 t2].
![Page 17: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/17.jpg)
We define a K -algebra homomorphism
π : S → T with xj 7→ taj .
The image of π is the K -subalgebra K [ta1 , . . . , tan ] of T , denoted
K [A]. We call K [A] the toric ring of A.
For the configuration matrix A of the above example we haveK [A] = K [t1, t
31 t
22 , t
21 t2].
The kernel of π is denoted by IA and is called the toric ideal of A.In our example, we have IA = (x1x2 − x23 ).
![Page 18: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/18.jpg)
We define a K -algebra homomorphism
π : S → T with xj 7→ taj .
The image of π is the K -subalgebra K [ta1 , . . . , tan ] of T , denoted
K [A]. We call K [A] the toric ring of A.
For the configuration matrix A of the above example we haveK [A] = K [t1, t
31 t
22 , t
21 t2].
The kernel of π is denoted by IA and is called the toric ideal of A.In our example, we have IA = (x1x2 − x23 ).
Proposition: Let A ∈ Zd×n. Then dimK [A] = rankA.
![Page 19: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/19.jpg)
Proof. Let K (A) be the quotient field of K [A]. Then the Krulldimension of K [A] is equal to the transcendence degreetrdeg(K (A)/K ) of K (A) over K .
![Page 20: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/20.jpg)
Proof. Let K (A) be the quotient field of K [A]. Then the Krulldimension of K [A] is equal to the transcendence degreetrdeg(K (A)/K ) of K (A) over K .
Let V ⊂ Qd be the Q-subspace of Qd generated by the columnvectors of A. Then rankA = dimQ V .
![Page 21: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/21.jpg)
Proof. Let K (A) be the quotient field of K [A]. Then the Krulldimension of K [A] is equal to the transcendence degreetrdeg(K (A)/K ) of K (A) over K .
Let V ⊂ Qd be the Q-subspace of Qd generated by the columnvectors of A. Then rankA = dimQ V .
Let b1, . . . ,bm be a Q-basis of integer vectors of V . Thenm = rankA and K (A) = K (tb1 , . . . , tbm). The desired result willfollow once we have shown that the elements tb1 , . . . , tbm arealgebraically independent over K .
![Page 22: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/22.jpg)
To see this, let F ∈ K [y1, . . . , ym] be a polynomial withF (tb1 , . . . , tbm) = 0. Say, F =
∑
c acyc with ac ∈ K .
Then0 =
∑
c
actc1b1+···+cmbm .
Since the vectors b1, . . . ,bm are linearly independent it followsthat the monomials tc1b1+···cmbm are pairwise distinct. This impliesthat F = 0. �
![Page 23: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/23.jpg)
Given a column vector
b =
b1b2...bn
belonging to Zn, we introduce the binomial fb ∈ S , defined by
fb =∏
bi>0
xbii −
∏
bj<0
x−bjj .
![Page 24: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/24.jpg)
Given a column vector
b =
b1b2...bn
belonging to Zn, we introduce the binomial fb ∈ S , defined by
fb =∏
bi>0
xbii −
∏
bj<0
x−bjj .
Note that fb = xb+ − xb
−, where
b+i =
{
bi , if bi ≥ 0,0, if bi < 0,
and b−i =
{
0, if bi > 0,−bi , if bi ≤ 0.
![Page 25: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/25.jpg)
For example, if b = (1,−1, 0, 2), then fb = x1x24 − x2.
![Page 26: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/26.jpg)
For example, if b = (1,−1, 0, 2), then fb = x1x24 − x2.
If b = (1, 2, 3, 1), then fb = x1x22 x
33 x4 − 1.
![Page 27: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/27.jpg)
For example, if b = (1,−1, 0, 2), then fb = x1x24 − x2.
If b = (1, 2, 3, 1), then fb = x1x22 x
33 x4 − 1.
If f = x21 x2 − x1x22x
33 x4, then f = x1x2fb with b = (1,−1,−3,−1).
![Page 28: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/28.jpg)
For example, if b = (1,−1, 0, 2), then fb = x1x24 − x2.
If b = (1, 2, 3, 1), then fb = x1x22 x
33 x4 − 1.
If f = x21 x2 − x1x22x
33 x4, then f = x1x2fb with b = (1,−1,−3,−1).
Theorem. Any toric ideal is a binomial ideal. More precisely, letA ∈ Zd×n. Then IA is generated by the binomials fb with b ∈ Zn
and Ab = 0.
![Page 29: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/29.jpg)
Proof. We first show that IA is a binomial ideal. Let f ∈ Ker πwith f =
∑
u λuu, λu ∈ K and each u a monomial in S . We writef =
∑
c f(c), where f (c) =
∑
u, π(u)=tc λuu.
![Page 30: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/30.jpg)
Proof. We first show that IA is a binomial ideal. Let f ∈ Ker πwith f =
∑
u λuu, λu ∈ K and each u a monomial in S . We writef =
∑
c f(c), where f (c) =
∑
u, π(u)=tc λuu.
It follows that
0 = π(f ) =∑
c
π(f (c)) =∑
c
(∑
u, π(u)=tc
λu)tc,
and hence∑
u, π(u)=tc λu = 0 for all c. Thus if f (c) 6= 0 and
u ∈ supp(f (c)), then f (c) =∑
v∈supp(f (c)) λv (v − u).
![Page 31: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/31.jpg)
Proof. We first show that IA is a binomial ideal. Let f ∈ Ker πwith f =
∑
u λuu, λu ∈ K and each u a monomial in S . We writef =
∑
c f(c), where f (c) =
∑
u, π(u)=tc λuu.
It follows that
0 = π(f ) =∑
c
π(f (c)) =∑
c
(∑
u, π(u)=tc
λu)tc,
and hence∑
u, π(u)=tc λu = 0 for all c. Thus if f (c) 6= 0 and
u ∈ supp(f (c)), then f (c) =∑
v∈supp(f (c)) λv (v − u).
Finally, let fb ∈ S . Then π(fb) = tAb+ − tAb
−. Hence fb ∈ Ker π if
and only if Ab+ = Ab−, and this is the case if and only if Ab = 0.�
![Page 32: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/32.jpg)
What are configuration matrices good for?
![Page 33: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/33.jpg)
What are configuration matrices good for?
Proposition. Let A ∈ Zd×n. The following conditions areequivalent:
![Page 34: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/34.jpg)
What are configuration matrices good for?
Proposition. Let A ∈ Zd×n. The following conditions areequivalent:
(a) A is a configuration matrix;
![Page 35: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/35.jpg)
What are configuration matrices good for?
Proposition. Let A ∈ Zd×n. The following conditions areequivalent:
(a) A is a configuration matrix;
(b) for all b = (b1, . . . , bn)t ∈ Zn with Ab = 0 we have
∑ni=1 bi = 0;
![Page 36: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/36.jpg)
What are configuration matrices good for?
Proposition. Let A ∈ Zd×n. The following conditions areequivalent:
(a) A is a configuration matrix;
(b) for all b = (b1, . . . , bn)t ∈ Zn with Ab = 0 we have
∑ni=1 bi = 0;
(c) IA is a graded ideal.
![Page 37: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/37.jpg)
What are configuration matrices good for?
Proposition. Let A ∈ Zd×n. The following conditions areequivalent:
(a) A is a configuration matrix;
(b) for all b = (b1, . . . , bn)t ∈ Zn with Ab = 0 we have
∑ni=1 bi = 0;
(c) IA is a graded ideal.
Proof. We only proof (b)⇔ (c). The binomials fb with Ab = 0generate IA. Thus IA is graded if and only if all fb arehomogeneous. This is the case if and only if
∑ni=1 bi = 0 for all b
with Ab = 0. �.
![Page 38: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/38.jpg)
Lattice ideals
We now give another interpretation of toric ideals.
![Page 39: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/39.jpg)
Lattice ideals
We now give another interpretation of toric ideals.
A subgroup L of Zn is called a lattice. Recall from basic algebrathat L is a free abelian group of rank m ≤ n. The binomial idealIL ⊂ S generated by the binomials fb with b ∈ L is called thelattice ideal of L.
![Page 40: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/40.jpg)
Lattice ideals
We now give another interpretation of toric ideals.
A subgroup L of Zn is called a lattice. Recall from basic algebrathat L is a free abelian group of rank m ≤ n. The binomial idealIL ⊂ S generated by the binomials fb with b ∈ L is called thelattice ideal of L.
Consider for example, the lattice L ⊂ Z3 with basis(1, 1, 1), (1, 0,−1). Then b ∈ L if and only if Ab = 0 whereA = (1− 2, 1). Thus in this case we have that IL is a toric ideal.
![Page 41: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/41.jpg)
Lattice ideals
We now give another interpretation of toric ideals.
A subgroup L of Zn is called a lattice. Recall from basic algebrathat L is a free abelian group of rank m ≤ n. The binomial idealIL ⊂ S generated by the binomials fb with b ∈ L is called thelattice ideal of L.
Consider for example, the lattice L ⊂ Z3 with basis(1, 1, 1), (1, 0,−1). Then b ∈ L if and only if Ab = 0 whereA = (1− 2, 1). Thus in this case we have that IL is a toric ideal.
On the other hand, any toric ideal is a lattice ideal.
![Page 42: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/42.jpg)
Proposition. Let A ∈ Zd×n. Then the toric ideal IA is equal to thelattice ideal IL, where L = {b : Ab = 0}.
![Page 43: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/43.jpg)
Proposition. Let A ∈ Zd×n. Then the toric ideal IA is equal to thelattice ideal IL, where L = {b : Ab = 0}.Proof: We know that IA is generated by the binomials fb withAb = 0. �
![Page 44: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/44.jpg)
Proposition. Let A ∈ Zd×n. Then the toric ideal IA is equal to thelattice ideal IL, where L = {b : Ab = 0}.Proof: We know that IA is generated by the binomials fb withAb = 0. �
Not all lattice ideals are toric ideals. The simplest such example isthe ideal IL for L = 2Z ⊂ Z. Here IL = (x2 − 1). If IL would be atoric ideal it would be a prime ideal. But x2 − 1 = (x + 1)(x − 1),and so IL is not a prime ideal.
![Page 45: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/45.jpg)
Proposition. Let A ∈ Zd×n. Then the toric ideal IA is equal to thelattice ideal IL, where L = {b : Ab = 0}.Proof: We know that IA is generated by the binomials fb withAb = 0. �
Not all lattice ideals are toric ideals. The simplest such example isthe ideal IL for L = 2Z ⊂ Z. Here IL = (x2 − 1). If IL would be atoric ideal it would be a prime ideal. But x2 − 1 = (x + 1)(x − 1),and so IL is not a prime ideal.
We have the following general result:
Theorem. Let L ⊂ Zn be a lattice. The following conditions areequivalent:
(a) the abelian group Zn/L is torsionfree;
(b) IL is a prime ideal;
The equivalent conditions hold, if and only if IL is a toric ideal.
![Page 46: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/46.jpg)
We only indicate the proof of (a) =⇒ (b): Since Zn/L is torsionfree, there exists an embedding Zn/L ⊂ Zd for some d . Lete1, . . . , en be the canonical basis of Zn. Then for i = 1, . . . , n,ei + L is mapped to ai ∈ Zd via this embedding. It follows that∑n
i=1 biai = 0 if and only if b = (b1, . . . , bn)t ∈ L. In other words,
b ∈ L if and only if Ab = 0, where A is the matrix whose columnvectors are a1, . . . , an. Therefore, IL is the toric ideal of A, andhence a prime ideal. �
![Page 47: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/47.jpg)
We only indicate the proof of (a) =⇒ (b): Since Zn/L is torsionfree, there exists an embedding Zn/L ⊂ Zd for some d . Lete1, . . . , en be the canonical basis of Zn. Then for i = 1, . . . , n,ei + L is mapped to ai ∈ Zd via this embedding. It follows that∑n
i=1 biai = 0 if and only if b = (b1, . . . , bn)t ∈ L. In other words,
b ∈ L if and only if Ab = 0, where A is the matrix whose columnvectors are a1, . . . , an. Therefore, IL is the toric ideal of A, andhence a prime ideal. �
Let I and J be two ideals. The saturation of I with respect to J isthe ideal I : J∞, where by definition I : J∞ =
⋃
k(I : Jk).
![Page 48: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/48.jpg)
Proposition. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞
is also a binomial ideal.
![Page 49: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/49.jpg)
Proposition. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞
is also a binomial ideal.
Proof: We set x =∏n
i=1 xi . Then
I : (
n∏
i=1
xi)∞ = ISx ∩ S .
![Page 50: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/50.jpg)
Proposition. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞
is also a binomial ideal.
Proof: We set x =∏n
i=1 xi . Then
I : (
n∏
i=1
xi)∞ = ISx ∩ S .
Consider the polynomial ring T = K [x1, . . . , xn, y1, . . . , yn] over K
in the variables x1, . . . , xn, y1, . . . , yn. Then
T/(x1y1 − 1, . . . , xnyn − 1) ' Sx , and henceT/(I , x1y1 − 1, . . . , xnyn − 1)T ' Sx/ISx .
Therefore, ISx ∩ S = (I , x1y1 − 1, . . . , xnyn − 1)T ∩ S , from whichit follows that I : (
∏ni=1 xi)
∞ is a binomial ideal. �
![Page 51: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/51.jpg)
Theorem. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞ is a
lattice ideal.
Proof. Let
L = {b ∈ Zn : ufb ∈ I for some monomial u}.
![Page 52: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/52.jpg)
Theorem. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞ is a
lattice ideal.
Proof. Let
L = {b ∈ Zn : ufb ∈ I for some monomial u}.
We claim that L ⊂ Zn is a lattice. Indeed, if b ∈ L then ufb ∈ I forsome monomial u and hence uf−b = −ufb ∈ I . This shows that−b ∈ L.
![Page 53: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/53.jpg)
Theorem. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞ is a
lattice ideal.
Proof. Let
L = {b ∈ Zn : ufb ∈ I for some monomial u}.
We claim that L ⊂ Zn is a lattice. Indeed, if b ∈ L then ufb ∈ I forsome monomial u and hence uf−b = −ufb ∈ I . This shows that−b ∈ L. Now let c ∈ L be another vector. Then there exists amonomial v such that vfc ∈ I . We get
(ufb)(vfc) = uv(wfb+c − xb−
fc − xc−
fb)
= uvwfb+c − xb−
u(vfc)− xc−
v(ufb).
It follows from this equation that b+ c ∈ L. This proves the claim.
![Page 54: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/54.jpg)
Theorem. Let I ⊂ S be a binomial ideal. Then I : (∏n
i=1 xi )∞ is a
lattice ideal.
Proof. Let
L = {b ∈ Zn : ufb ∈ I for some monomial u}.
We claim that L ⊂ Zn is a lattice. Indeed, if b ∈ L then ufb ∈ I forsome monomial u and hence uf−b = −ufb ∈ I . This shows that−b ∈ L. Now let c ∈ L be another vector. Then there exists amonomial v such that vfc ∈ I . We get
(ufb)(vfc) = uv(wfb+c − xb−
fc − xc−
fb)
= uvwfb+c − xb−
u(vfc)− xc−
v(ufb).
It follows from this equation that b+ c ∈ L. This proves the claim.
In the next step one shows that I : (∏n
i=1 xi )∞ = IL. �
![Page 55: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/55.jpg)
Lattice ideals are saturated, as the following result shows
![Page 56: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/56.jpg)
Lattice ideals are saturated, as the following result shows
Theorem. Let L ⊂ Zn be a lattice. Then IL : (∏n
i=1 xi)∞ = IL.
![Page 57: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/57.jpg)
Lattice ideals are saturated, as the following result shows
Theorem. Let L ⊂ Zn be a lattice. Then IL : (∏n
i=1 xi)∞ = IL.
Proof. We only need to show that IL : (∏n
i=1 xi )∞ ⊂ IL. Let
f ∈ IL : (∏n
i=1 xi )∞. We may assume that f is a binomial, and we
may further assume that f = fb for some b ∈ Zn.
![Page 58: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/58.jpg)
Lattice ideals are saturated, as the following result shows
Theorem. Let L ⊂ Zn be a lattice. Then IL : (∏n
i=1 xi)∞ = IL.
Proof. We only need to show that IL : (∏n
i=1 xi )∞ ⊂ IL. Let
f ∈ IL : (∏n
i=1 xi )∞. We may assume that f is a binomial, and we
may further assume that f = fb for some b ∈ Zn.
We want to show that b ∈ L. Since fb ∈ IL : (∏n
i=1 xi )∞, it follows
that 1− xb ∈ ILSx where x =∏n
i=1 xi . Observe that ILSx isgenerated by the binomials 1− xc with c ∈ L. Therefore,Sx/ILSx isisomorphic to the group ring K [Zn/L] which admits the K -basisconsisting of the elements of the group G = Zn/L.
![Page 59: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/59.jpg)
Lattice ideals are saturated, as the following result shows
Theorem. Let L ⊂ Zn be a lattice. Then IL : (∏n
i=1 xi)∞ = IL.
Proof. We only need to show that IL : (∏n
i=1 xi )∞ ⊂ IL. Let
f ∈ IL : (∏n
i=1 xi )∞. We may assume that f is a binomial, and we
may further assume that f = fb for some b ∈ Zn.
We want to show that b ∈ L. Since fb ∈ IL : (∏n
i=1 xi )∞, it follows
that 1− xb ∈ ILSx where x =∏n
i=1 xi . Observe that ILSx isgenerated by the binomials 1− xc with c ∈ L. Therefore,Sx/ILSx isisomorphic to the group ring K [Zn/L] which admits the K -basisconsisting of the elements of the group G = Zn/L.
Let g = b+ L. Then 1− g = 0 in K [Zn/L] because 1− xb ∈ ILSx .This implies that b+ L = 0 + L, and hence b ∈ L, as desired. �
![Page 60: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/60.jpg)
Lattice ideals are saturated, as the following result shows
Theorem. Let L ⊂ Zn be a lattice. Then IL : (∏n
i=1 xi)∞ = IL.
Proof. We only need to show that IL : (∏n
i=1 xi )∞ ⊂ IL. Let
f ∈ IL : (∏n
i=1 xi )∞. We may assume that f is a binomial, and we
may further assume that f = fb for some b ∈ Zn.
We want to show that b ∈ L. Since fb ∈ IL : (∏n
i=1 xi )∞, it follows
that 1− xb ∈ ILSx where x =∏n
i=1 xi . Observe that ILSx isgenerated by the binomials 1− xc with c ∈ L. Therefore,Sx/ILSx isisomorphic to the group ring K [Zn/L] which admits the K -basisconsisting of the elements of the group G = Zn/L.
Let g = b+ L. Then 1− g = 0 in K [Zn/L] because 1− xb ∈ ILSx .This implies that b+ L = 0 + L, and hence b ∈ L, as desired. �
Corollary. Let I ⊂ S be a binomial ideal. Then I is a lattice idealif and only if I : (
∏ni=1 xi )
∞ = I .
![Page 61: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/61.jpg)
We have seen above that IL is not always a prime ideal. The latticeideal IL need to be even a radical ideal if char(K ) = p > 0.
![Page 62: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/62.jpg)
We have seen above that IL is not always a prime ideal. The latticeideal IL need to be even a radical ideal if char(K ) = p > 0.
Indeed if L = (p,−p) ⊂ Z2, then IL = (xp − yp), and we havef = x − y 6∈ IL but f p ∈ IL.
![Page 63: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/63.jpg)
We have seen above that IL is not always a prime ideal. The latticeideal IL need to be even a radical ideal if char(K ) = p > 0.
Indeed if L = (p,−p) ⊂ Z2, then IL = (xp − yp), and we havef = x − y 6∈ IL but f p ∈ IL.
However, if char(K ) = 0 or char(K ) = p > 0 and p is big enough,then IL is a radical ideal. More precisely, we have
![Page 64: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/64.jpg)
We have seen above that IL is not always a prime ideal. The latticeideal IL need to be even a radical ideal if char(K ) = p > 0.
Indeed if L = (p,−p) ⊂ Z2, then IL = (xp − yp), and we havef = x − y 6∈ IL but f p ∈ IL.
However, if char(K ) = 0 or char(K ) = p > 0 and p is big enough,then IL is a radical ideal. More precisely, we have
Theorem Let L ⊂ Zn be a lattice and let t be the maximal orderof a torsion element of Zn/L. If char(K ) = 0 or char(K ) > t, thenIL is a radical ideal.
![Page 65: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/65.jpg)
Lattice basis ideals
Let L ⊂ Zn be a lattice and let B = b1, . . . ,bm be a Z-basis of L.The ideal IB is called a lattice basis ideal of L.
![Page 66: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/66.jpg)
Lattice basis ideals
Let L ⊂ Zn be a lattice and let B = b1, . . . ,bm be a Z-basis of L.The ideal IB is called a lattice basis ideal of L.
In general, IB 6= IL. Consider for example, A = (3, 4, 5) ∈ Z1×3.The toric ideal IA is the lattice ideal of the lattice L with basisB = (2, 1,−2), (1,−2, 1). Then IB = (x2y − z2, xz − y2), while ILalso contains the binomial x3 − yz .
![Page 67: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/67.jpg)
Lattice basis ideals
Let L ⊂ Zn be a lattice and let B = b1, . . . ,bm be a Z-basis of L.The ideal IB is called a lattice basis ideal of L.
In general, IB 6= IL. Consider for example, A = (3, 4, 5) ∈ Z1×3.The toric ideal IA is the lattice ideal of the lattice L with basisB = (2, 1,−2), (1,−2, 1). Then IB = (x2y − z2, xz − y2), while ILalso contains the binomial x3 − yz .
However one has
Corollary. Let B be a basis of the lattice L. ThenIB : (
∏ni=1 xi )
∞ = IL.
![Page 68: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/68.jpg)
Proof. There exists a lattice L′ ⊂ Zn such thatIB : (
∏ni=1 xi )
∞ = IL′ . Since B ⊂ L′ it follows that L ⊂ L′.
![Page 69: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/69.jpg)
Proof. There exists a lattice L′ ⊂ Zn such thatIB : (
∏ni=1 xi )
∞ = IL′ . Since B ⊂ L′ it follows that L ⊂ L′.
On the other hand, IB ⊂ IL. Thus,IL′ = IB : (
∏ni=1 xi)
∞ ⊂ IL : (∏n
i=1 xi )∞ = IL. This shows that
L′ ⊂ L, and hence L′ = L. �
![Page 70: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/70.jpg)
Proof. There exists a lattice L′ ⊂ Zn such thatIB : (
∏ni=1 xi )
∞ = IL′ . Since B ⊂ L′ it follows that L ⊂ L′.
On the other hand, IB ⊂ IL. Thus,IL′ = IB : (
∏ni=1 xi)
∞ ⊂ IL : (∏n
i=1 xi )∞ = IL. This shows that
L′ ⊂ L, and hence L′ = L. �
We fix a field K , and let X = (xij) be an (m × n)-matrix ofindeterminates. The ideal of all 2-minors of X is a prime ideal andand hence may be viewed as a toric ideal, or as a lattice ideal IL forthe lattice L ⊂ Zm×n with lattice basis B consisting of the vectors
eij + ei+1,j+1 − ei ,j+1 − ei+1,j , 1 ≤ i ≤ m − 1, 1 ≤ j ≤ n − 1.
![Page 71: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/71.jpg)
Proof. There exists a lattice L′ ⊂ Zn such thatIB : (
∏ni=1 xi )
∞ = IL′ . Since B ⊂ L′ it follows that L ⊂ L′.
On the other hand, IB ⊂ IL. Thus,IL′ = IB : (
∏ni=1 xi)
∞ ⊂ IL : (∏n
i=1 xi )∞ = IL. This shows that
L′ ⊂ L, and hence L′ = L. �
We fix a field K , and let X = (xij) be an (m × n)-matrix ofindeterminates. The ideal of all 2-minors of X is a prime ideal andand hence may be viewed as a toric ideal, or as a lattice ideal IL forthe lattice L ⊂ Zm×n with lattice basis B consisting of the vectors
eij + ei+1,j+1 − ei ,j+1 − ei+1,j , 1 ≤ i ≤ m − 1, 1 ≤ j ≤ n − 1.
The ideal IB is called the ideal of adjacent minors of X . It has firstbeen studied by Hosten and Sullivant.
![Page 72: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/72.jpg)
In general, IB is not a radical ideal.
![Page 73: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/73.jpg)
In general, IB is not a radical ideal.
Let I be an ideal in a Noetherian ring. Then there exists an integerk such that (
√I )k ⊂ I . The smallest integer with this property is
called the index of nilpotency, denoted nilpot(I ).
![Page 74: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/74.jpg)
In general, IB is not a radical ideal.
Let I be an ideal in a Noetherian ring. Then there exists an integerk such that (
√I )k ⊂ I . The smallest integer with this property is
called the index of nilpotency, denoted nilpot(I ).
Theorem. (Ene, H, Hibi and Qureshi) I be the ideal of adjacent2-minors of the generic (m × n)-matrix X , and let n = 4k + p andn = 4l + q with 0 ≤ p, q < 4. Then
nilpot(I ) ≥ (k + bp3c)(l + bq
3c) + 1 ≈ mn
16.
![Page 75: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/75.jpg)
Problems
Problem 1. Show that
A =
2 0 3 41 −2 1 −13 0 5 17 −1 12 5
is a configuration matrix.
Problem 2. Let A ∈ Zd×n. Then IA is a principal ideal if and onlyif rankA = n − 1.
Problem 2. Let A = (3, 4, 5) ∈ Z1×3. Compute IA.
![Page 76: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/76.jpg)
Problem 3. Let I ⊂ K [x1, . . . , xn, y1, . . . , yn] be the idealgenerated by a set S of 2-minors of the 2× n-matrix
X =
(
x1 · · · xny1 · · · yn
)
. We denote by [i , j] a 2-minor with rows i
and j . Show that I is a prime ideal if and only [n] is the disjointunion of sets S1, . . . ,Sk such that S =
⋃ki=1{[i , j] : {i , j} ⊂ Sk}.
Problem 4. Let char(K ) = 0 and let b ∈ Zn. Then I = (fb) ⊂ S
is a radical ideal.
Problem 5. Let b1, . . . ,br ∈ Zn be Q-linearly independentvectors. Then fb1 , . . . , fbr is a regular sequence.
![Page 77: Ju¨rgen Herzog Universitat Duisburg-Essen August 17-24 ...](https://reader031.fdocuments.in/reader031/viewer/2022012514/618d8c56dfe6e435937c260e/html5/thumbnails/77.jpg)
Problem 6. Show that (xk − yk , x l − y l) : (xy)∞ = (x − y).Which is the smallest integer m with the property that(xk − yk , x l − y l) : (xy)m = (x − y)?
Problem 7. Let L ⊂ Zn be a lattice. Prove that height IL = rankL.
Problem 6. Let B be a basis of a lattice L for which Zn/L istorsionfree. Then IB = IL if and only if IB is a prime ideal.
Problem 7. Let I ⊂ S be the ideal of adjacent 2-minors of am × n-matrix of indeterminates.(a) Show that I is a radical ideal if and only if m ≤ 2 or n ≤ 2.(b) Find a polynomial f ∈ S \ I with f 2 ∈ I , if m = n = 3.