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Analogous to the notation I(x; µ) introduced above we may write

J(x; δ) = J(X; δ)/J(X; γ)

when a ring G(X; γ) is investigated. It follows directly from (1.4) that for γ = [a 1 , . . . , am ]
the ring G(X; γ) is the Schubert cycle associated with

„¦(n ’ am + 1, n ’ am’1 + 1, . . . , n ’ a1 + 1).
53
B. Syzygies and the Straightening Law

(5.4) Theorem. (a) The rings G(X; γ) are exactly the Schubert cycles.
(b) G(X; γ) is a graded ASL on “(X; γ).
(c) “(X; γ) is a distributive lattice.
All this is evident now. In 4.D we have extended the matrix X in order to get the
representation

µ = (’1)m(m’1)/2 .
B[X] = G(X)/G(X)([n + 1, . . . , n + m] ’ µ),

For δ ∈ ∆(X), δ = [a1 , . . . , ar |b1 , . . . , br ], we choose

δ = [b1 , . . . , br , n + m + 1 ’ am’r , . . . , n + m + 1 ’ a1 ],

{a1 , . . . , am’r } being complementary to {a1 , . . . , ar } in {1, . . . , m}. Then, by virtue of
(4.9), the epimorphism G(X) ’’ B[X] maps a generating set of J(X; δ) onto a set of
generators of I(X; δ). Therefore one obtains immediately:
(5.5) Theorem. With the notations just introduced, R(X; δ) is the dehomogeniza-
tion of G(X; δ) with respect to µ[n + 1, . . . , n + m].
The geometric signi¬cance of (5.5) has been indicated brie¬‚y in 1.D. We will use
(5.5) mainly to transfer information from G(X; δ) to R(X; δ).

B. Syzygies and the Straightening Law

As we have seen in (4.2), an ASL A is de¬ned in terms of generators and relations
by its underlying poset Π and the straightening relations. In a very similar way the
module of syzygies of an ideal AΨ, Ψ an ideal in Π, is determined by the straightening
relations involving elements of Ψ (and the “Koszul” relations corresponding to pairs of
comparable elements of Ψ).
To have a compact notation in the following proposition we denote the smallest
factor of a standard monomial µ by ±(µ) and the product of the remaining factors by
ω(µ). Note that for every standard monomial µ in a straightening relation ξψ = aµ µ
one has ±(µ) ∈ Ψ if Ψ is a poset ideal and ψ ∈ Ψ.
(5.6) Proposition. Let A be a graded ASL on Π over B, Ψ ‚ Π an ideal, and e ψ ,
ψ ∈ Ψ, denote the canonical basis of the free A-module AΨ .
(a) Then the kernel U of the natural epimorphism

AΨ ’’ AΨ, eψ ’’ ψ,

is generated by the elements

•eψ ’ ψe• , •, ψ ∈ Ψ, • < ψ,

and the elements

ξeψ ’ ξ ∈ Π, ψ ∈ Ψ, ξ and ψ incomparable,
aµ ω(µ)e±(µ) ,

corresponding to the straightening relations ξψ = aµ µ.
54 5. The Structure of an ASL

(b) Suppose that the submodule U ‚ U contains elements

ξeψ ’ cξψ• ∈ A,
cξψ• e• ,
•<ψ


for all elements ξ ∈ Π, ψ ∈ Ψ such that ξ ≥ ψ. Then U = U .
Proof: Since, with the notations of part (a), ±(µ) < ψ, (b) is a generalization of
(a). Let
dψ ∈ A,
u= dψ eψ ,
ψ∈Ψ

be an element of U . Each dψ has a standard representation

aµψ ∈ B, aµψ = 0.
dψ = aµψ µ,
µ∈Mψ

Suppose that ψ ¤ ±(µ) for all the standard monomials µ ∈ Mψ . Then, by the linear
independence of the standard monomials, one has u = 0. This observation is the base
of an inductive proof: Modulo U every term aµψ µeψ with ±(µ) ≥ ψ can be replaced
by a linear combination of the elements e• , • < ψ, and after ¬nitely many iterations
one obtains the zero element of U , as has just been seen. In other words, one creates a
sequence

ui+1 ≡ ui mod U , i = 0, . . . , n ’ 1. ”
u = un , un’1 , . . . , u0 = 0,

In order to prove that a given set of relations of the elements ψ ∈ Ψ generates U
one will of course show that all the relations required in (b) can be obtained as linear
combinations of the given ones.

C. Nilpotents, Regular Elements and Dimension
A general and extremely important property of ASLs is that they have no nilpotent
elements, provided B has no nilpotents:
(5.7) Proposition. Let A be a graded ASL on Π over B. Then A is reduced if
(and only if ) B is reduced.
Proof: The proof of the nontrivial statement is by induction on |Π|. In case |Π| = 1,
A is the polynomial ring in one indeterminate over B. Let |Π| > 1. Let x ∈ A be nilpotent
and suppose x = 0. We choose a minimal element ξ ∈ Π. By induction x ∈ Aξ, so x = ξ d y
such that y ∈ Aξ and d ≥ 1, simply from consideration of the standard representation.
/
If ξ is the single minimal element of Π, then it is not a zero-divisor obviously, and y is
nilpotent, contradicting the inductive hypothesis. Otherwise there is a second minimal
element …, and by inductive reasoning again x = ξ d … e z, e ≥ 1. However, ξ… = 0 from
(H2 ). ”
It follows easily from (5.7) that for general B the nilradical of A is the extension of
the nilradical of B.
The argument that ξ… = 0 for a minimal element ξ ∈ Π and an element … ∈ Π not
comparable to it, will be used several times below.
55
C. Nilpotents, Regular Elements and Dimension

(5.8) Corollary. If the ring B of coe¬cients is reduced, then all the rings R(X; δ)
and G(X; γ) are reduced.
In particular, they are the coordinate rings of the varieties associated with them
(under a suitable hypothesis on B). Our next goal is the computation of their dimensions.
For this purpose we exhibit a natural candidate for a system of parameters in a certain
localization (which under special circumstances can also serve as a maximal regular
sequence). For an element ξ ∈ Π we de¬ne its rank by:

⇐’ there is a chain ξ = ξk > ξk’1 > · · · > ξ1 , ξi ∈ Π,
rk ξ = k
and no such chain of greater length exists.

For a subset „¦ ‚ Π let
rk „¦ = max{rk ξ : ξ ∈ „¦}.
(The preceding de¬nition of rank di¬ers from the usual one in combinatorics which gives
a result smaller by 1. In order to reconcile the two de¬nitions the reader should imagine
an element ’∞ added to Π, vaguely representing 0 ∈ A.)
(5.9) Lemma. Let „¦ ‚ Π be an ideal, k = rk „¦, and xi = ξ. Then
rk ξ=i
ξ∈„¦
k
Rad A„¦ = Rad Axi .
i=1
k
Proof: Let ζ1 , . . . , ζm be the minimal elements of „¦, and J = Rad i=1 Axi . Then,
2
since ζu ζv = 0 for u = v and ζ1 + · · · + ζm ∈ J, ζu ∈ J, hence ζu ∈ J, u = 1, . . . , m. The
rest is induction. ”
A particularly simple example is „¦ = Π = ∆(X; [1|1]), the poset underlying R2 (X):
rk „¦ = m + n ’ 1 and

i = 1, . . . , m + n ’ 1;
xi = [u|v],
u+v=i+1


the xi are the sums over the diagonals of the matrix.
Especially for „¦ = Π one has Rad rk Π Axi = Rad AΠ. When B = K is a ¬eld, AΠ
i=1
is the irrelevant maximal ideal, so dim A ¤ rk Π. This bound turns out to be precise, and
the general case regarding B can be treated via the dimension formula for ¬‚at extensions.
(5.10) Proposition. Let A be a graded ASL on Π over the noetherian ring B.
Then
dim A = dim B + rk Π and ht AΠ = rk Π.

Proof: Let P be a prime ideal of A, and Q = B © P . Since AP is a localization of
A — BQ , it is a ¬‚at local extension of BQ , so

dim AP = dim BQ + dim AP — (BQ /QBQ ).

The ring AP — (BQ /QBQ ) is just (A/QA)P/QA , hence a localization of A — (BQ /QBQ )
which is a graded ASL over the ¬eld BQ /QBQ on Π. From what has been said above

dim AP ¤ dim B + rk Π,
56 5. The Structure of an ASL

so
dim A ¤ dim B + rk Π.
Now it is enough to show that ht AΠ ≥ rk Π. Let „¦ be the set of minimal elements of Π.
As we shall see in Lemma (5.11), the element ξ∈„¦ ξ is not a zero-divisor of A, thus

ht(Π\„¦)(A/A„¦) ¤ ht AΠ ’ 1.

On the other hand, arguing by induction,

ht(Π\„¦)(A/A„¦) = rank(Π\„¦) = rankΠ ’ 1. ”

(5.11) Lemma. Let „¦ ‚ Π consist of pairwise incomparable elements, and suppose
that every maximal chain (linearly ordered subset) of Π intersects „¦. Then x = ξ∈„¦ ξ
is not a zero-divisor of A.
Proof: Suppose that yx = 0, y = 0. Let Min Π be the set of minimal elements of
Π. By induction on |Π| one immediately obtains the following auxiliary claim: (—) Let
π ∈ Min Π, π ∈ „¦; then y ∈ Aπ, y = π d y , d ≥ 1, y ∈ Aπ and y x = 0.
/ /
Case 1: „¦ = Min Π. We pick a standard monomial µ0 in the standard representation
aµ µ. There is an ω0 ∈ „¦ such that ω0 µ0 is a standard monomial, and ω0 µ0 can
y=
not appear in the standard representations of any of the products ωµ, ω ∈ „¦, ω = ω0
or µ = µ0 ! Therefore ω0 µ0 occurs in the standard representation of yx with a nonzero
coe¬cient. Contradiction.
Case 2: | Min Π| = 1, Min Π = {π}. Then either „¦ = {π}, a case covered already, or
π ∈ „¦. In the latter case the contradiction results from (—), since π is not a zero-divisor.
/
Case 3: |(Min Π) \ „¦| ≥ 2, π1 , π2 ∈ (Min Π) \ „¦, π1 = π2 . Then, by (—), y ∈
Aπ1 © Aπ2 = 0.
Case 4: |(Min Π) \ „¦| = 1. Let π ∈ Min Π, π ∈ „¦. Excluding case 2, we may assume
/
that there is a σ ∈ (Min Π) © „¦. Write

x =x +x , x= ξ.
ξ∈(Min Π)©„¦

We want to construct a subset „¦ of Π which satis¬es the hypothesis of the lemma modulo
Aσ. Let

„¦ = („¦ \ {σ}) ∪ „¦ , „¦ = {„ : „ an upper neighbour of σ
not comparable to any ω ∈ „¦, ω = σ}.

Then „¦ consists of incomparable elements. A maximal chain “ in Π \ {σ} which does
not intersect „¦ \ {σ}, passes through an upper neighbour ρ of σ, ρ ∈ Min(Π \ {σ}). If
ρ ∈ „¦ , “ © „¦ = …. Otherwise ρ is comparable to an ω ∈ „¦ \ {σ}. Since ρ is minimal in
Π \ {σ}, ρ < ω, a fortiori σ < ω, in contradiction to the hypothesis on „¦.
Let „ ∈ „¦ and suppose „ ≥ π. Then there is a maximal chain in Π \ {σ} starting
from π and passing through „ . This chain has to intersect „¦ \ {σ} which is impossible
by de¬nition of „¦ . So π and „ ∈ „¦ are incomparable. Let x = ξ∈„¦ ξ. Then

x=x ’σ+x + „
„ ∈„¦
57
C. Nilpotents, Regular Elements and Dimension

and
yx = yx ’ y π d σ + y π d „ = 0.
„ ∈„¦

„¦ satis¬es the hypothesis of the lemma modulo Aσ. Therefore y ∈ Aσ © Aπ = 0, a
contradiction settling case 4.
Since Min Π ‚ „¦ implies Min Π = „¦, all the possible cases have been covered. ”
A subset „¦ of Π which satis¬es the hypothesis of (5.11) is of course maximal with
respect to having pairwise incomparable elements. This weaker property is not su¬cient
for x to be not a zero-divisor. For
ξs sρ
@
Π= and ρξ = ρπ = σπ = 0
@
π s @s σ
one has σξ(ρ + π) = 0.
After Proposition (5.10) the computation of dim A is a purely combinatorial problem.
Let again X be an m — n matrix of indeterminates over B, γ = [a1 , . . . , am ] ∈ “(X).
Any maximal chain in “(X; γ) starts at γ, and one moves to an upper neighbour raising
exactly one index by 1. Therefore
m m
m(m + 1)
(n ’ m + i ’ ai ) = m(n ’ m) + ’
rk “(X; γ) = ai + 1.
2
i=1 i=1

The rank of ∆(X; δ) can most conveniently be computed by relating it to “(X; δ) as in
Theorem (5.5): For δ = [a1 , . . . , ar |b1 , . . . , br ] one has
δ = [b1 , . . . , br , (n + m + 1) ’ am’r , . . . , (n + m + 1) ’ a1 ],
{a1 , . . . , am’r } being complementary to {a1 , . . . , ar } in {1, . . . , m}. An easy computation
gives
r
rk ∆(X; δ) = rk “(X; δ) ’ 1 = (m + n)r ’ (ai + bi ) + r.
i=1

(5.12) Corollary. Let B be a noetherian ring, X an m—n matrix of indeterminates
over B.
(a) Let δ = [a1 , . . . , ar |b1 , . . . , br ] ∈ ∆(X). Then
r
dim R(X; δ) = dim B + (m + n)r ’ (ai + bi ) + r,
i=1

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