Vol. 82, Nos. 1–4, (2013), 01-15
R. P. STANLEY’S SOLUTION OF
ANAND-DUMIR-GUPTA CONJECTURES
ABOUT MAGIC SQUARES*
J. K. VERMA
Abstract. In 1966 Anand-Dumir-Gupta formulated a number of conjectures concerning enumeration of doubly stochastic matrices with given line sum. In 1973 Stanley solved most of these conjectures using techniques from commu-tative algebra. We sketch Stanley’s solution after recalling necessary results from commutative algebra.
1. THE ANAND-DUMIR-GUPTA CONJECTURES
The objective of this exposition is to outline Stanley’s solution of the Anand-Dumir-Gupta (ADG) conjectures. The ADG conjectures concern enumeration of doubly stochastic matrices or magic squares. Let
N
denote the set of nonnegative integers and letP
denote the set of positive integers. Ann×nmatrixM is called amagic squareif its entries are inN
and the sum of entries in any row or column is a given integerr. The number ris called the line sum of M.LetHn(r) denotethe number ofn×nmagic squares with line sumr.It is clear that
H1(r) = 1 and H2(r) =r+ 1.
MacMahon [5] and independently Anand-Dumir-Gupta [1] proved that
H3(r) =
(
r+ 4 4
)
+
(
r+ 3 4
)
+
(
r+ 2 4
)
.
Inspired by these formulas H. Anand, V. C. Dumir and Hansraj Gupta pro-posed the following conjectures in [1]:
Conjecture 1.1 (Anand-Dumir-Gupta (1966)). Fixn≥1.Then (1) Hn(r)∈
C
[r].(2) degHn(r) = (n−1)2.
2010 AMS Subject Classification: 13D40, 13D45,05A15.
keywords:magic square, Hilbert series, Cohen-Macaulay rings, Gorenstein rings, canonical module, local cohomology modules, ring of invariants.
* The expanded version of the text of the 23rd Srinivasa Ramanujan Memorial Award lecture delivered at the 78th Annual Conference of the Indian Mathematical Society held at Banaras Hindu University, Varanasi - 221 005, UP, during the period January 22 - 25, 2013.
c
⃝
Indian Mathematical Society, 2013
.(3) Hn(i) = 0 fori=−1,−2, . . . ,−(n−1).
(4) Hn(−n−r) = (−1)n−1Hn(r).
Quite often integer valued sequences are studied effectively by understanding the properties of their generating functions. Stanley translated the ADG conjec-tures in terms the generating function ofHn(r).He showed that the four assertions aboutHn(r) are equivalent to the following:
∞
∑
r=0
Hn(r)λr=
h0+h1λ+· · ·+hdλd
(1−λ)(n−1)2+1 ,
whereh0, h1, . . . , hdare integers,d= (n−1)2+ 1−n, h0+h1+· · ·+hd̸= 0 and
hd−i=hi fori= 0,1, . . . , d.Stanley made the additional conjectures that
(5)hi≥0 for alliand (6)h0≤h1≤ · · · ≤h[d/2].
Stanley settled (1)-(5) in 1973 [7]. A geometric proof based on Ehrhart poly-nomials of integral polytopes appears in Stanley’s Green Book [8]. The conjecture (6) has been proved by Athanasiadis [2].
Stanley’s solution used several techniques from commutative algebra to solve the ADG conjectures such as Hilbert series of graded modules, Cohen-Macaulay and Gorenstein rings, invariant theory, local cohomology modules and canonical modules of graded Cohen-Maculay rings. We will describe these techniques first and then return to a sketch of Stanley’s solution. We have tried to simplify the proof by using the Grothendieck-Serre formula for the difference of Hilbert function and the Hilbert polynomial of a finitely generated graded module over a graded Noetherian ring. We refer the reader to [11] and an excellent survey article by W. Bruns [3] on Stanley’s solution in which the techniques required from commu-tative algebra have been presented in full details. We also refer the reader to an expository account, by Anurag Singh [12], of a solution of the parts (1) and (2) of the ADG conjectures.
Acknowledgements: Thanks are due to Manoj Kummini and Anurag Singh for a careful reading of the paper.
2. HILBERT SERIES OF GRADED ALGEBRAS
Letkbe a field. An
N
-gradedk-algebraRis a commutative ring containing a sequence ofk-vector spacesR0=k, R1, R2, . . . ,such thatR=⊕∞i=0Ri and themultiplication inRobeys the rule RiRj ⊆Ri+j for alli, j.The elments of Ri are
called homogeneous of degreei.IfRis Noetherian then there exist homogeneous el-ementsa1, a2, . . . , aswith degreeai=ei,for certain positive integerse1, e2, . . . , es
ringA =k[y1, y2, . . . , ys] where degyi = ei for i = 1,2, . . . , s. The algebra R is
calledstandardif finitely many elements ofR1generateR as ak-algebra.
A
Z
-gradedR-moduleM is anR-module having a direct sum decompositionM =⊕∞i=0Mi where eachMi is ak-vector space and RiMj ⊆Mi+j for alli, j. A
mapf :M −→Nof gradedR-modulesMandNis called graded iff(Mi)⊆Nifor alli.A submoduleNofM is called homogeneous if it is generated by homogeneous elements. IfN is homogeneous submodule ofM,thenN =⊕
n∈
Z
N∩[M]n.Here[M]n denotes theR0-submodule ofM generated by elements of degreenin M.If
N is homogeneous submodule ofM, then the quotient moduleM/N is a graded
R-module with [M/N]n=Mn/Nn for alln.
It is clear that dimMi < ∞ for all i for a Noetherian
Z
-graded R-module. TheHilbert seriesofM is the generating functionF(M, λ) = ∞
∑
n=0
dimkMnλn.
The function H(M, n) = dimkMn is called the Hilbert function of M. Recall
that thedimensionofR, dimR,is the maximum number of algebraically indepen-dent elements inRoverk.Recall that the degree of a rational functionp(x)/q(x) is defined to be degp(x)−degq(x).Finally the dimension of anR-moduleM is defined by dimM = dimR/ann(M),where ann(M) ={r∈R:rM= 0}.
Theorem 2.1 (Hilbert-Serre Theorem). Let R be a Noetherian graded k-algebra generated by elements of positive degreese1, e2, . . . , esandM, a Noetherian
gradedR-module. Then
(1) There existsh(λ, λ−1)∈
Z
[λ, λ−1] such thatF(M, λ) = (1−λhe(1λ,λ)···−(11−)λes).(2) The order of the pole atλ= 1 ofF(M, λ) = dimM.
(3) If R is standard then H(M, n) is given by a polynomial P(M, n) with rational coefficients for large n. MoreoverdegP(M, n) = dimM−1. (4) max{n:H(M, n)̸=P(M, n)}= degF(M, λ).
3. COHEN-MACAULAY GRADED RINGS AND MODULES
In this section, we review the basics of Cohen-Macaulay rings and modules for Noetherian graded algebras over a field. Throughout this sectionRwill denote an
N
-graded Noetheriank-algebra wherekis a field andM will be a gradedZ
-module overR.Definition 3.1. A partial homogeneous system of parameters (hsop) forM is a sequence of homogenous elementsa=a1, a2, . . . , ar of positive degree in R such
that dimM/aM = dimM−r. We saya is an hsop forM ifr= dimM.
Lemma 3.1. Let dimM = d. A sequence of elements a = a1, a2, . . . , ad is an
Homogeneous systems of parameters forM exist and ifkis infinite, we can choose an hsop forM fromR1.
Definition 3.2. A sequence a = a1, a2, . . . , ag of homogeneous elements of R
of positive degree is called an M-sequence if for i = 0,1, . . . , g −1, ai+1 is a
nonzerodivisor onM/(a1, a2, . . . , ai)M.
Theorem 3.1. (1) All maximal M-sequences in ideal I with IM ̸=M have equal length.
(2) The sequence a1, a2, . . . , ag is anM-sequence if and only if a1, a2, . . . , ag are algebraically independent over k and M is a free k[a1, a2, . . . , ag]
-module.
(3) The sequence a=a1, a2, . . . , ag is an M-sequence if and only if
F(M, λ) =∏Fr(M/aM), λ)
1=1(1−λdegai)
.
(4) depthM ≤dimM.
Definition 3.3. The length of longestM-sequence in an idealI,(resp. the max-imal homogeneous ideal of R) denoted by depthIM, (resp. depthM) is called the I−depth of M. The module M is called Cohen-Macaulay R-module if depthM = dimM.
Definition 3.4. Let M be a finitely generated graded module over a graded Noetherian ring R. By projective dimension pd(M) of M over R, we mean the smallest nonnegative integerpso that there is an exact sequence of freeR-modules of finite rank
0−→Fp−→ϕp Fp−1
ϕp−1
−→ · · · ϕ1
−→F0−→0
so that cokerϕ1=M.
Theorem 3.2(Auslander-Buchsbaum Formula). Let M be a finitely gener-ated graded module of finite projective dimension over a gradedk-algebra. Then
depthM + pdM = depthR. Theorem 3.3. The following are equivalent:
(1) M is Cohen-Macaulay.
(2) Every hsop for M is anM-sequence.
(3) M is a finitely generated free module over k[a] for some ( hence every )
hsop a.
Theorem 3.4. Let I be a proper ideal of a Cohen-Macaulay graded algebra R. Then
4. MACAULAY’S THEOREM ABOUT GORENSTEIN GRADED RINGS
The purpose of this section is to recall the basic definitions and facts about Gorenstein graded rings and provide a proof of Macaulay’s theorem concerning Hilbert series of Gorenstein graded rings. We begin by describing
The category of graded modules
Let R be an
N
-graded ring. Let M be the category ofZ
-graded R-modules. Consider theZ
-graded modulesM =⊕
n∈Z
Mn andN=
⊕
n∈Z
Nn ∈ M.
AnR-linear mapf :M −→N is a morphism inMiff(Mn)⊆Nn for alln∈
Z
.ByM(n) we mean the moduleM with grading defined by [M(n)]m=Mm+n for
allm∈
Z
.Put∗Hom(M, N)
n={f :M −→N(n)|f is degree preserving}.
and∗Hom(M, N) =⊕n∈Z∗Hom(M, N)n.It is easy to check that ifM is finitely
generated then∗Hom(M, N) = Hom(M, N).LetM be a graded R-module. Con-sider a projective resolution ofM in the category M:
P :· · · −→P2
ϕ2
−→P1
ϕ1
−→P0
ϕ0
−→M −→0.
Apply the contravariant functor ∗Hom(−, N) to the above resolution to get the complex
0 −−−−→ *Hom(P 0, N)
ϕ1⊗id
−−−−→ *Hom(P 1, N)
ϕ2⊗id
−−−−→ *Hom(P
2, N) −−−−→ · · ·.
Then∗Exti(M, N) = Kernel(ϕi+1⊗1)/Image(ϕi⊗1).
Proposition 4.1. Let A=k[x1, x2, . . . , xs]be polynomial ring over a field k. Let
I be a homogeneous ideal ofA. Let A/I be Cohen-Macaulay. Then
Exti(A/I, A)̸= 0 ⇐⇒ i=h=ht(I). ProofBy Auslander-Buchsbaum formula
pd(A/I) = depthA−dimA/I =s−(s−h) =h.
Write a graded minimal resolution ofA/I as anA-module:
0 −−−−→ Aβh −−−−→ Aβh−1 −−−−→ · · · −−−−→ Aβ1 −−−−→ A −−−−→ 0. Thus Exti(A/I, A) = 0 fori > h.SinceAis Cohen-Macaulay ,
depthI(A) = min{i: Exti(A/I, A)̸= 0}=h.
Definition 4.1. The A-module KA/I = Ext h
(A/I, A) is called the canonical moduleofA/I.The ringA/I is calledGorensteinifKA/I ≃A/I(a). for some a∈
Z
.The integerais called the a-invariant ofA/I.Macaulay’s theorem for Hilbert series of Gorenstein graded rings
Theorem 4.1. Put R = A/I and d = dim(R). Suppose degxi = ei ∈
P
for i= 1,2, . . . , s. Then as rational functions ofλF(KR, λ) = (−1)d F(R,1/λ) λ− ∑s
i=1ei.
Proof (Stanley)Write a minimal free resolution ofR as anA-module: 0 −−−−→ Mh −−−−→ϕh
Mh−1
ϕh−1
−−−−→ · · · −−−−→ M1
ϕ1
−−−−→ M0 −−−−→ 0.
Apply Hom(−, A) to the above resolution to get the complex: 0 −−−−→ Hom(M0, A)
ϕ∗0
−−−−→ · · · −−−−→ϕ∗h
Hom(Mh, A) −−−−→ 0.
ThusKR≃Hom(Mh, A)/Im(ϕ∗h).Hence we have the following minimal free
reso-lution forKR as anA-module: 0 −−−−→ Hom(M0, A)
ϕ∗0
−−−−→ · · · ϕ∗h
−−−−→ Hom(Mh, A) −−−−→ KR −−−−→ 0.
It is easy to see that for integersmandn,
F(M(n), λ) =λ−nF(M, λ) and Hom(A(m), A)≃A(−m).
Let rank(Mi) = βi, and Mi = ⊕βi
j=1A(−gij) for i = 0,1, . . . , h. Put D(λ) =
∏s
p=1(1−λ
ep) and Ni(λ) =∑βi
j=1λ
gij.Now we calculate the Hilbert series of R
andKR from their minimal free resolutions written above.
F(Mi, λ) =
βi ∑
j=1
F(A(−gij), λ) =
∑βi
j=1λ
gij ∏s
p=1(1−λep)
=Ni(λ)
D(λ). HenceF(R, λ) =∑hi=0Ni(λ)/D(λ)(−1)i.To findF(KR, λ),note that
F(KR, λ) = h
∑
i=0
(−1)i+hF(Mi∗, λ) =
h
∑
i=0
(−1)i+hNi(λ−1)/D(λ).
SinceD(λ−1) = (−1)sD(λ)λ−∑sp=1ep,we get
F(R,1/λ) =
h
∑
i=0
(−1)iNi(λ −1)
D(λ−1)
= (−1)s−hλ∑si=1eiF(KR, λ)
= (−1)dλ∑eiF(KR, λ).
Corollary 4.1 (Macaulay’s Theorem). If the ringR =A/I is Gorenstein of dimensiondthen for some σ∈
Z
,F(R,1/λ) = (−1)dλσF(R, λ).
IfR is standard Gorenstein with F(R, λ) = h0+h1λ+···+hgλg
(1−λ)d , and hg ̸= 0,then (1) hi=hg−i, for alli= 0,1, . . . , g.
(2) σ=d−g.
(3) Ifσ≥1,thenH(n) = dimRn is a polynomial for alln, (a) H(−i) = 0 for all i= 1,2, . . . ,(σ−1), and (b) H(n) = (−1)d−1H(−σ−n) for all n∈
Z
.Proof(1) and (2): Pute=∑si=1ei.SupposeRis Gorenstein. ThenKR≃R(a),
for somea∈
Z
.HenceF(KR, λ) =λ−aF(R, λ) = (−1)dλ−eF(R,1/λ).
HenceF(R, λ) =λa−e(−1)dF(R,1/λ).Now letR be standard Gorenstein. Write
F(R, λ) = (h0+h1λ+h2λ2+· · ·+hgλg)/(1−λ)d
wherehg̸= 0.Then
F(R,1/λ) = (−1)dλd−g(h0λg+h1λg−1+· · ·+hg)/(1−λ)d=λe−a(−1)dF(R, λ).
Henced−g=e−a=σandhi=hg−i for alli= 0,1, . . . , g.
(3) By theorem 2.1, it is clear that ifσ ≥1, then H(n) is a polynomial for all
n∈
Z
and as dimRn= 0 for alln <0, H(n) = 0 for alln=−1,−2, . . . ,−(σ−1),andH(−σ)̸= 0.We have for alln≥ −(σ−1),
H(n) =h0
(
n+d−1
d−1
)
+h1
(
n+d−2
d−1
)
+· · ·+hg
(
d−1 +n−g d−1
)
Now use the fact thathi =hg−i for all i = 1,2, . . . , g, and
(n p
)
= (−1)p(p−np−1),
as polynomials,
H(n) =
g
∑
i=0
hi
(
d−1 +n−i d−1
)
=
g
∑
i=0
hg−i
(
d−1−(d−1 +n−i)−1
d−1
)
(−1)d−1
= g ∑ i=0 hi (
g−i−n−1
d−1
)
(−1)d−1
= g ∑ i=0 hi (
d−σ−i−n−1
d−1
)
(−1)d−1 = (−1)d−1H(−σ−n).
Definition 4.2. The vector (h0, h1, . . . , hg) is called theh-vectorof the standard
graded algebraR.If the conditionhi =hg−i is satisfied for alli= 0, i, . . . , g then
we say that theh-vector ofRissymmetric.
Example4.1. The symmetry of theh-vector of a standard graded Cohen-Macaulay algebraRdoes not imply thatRis Gorenstein. We construct an example. Consider the idealI= (xyz, xw, zw) of the polynomial ringA=k[x, y, z, w].The idealI is generated by the maximal minors of the matrix
M =
[
0 −z x
w −yz 0
]
A resolution ofR=A/I as anA-module is:
0 −−−−→ A(−3)⊕A(−4) −−−−→f A(−3)⊕A(−2)2 −−−−→g A −−−−→ 0,
where the mapsf and gare defined as
f([r, s]) = [r, s]
[
0 −z x
w −yz 0
]
and g([r, s, t]) =rxyz+sxw+tzw.
It can be shown easily that the above sequence is a minimal resolution ofR.Hence by Auslander-Buchsbaum formula, depthR= depthA−pdR= 4−2 = 2 = dimR.
HenceRis Cohen-Macaulay. However it is not Gorenstein as the above resolution shows that rankKR= 2.The Hilbert series ofRcan be found from the resolution and it turns out to be (1 + 2λ+λ2)/(1−λ)2.Hence theh-vector ofRis symmetric
although it is not Gorenstein.
5. LOCAL COHOMOLOGY OF GRADED MODULES
We will need an alternative characterization of canonical modules of Cohen-Macaulay graded rings over fields in terms of local cohomology modules. Therefore we will introduce these modules and summarise their fundamental properties. For detailed proofs we refer the reader to [4]. We will also derive a formula for the difference between Hilbert function and Hilbert polynomial of a graded algebra over a field. This formula plays a crucial role in the solution.
LetR be a standard graded ring over a fieldk. Letm be the maximal homo-geneous ideal ofR.Letx=x1, x2, . . . , xrbe a sequence of homogeneous elements
ofRof positive degree. LetRf denote the localization ofRat the multiplicatively
closed set generated byf.For a
Z
-gradedR-moduleM, LetK(x∞, M) denote the complexr
⊗
i=1
(0−→R−→Rxi −→0)⊗M.
This is called thelocalisation complexofM. The components of this complex are displayed in the sequence:
0−→δ0
M −→δ1
r
∏
i=1
Mxi−→δ2 ∏
i<j Mxixj
δ3
−→ · · ·−→δr
Mx1x2...xr −→0.
The mapδj+1is defined as follows: Letm∈N =Mxi1xi2...xij.Put
[r]\ {i1, i2, . . . , ij}={l1, l2, . . . , lr−j}
withl1< l2<· · ·< lr−j.Letϕls :N −→Nxlsbe the natural map for 1≤s≤r−j.
Then
δj+1(m) =ϕl1(m)−ϕl2(m) +· · ·+ (−1)
r−jϕl
r−j(m).
The ith local cohomology module Hxi(M) is the ith cohomology module of the complexK(x∞, M).Hence
Hxi(M) =Hi(K(x∞, M)) = kerδi+1/imδi.
The local cohomology modules are
Z
-graded modules. These modules are depth sensitive.Lemma 5.1. If the sequence of finite gradedR-modules
0 −−−−→ M1 −−−−→ M2 −−−−→ M3 −−−−→ 0,
is exact, then we have the long exact sequences of local cohomology modules
0 −−−−→ Hx0(M1) −−−−→ Hx0(M2) −−−−→ Hx0(M2)
−−−−→ H1
x(M1) −−−−→ Hx1(M2) −−−−→ Hx1(M2) −−−−→ · · ·
Theorem 5.1. Let R be a standard graded ring with maximal graded ideal m
generated by homogeneous elements x1, x2, . . . , xr. Let M be a finitely generated gradedR-module. Then
depth(M) = min{i:Hxi(M)̸= 0}.
Corollary 5.1. The moduleM is Cohen-Macaulay if and only ifHi
x(M) = 0for alli <dimM.
Let ℓ(N) denote length of a module N. Let R = ⊕∞n=0Rn be a standard Noetherian graded ring over an artinian local ring R0. Let R+ denote the ideal
ofR generated by elements of positive degrees. LetM =⊕
n∈
Z
Mn be a finitelygenerated gradedR-module. LetHM(n) = ℓ(Mn) be the Hilbert function of M
andPM(n) be the corresponding Hilbert polynomial. The Hilbert series ofM is the formal power seriesF(M, λ) =∑
n∈
Z
ℓ(Mn)λn. We know that HM(n) = PM(n)
for all n >degF(M, λ). The graded local cohomology modules Hi
R+(M) can be used to find the differenceHM(n)−PM(n).Define the functionχM :
Z
−→Z
byχM(n) =
∞
∑
i=0
(−1)iℓ(HRi
+(M)n)
We have the following more precise result of Grothendieck and Serre.
Theorem 5.2. [Grothendieck-Serre] Let R and M be as above. Then for all
n∈
Z
,HM(n)−PM(n) =χM(n). Proof: First observe that if
0 −−−−→ M −−−−→ N −−−−→ P −−−−→ 0
is an exact sequence of
Z
-gradedR-modules, then by using the long exact sequence of local cohomology modules and additivity of length, we getχN =χM +χP.Weapply induction on d= dimM. If d= 0, then M =H0
R+(M), and PM(n) = 0. Hence the formula is true. Now let d >0. We may replace M by M/H0
R+(M) without any harm. Thus we have H0
R+(M) = 0. Hence there is an M-regular elementaof degree one inR.Consider the exact sequence
0 −−−−→ M(−1) µa
−−−−→ M −−−−→ M/aM −−−−→ 0.
Hence for alln∈
Z
,χM/aM(n) =χM(n)−χM(n−1) and HM/aM(n) =HM(n)−HM(n−1) ThusPM/aM(n) =PM(n)−PM(n−1).Putf(n) =HM(n)−PM(n).Then these equations show that the first difference ∆f(n) =f(n)−f(n−1) coincides with ∆(χM)(n).Sincef(n) =χM(n) = 0,for all largen,we havef(n) =χM(n) for all
6. LINEAR HOMOGENEOUS DIOPHANTINE EQUATIONS
Let xij; i, j = 1,2, . . . , n be indeterminates. The entries of an n×n magic
square are solutions to the following system of linear Diophantine equations:
x11+x12+· · ·+x1n= n
∑
j=1
xij fori= 2,3. . . , n. (6.1)
x11+x12+· · ·+x1n= n
∑
i=1
xij forj= 2,3. . . , n.
Thus the problem of counting magic squares is a special case of counting nonnegative integer solutions of a system of linear Diophantine equations. Let Φ be an r×n
Z
-matrix. Let x1, x2, . . . , xn be indeterminates. Let X denote thecolumn vector (x1, x2, . . . , xn)t.We are interested in the
N
-solutions to the systemΦX = 0.We gather all the solutions in the monoid
EΦ={β∈
N
n
: Φβ= 0}.
Letkbe any field. Putxβ=xβ1
1 x
β2
2 · · ·x
βn
n .WithEΦwe can associate the monoid
algebra
RΦ=k[xβ:β ∈EΦ].
Stanley studied the monoid algebra Rµ where µ is the (2n−2)×n2 coefficient
matrix of the system (6.1). In particular he showed that the ringRµis a Gorenstein and calculated its canonical module and thus itsa-invariant. We shall see that these observations are enough to settle the conjectures (1)-(5). Let us begin by observing the
Theorem 6.1. The monoid algebraRΦis a finitely generatedk-algebra.
Proof: LetI denote the ideal inR=k[x1, x2, . . . , xn] generated by the set
P ={xβ : 0̸=β ∈EΦ}.
SinceRis Noetherian,I is generated by a finite subsetG={xδ1, xδ2, . . . , xδt} of
P. We claim that
RΦ=k[xδ :xδ ∈G].
Indeed, anyxβ∈R
Φ can be written as xβ =xδixγ for someiand xγ ∈R.Thus
γ =β −δi ∈ EΦ. The argument can be repeated for xγ, eventually yielding an
expression forxβ in terms ofxδi fori= 1,2, . . . , t.
As far as the structure of Rµ is concerned, we have more precise information due to
Thus Rµ is generated by n! degree n monomials. Let [Rµ]r denote the k
-subspace of Rµ generated by the monomials of degree nr. These monomials are
in one-to-one correspondence with magic squares of line sumr. Moreover, Rµ =
⊕∞
r=0[Rµ]r.Thus
H(Rµ, r) = dimk[Rµ]r=Hn(r).
This observation will eventually lead to the conclusion thatHn(r) is a polynomial inrfor allr.But for the time being we can see that it is so for all large values of
rin view of the Hilbert-Serre theorem.
Lemma 6.1. IfΦX = 0has a positive solution, thendimRΦ=n−rank Φ.
Proof: We show that the vectorsβ1, β2, . . . , βd∈EΦ are
Q
-linearly independentif and only ifxβ1, xβ2, . . . , xβdare algebraically independent over k.Suppose that
β1, β2, . . . , βd∈EΦ are linearly independent over
Q
.Let∑
α
aα(xβ1)α1(xβ2)α2. . .(xβd)αd= 0,
for certainaα ∈k and distinct vectors α= (α1, . . . , αd)∈
N
d
. Asβ1, β2, . . . , βd
are linearly independent over
Q
,the vectorsα1β1+· · ·+αdβd are distinct. Hence aα= 0 for allα.Conversely let xβ1, xβ2, . . . , xβd be algebraically independent over k.Suppose
thatα1, . . . , αd ∈
Q
such that α1β1+· · ·+αdβd = 0.Without loss of generalitywe may assume thatα1, α2, . . . , αp>0 andαp+1, . . . , αd<0.Then α1β1+· · ·+αpβp=αp+1βp+1+· · ·+αdβd.
This yields the relationxα1β1· · ·xαpβp =xαp+1βp+1· · ·xαdβd.
Let α ∈
P
n∩EΦ. Let d = n−rank Φ. Pick linearly independent solutionsβ1, β2, . . . , βd ∈
Z
nof ΦX = 0. Let t ∈
P
. If α−tβ1, α−tβ2, . . . , α−tβd arelinearly dependent over
Q
,there exista1, a2, . . . , ad∈Z
,not all zero such thata1(α−tβ1) +a2(α−tβ2) +· · ·+ad(α−tβd) = 0.
We haveb1, b2, . . . , bd∈
Q
such thatα=b1β1+b2β2+· · ·+bdβd.Hencea1(b1−t)β1+a2(b2−t)β2+· · ·+ad(bd−t)βd= 0.
Hence by selectingt ∈
P
sufficiently small, we get ai = 0, for alli = 1,2. . . , d.This contradiction proves thatα−tβ1, α−tβ2, . . . , α−tβdare linearly independent
7. STANLEY’S SOLUTION OF THE ADG CONJECTURES Theorem 7.1. Hn(r)is a polynomial of degree (n−1)2 for larger.
Proof: The ringRµis a standard gradedk-algebra. Therthgraded component of
it is generated by monomials of degreerncorresponding to magic squares of line sumr.Hence by Theorem 2.1, degHn(r) = dimRµ−1 andHn(r) is a polynomial for larger.We show that
dimRµ= (n−1)2+ 1.
By the above lemma, dimRµ= nullity Φ.Note that to construct a magic square,
we may assign any nonnegative values to the variablesxij,fori, j= 1,2, . . . ,(n−1) and a value forx1n will determine the rest of the entries of it. Thus nullity Φ =
(n−1)2+ 1.
The next important step in the solution is the observation that the graded ringRΦis Cohen-Macaulay.
This is proved by showing that it is a ring of invariants of an algebraic torus acting linearly on a polynomial ring. A well-known theorem of Hochster then implies that it is Cohen-Macaulay.
Letk∗ denote the multiplicative group ofk.Consider the algebraic torus
T ={diag(uγ1, uγ2, . . . , uγn) :u= (u
1, u2, . . . , ur)∈(k∗)r}.
Let us write ther×nmatrix Φ = [γ1, γ2, . . . , γn] whereγiis theith column vector
of Φ. T acts onR=k[x1, x2, . . . , xn] via the automorphisms
τu:xi −→uγixi, i= 1,2, . . . , n.
Letβ ∈
N
n.Thenτu(xβ) = (uγ1x
1)β1(uγ2x2)β2· · ·(uγnxn)βn
=uβ1γ1+β2γ2+···βnγnxβ
Henceτu(xβ) =xβ if and only ifβ∈E
Φ.HenceRΦis the ring of invariants of the
torusT acting linearly onR.By Hochster’s theorem [6] RΦis Cohen-Macaulay.
We can now dispose the conjecture (5) of Stanley. SinceRµis Cohen-Macaulay homogenous ring of dimensiond= (n−1)2+ 1,there exists an hsop aforR
µ of
elements of degree one. Hence
F(Rµ, λ) =F(Rµ/(a), λ) (1−λ)d .
Finally we sketch the solutions of the parts (1)-(4) of the ADG-Conjecture By Corollary 4.1, we need to show that the degree of the Hilbert series ofRµ is−n
and it is Gorenstein. This is done by computing its canonical module. Stanley showed that the canonical module ofRΦis preciselyk[xβ:β∈
P
n
∩EΦ].Thus if
γ= (1,1, . . . ,1)∈EΦ, then
KRΦ =xγRΦ.
Hence in this case, RΦ is Gorenstein. For the case of magic squares, then×n
magic square whose each entry is 1 is the smallest positive solution. ThusKRµ = ∏
1≤i,j≤nxijRµ. Since the degree of
∏
1≤i,j≤nxij as an element of Rµ is n, the
degree of its Hilbert series is −n. It proves that Hn(r) is a polynomial for all r >−n.MoreoverHn(−n)̸= 0 and
Hn(−1) =Hn(−2) =· · ·=Hn(−(n−1)) = 0.
Corollary 4.1, we conclude that
Hn(r) = (−1)(n−1)2Hn(−r−n),
for alln.But (n−1)2 and (n−1) have same parity. Thus for allr
Hn(r) = (−1)n−1Hn(−r−n),
References
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J. K. VERMA.
DEPARTMENT OF MATHEMATICS,
INDIAN INSTITUTE OF TECHNOLOGY BOMBAY, MUMBAI - 400 076, INDIA.