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Article

Cohomology Algebras of a Family of DG Skew Polynomial Algebras

by 1,* and 2
1
Department of Mathematics, Shanghai University, Shanghai 200444, China
2
School of Economics, Shanghai University, Shanghai 200444, China
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(7), 1617; https://doi.org/10.3390/math11071617
Received: 27 January 2023 / Revised: 18 March 2023 / Accepted: 22 March 2023 / Published: 27 March 2023
(This article belongs to the Special Issue Algebra and Discrete Mathematics 2023)

Abstract

:
Let A be a connected cochain DG algebra such that its underlying graded algebra A # is the graded skew polynomial algebra k x 1 , x 2 , x 3 / x 1 x 2 + x 2 x 1 x 2 x 3 + x 3 x 2 x 3 x 1 + x 1 x 3 , | x 1 | = | x 2 | = | x 3 | = 1 . Then the differential A is determined by A ( x 1 ) A ( x 2 ) A ( x 3 ) = M x 1 2 x 2 2 x 3 2 for some M M 3 ( k ) . When the rank r ( M ) of M belongs to { 1 , 2 , 3 } , we compute H ( A ) case by case. The computational results in this paper give substantial support for the research of the various homological properties of such DG algebras. We find some examples, which indicate that the cohomology graded algebras of such kind of DG algebras may be not left (right) Gorenstein.

1. Introduction

In the literature, Koszul, homologically smooth, Gorenstein and Calabi-Yau properties of cochain DG algebras have been frequently studied. In general, these homological properties are difficult to detect. For a non-trivial DG algebra A , the trivial DG algebra H ( A ) is much simpler to study since it has zero differential. There have been some attempts to judge the various homological properties of A from H ( A ) . It is shown in [1,2,3] that a connected cochain DG algebra A is a Kozul Calabi-Yau DG algebra if H ( A ) belongs to one of the following cases:
( a ) H ( A ) k ; ( b ) H ( A ) = k [ z ] , z ker ( A 1 ) ; ( c ) H ( A ) = k z 1 , z 2 ( z 1 z 2 + z 2 z 1 ) , z 1 , z 2 ker ( A 1 ) .
A more general result is proved in [4] that A is Calabi-Yau if the trivial DG algebra ( H ( A ) , 0 ) is Calabi-Yau. In particular, A is a Calabi-Yau DG algebra if
H ( A ) = k x , y , z / a y z + b z y + c x 2 a z x + b x z + c y 2 a x y + b y x + c z 2 ,
where ( a , b , c ) P k 2 D and x , y , z ker ( A 1 ) . By [5] (Proposition 6.2 ), A is not a Gorenstein DG algebra but a Koszul and homologically smooth DG algebra if H ( A ) = k y 1 , , y n , for some degree 1 cocycle elements y 1 , , y n in A . In addition, [6] (Proposition 6.5 ) indicates that A is Calabi-Yau if H ( A ) = k [ z 1 , z 2 ] , where z 1 ker ( A 1 ) and z 2 ker ( A 2 ) . In [7], it is proved that A is a Koszul homologically smooth DG algebra if H ( A ) = k [ y 1 , , y m ] , for some central, cocycle and degree 1 elements y 1 , , y m in A . Moreover, A is 0-Calabi-Yau if and only if m is an odd integer. It is proved in [1] (Proposition 4.3) that A is a Koszul and Calabi-Yau DG algebra if
H ( A ) = k y 1 , y 2 / ( t 1 y 1 2 + t 2 y 2 2 + t 3 ( y 1 y 2 + y 2 y 1 ) )
with y 1 , y 2 Z 1 ( A ) and ( t 1 , t 2 , t 3 ) P k 2 { ( t 1 , t 2 , t 3 ) | t 1 t 2 t 3 2 0 } . These results indicate that it is worthwhile to compute the cohomology algebra of a given DG algebra if one wants to study its homological properties.
Recently, the constructions and studies on some specific family of connected cochain DG algebras have attracted much attention. In [5,6,7], DG down–up algebras, DG polynomial algebras and DG-free algebras are introduced and systematically studied, respectively. It is exciting to discover that non-trivial DG down–up algebras and DG free algebras with 2 degree 1 variables are Calabi-Yau DG algebras. It seems to be a good way to construct some interesting homologically smooth DG algebras on AS-regular algebras. The notion of AS-regular algebras was introduced by Artin-Schelter in [8]. AS-regular algebras are thought to be the coordinate rings of the corresponding non-commutative projective spaces in the non-commutative projective geometry (cf. [9,10,11]). One of the central questions in non-commutative projective geometry is to classify non-commutative projective spaces, or equivalently, to classify the corresponding Artin–Schelter regular algebras. In the last twenty years, they have been intensively studied in the literature (cf. [12,13,14,15,16,17,18,19,20]).
Let D be the subset of the projective plane P k 2 consisting of the 12 points:
D : = { ( 1 , 0 , 0 ) , ( 0 , 1 , 0 ) , ( 0 , 0 , 1 ) } { ( a , b , c ) | a 3 = b 3 = c 3 } .
Recall that the points ( a , b , c ) P k 2 D parametrize the 3-dimensional Sklyanin algebras,
S a , b , c = k x 1 , x 2 , x 3 ( f 1 , f 2 , f 3 ) ,
where
f 1 = a x 2 x 3 + b x 3 x 2 + c x 1 2 f 2 = a x 3 x 1 + b x 1 x 3 + c x 2 2 f 3 = a x 1 x 2 + b x 2 x 1 + c x 3 2 .
The 3-dimensional Sklyanin algebras form the most important class of Artin-Schelter regular algebras of global dimension 3 (cf. [21,22,23,24,25]). We say that a cochain DG algebra A is a 3-dimensional Sklyanin DG algebra if its underlying graded algebra A # is a 3-dimensional Sklyanin algebra S a , b , c for some ( a , b , c ) P k 2 D . In [2], all possible differential structures on 3-dimensional DG Sklyanin algebras are classified. By [2] (Theorem A), A = 0 when | a | | b | or c 0 . Note that A 0 only if either a = b , c = 0 or a = b , c = 0 . When a = b , c = 0 , the 3-dimensional DG Sklyanin algebras A is just a DG polynomial algebra, which is systematically studied in [7]. For the case a = b , c = 0 , the differential A is defined by
A ( x 1 ) A ( x 2 ) A ( x 3 ) = M x 1 2 x 2 2 x 3 2 , for some M M 3 ( k ) .
In this case, the 3-dimensional DG Sklyanin algebra is just A O 1 ( k 3 ) ( M ) in [1]. Note that such 3-dimensional DG Sklyanin algebras are actually a family of cochain DG skew polynomial algebras. The motivation of this paper is to compute H ( A ) when the rank r ( M ) of M belongs to { 1 , 2 , 3 } .
For any M M 2 ( k ) , one sees that H [ A O 1 ( k 2 ) ( M ) ] is always AS-Gorenstein by [26]. In addition, each DG algebra A O 1 ( k 2 ) ( M ) is a Koszul Calabi-Yau DG algebra by [3] (Theorem C). It is natural for us to put forward the following conjecture.
Conjecture 1. 
For any M M 3 ( k ) , H ( A O 1 ( k 3 ) ( M ) ) is a left (right) Gorenstein graded algebra.
Finally, we give a concrete counterexample to disprove Conjecture 1 (see Example 1). More generally, we have the following theorem (see Theorem 2).
Theorem 1. 
Let A be a connected cochain DG algebra such that
A # = k x 1 , x 2 , x 3 / x 1 x 2 + x 2 x 1 x 2 x 3 + x 3 x 2 x 3 x 1 + x 1 x 3 , | x 1 | = | x 2 | = | x 3 | = 1 ,
and A is determined by
A ( x 1 ) A ( x 2 ) A ( x 3 ) = N x 1 2 x 2 2 x 3 2 .
Then, the graded algebra H ( A ) is not left (right) Gorenstein if and only if there exists some C = ( c i j ) 3 × 3 QPL 3 ( k ) satisfying N = C 1 M ( c i j 2 ) 3 × 3 , where
M = 1 1 0 1 1 0 1 1 0 or M = m 11 m 12 m 13 l 1 m 11 l 1 m 12 l 1 m 13 l 2 m 11 l 2 m 12 l 2 m 13
with m 12 l 1 2 + m 13 l 2 2 m 11 , l 1 l 2 0 and 4 m 12 m 13 l 1 2 l 2 2 = ( m 12 l 1 2 + m 13 l 2 2 m 11 ) 2 .
Here, QPL n ( k ) is the set of quasi-permutation matrixes in GL n ( k ) . Recall that a square matrix is called a quasi-permutation matrix if each row and each column has at most one non-zero element (cf. [27]). By [1] (Lemma 3.3), a matrix M = ( m i j ) n × n in GL n ( k ) is a quasi-permutation if and only if m i r m j r = 0 , for any 1 i < j n and r { 1 , 2 , , n } .

2. Preliminaries

2.1. Notations and Conventions

Throughout this paper, k is an algebraically closed field of characteristic 0. For any k-vector space V, we write V = Hom k ( V , k ) . Let { e i | i I } be a basis of a finite dimensional k-vector space V. We denote the dual basis of V by { e i * | i I } , i.e., { e i * | i I } is a basis of V such that e i * ( e j ) = δ i , j . For any graded vector space W and j Z , the j-th suspension Σ j W of W is a graded vector space defined by ( Σ j W ) i = W i + j .
A cochain DG algebra is a graded k-algebra A together with a differential A : A A of degree 1 such that
A ( a b ) = ( A a ) b + ( 1 ) | a | a ( A b )
for all homogeneous elements a , b A . We write A o P for its opposite DG algebra, whose multiplication is defined as a · b = ( 1 ) | a | · | b | b a for all homogeneous elements a and b in A . Let A be a cochain DG algebra. We denote by A i its i-th homogeneous component. The differential A is a sequence of linear maps A i : A i A i + 1 such that A i + 1 A i = 0 , for all i Z . If A 0 , A is called non-trivial. The cohomology graded algebra of A is the graded algebra
H ( A ) = i Z ker ( A i ) im ( A i 1 ) .
Let z ker ( A i ) be a cocycle element of degree i. We write z for the cohomology class in H ( A ) represented by z. If A 0 = k and A i = 0 , i < 0 , then we say that A is connected. One sees that H ( A ) is a connected graded algebra if A is a connected cochain DG algebra. Let A be a connected cochain DG k-algebra. We write m as the maximal DG ideal A > 0 of A . Via the canonical surjection ε : A k , k is both a DG A -module and a DG A o P -module. It is easy to check that the enveloping DG algebra A e = A A o P of A is also a connected cochain DG algebra with H ( A e ) H ( A ) e , and
m A e = m A A o P + A m A o P .
The derived category of left DG modules over A (DG A -modules for short) is denoted by D ( A ) . A DG A -module M is compact if the functor Hom D ( A ) ( M , ) preserves all coproducts in D ( A ) [28,29,30,31]. By [32] (Proposition 3.3), a DG A -module is compact if and only if it admits a minimal semi-free resolution with a finite semi-basis. The full subcategory of D ( A ) consisting of compact DG A -modules is denoted by D c ( A ) . The right derived functor of Hom is denoted by R Hom , and the left derived functor of ⊗ is denoted by L . They can be computed via K-projective, K-injective and K-flat resolution of the DG modules. For any M , N D ( A ) and L D ( A o P ) , let F M , N I and P L be a K-projective resolution of M, K-injective resolution of N and K-flat resolution of L, respectively. Then, we have R Hom A ( M , N ) = Hom A ( F , N ) Hom A ( M , I ) and L L A M = P A M (cf. [33,34,35,36]).
In the rest of this subsection, we review some important homological properties for DG algebras.
Definition 1. 
Let  A be a connected cochain DG algebra.
1 . 
If dim k H ( R Hom A ( k , A ) ) = 1 (resp. dim k H ( R Hom A o P ( k , A ) ) = 1 ) , then A is called the left (resp. right) Gorenstein (cf. [37]);
2 . 
If A k , or equivalently A e A , has a minimal semi-free resolution with a semi-basis concentrated in degree 0, then A is called Koszul (cf. [38]);
3 . 
If A k , or equivalently the DG A e -module A is compact, then A is called homologically smooth (cf. [39] (Corollary 2.7));
4 . 
If A is homologically smooth and R Hom A e ( A , A e )   Σ n A in the derived category D ( ( A e ) o P ) of right DG A e -modules, then A is called an n-Calabi-Yau DG algebra (cf. [40,41]).
Note that the DG algebras considered in this paper are not graded commutative in general. We should distinguish between left and right Gorenstein properties. To extend the rich theory of commutative Gorenstein rings to DG algebras, people have completed a lot of work. We refer to [33,35,42,43,44] for more details on them.

2.2. AS-Gorenstein (AS-Regular) Graded Algebras

In this subsection, we let A be a connected graded algebra. We have the following definitions on AS-Gorenstein graded algebras and AS-regular graded algebras [45,46,47].
Definition 2. 
We say that A is left (resp. right) Gorenstein if dim k Ext A * ( k , A ) = 1 (resp. dim k Ext A o P * ( k , A ) = 1 ), where Ext A * ( k , A ) = i Z Ext A i ( k , A ) . For a left Gorenstein graded algebra A, there is some integer l such that
Ext A i ( k , A ) = 0 , i depth A A , k ( l ) , i = depth A A .
A left (resp. right) Gorenstein graded algebra A is called left (resp. right) AS-Gorenstein (AS stands for Artin-Schelter) if its left injective dimension id A A < (resp. right injective dimension id A o P A < ). If further, its global dimension gl . dim A < , then we say A is left (resp. right) AS-regular.
Lemma 1. 
Let A be a Noetherian and AS-Gorenstein graded algebra. Then, the graded algebra B = A [ x ] with | x | = 2 is also a Noetherian and AS-Gorenstein graded algebra.
Proof. 
By the well-known `Hilbert basis Theorem’, one sees that B is Noetherian. We have B = A k [ x ] . Let P and Q be the finitely generated minimal free resolutions of A k and k [ x ] k , respectively. Then, P Q is a finitely generated minimal free resolution of B k . We have
H ( Hom B ( P Q , B ) ) = H ( Hom A k [ x ] ( P Q , A k [ x ] ) ) H ( Hom A ( P , Hom k [ x ] ( Q , A k [ x ] ) ) ) H ( Hom A ( P , A Hom k [ x ] ( Q , k [ x ] ) ) H ( Hom A ( P , A ) Hom k [ x ] ( Q , k [ x ] ) ) H ( Hom A ( P , A ) ) H ( Hom k [ x ] ( Q , k [ x ] ) ) .
Since A and k [ x ] are both AS-Gorenstein, we have
dim k Ext B * ( k , B ) = dim k H ( Hom B ( P Q , B ) ) = 1 .
Thus, B = A [ x ] is left AS-Gorenstein. We can similarly show that B = A [ x ] is right AS-Gorenstein. □
Lemma 2. 
Let A be a connected graded algebra such that
A = k x , y ( a x 2 + a b ( x y + y x ) + b y 2 ) , a b > 0 , | x | = | y | = 1 .
Then, A is not left (right) Gorenstein.
Proof. 
The trivial module A k admits a finitely generated minimal free resolution
d n + 1 F n d n F n 1 d n 1 d 3 F 2 d 2 F 1 = A e x A e y d 1 A ε A k 0 ,
where
F n 1 = A e n 1 , d n ( e n ) = ( a x + a b y ) e n 1 , n 3 ; d 2 ( e 2 ) = ( a x + a b y ) e x + ( a b x + b y ) e y , d 1 ( e x ) = x , d 1 ( e y ) = y .
Acting the functor Hom A ( , A ) on the deleted complex of the minimal free resolution above, we obtain the complex
0 1 * A d 1 * e x * A e y * A d 2 * e 2 * A d 3 * e 3 * A d 4 * d n * e n * A d n + 1 * ,
where
d 1 * ( 1 * ) = e x * x + e y * y ; d 2 * ( e x * ) = e r * ( a x + a b y ) , d 2 * ( e y * ) = e r * ( a b x + b y ) ; d i + 1 * ( e i * ) = e i + 1 * ( a x + a b y ) , i 2 .
We have
Ext A 0 ( k , A ) = ker ( d 1 * ) = 0 ; Ext A 1 ( k , A ) = ker ( d 2 * ) im ( d 1 * ) = ( b a e x * e y * ) A ( e x * x + e y * y ) A ( e x * x + e y * y ) A ( b a e x * e y * ) A ; Ext A i ( k , A ) = ker ( d i + 1 * ) im ( d i * ) = e i * ( a x + a b y ) A e i * ( a x + a b y ) A = 0 , i 2 .
Obviously, dim k Ext A * ( k , A ) 1 and hence A is not left Gorenstein, similarly, we can show that A is not right Gorenstein. □
Lemma 3. 
Let A be a connected graded algebra such that
A = k x , y ( a x 2 + b y 2 ) , a b = 0 , ( a , b ) ( 0 , 0 ) , | x | = | y | = 1 .
Then, A is not left (right) Gorenstein.
Proof. 
Without the loss of generality, we assume that a = 0 , b 0 . The trivial module A k admits a finitely generated minimal free resolution
d n + 1 F n d n F n 1 d n 1 d 3 F 2 d 2 F 1 = A e x A e y d 1 A ε A k 0 ,
where
F n = A e n , d n ( e n ) = ( b y ) e n 1 , n 3 ; d 2 ( e 2 ) = ( b y ) e y , d 1 ( e x ) = x , d 1 ( e y ) = y .
Acting the functor Hom A ( , A ) on the deleted complex of the minimal free resolution above, we obtain the complex
0 1 * A d 1 * e x * A e y * A d 2 * e 2 * A d 3 * e 3 * A d 4 * d n * e n * A d n + 1 * ,
where
d 1 * ( 1 * ) = e x * x + e y * y ; d 2 * ( e x * ) = 0 , d 2 * ( e y * ) = e r * ( b y ) ; d i + 1 * ( e i * ) = e i + 1 * ( b y ) , i 2 .
Ext A 0 ( k , A ) = ker ( d 1 * ) = 0 ; Ext A 1 ( k , A ) = ker ( d 2 * ) im ( d 1 * ) = e x * A ( e x * x + e y * y ) A ( e x * x + e y * y ) A e x * A ; Ext A i ( k , A ) = ker ( d i + 1 * ) im ( d i * ) = e i * ( b y ) A e i * ( b y ) A = 0 , i 2 .
Since dim k Ext A * ( k , A ) 1 , A is not left Gorenstein. Similarly, we can show that A is not right Gorenstein. □

3. Some Basic Lemmas

In this section, we give some simple lemmas, which will be used in the subsequent computations. If no special assumption is emphasized, we let A be a DG Sklyanin algebra with A # = S a , a , 0 , and A is determined by a matrix M in M 3 ( k ) .
Lemma 4. 
For any t N , x 1 2 t , x 2 2 t , x 3 2 t are cocycle central elements of A .
Proof. 
One sees that x i 2 is a central element of A since
x i 2 x j = x i x i x j = x i x j x i = x j x i 2 ,
when i j . This implies that each x i 2 t is a central element of A . We have
A ( x i 2 ) = A ( x i ) x i x i A ( x i ) = j = 1 n m i j x j 2 x i x i j = 1 n m i j x j 2 = j = 1 n m i j ( x j 2 x i x i x j 2 ) = 0 .
Using this, we can inductively prove A ( x i 2 t ) = 0 . □
Lemma 5. 
Let Ω be a coboundary element in A of degree d 3 .
( 1 ) If d = 2 l + 1 is odd, then Ω = A [ x 1 x 2 f + x 1 x 3 g + x 2 x 3 h ] , where f , g and h are all linear combinations of monomials with non-negative even exponents.
( 2 ) If d = 2 l is even, then Ω = A [ x 1 f + x 2 g + x 3 h + x 1 x 2 x 3 u ] , where f , g , h and u are all linear combinations of monomials with non-negative even exponents.
Proof. 
By the assumption, we have
Ω = A [ l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 ] .
If d = 2 l + 1 is odd, then d = 2 l is even. Since
l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 are odd , l 3 is even C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 + l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 3 are odd , l 2 is even C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 + l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 2 , l 3 are odd , l 1 is even C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 + l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 , l 3 are even C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 ,
we have
Ω = A [ l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 ] = A [ x 1 x 2 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 are odd , l 3 is even C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 1 x 3 l 3 ] + A [ x 1 x 3 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 3 are odd , l 2 is even C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 x 3 l 3 1 ] + A [ x 2 x 3 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 2 , l 3 are odd , l 1 is even C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 1 x 3 l 3 1 ]
by Lemma 4. Let
f = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 are odd , l 3 is even C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 1 x 3 l 3 , g = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 3 are odd , l 2 is even C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 x 3 l 3 1 , h = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 2 , l 3 are odd , l 1 is even C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 1 x 3 l 3 1 .
This proves (1).
If d = 2 l is even, then d 1 = 2 l 1 is odd. Since
l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 are even , l 3 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 + l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 3 are even , l 2 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 + l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 2 , l 3 are even , l 1 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 + l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 , l 3 are odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 ,
we have
Ω = A [ l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 ] = A [ x 3 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 are even , l 3 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 1 ] + A [ x 2 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 3 are even , l 2 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 1 x 3 l 3 ] + A [ x 1 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 2 , l 3 are even , l 1 is odd C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 x 3 l 3 ] + A [ x 1 x 2 x 3 l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 , l 3 are odd C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 1 x 3 l 3 1 ] .
Let
f = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 2 , l 3 are even , l 1 is odd C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 x 3 l 3 , g = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 3 are even , l 2 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 1 x 3 l 3 , h = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 are even , l 3 is odd C l 1 , l 2 , l 3 x 1 l 1 x 2 l 2 x 3 l 3 1 , u = l 1 + l 2 + l 3 = d 1 l 1 , l 2 , l 3 0 l 1 , l 2 , l 3 are odd C l 1 , l 2 , l 3 x 1 l 1 1 x 2 l 2 1 x 3 l 3 1 .
This proves ( 2 ) . □
Lemma 6. 
Let M = ( m i j ) 3 × 3 be a matrix in GL 3 ( k ) . Then, x 1 2 , x 2 2 , x 3 2 are coboundary elements in A .
Proof. 
For a 1 , a 2 , a 3 k , we have
A ( c 1 x 1 + c 2 x 2 + c 3 x 3 ) = a 1 ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) + a 2 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) + a 3 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) = ( a 1 m 11 + a 2 m 21 + a 3 m 31 ) x 1 2 + ( a 1 m 12 + a 2 m 22 + a 3 m 32 ) x 2 2 + ( a 1 m 13 + a 2 m 23 + a 3 m 33 ) x 3 2 .
So, A ( a 1 x 1 + a 2 x 2 + a 3 x 3 ) = x 1 2 if and only if
a 1 m 11 + a 2 m 21 + a 3 m 31 = 1 a 1 m 12 + a 2 m 22 + a 3 m 32 = 0 a 1 m 13 + a 2 m 23 + a 3 m 33 = 0 M T a 1 a 2 a 3 = 1 0 0 .
Since r ( M ) = 3 , there exists
a 1 = m 22 m 33 m 23 m 32 | M | a 2 = m 13 m 32 m 12 m 33 | M | a 3 = m 12 m 23 m 13 m 22 | M |
such that A ( a 1 x 1 + a 2 x 2 + a 3 x 3 ) = x 1 2 . Similarly, we can show there exist
b 1 = m 23 m 31 m 21 m 33 | M | b 2 = m 11 m 33 m 13 m 31 | M | b 3 = m 13 m 21 m 11 m 23 | M | and c 1 = m 21 m 32 m 22 m 31 | M | c 2 = m 12 m 31 m 11 m 32 | M | c 3 = m 11 m 22 m 12 m 21 | M |
such that A ( b 1 x 1 + b 2 x 2 + b 3 x 3 ) = x 2 2 and A ( c 1 x 1 + c 2 x 2 + c 3 x 3 ) = x 3 2 , respectively. □
Lemma 7. 
Let M = ( m i j ) 3 × 3 be a matrix in GL 3 ( k ) and m 22 m 33 m 23 m 32 0 . If g ( x 2 ¯ , x 3 ¯ ) Z 2 l + 1 [ A / ( x 1 2 ) ] and h ( x 2 ¯ , x 3 ¯ ) Z 2 l [ A / ( x 1 2 ) ] are sum of monomials in variables x 2 ¯ and x 3 ¯ with l 1 . Then
h ( x 2 ¯ , x 3 ¯ ) = i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i with r 2 i k , 0 i l .
Furthermore, there exist u ( x 2 , x 3 ) and v ( x 2 , x 3 ) , which are sums of monomials in variables x 2 and x 3 , such that
g ( x 2 ¯ , x 3 ¯ ) = A [ u ( x 2 , x 3 ) ] ¯ , h ( x 2 ¯ , x 3 ¯ ) = A [ v ( x 2 , x 3 ) ] ¯ .
Proof. 
Let g ( x 2 ¯ . x 3 ¯ ) = j = 0 2 l + 1 t j x 2 ¯ 2 l + 1 j x 3 ¯ j and h ( x 2 ¯ , x 3 ¯ ) = j = 0 2 l r j x 2 ¯ 2 l j x 3 ¯ j , where each t j , r j k . Then
0 = A ( j = 0 2 l + 1 t j x 2 2 l + 1 j x 3 j ) ¯ = A ( i = 0 l t 2 i x 2 2 l 1 2 i x 3 2 i + i = 1 l + 1 t 2 i 1 x 2 2 l 2 i x 3 2 i 1 ) ¯ = i = 0 l [ t 2 i ( m 22 x 2 ¯ 2 + m 23 x 3 ¯ 2 ) x 2 ¯ 2 l 2 i 2 x 3 ¯ 2 i + t 2 i + 1 x 2 ¯ 2 l 2 i 2 x 3 ¯ 2 i ( m 32 x 2 ¯ 2 + m 33 x 3 ¯ 2 ) ] = i = 0 l [ ( t 2 i m 22 + t 2 i + 1 m 32 ) x 2 ¯ 2 l 2 i x 3 ¯ 2 i + ( t 2 i m 23 + t 2 i + 1 m 33 ) x 2 ¯ 2 l 2 i 2 x 3 ¯ 2 i + 2 ]
and
0 = A ( j = 0 2 l r j x 2 2 l j x 3 j ) ¯ = i = 1 l r 2 i 1 [ ( m 22 x 2 ¯ 2 + m 23 x 3 ¯ 2 ) x 2 ¯ 2 l 2 i x 3 ¯ 2 i 1 x 2 ¯ 2 l 2 i + 1 x 3 ¯ 2 i 2 ( m 32 x 2 ¯ 2 + m 33 x 3 ¯ 2 ) ] .
They imply
t 0 m 22 + t 1 m 32 = 0 t 2 m 22 + t 3 m 32 + t 0 m 23 + t 1 m 33 = 0 t 4 m 22 + t 5 m 32 + t 2 m 23 + t 3 m 33 = 0 . . . . . . . . t 2 l 2 m 22 + t 2 l 1 m 32 + t 2 l 4 m 23 + t 2 l 3 m 33 = 0 t 2 l m 22 + t 2 l + 1 m 32 + t 2 l 2 m 23 + t 2 l 1 m 33 = 0 t 2 l m 23 + t 2 l + 1 m 33 = 0
and
r 1 m 32 = 0 r 1 m 22 = 0 r 1 m 33 + r 3 m 32 = 0 r 1 m 23 + r 3 m 22 = 0 . . . . . . . . r 2 l 3 m 33 + r 2 l 1 m 32 = 0 r 2 l 3 m 23 + r 2 l 1 m 22 = 0 r 2 l 1 m 33 = 0 r 2 l 1 m 23 = 0 .
Since m 22 m 33 m 23 m 32 0 , the rank of the system matrix
m 22 m 32 0 0 0 0 0 0 0 0 0 0 0 0 m 23 m 33 m 22 m 32 0 0 0 0 0 0 0 0 0 0 0 0 m 23 m 33 m 22 m 32 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 m 23 m 33 m 22 m 32 0 0 0 0 0 0 0 0 0 0 0 0 m 23 m 33 m 22 m 32 0 0 0 0 0 0 0 0 0 0 0 0 m 23 m 33
of (2) is l + 2 . Hence, the space of the solutions of (2) is of dimension l. On the other hand, for any 1 i l , A ( x 2 2 l 2 i + 1 x 3 2 i 1 ) ¯ is
m 32 x 2 ¯ 2 l 2 i + 3 x 3 ¯ 2 i 2 + m 22 x 2 ¯ 2 l 2 i + 2 x 3 ¯ 2 i 1 m 33 x 2 ¯ 2 l 2 i + 1 x 3 ¯ 2 i + m 23 x 2 ¯ 2 l 2 i x 3 ¯ 2 i + 1 .
Therefore, m 32 m 22 m 33 m 23 0 0 0 0 0 0 0 0 0 0 , 0 0 m 32 m 22 m 33 m 23 0 0 0 0 0 0 0 0 , , 0 0 0 0 0 0 0 0 m 32 m 22 m 33 m 23 0 0 , 0 0 0 0 0 0 0 0 0 0 m 32 m 22 m 33 m 23 is a k-basis of the space of the solutions of system (2). So, there exists { s 2 i 1 k | 1 i l } such that A ( i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 ) ¯ = g ( x 2 ¯ , x 3 ¯ ) . Take u ( x 2 , x 3 ) = i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 .
Since m 22 m 23 m 32 m 33 0 , we can conclude r 1 = r 3 = = r 2 l 1 = 0 from the system of Equation (3). So, h ( x 2 ¯ , x 3 ¯ ) = i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i . Since
A [ m 33 m 22 m 33 m 23 m 32 x 2 m 23 m 22 m 33 m 23 m 32 x 3 ] ¯ = x 2 ¯ 2 A [ m 32 m 22 m 33 m 23 m 32 x 2 + m 22 m 22 m 33 m 23 m 32 x 3 ] ¯ = x 3 ¯ 2 ,
we have
h ( x 2 ¯ , x 3 ¯ ) = i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i = A [ i = 0 l 1 r 2 i ( m 33 x 2 m 22 m 33 m 23 m 32 m 23 x 3 m 22 m 33 m 23 m 32 ) x 2 2 l 2 i 2 x 3 2 i ] ¯ + A [ r 2 l ( m 32 x 2 m 22 m 33 m 23 m 32 + m 22 x 3 m 22 m 33 m 23 m 32 ) x 3 2 l 2 ] . ¯
Take
v ( x 2 , x 3 ) = i = 0 l 1 r 2 i ( m 33 x 2 m 22 m 33 m 23 m 32 m 23 x 3 m 22 m 33 m 23 m 32 ) x 2 2 l 2 i 2 x 3 2 i + r 2 l ( m 32 x 2 m 22 m 33 m 23 m 32 + m 22 x 3 m 22 m 33 m 23 m 32 ) x 3 2 l 2 .
Then, we are finished. □
Remark 1. 
Since x 2 2 and x 3 2 are cocycle elements in A , one sees that u ( x 2 , x 3 ) in Lemma 7 can be chosen as u ( x 2 , x 3 ) = i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 with s 2 i 1 k , 1 i l .
Lemma 8. 
Let M = ( m i j ) 3 × 3 be a matrix in GL 3 ( k ) with m 22 m 33 m 23 m 32 0 and m 33 0 . Assume that I 1 = ( x 1 2 ) , I 2 = ( x 1 2 , x 2 2 ) and I 3 = ( x 1 2 , x 2 2 , x 3 2 ) are the three DG ideals generated by the subsets { x 1 2 } , { x 1 2 , x 2 2 } and { x 1 2 , x 2 2 , x 3 2 } of the DG algebra A , respectively. Then,
H i ( I 2 / I 1 ) = k x 2 ¯ 2 , if i = 2 k x 1 ¯ x 2 ¯ 2 + x 2 ¯ 2 ( m 13 m 32 m 12 m 33 m 22 m 33 m 23 m 32 x 2 ¯ + m 12 m 23 m 13 m 22 m 22 m 33 m 23 m 32 x 3 ¯ ) , if i = 3 0 , if i 4
and
H i ( I 3 / I 2 ) = k x 3 ¯ 2 , if i = 2 k m 33 x 1 ¯ x 3 ¯ 2 + m 13 x 3 ¯ 3 k m 33 x 2 ¯ x 3 ¯ 2 + m 23 x 3 ¯ 3 , if i = 3 k m 23 x 1 ¯ x 3 ¯ 3 m 13 x 2 ¯ x 3 ¯ 3 m 33 x 1 ¯ x 2 ¯ x 3 ¯ 2 , if i = 4 0 , if i 5 .
Proof. 
By Lemma 4, each x i 2 is a central cocycle element of A . So, I 1 , I 2 and I 3 are indeed DG ideals of A . Then, H 2 ( I 2 / I 1 ) = k x 2 2 and H 2 ( I 3 / I 2 ) = k x 3 2 since I 2 / I 1 and I 3 / I 2 are concentrated in degrees 2 , ( I 2 / I 1 ) 2 = k x 2 2 and ( I 3 / I 2 ) 2 = k x 3 2 .
Any graded cocycle element Ω of degree d in I 2 / I 1 can be written as
Ω = x 1 ¯ x 2 ¯ 2 f ( x 2 ¯ , x 3 ¯ ) + x 2 ¯ 2 g ( x 2 ¯ , x 3 ¯ ) ,
where f ( x 2 ¯ , x 3 ¯ ) and g ( x 2 ¯ , x 3 ¯ ) are sums of monomials in variables x 2 ¯ and x 3 ¯ . We have
0 = I 2 / I 1 ( z ) = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) x 2 ¯ 2 f ( x 2 ¯ , x 3 ¯ ) x 1 ¯ x 2 ¯ 2 A [ f ( x 2 , x 3 ) ] ¯ + x 2 ¯ 2 A [ g ( x 2 , x 3 ) ] ¯ = x 2 ¯ 2 { ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) f ( x 2 ¯ , x 3 ¯ ) + A [ g ( x 2 , x 3 ) ] ¯ } x 1 ¯ x 2 ¯ 2 A [ f ( x 2 , x 3 ) ] ¯ .
Thus
A [ f ( x 2 , x 3 ) ] ¯ = 0 A [ g ( x 2 , x 3 ) ] ¯ = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) f ( x 2 ¯ , x 3 ¯ ) .
When d = 3 , we have | f ( x 2 ¯ , x 3 ¯ ) | = 0 and | g ( x 2 ¯ , x 3 ¯ ) | = 1 . Let f ( x 2 ¯ , x 3 ¯ ) = c k and g ( x 2 ¯ , x 3 ¯ ) = c 1 x 2 ¯ + c 2 x 3 ¯ . Then
( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) c = A [ g ( x 2 , x 3 ) ] ¯ = A ( c 1 x 2 + c 2 x 3 ) ¯ = c 1 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) + c 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) ¯ = ( c 1 m 22 + c 2 m 32 ) x 2 ¯ 2 + ( c 1 m 23 + c 2 m 33 ) x 3 ¯ 2 .
This implies that
c 1 m 22 + c 2 m 32 = c m 12 c 1 m 23 + c 2 m 33 = c m 13 .
Hence
c 1 = m 12 m 32 m 13 m 33 m 22 m 32 m 23 m 33 c = c ( m 13 m 32 m 22 m 33 ) m 22 m 33 m 23 m 32 c 2 = m 22 m 12 m 23 m 13 m 22 m 32 m 23 m 33 c = c ( m 12 m 23 m 13 m 22 ) m 22 m 33 m 23 m 32
Then,
Ω = x 1 ¯ x 2 ¯ 2 c + x 2 ¯ 2 [ c ( m 13 m 32 m 22 m 33 ) m 22 m 33 m 23 m 32 x 2 ¯ + c ( m 12 m 23 m 13 m 22 ) m 22 m 33 m 23 m 32 x 3 ¯ ]
and
H 3 ( I 2 / I 1 ) = k x 1 ¯ x 2 ¯ 2 + x 2 ¯ 2 ( m 13 m 32 m 22 m 33 m 22 m 33 m 23 m 32 x 2 ¯ + m 12 m 23 m 13 m 22 m 22 m 33 m 23 m 32 x 3 ¯ )
since B 3 ( I 2 / I 1 ) = 0 .
When d = 4 , we have | f ( x 2 ¯ , x 3 ¯ ) | = 1 and | g ( x 2 ¯ , x 3 ¯ ) | = 2 . Let f ( x 2 ¯ , x 3 ¯ ) = l 1 x 2 ¯ + l 2 x 3 ¯ and g ( x 2 ¯ , x 3 ¯ ) = t 1 x 2 ¯ 2 + t 2 x 2 ¯ x 3 ¯ + t 3 x 3 ¯ 2 . Then, by (4), we have
0 = A [ f ( x 2 , x 3 ) ] ¯ = A ( l 1 x 2 + l 2 x 3 ) ¯ = l 1 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) + l 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) ¯ = ( l 1 m 22 + l 2 m 32 ) x 2 ¯ 2 + ( l 1 m 23 + l 2 m 33 ) x 3 ¯ 2 ,
which implies that
l 1 m 22 + l 2 m 32 = 0 l 1 m 23 + l 2 m 33 = 0 .
Since m 22 m 33 m 23 m 32 0 , we obtain l 1 = l 2 = 0 and hence f ( x 2 ¯ , x 3 ¯ ) = 0 . Then, by (4), we have
0 = A [ g ( x 2 , x 3 ) ] ¯ = A [ t 1 x 2 2 + t 2 x 2 x 3 + t 3 x 3 2 ] ¯ = t 2 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) x 3 t 2 x 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) ¯ = t 2 m 22 x 2 ¯ 2 x 3 ¯ + t 2 m 23 x 3 ¯ 3 t 2 m 32 x 2 ¯ 3 t 2 m 33 x 2 ¯ x 3 ¯ 2 .
Thus, t 2 m 22 = t 2 m 23 = t 2 m 32 = t 2 m 33 = 0 . Since m 22 m 23 m 32 m 33 0 , we obtain t 2 = 0 . So, Ω = x 1 ¯ x 2 ¯ 2 f ( x 2 ¯ , x 3 ¯ ) + x 2 ¯ 2 g ( x 2 ¯ , x 3 ¯ ) = x 2 ¯ 2 ( t 1 x 2 ¯ 2 + t 3 x 3 ¯ 2 ) . By the proof of Lemma 6, there exist
b 1 = m 23 m 31 m 21 m 33 | M | b 2 = m 11 m 33 m 13 m 31 | M | b 3 = m 13 m 21 m 11 m 23 | M | and c 1 = m 21 m 32 m 22 m 31 | M | c 2 = m 12 m 31 m 11 m 32 | M | c 3 = m 11 m 22 m 12 m 21 | M |
such that A ( b 1 x 1 + b 2 x 2 + b 3 x 3 ) = x 2 2 and A ( c 1 x 1 + c 2 x 2 + c 3 x 3 ) = x 3 2 , respectively. Then,
z = x 2 ¯ 2 ( t 1 x 2 ¯ 2 + t 3 x 3 ¯ 2 ) = x 2 ¯ 2 [ t 1 A ( b 1 x 1 + b 2 x 2 + b 3 x 3 ) ¯ + t 3 A ( c 1 x 1 + c 2 x 2 + c 3 x 3 ) ¯ ] = I 2 / I 1 { x 2 ¯ 2 [ t 1 ( b 1 x 1 ¯ + b 2 x 2 ¯ + b 3 x 3 ¯ ) + t 3 ( c 1 x 1 ¯ + c 2 x 2 ¯ + c 3 x 3 ¯ ) ] } .
Hence, H 4 ( I 2 / I 1 ) = 0 .
When d = 2 l + 3 , l 1 , we have | f ( x 2 ¯ , x 3 ¯ ) | = 2 l and | g ( x 2 ¯ , x 3 ¯ ) | = 2 l + 1 . Since A [ f ( x 2 , x 3 ) ] ¯ = 0 by (4), we obtain f ( x 2 ¯ , x 3 ¯ ) = i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i by Lemma 7, where r 2 i k , 0 i l . Then by (4), we have
A [ g ( x 2 , x 3 ) ] ¯ = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) f ( x 2 ¯ , x 3 ¯ ) = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) ( i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i ) = A [ ( m 12 m 33 m 22 m 33 m 23 m 32 x 2 + m 12 m 23 m 22 m 33 m 23 m 32 x 3 ) ( i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i ) ] ¯ + A [ ( m 13 m 32 m 22 m 33 m 23 m 32 x 2 m 13 m 22 m 22 m 33 m 23 m 32 x 3 ) ( i = 0 l r 2 i x 2 ¯ 2 l 2 i x 3 ¯ 2 i ) ] ¯ = A { i = 0 l r 2 i [ m 13 m 32 m 12 m 33 m 22 m 33 m 23 m 32 x 2 2 l 2 i + 1 x 3 2 i + m 12 m 23 m 13 m 22 m 22 m 33 m 23 m 32 x 2 2 l 2 i x 3 2 i + 1 ] } ¯
Then, by Lemma 7, we may let
g ( x 2 ¯ , x 3 ¯ ) = i = 0 l r 2 i [ m 13 m 32 m 12 m 33 m 22 m 33 m 23 m 32 x 2 ¯ 2 l 2 i + 1 x 3 ¯ 2 i + m 12 m 23 m 13 m 22 m 22 m 33 m 23 m 32 x 2 ¯ 2 l 2 i x 3 ¯ 2 i + 1 ] + A [ u ( x 2 , x 3 ) ] ¯ ,
where u ( x 2 , x 3 ) is a sum of monomials in variables x 2 and x 3 . Then,
Ω = x 1 ¯ x 2 ¯ 2 f ( x 2 ¯ , x 3 ¯ ) + x 2 ¯ 2 g ( x 2 ¯ , x 3 ¯ ) = i = 0 l r 2 i x 1 ¯ x 2 ¯ 2 l 2 i + 2 x 3 ¯ 2 i + x 2 ¯ 2 A [ u ( x 2 , x 3 ) ] ¯ + i = 0 l r 2 i [ m 13 m 32 m 12 m 33 m 22 m 33 m 23 m 32 x 2 ¯ 2 l 2 i + 3 x 3 ¯ 2 i + m 12 m 23 m 13 m 22 m 22 m 33 m 23 m 32 x 2 ¯ 2 l 2 i + 2 x 3 ¯ 2 i + 1 ] = i = 0 l r 2 i [ x 1 ¯ + ( m 13 m 32 m 12 m 33 ) x 2 ¯ + ( m 12 m 23 m 13 m 22 ) x 3 ¯ m 22 m 33 m 23 m 32 ] x 2 ¯ 2 l 2 i + 2 x 3 ¯ 2 i + x 2 ¯ 2 A [ u ( x 2 , x 3 ) ] ¯ .
One sees that ω = x 1 + ( m 13 m 32 m 12 m 33 ) x 2 + ( m 12 m 23 m 13 m 22 ) x 3 m 22 m 33 m 23 m 32 is a cocycle element in A . Hence,
z = A [ i = 0 l 1 r 2 i ω ( b 1 x 1 + b 2 x 2 + b 3 x 3 ) x 2 2 l 2 i x 3 2 i r 2 l ω x 2 2 x 3 2 l 2 ( c 1 x 1 + c 2 x 2 + c 3 x 3 ) ] ¯ + x 2 ¯ 2 A [ u ( x 2 , x 3 ) ] ¯ = I 2 / I 1 { [ i = 0 l 1 r 2 i ω ( b 1 x 1 ¯ + b 2 x 2 ¯ + b 3 x 3 ¯ ) x 2 ¯ 2 l 2 i 2 x 3 ¯ 2 i ] x 2 ¯ 2 } + I 2 / I 1 { [ r 2 l ω ( c 1 x 1 ¯ + c 2 x 2 ¯ + c 3 x 3 ¯ ) x 3 ¯ 2 l 2 + u ( x 2 ¯ , x 3 ¯ ) ] x 2 ¯ 2 } .
Thus, H 2 l + 3 ( I 2 / I 1 ) = 0 .
When d = 2 l + 4 , we have | f ( x 2 ¯ , x 3 ¯ ) | = 2 l + 1 and | g ( x 2 ¯ , x 3 ¯ ) | = 2 l + 2 . Since A [ f ( x 2 , x 3 ) ] ¯ = 0 by (4), we have
f ( x 2 ¯ , x 3 ¯ ) = A [ i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 ] ¯
by Lemma 7 and Remark 1, where s 2 i 1 k , 1 i l . Then, by (4), we have
A [ g ( x 2 , x 3 ) ] ¯ = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) f ( x 2 ¯ , x 3 ¯ ) = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) A [ i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 ] ¯ .
Then, by Lemma 7, we may let
g ( x 2 ¯ , x 3 ¯ ) = ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) [ i = 1 l s 2 i 1 x 2 ¯ 2 l 2 i + 1 x 3 ¯ 2 i 1 ] + A [ v ( x 2 , x 3 ) ] ¯ .
where v ( x 2 , x 3 ) is a sum of monomials in variables x 2 and x 3 . Then,
Ω = x 1 ¯ x 2 ¯ 2 f ( x 2 ¯ , x 3 ¯ ) + x 2 ¯ 2 g ( x 2 ¯ , x 3 ¯ ) = x 1 ¯ x 2 ¯ 2 A [ i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 ] ¯ ( m 12 x 2 ¯ 2 + m 13 x 3 ¯ 2 ) [ i = 1 l s 2 i 1 x 2 ¯ 2 l 2 i + 3 x 3 ¯ 2 i 1 ] + x 2 ¯ 2 A [ v ( x 2 , x 3 ) ] ¯ = A [ x 1 i = 1 l s 2 i 1 x 2 2 l 2 i + 1 x 3 2 i 1 v ( x 2 , x 3 ) ] ¯ x 2 ¯ 2 = I 2 / I 1 [ ( v ( x 2 ¯ , x 3 ¯ ) x 1 ¯ i = 1 l s 2 i 1 x 2 ¯ 2 l 2 i + 1 x 3 ¯ 2 i 1 ) x 2 ¯ 2 ]
and hence H 2 l + 4 ( I 2 / I 1 ) = 0 .
Since ( I 3 / I 2 ) 3 = k x 1 ¯ x 3 ¯ 2 k x 2 ¯ x 3 ¯ 2 k x 3 ¯ 3 , any cocycle element in ( I 3 / I 2 ) 3 can be denoted by c 1 x 1 ¯ x 3 ¯ 2 + c 2 x 2 ¯ x 3 ¯ 2 + c 3 x 3 ¯ 3 where c 1 , c 2 , c 3 k . Then,
0 = I 3 / I 2 [ c 1 x 1 ¯ x 3 ¯ 2 + c 2 x 2 ¯ x 3 ¯ 2 + c 3 x 3 ¯ 3 ] = c 1 m 13 x 3 ¯ 4 + c 2 m 23 x 3 ¯ 4 + c 3 m 33 x 3 ¯ 4 = ( c 1 m 13 + c 2 m 23 + c 3 m 33 ) x 3 ¯ 4
and hence c 1 m 13 + c 2 m 23 + c 3 m 33 = 0 , which has a basic solution system
m 33 0 m 13 , 0 m 33 m 23
So, Z 3 ( I 3 / I 2 ) = k ( m 33 x 1 ¯ x 3 ¯ 2 + m 13 x 3 ¯ 3 ) k ( m 33 x 2 ¯ x 3 ¯ 2 + m 23 x 3 ¯ 3 ) . Then,
H 3 ( I 3 / I 2 ) = k m 33 x 1 ¯ x 3 ¯ 2 + m 13 x 3 ¯ 3 k m 33 x 2 ¯ x 3 ¯ 2 + m 23 x 3 ¯ 3
since one sees easily that B 3 ( I 3 / I 2 ) = 0 . Any graded cocycle element z of degree d , d 4 in I 3 / I 2 can be written as
χ = x 1 ¯ x 3 ¯ 2 ϕ ( x 3 ¯ ) + x 2 ¯ x 3 ¯ 2 φ ( x 3 ¯ ) + x 1 ¯ x 2 ¯ x 3 ¯ 2 ψ ( x 3 ¯ ) + x 3 ¯ 2 λ ( x 3 ¯ ) .
We have
0 = I 3 / I 2 ( χ ) = I 3 / I 2 [ x 1 ¯ x 3 ¯ 2 ϕ ( x 3 ¯ ) + x 2 ¯ x 3 ¯ 2 φ ( x 3 ¯ ) + x 1 ¯ x 2 ¯ x 3 ¯ 2 ψ ( x 3 ¯ ) + x 3 ¯ 2 λ ( x 3 ¯ ) ] = m 13 x 3 ¯ 4 ϕ ( x 3 ¯ ) x 1 ¯ x 3 ¯ 2 A [ ϕ ( x 3 ) ] ¯ + m 23 x 3 ¯ 4 φ ( x 3 ¯ ) x 2 ¯ x 3 ¯ 2 A [ φ ( x 3 ) ] ¯ + m 13 x 2 ¯ x 3 ¯ 4 ψ ( x 3 ¯ ) m 23 x 1 ¯ x 3 ¯ 4 ψ ( x 3 ¯ ) + x 1 ¯ x 2 ¯ x 3 ¯ 2 A [ ψ ( x 3 ¯ ) ] ¯ + x 3 ¯ 2 A [ λ ( x 3 ) ] ¯ = x 3 ¯ 2 [ m 13 x 3 ¯ 2 ϕ ( x 3 ¯ ) + m 23 x 3 ¯ 2 φ ( x 3 ¯ ) + A [ λ ( x 3 ) ] ¯ ] + x 1 ¯ x 2 ¯ x 3 ¯ 2 A [ ψ ( x 3 ¯ ) ] ¯ x 1 ¯ [ x 3 ¯ 2 A [ ϕ ( x 3 ) ] ¯ + m 23 x 3 ¯ 4 ψ ( x 3 ¯ ) ] + x 2 ¯ [ m 13 x 3 ¯ 4 ψ ( x 3 ¯ ) x 3 ¯ 2 A [ φ ( x 3 ) ] ¯ ] .
Hence,
m 13 x 3 ¯ 2 ϕ ( x 3 ¯ ) + m 23 x 3 ¯ 2 φ ( x 3 ¯ ) + A [ λ ( x 3 ) ] ¯ = 0 x 3 ¯ 2 A [ ϕ ( x 3 ) ] ¯ + m 23 x 3 ¯ 4 ψ ( x 3 ¯ ) = 0 m 13 x 3 ¯ 4 ψ ( x 3 ¯ ) x 3 ¯ 2 A [ φ ( x 3 ) ] ¯ = 0 A [ ψ ( x 3 ¯ ) ] ¯ = 0 .
When d = 4 , we have | ψ ( x 3 ¯ ) | = 0 , | φ ( x 3 ¯ ) | = | ϕ ( x 3 ¯ ) | = 1 and | λ ( x 3 ¯ ) | = 2 . Let ψ ( x 3 ¯ ) = c k . Then, by (5), we obtain φ ( x 3 ) = c m 13 m 33 x 3 , ϕ ( x 3 ) = c m 23 m 33 x 3 and λ ( x 3 ¯ ) = c x 3 ¯ 2 , for some c k . So,
Z 4 ( I 3 / I 2 ) = k ( m 23 x 1 ¯ x 3 ¯ 3 + m 13 x 2 ¯ x 3 ¯ 3 + m 33 x 1 ¯ x 2 ¯ x 2 ¯ 2 ) k x 3 ¯ 4 .
Then, H 4 ( I 3 / I 2 ) = k m 23 x 1 ¯ x 3 ¯ 3 m 13 x 2 ¯ x 3 ¯ 3 m 33 x 1 ¯ x 2 ¯ x 2 ¯ 2 since B 4 ( I 3 / I 2 ) = k x 3 ¯ 4 . When d = 2 l 1 5 , we have | ϕ ( x 3 ¯ ) | = 2 l 4 , | φ ( x 3 ¯ ) | = 2 l 4 , | ψ ( x 3 ¯ ) | = 2 l 5 and | λ ( x 3 ¯ ) | = 2 l 3 . Let ψ ( x 3 ¯ ) = q x 3 ¯ 2 l 5 for some q k . Then 0 = A [ ψ ( x 3 ¯ ) ] ¯ = q m 33 x 3 ¯ 2 l 4 by (5). So, q = 0 and ψ ( x 3 ¯ ) = 0 . Then, we obtain A [ ϕ ( x 3 ) ] ¯ = A [ φ ( x 3 ) ] ¯ = 0 by (5). Let ϕ ( x 3 ) = P x 3 2 l 4 and φ ( x 3 ) = r x 3 2 l 4 , P , r k . Then,
A [ λ ( x 3 ) ] ¯ = m 13 P x 3 ¯ 2 l 2 m 23 r x 3 ¯ 2 l 2 .
So, λ ( x 3 ¯ ) = ( m 13 P + m 23 r ) x 3 ¯ 2 l 3 m 33 . Then,
χ = x 1 ¯ x 3 ¯ 2 ϕ ( x 3 ¯ ) + x 2 ¯ x 3 ¯ 2 φ ( x 3 ¯ ) + x 1 ¯ x 2 ¯ x 3 ¯ 2 ψ ( x 3 ¯ ) + x 3 ¯ 2 λ ( x 3 ¯ ) = P x 1 ¯ x 3 ¯ 2 l 2 + r x 2 ¯ x 3 ¯ 2 l 2 ( m 13 P + m 23 r ) x 3 ¯ 2 l 1 m 33 = [ m 33 ( P x 1 ¯ + r x 2 ¯ ) ( P m 13 + r m 23 ) x 3 ¯ ] m 33 x 3 ¯ 2 l 2 = I 3 / I 2 { [ m 33 ( P x 1 ¯ + r x 2 ¯ ) + ( P m 13 + r m 23 ) x 3 ¯ ] m 33 2 x 3 ¯ 2 l 3 } .
Thus, H 2 l 1 ( I 3 / I 2 ) = 0 , for any l 3 . When d = 2 l 6 , we have | ϕ ( x 3 ¯ ) | = 2 l 3 , | φ ( x 3 ¯ ) | = 2 l 3 , | ψ ( x 3 ¯ ) | = 2 l 4 and | λ ( x 3 ¯ ) | = 2 l 2 . So, A [ λ ( x 3 ) ] ¯ = 0 and A [ ψ ( x 3 ) ] ¯ = 0 . Then, (5) is equivalent to
m 13 x 3 ¯ 2 ϕ ( x 3 ¯ ) + m 23 x 3 ¯ 2 φ ( x 3 ¯ ) = 0 x 3 ¯ 2 A [ ϕ ( x 3 ) ] ¯ + m 23 x 3 ¯ 4 ψ ( x 3 ¯ ) = 0 m 13 x 3 ¯ 4 ψ ( x 3 ¯ ) x 3 ¯ 2 A [ φ ( x 3 ) ] ¯ = 0 .
Let λ ( x 3 ¯ ) = s x 3 ¯ 2 l 2 and ψ ( x 3 ) = t x 3 ¯ 2 l 4 . Then, by the system of equations above, we obtain ϕ ( x 3 ¯ ) = m 23 t m 33 x 3 ¯ 2 l 3 and φ ( x 3 ¯ ) = m 13 t m 33 x 3 ¯ 2 l 3 . Then
χ = x 1 ¯ x 3 ¯ 2 ϕ ( x 3 ¯ ) + x 2 ¯ x 3 ¯ 2 φ ( x 3 ¯ ) + x 1 ¯ x 2 ¯ x 3 ¯ 2 ψ ( x 3 ¯ ) + x 3 ¯ 2 λ ( x 3 ¯ ) = m 23 t m 33 x 1 ¯ x 3 ¯ 2 l 1 + m 13 t m 33 x 2 ¯ x 3 ¯ 2 l 1 + t x 1 ¯ x 2 ¯ x 3 ¯ 2 l 2 + s x 2 ¯ 2 l = [ m 23 x 1 ¯ x 3 ¯ + m 13 x 2 ¯ x 3 ¯ + m 33 x 1 ¯ x 2 ¯ m 33 ] t x 3 ¯ 2 l 2 + s x 3 ¯ 2 l = I 3 / I 2 { [ m 23 x 1 ¯ x 3 ¯ + m 13 x 2 ¯ x 3 ¯ + m 33 x 1 ¯ x 2 ¯ m 33 2 ] t x 3 ¯ 2 l 1 + s m 33 x 3 ¯ 2 l 1 }
Hence, H 2 l ( I 3 / I 2 ) = 0 for any l 3 . □
Lemma 9. 
Let M = ( m i j ) 3 × 3 and r ( M ) = 2 . Then, r ( X ) = 5 , where
X = m 11 m 21 m 31 0 0 0 0 0 0 m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 0 0 0 m 12 m 22 m 32 0 0 0 0 0 0 0 0 0 m 13 m 23 m 33 .
Proof. 
Since r ( M ) = 2 , there exists ( s 1 , s 2 , s 3 ) T 0 such that M ( s 1 , s 2 , s 3 ) T = 0 , which is equivalent to
s 1 m 11 m 21 m 31 + s 2 m 12 m 22 m 32 + s 3 m 13 m 23 m 33 = 0 .
Without the loss of generality, let s 1 0 . Then, m 12 m 22 m 32 , m 13 m 23 m 33 are linearly independent and
( m 11 , m 21 , m 31 ) + s 2 s 1 ( m 12 , m 22 , m 32 ) + s 3 s 1 ( m 13 , m 23 , m 33 ) = 0 .
For X, we can perform the following elementary row transformations
X r 1 + s 3 s 1 × r 3 r 1 + s 2 s 1 × r 2 0 0 0 s 2 s 1 m 11 s 2 s 1 m 21 s 2 s 1 m 31 s 3 s 1 m 11 s 3 s 1 m 21 s 3 s 1 m 31 m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 0 0 0 m 12 m 22 m 32 0 0 0 0 0 0 0 0 0 m 13 m 23 m 33 r 1 + s 2 s 3 s 1 2 × r 4 r 1 + s 2 2 s 1 2 × r 5 0 0 0 0 0 0 s 3 2 s 1 2 m 13 s 3 2 s 1 2 m 23 s 3 2 s 1 2 m 33 m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 0 0 0 m 12 m 22 m 32 0 0 0 0 0 0 0 0 0 m 13 m 23 m 33 r 1 + s 3 2 s 1 2 × r 6 0 0 0 0 0 0 0 0 0 m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 0 0 0 m 12 m 22 m 32 0 0 0 0 0 0 0 0 0 m 13 m 23 m 33 .
This indicates r ( X ) 5 and
r ( X ) = r m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 0 0 0 m 12 m 22 m 32 0 0 0 0 0 0 0 0 0 m 13 m 23 m 33 .
Let
l 1 m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 + l 2 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 + l 3 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 + l 4 0 0 0 m 12 m 22 m 32 0 0 0 + l 5 0 0 0 0 0 0 m 13 m 23 m 33 = 0 .
Then,
l 1 m 12 + l 2 m 13 = 0 l 1 m 22 + l 2 m 23 = 0 l 1 m 32 + l 2 m 33 = 0 l 1 m 11 + l 3 m 13 + l 4 m 12 = 0 l 1 m 21 + l 3 m 23 + l 4 m 22 = 0 l 1 m 31 + l 3 m 33 + l 4 m 32 = 0 l 2 m 11 + l 3 m 12 + l 5 m 13 = 0 l 2 m 21 + l 3 m 22 + l 5 m 23 = 0 l 2 m 31 + l 3 m 32 + l 5 m 33 = 0 ,
which implies l 1 = l 2 = l 3 = l 4 = l 5 = 0 since m 12 m 22 m 32 , m 13 m 23 m 33 are linearly independent. Thus,
r ( X ) = r m 12 m 22 m 32 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 0 0 0 m 11 m 21 m 31 0 0 0 m 13 m 23 m 33 m 12 m 22 m 32 0 0 0 m 12 m 22 m 32 0 0 0 0 0 0 0 0 0 m 13 m 23 m 33 = 5 .
Similarly, we can show r ( X ) = 5 when s 2 0 or s 3 0 . □
Lemma 10. 
Let M = ( m i j ) 3 × 3 be a matrix in M 3 ( k ) with r ( M ) = 2 . If
r 1 = m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 , r 2 = m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 , r 3 = m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ,
then the graded ideal ( r 1 , r 2 , r 3 ) is a prime graded ideal of the polynomial graded algebra k [ x 1 2 , x 2 2 , x 3 2 ] .
Proof. 
Since r ( M ) = 2 , there exist a non-zero solution vector ( t 1 , t 2 , t 3 ) T of the homogeneous linear equations M T X = 0 . We have
t 1 r 1 + t 2 r 2 + t 3 r 3 = ( t 1 , t 2 , t 3 ) r 1 r 2 r 3 = ( t 1 , t 2 , t 3 ) M x 1 2 x 2 2 x 3 2 = 0 .
Since ( t 1 , t 2 , t 3 ) T 0 , we may as well let t 3 0 . Then, r 3 = t 1 t 3 r 1 t 2 t 3 r 2 and hence ( r 1 , r 2 , r 3 ) = ( r 1 , r 2 ) . Since
m 31 m 32 m 33 = t 1 t 3 m 11 m 12 m 13 t 2 t 3 m 21 m 22 m 23 ,
we have
r m 11 m 12 m 13 m 21 m 22 m 23 = 2 ,
this indicates that there at least one non-zero minor among
m 11 m 12 m 21 m 22 , m 11 m 13 m 21 m 23 , m 12 m 13 m 22 m 23 .
We may as well let m 11 m 12 m 21 m 22 0 . Then, one sees that
k [ x 1 2 , x 2 2 , x 3 2 ] / ( r 1 , r 2 ) k [ x 3 2 ]
is a domain. So, ( r 1 , r 2 , r 3 ) = ( r 1 , r 2 ) is a graded prime ideal of k [ x 1 2 , x 2 2 , x 3 2 ] . □
Lemma 11. 
Assume that M = ( m i j ) 3 × 3 M 3 ( k ) with r ( M ) = 2 , k ( s 1 , s 2 , s 3 ) T and k ( t 1 , t 2 , t 3 ) T are the solution spaces of homogeneous linear equations M X = 0 and M T X = 0 , respectively. We have the following statements.
(1) If s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 0 , then k [ t 1 x 1 + t 2 x 2 + t 3 x 3 ] is a subalgebra of H ( A ) ;
(2) If s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 = 0 , then
k [ t 1 x 1 + t 2 x 2 + t 3 x 3 , s 1 x 1 2 + s 2 x 2 2 + s 3 x 3 2 ] / ( t 1 x 1 + t 2 x 2 + t 3 x 3 2 )
is a subalgebra of H ( A ) .
Proof. 
Clearly, we have H 0 ( A ) = k . Since r ( M T ) = 2 < 3 , there is a non-zero solution vector ( t 1 , t 2 , t 3 ) T of the homogeneous linear equations M T X = 0 . For any c 1 x 1 + c 2 x 2 + c 3 x 3 Z 1 ( A ) , we have
0 = A ( c 1 x 1 + c 2 x 2 + c 3 x 3 ) = ( c 1 , c 2 , c 3 ) A ( x 1 ) A ( x 2 ) A ( x 3 ) = ( c 1 , c 2 , c 3 ) M x 1 2 x 2 2 x 3 2 ,
which implies that ( c 1 , c 2 , c 3 ) M = 0 or equivalently M T c 1 c 2 c 3 = 0 . Thus, H 1 ( A ) = k t 1 x 1 + t 2 x 2 + t 3 x 3 .
For any l 11 x 1 2 + l 12 x 1 x 2 + l 13 x 1 x 3 + l 22 x 2 2 + l 23 x 2 x 3 + l 33 x 3 2 ker ( A 2 ) , we have
0 = A [ l 11 x 1 2 + l 12 x 1 x 2 + l 13 x 1 x 3 + l 22 x 2 2 + l 23 x 2 x 3 + l 33 x 3 2 ] = l 12 ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) x 2 l 12 x 1 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) + l 13 ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) x 3 l 13 x 1 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) + l 23 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) x 3 l 23 x 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) = ( l 12 m 21 + l 13 m 31 ) x 1 3 + ( l 12 m 11 l 23 m 31 ) x 1 2 x 2 + ( l 13 m 11 + l 23 m 21 ) x 1 2 x 3 ( l 12 m 22 + l 13 m 32 ) x 1 x 2 2 + ( l 13 m 12 + l 23 m 22 ) x 2 2 x 3 + ( l 12 m 12 l 23 m 32 ) x 2 3 ( l 12 m 23 + l 13 m 33 ) x 1 x 3 2 + ( l 12 m 13 l 23 m 33 ) x 2 x 3 2 + ( l 13 m 13 + l 23 m 23 ) x 3 3 .
Hence,
l 12 m 21 + l 13 m 31 = 0 l 12 m 11 l 23 m 31 = 0 l 13 m 11 + l 23 m 21 = 0 l 12 m 22 + l 13 m 32 = 0 l 13 m 12 + l 23 m 22 = 0 l 12 m 12 l 23 m 32 = 0 l 12 m 23 + l 13 m 33 = 0 l 12 m 13 l 23 m 33 = 0 l 13 m 13 + l 23 m 23 = 0 l 12 m 21 + l 13 m 31 = 0 l 12 m 22 + l 13 m 32 = 0 l 12 m 23 + l 13 m 33 = 0 l 12 m 11 l 23 m 31 = 0 l 12 m 12 l 23 m 32 = 0 l 12 m 13 l 23 m 33 = 0 l 13 m 11 + l 23 m 21 = 0 l 13 m 12 + l 23 m 22 = 0 l 13 m 13 + l 23 m 23 = 0 ,
which is equivalent to
m 11 m 21 m 31 m 12 m 22 m 32 m 13 m 23 m 33 0 l 12 l 13 l 12 0 l 23 l 13 l 23 0 = 0 3 × 3 .
We claim that l 12 = l 23 = l 13 = 0 . Indeed, if any one of l 12 , l 23 , l 13 is non-zero, then there are at least two non-zero linear independent vectors among
0 l 12 l 13 , l 12 0 l 23 , l 13 l 23 0 ,
which are all solutions of M X = 0 . This contradicts with r ( M ) = 2 . Hence, ker ( A 2 ) = k x 1 2 k x 2 2 k x 3 2 . In A , we have
( t 1 x 1 + t 2 x 2 + t 3 x 3 ) 2 = t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 .
(1) If s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 0 , we claim that t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 B 2 ( A ) . Indeed, if there exist q 1 x 1 + q 2 x 2 + q 3 x 3 A 1 such that A ( q 1 x 1 + q 2 x 2 + q 3 x 3 ) = t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 , then
( q 1 , q 2 , q 3 ) M x 1 2 x 2 2 x 3 2 = A ( q 1 x 1 + q 2 x 2 + q 3 x 3 ) = t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 = ( t 1 2 , t 2 2 , t 3 2 ) x 1 2 x 2 2 x 3 2 ,
which implies that ( q 1 , q 2 , q 3 ) M = ( t 1 2 , t 2 2 , t 3 2 ) and hence
0 = ( q 1 , q 2 , q 3 ) M s 1 s 2 s 3 = ( t 1 2 , t 2 2 , t 3 2 ) s 1 s 2 s 3 = s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 .
This contradicts with the assumption that s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 0 . Then, we obtain that t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 B 2 ( A ) if s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 0 . On the other hand, we have dim k B 2 ( A ) = 2 since r ( M ) = 2 . Therefore, dim k H 2 ( A ) = 1 and
H 2 ( A ) = k t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 = k t 1 x 1 + t 2 x 2 + t 3 x 3 2 .
In order to show k [ t 1 x 1 + t 2 x 2 + t 3 x 3 ] is a subalgebra of H ( A ) , we need to show ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n B n ( A ) for any n 3 . If this not the case, we have
( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n = A [ x 1 x 2 f + x 1 x 3 g + x 2 x 3 h ] , if n = 2 j + 1 is odd A [ x 1 f + x 2 g + x 3 h + x 1 x 2 x 3 u ] , if n = 2 j is even
where f , g , h and u are all linear combinations of monomials with non-negative even exponents. When n = 2 j is even, we have
( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n = A [ x 1 f + x 2 g + x 3 h + x 1 x 2 x 3 u ] = ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) f + ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) g + ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) h + ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) x 2 x 3 u x 1 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) x 3 g + x 1 x 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) u .
Considering the parity of exponents of the monomials that appear on both sides, the equation above implies that
( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) f + ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) g + ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) h = A ( x 1 ) f + A ( x 2 ) g + A ( x 3 ) h
and
A ( x 1 x 2 x 3 u ) = ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) x 2 x 3 u x 1 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) x 3 g + x 1 x 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) u = 0 .
Therefore, ( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j is in the graded ideal ( A ( x 1 ) , A ( x 2 ) , A ( x 3 ) ) of k [ x 1 2 , x 2 2 , x 3 2 ] . By Lemma 10, ( A ( x 1 ) , A ( x 2 ) , A ( x 3 ) ) is a graded prime ideal of k [ x 1 2 , x 2 2 , x 3 2 ] . So, t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ( A ( x 1 ) , A ( x 2 ) , A ( x 3 ) ) . Hence, there exist a 1 , a 2 and a 3 in k such that
t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 = a 1 A ( x 1 ) + a 2 A ( x 2 ) + a 3 A ( x 3 ) = A ( a 1 x 1 + a 2 x 2 + a 3 x 3 ) .
However, this contradicts with the fact that t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 B 2 ( A ) , which we have proved above. Thus, ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n B n ( A ) when n is even.
When n = 2 j + 1 is odd, we have
( t 1 x 1 + t 2 x 2 + t 3 x 3 ) ( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n = A [ x 1 x 2 f + x 1 x 3 g + x 2 x 3 h ] = ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) x 2 f x 1 ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) f + ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) x 3 g x 1 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) g + ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) x 3 h x 2 ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) h = x 1 [ ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) f + ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) g ] + x 2 [ ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) f ( m 31 x 1 2 + m 32 x 2 2 + m 33 x 3 2 ) h ] + x 3 [ ( m 21 x 1 2 + m 22 x 2 2 + m 23 x 3 2 ) h + ( m 11 x 1 2 + m 12 x 2 2 + m 13 x 3 2 ) g ] = x 1 [ A ( x 2 ) f A ( x 3 ) g ] + x 2 [ A ( x 1 ) f A ( x 3 ) h ] + x 3 [ A ( x 2 ) h + A ( x 1 ) g ] .
This implies that
t 1 ( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = A ( x 2 ) f A ( x 3 ) g = A [ x 2 f x 3 g ] t 2 ( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = A ( x 1 ) f A ( x 3 ) h = A [ x 1 f x 3 h ] t 3 ( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = A ( x 2 ) h + A ( x 1 ) g = A [ x 2 h + x 1 g ] .
Since ( t 1 , t 2 , t 3 ) T 0 , there is at least one non-zero t i , i { 1 , 2 , 3 } . Then, we obtain ( t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 ) j = ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) 2 j B 2 j ( A ) , which contradicts with the proved fact that ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n B n ( A ) when n is even. Therefore, ( t 1 x 1 + t 2 x 2 + t 3 x 3 ) n B n ( A ) when n is odd.
Then, we reach a conclusion that k [ t 1 x 1 + t 2 x 2 + t 3 x 3 ] is a subalgebra of H ( A ) when s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 0 .
(2) When s 1 t 1 2 + s 2 t 2 2 + s 3 t 3 2 = 0 , we should show t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 B 2 ( A ) and s 1 x 1 2 + s 2 x 2 2 + s 3 x 3 2 B 2 ( A ) first. In order to prove t 1 2 x 1 2 + t 2 2 x 2 2 + t 3 2 x 3 2 B 2 ( A ) , we need to show the existence of an element q 1 x 1 + q 2 x 2 + q 3 x 3 A 1 such that