RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS. 1. Introduction

Transcription

1 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS 1 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS AUTHOR: YUZHE BAI SUPERVISOR: DR. EUGENE GORSKY Abstract. In this thesis, we will provide a basic recursive formula of the Hilbert Series of a quotient ring R n/i n where R n C[x 0,..., x ] and I n is the quadratic defining ideal of the n th jet scheme of a double point. Knowing the Hilbert Series of the quotient will give us a way to derive the Hilbert Series of the ideal. To achieve our goal, we will study the first syzygy module of the quotient ring and describe its generators. In fact, by learning the structure of the first syzygy module, we can derive a recursive formula for the minimal free resolution of the quotient ring as R n module which will provide an explicit form for the Hilbert Series. Consider the ring and the ideal I n (f 1, f 2..., f n ) R n, where 1. Introduction R n C[x 0, x 1, x 2..., x ] f n x k x k. For example, when n 1, we have I 1 (x 2 0), and when n 2, we have I 2 (x 2 0, 2x 0 x 1 ). It is easy to see that R n is an infinite dimensional complex vector space where the basis consists of all monomials of x 0, x 1..., x. Now, we define the bi-degree deg(m) (a 1, a 2 ) to be the degree of a monomial m, where deg(x i ) (i, 1) and deg(m 1 m 2 ) deg(m 1 ) + deg(m 2 ). Then obviously, R n is a commutative bi-graded C algebra, and f i are homogeneous with respect to this bi-grading. Definition 1.1. For any field k and any bi-graded k vector space V, the Hilbert Series H V (q, t) of V over k is defined as H V (q, t) q a1 t a2 dim V (a 1, a 2 ), where dim V (a 1, a 2 ) is the dimension of the subspace of V that has bi-degree (a 1, a 2 ). Now, let s consider the Hilbert series of R n over C H(q, t) n q a1 t a2 d Rn (a 1, a 2 ) where d Rn (a 1, a 2 ) dim Rn (a 1, a 2 ) is the dimension of the subspace with bi-degree (a 1, a 2 ) in R n ; in other word, it is the number of monomials in R n having bi-degree (a 1, a 2 ). For example, when n 1, R 1 C[x 0 ], the standard basis for R 1 as a complex vector space is B 1 {1, x 0, x }. Then a 1 (x k 0) 0, a 2 (x k 0) k, and that implies when a 1 0 and a 2 Z 0, d R1 (a 1, a 2 ) 1, and when a 1 0, d R1 (a 1, a 2 ) 0. Thus, the Hilbert series of R 1 is. H(q, t) 1 q a1 t a2 d R1 (a 1, a 2 ) a 20 t a2 1 1 t Proposition 1.2. The Hilbert series H(q, t) n of R n is given by the equation H(q, t) n 1 1 q k t.

2 2 AUTHOR: YUZHE BAI SUPERVISOR: DR. EUGENE GORSKY Proof. We are going to prove this proposition by using mathematical induction. For the base case, as we stated earlier, when n 1, Suppose for some m Z >0, H(q, t) t 1 1 q k t. H(q, t) m m q k t, Let s consider n m + 1. Let B m be the standard basis of R m, then the standard basis of R m+1 is B m+1 xk mb m. Moreover, we know deg(x k m) (km, k). Since deg(m 1 m 2 ) deg(m 1 ) + deg(m 2 ), we can convert the Hilbert series H(q, t) m+1 of R m+1 as following H(q, t) m+1 q a1 t a2 d Rm+1 (a 1, a 2 ) q jm t j q a1 t a2 d Rm (a 1, a 2 ) q jm t j H(q, t) m 1 1 q m t H(q, t) m m 1 1 q k t. Therefore, by mathematical induction, for n Z >0, the Hilbert series H(q, t) n of R n is H(q, t) n 1 1 q k t. Remark 1.3. For any commutative bi-graded C algebra R and any bi-graded R module M, denote the Hilbert series of M as H(M). We know that for any bi-graded submodule S M, the Hilbert Series over C satisfy H(M) H(S) + H(M/S). Thus, instead of calculate the Hilbert Series of I n directly, we can first calculate H(R n /I n ) and subtract it from H(R n ) which is known. Because of the relation, we can calculate the Hilbert Series of I n over C by subtracting the Hilbert Series of R n /I n from the Hilbert Series of R n. Since we already have the formula for H(R n ) for n N, once we know H(R n /I n ), we can easily deduce H(I n ). In this sense, the main goal of this thesis is to prove the following theorem. Main Theorem. recursion: Let H n (q, t) H(R n /I n ), then when n 4, it can be derived by the following H n (q, t) th n 3(q, q 2 t) + H n 2 (q, qt) 1 q. t When n is approching to, an explicit formula for the Hilbert Series of R n /I n is proven by Bruschek, Mourtada and Schepers[1], which relates the Hilbert series of the arc space for the double point to the Rogers-Ramanujan identity. A similar result for n was obtained by Feigin-Stoyanovsky [2, 3], Lepowsky et al. [4, 5], and Gorsky, Oblomkov and Rasmussen in [6]. Moreover, the result of this thesis, together with the explicit formula for the Hilbert Series of R n /I n can be found in [7].

3 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS 3 2. Build up for recursive relation Basis of R n /I n for small n In order to calculate the Hilbert Series of R n /I n, we need to know the basis of R n /I n over C. Now, let s consider the basis B n of R n /I n for n 1, 2, 3. When n 1, R 1 C[x 0 ] and I 1 (x 2 0), then B 1 {1, x 0 }; When n 2, R 2 C[x 0, x 1 ] and I 2 (x 2 0, 2x 0 x 1 ), then B 2 {1, x 0, x k 1 k N}; When n 3, R 3 C[x 0, x 1, x 2 ] and I 3 (x 2 0, 2x 0 x 1, 2x 0 x 2 + x 2 1), we can see that by choosing the reversed degree lexicographic order as the monomial order, {x 2 0, 2x 0 x 1, 2x 0 x 2 + x 2 1} satisfies Buchberger s Criterion and is indeed a reduced Groebner basis for I 3. Therefore, B 3 {x i 2, x 0 x j 2, x 1x k 2 i, j, k Z 0 } which consists of all monomials which are not divisible by any of the leading terms of the elements in the Groebner basis. For bigger n, we will get the basis from a recursive relation, but before we do that, we need some tools. Definition 2.1. A shift of a polynomial f(x 1,..., x n ) R n is f(x 2,..., x n+1 ), which is in R m for all m > n, we denote it as S(f). Moreover, for any subset M R n or quotient of R n, we denote the shift of M as S(M) {S(f) : f M}. If we consider S as a linear function from R n to R m where m n, then S is an injection as its kernel is trivial. Lemma 2.2. For any α 1,..., α n R n, n i1 S(α i)s(f i ) 0 if and only if n i1 α if i 0. Proof. Since n i1 S(α i)s(f i ) S( n i1 α if i ). Because S is an injection, then S(g) 0 if and only if g 0. Thus, n i1 S(α i)s(f i ) 0 if and only if n i1 α if i 0. Recursive relation between the bases of R n /I n Now, we claim that for n > 3, B n xk [x 0 S 2 (B n 3 ) S(B n 2 )]. The way we consider this is to construct two subsets of R n /I n, which we name as Q n and P n, where Q n consists of all elements in R n /I n that are divisible by x 0 and P n consists of all elements in R n /I n which contain no term divisible by x 0. We can see that Q n is an ideal in R n /I n and P n is a subring of R n /I n which is isomorphic to [R n /I n ]/Q n. Consequently, we have H(R n /I n ) H(Q n ) + H(P n ). In fact, Q n can be considered as x 0 R n /[I n (x0 )] and P n can be considered as R n /[I n + (x 0 )]. Now, we need to study their structure respectively. First, let s consider the quotient ring R n /[I n + (x 0 )]. Lemma 2.3. For the subring P n of R n /I n which consists of elements having no term divisible by x 0, we have ( ) R n P n I n + (x 0 ) S Rn 2 [x ]. I n 2 Proof. By the Third Isomorphism Theorem, it s isomorphic to R n /(x 0 ) [I n + (x 0 )]/(x 0 ), where R n /(x 0 ) S(R n 2 )[x ]. Since I n + (x 0 ) is generated by f 1, f 2,..., f n, x 0, by the natural map, the quotient [I n + (x 0 )]/(x 0 ) is generated by the residue of the generator of I n + (x 0 ), where the residue of f 1, f 2 is 0 since they contain x 0, and residue of f n is S(f n 2 ) for n 3 as f n S(f n 2 ) + 2x 0 x. As a result, R n I n + (x 0 ) S ( Rn 2 I n 2 ) [x ]. Thus, the basis of P n is xk S(B n 2 ). In order to study Q n, we first need to know the structure of I n (x0 ), which requires us to study the module of first syzygy of f 1,..., f n. 3. Study of the first syzygy Definition 3.1. Let R be a commutative ring and M be an R-module generated by m 1, m 2,...m n, where n <. Let F be an rank n free R-module with basis B {e 1,..., e n } and there is a R-module homomorphism φ : F M such that φ(e i ) m i for i 1,..., n. The first syzygy module of m 1,...m n is ker φ and an element of ker φ is a syzygy of m 1,..., m n. Moreover, since F is a free module, every

4 4 AUTHOR: YUZHE BAI SUPERVISOR: DR. EUGENE GORSKY element in F can be written as a unique R combination of e 1,..., e n. As a result, we can write the element in ker φ in n-tuple form. In our case, I n is an R n -module and {f 1,..., f n } is a generating set of I n. It is important for us to learn the first syzygy module of f 1,..., f n as it will help us understand the structure of I n (x0 ). Lemma 3.2. For f 1, f 2,...f n, the following relation holds (n 1 3i)x i f n i 0. i0 Proof. It is easy to see that all terms in i0 (n 1 3i)x if n i 0 are scalar multiples of monomials x a x b x c such that a + b + c n 1 since f n is the summation of monomials x a x b such that a + b n 1. If a b c, then the coefficient of x a x b x c is 2(n 1 3a) + 2(n 1 3b) + (n 1 3c) 6(n 1) 6a 6b 6c If a b c, then the coefficient of x a x b x c is 6(n 1) 6(a + b + c) 6(n 1) 6(n 1) 0; 2(n 1 3a) + (n 1 3c) 3(n 1) (6a + 3c) 3(n 1) 3(a + b + c) 3(n 1) 3(n 1) If a b c, then 3a n 1 and a b c (n 1 a)/2. Thus, the coefficient of x a x b x c is 0; 3(n 1 a) n 1 3a n 1 2 3a (n 1) 2 0. As a result, i0 (n 1 3i)x if n i 0. From Lemma 1, we can recognize some syzygies of f 1, f 2,..., f n. Proposition 3.3. Let µ i ( 2ix i, ( 2i + 3)x i 1,..., ix 0, 0,..., 0), ν j,k f k e j + f j e k. Then µ i and ν j,k are syzygies of f 1,..., f n for i n 1 and j, k n such that j < k. Proof. For µ i, it directly follows from Lemma 3.2. For ν j,k, we realize that f j f k f k f j 0 when j > k, the preimage of this relation is ν j,k. As a result, µ i and ν j,k are in the first syzygy module of f 1,..., f n. In fact, they are the generators of the first syzygy module of f 1,.., f n and we will prove this later. If we have already had the conclusion in hand, we will have enough information to discover the structure of I n (x0 ), which gives us an explicit form of Q n. Proposition 3.4. When n 3, (n 3i)x i f n+1 i nx 0 S(f ) n[x 0 x 1 x + x 0 S 2 (f n 3 )], i1 where the second equation holds when n 4. Proof. Similar to Lemma 1, we can see that each term of the sum is a scalar multiple of monomial x a x b x c where a + b + c n. When a, b, c 0, if a b c, then the coefficient of x a x b x c is 2(n 3a) + 2(n 3b) + 2(n 3c) 6n 6(a + b + c) 0;

5 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS 5 if a b c, then the coefficient of x a x b x c is 2(n 3a) + (n 3c) 3n 6a 3c 3n 3(a + b + c) 0; if a b c, then a b c n/3. The coefficient of x a x b x c is n 3a 0. Thus, if a, b, c 0, then the coefficient of x a x b x c is 0. As a result, x 0 i1 (n 3i)x if n+1 i. Now, let s consider the coefficient of x 0 x a x b. From the observation above, we have a + b n, if a b, then the coefficient of x 0 x a x b is if a b, then the coefficient of x 0 x a x b is 2(n 3a) + (2n 3b) 4n 6(a + b) 4n 6n 2n; 2(n 3a) 2n 6a 2n 3(a + b) 2n 3n n. Therefore, we can rewrite i1 (n 3i)x if n+1 i as nx 0 n 2 x i x n i nx 0 x i+1 x (n 2 i)+1 nx 0 S(f ). i1 i0 Moreover, because f i 2x 0 x i 1 + S(f i 2 ) when i 3, then S(f i ) 2x 1 x i + S 2 (f i 2 ) when i 3. Hence, when n 4, nx 0 S(f ) nx 0 (2x 1 x + S 2 (f n 3 )) n[x 0 x 1 x + x 0 S 2 (f n 3 )]. By Proposition 3.4, we notice that x 0 S 2 (f n 3 ) I n when n 4, which implies that. I n (x0 ) (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ] 4. Main theorems Theorem 4.1. If the first syzygy module of f 1,..., f n is generated by µ i and ν j,k where i n 1 and j, k n such that j k,then for n 4, x 0 R n I n (x0 ) x 0S 2 ( R n 3 )[x ]. I n 3 Proof. In order to prove Theorem 4.1, we first of all need to show that I n (x0 ) (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ] when the first syzygy module is generated by µ and ν. One direction is already derived by Proposition 2, we just need to show the other direction. For all g I n, there exists α i R n such that g n i1 α if i. For each α i, we can separate it into two parts, one is divisible by x 0 and the other contains no term divisible by x 0 : Do the similar thing to f i, we get f 1 x 2 0 f 2 2x 0 x 1 α i x 0 α i + S(β i ). f i 2x 0 x i 1 + S(f i 2 ) for i 3

6 6 AUTHOR: YUZHE BAI SUPERVISOR: DR. EUGENE GORSKY Thus, we can rewrite g into the following form g x 2 0(x 0 α 1 + S(β 1 )) + 2x 0 x 1 (x 0 α 2 + S(β 2 )) + 2x 2 0x i 1 α i + x 0 α is(f i 2 ) + 2x 0 x i 1 S(β i ) + S(β i )S(f i 2 ) f 1 (x 0 α 1 + S(β 1 )) + f 2 (x 0 α 2 + S(β 2 )) + + α 4x 2 f 2 + 2f 1 x i 1 α i α 3x 1 f 2 x 0 α i(s 2 (f i 4 ) + 2x 1 x i 2 ) + f 1 (x 0 α 1 + S(β 1 )) + f 2 (x 0 α 2 + S(β 2 )) + + α 4x 2 f 2 + 2x 0 x i 1 S(β i ) + S(β i )S(f i 2 ) 2f 1 x i 1 α i α 3x 1 f 2 [x i 2 α if 2 + α ix 0 S 2 (f i 4 )] + 2x 0 x i 1 S(β i ) + S(β i )S(f i 2 ) The only thing that may not in (x 0 S 2 (I n 3 )+(f 1, f 2 ))[x ] is n 2x 0x i 1 S(β i )+ n S(β i)s(f i 2 ) and it s totally depends on β i. If g I n x0 R n, we know n S(β i)s(f i 2 ) n 2 i1 S(β i+2)s(f i ) 0, which implies n 2 i1 β i+2f i 0. By assumptions, (β 3,..., β n ) is a combination of µ i and ν j,k where i n 3, and j < k n 2. For µ i ( 2ix i, ( 2i+3)x i 1,..., ix 0, 0,..., 0), we have S(µ i ) (0, 0, 2ix i+1, ( 2i+3)x i,..., ix 1, 0,..., 0), and the image of S(µ i ) onto R n is i (i 3j)x j+1 f i j+3 i (i 3j)x j+1 (S(f i j+1 ) 2x 0 x i j+2 ) i (i 3j)S(x j )S(f i j+1 ) i 2x 0 (i 3j)x j+1 x i j+2 (3 i)x 0 S(f i+1 ) 6x 0 x 1 x i+2 i (i 3j)2x 0 x j+1 x i j+2 2(3 i)x 0 x 1 x i+2 + (3 i)x 0 S 2 (f i 1 ) 6x 0 x 1 x i+2 (3 i)x 0 S 2 (f i 1 ) 2if 2 x i+2 Since i n 3, then the image is inside (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ]. For ν j,k f k e j + f j e k, we have S(ν j,k ) S(f k )e j+2 + S(f j )e k+2, and the image of S(ν j,l ) onto R n is S(f k )f j+2 + S(f j )f k+2 S(f k )(S(f j ) + 2x 0 x j 1 ) + S(f j )(S(f k ) + 2x 0 x k 1 ) S(f k )S(f j ) + S(f j )S(f k ) 2S(f k )x 0 x j 1 + 2S(f j )x 0 x k 1 2(S 2 (f j 2 ) + 2x 1 x j )x 0 x k 1 2(S 2 (f k 2 ) + 2x 1 x k )x 0 x j 1 2x 0 (S 2 (f j 2 )x k 1 S 2 (f k 2 )x j 1 ) + 4f 2 (x j x k 1 x k x j 1 ) Since j < k n 2, the image is in (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ]. Therefore, by our assumption on the generators of the first syzygy module, we have for all g I n x0 R n, g (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ]. As a result, we have x 0 R n I n (x0 ) x 0 R n (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ].

7 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS 7 Since R n C[x 0, x 1,..., x ], we can consider the ideal x 0 R n as C[x 2 0, x 0 x 1,..., x 0 x ]. Then x 0 R n (x 0 S 2 (I n 3 ) + (f 1, f 2 ))[x ] C[x 2 0, x 0 x 1,..., x 0 x ] (x 0 S 2 (I n 3 ) + (x 2 0, 2x 0x 1 ))[x ] C[x 0x 2, x 0 x, 3,..., x 0 x ] x 0 S 2 (I n 3 )[x ] x 0C[S 2 (x 0 ), S 2 (x 1 ),..., S 2 (x n 4 )][x ] x 0 S 2 (I n 3 )[x ] x 0S 2 (R n 3 )[x ] x 0 S 2 (I n 3 )[x ] x 0 S 2 ( Rn 3 I n 3 ) [x ] Now, since we already have the recursive form for the ideal Q n of R n /I n which consists of all elements divisible by x 0 and the quotient P n (R n /I n )/Q n, we are ready to calculate the recursive relation of the Hilbert Series of R n /I n. However, before we start the calculation, we still need to know how the shift map affects the Hilbert Series. Lemma 4.2. For any bi-graded M where M consists of polynomials of x 1, x 2,..., x n, denote the Hilbert Series of M as H M (q, t). Then the Hilbert Series of S(M), which is denoted as H S(M) (q, t), satisfies H S(M) (q, t) H M (q, qt). Proof. By definition, we know that then we have H M (q, t) H S(M) (q, t) q a1 t a2 dim M (a 1, a 2 ), q a1 t a2 dim S(M) (a 1, a 2 ). Since S(x i ) x i+1, which shift the first degree for all x i. Then because deg(m 1 m 2 ) deg(m 1 ) + deg(m 2 ) for all monomials m 1 and m 2, a 1 [S(m)] a 1 (m) + a 2 (m) and a 2 [S(m)] a 2 (m) for any monomial m as a 2 denotes the total degree of a monomial and that won t be changed by shifting. Hence, dim M (a 1, a 2 ) dim S(M) (a 1 + a 2, a 2 ) for all a 1, a 2 Z 0, and we have H S(M) (q, t) q a1 t a2 dim S(M) (a 1, a 2 ) q a1+a2 t a2 dim S(M) (a 1 + a 2, a 2 ) q a1 (qt) a2 dim M (a 1, a 2 ) H M (q, qt) Proposition 4.3. For all n N, H Sn (M)(q, t) H M (q, q n t). We can derive this result by repeating the process n times. Notice that H S2 (M)(q, t) H M (q, q 2 t). Now, we have enough tools to prove our main theorem. Theorem 4.4. Denote H n (q, t) to be the Hilbert Series of R n /I n, then for n 4, it follows the following recursive relation: H n (q, t) th n 3(q, q 2 t) + H n 2 (q, qt) 1 q. t Proof. Notice that x 0 R n /[I n (x0 )] is an ideal of R n /I n and their quotient yields R n / x 0 R n I n I n (x0 ) R n / R n (x0 ) I n I n + (x 0 ).

8 8 AUTHOR: YUZHE BAI SUPERVISOR: DR. EUGENE GORSKY Then, because H(M) H(M/S) + H(S) for all bi-graded C space M and bi-graded subspace S M. we have ( ) ( ) x0 R n Rn H n (q, t) H + H. I n (x0 ) I n + (x 0 ) Because we know that we can derive as well as x 0 R n I n (x0 ) x 0S 2 ( Rn 3 I n 3 ( ) x0 R n H I n (x0 ) t ( ) Rn H I n + (x 0 ) ) [x ] and R n I n + (x 0 ) S ( Rn 2 I n 2 H n 3 (q, q 2 t)(q t) k th n 3(q, q 2 t) 1 q, t H n 2 (q, qt)(q t) k H n 2(q, qt) 1 q t. As a result, we will have the recursive formula for H n (q, t) as By Theorem 4.4, we will have H(I n ) H n (q, t) th n 3(q, q 2 t) + H n 2 (q, qt) 1 q. t m 1 1 q k t th n 3(q, q 2 t) + H n 2 (q, qt) 1 q. t ) [x ], For the base cases of H n (q, t), we have: H 0 (q, t) 0; H 1 (q, t) 1 + t as R 1 /I 1 has basis {1, x 0 }; H 2 (q, t) t + 1/(1 qt) as R 2 /I 2 has basis {x 0, x k 1 k Z 0 }. Although from Theorem 4.1, we can derive the recursive formula of Hilbert Series in Theorem 4.4, the proof of Theorem 4.1 is based on the assumption that the first syzygy module of f 1,..., f n is generated by certain relations. Now, we are going to prove that it is true. Theorem 4.5. For n N, the first syzygy module of f 1, f 2,...f n, namely M n, is generated by µ i and ν j,k where i n 1, j k where j, k n. Before we start the proof of the theorem, let s first see some examples of cases with small n. Example 4.6. (when n 3) In I 3, consider 3 i1 α if i 0, write α i x 0 α i + α i where x 0 α i and j 1, 2, 3. Then, 3 i1 and any term of α j α i f i (x 0 α 1 + α 1)x (x 0 α 2 + α 2)2x 0 x 1 + (x 0 α 3 + α 3)(2x 0 x 2 + x 2 1). Since α 3x 2 1 is the only term not divisible by x 0, we must have α 3 0, what remains is (x 0 α 1 + α 1)x (x 0 α 2 + α 2)2x 0 x 1 + x 0 α 3(2x 0 x 2 + x 2 1). for i 2, 3 Similarly, 2α 2x 0 x 1 and x 0 α 3x 2 1 are the only two terms not divisible by x 2 0, so 2α 2x 0 x 1 +x 0 α 3x 2 1 0, which implies α 2 x 1 α 3/2. Moreover, since 2x 0 (2x 0 x 2 + x 2 1) x 1 (2x 0 x 1 ) 4x 2 (x 2 0) 0 is a standard syzygy for f 1, f 2, f 3, which is µ 2, then α 2(2x 0 x 1 ) + x 0 α 3(2x 0 x 2 + x 2 1) x 1α 3f 2 + x 0 α 2 3f 3 2x 2 α 3f 1 x 1α 3f 2 2 α 3µ x 2 α 3f 1 + x 0 α 3f 3 + 2x 2 α 3f 1

9 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS 9 Therefore, we can rewrite the relation as (x 0 α 1 + α 1 + 2x 2 α 3)f 1 + x 0 α 2f 2 + α 3µ 2 2. Notice that µ 2 0, then (x 0 α 1 + α 1 + 2x 2 α 3)f 1 + x 0 α 2f 2 0 as well which is a relation between f 1 and f 2 and we know it must be a multiple of µ 1. As a result, any arbitrary syzygy of f 1, f 2, f 3 is generated by µ 1 and µ 2. Example 4.7. (when n 4) Similar to the case in I 3, write the arbitrary syzygy as 4 i1 α if i as (x 0 α 1 + α 1)x (x 0 α 2 + α 2)2x 0 x 1 + (x 0 α 3 + α 3)(2x 0 x 2 + x 2 1) + (x 0 α 4 + α 4)(2x 0 x 3 + 2x 1 x 2 ). Since the only terms not divisible by x 0 are α 3x 2 1 and α 4(2x 1 x 2 ) where x 2 1 S(f 1 ) and 2x 1 x 2 S(f 2 ), then there β C[x 1, x 2, x 3 ] such that α 3 2x 2 β and α 4 x 1 β. Rewrite the relation as (x 0 α 1 + α 1)x (x 0 α 2 + α 2 α α 2 )2x 0x 1 + x 0 α 3(2x 0 x 2 + x 2 1) + x 0 α 4(2x 0 x 3 + 2x 1 x 2 ) 4βx 0 x 2 2, 2 + x 3 β. Then the terms not divisible by x 2 0 are α 2 (2x 0x 1 ), α 3x 0 x 2 1, 2α 4x 0 x 1 x 2, and where 4βx 0 x 2 2. If α 2, α 3, and α 4 are all 0, then β 0 and we can precede to the next step. Otherwise, we have first three terms are divisible by x 1, then the last one must be divisible by x 1 as well. Therefore, there γ C[x 1, x 2, x 3 ] such that β x 1 γ. As a result, α 3 2x 1 x 2 γ γf 4 + 2x 0 x 3 γ, and α 4 x 2 1γ γf 3 2x 0 x 2 γ. Then, the relation becomes (x 0 α 1 + α 1)x (x 0 α 2 + α 2 + x 1x 3 γ 2x 2 2γ)2x 0 x 1 + x 0 α 3(2x 0 x 2 + x 2 1) + x 0 α 4(2x 0 x 3 + 2x 1 x 2 ) γν 3,4. Then, by taking out a α 4µ 3, we get (x 0 α 1 + α 1 x 3 α 4)f 1 + (x 0 α 2 + α 2 + x 1x 3 γ 2x 2 2γ x 2 α 4)f 2 + (x 0 α 3 x 1 α 4)f 3 + α 4µ 3 γν 3,4. Because µ 3 0 and ν 3,4 0, we have (x 0 α 1 + α 1 x 3 α 4)f 1 + (x 0 α 2 + α 2 + x 1x 3 γ 2x 2 2γ x 2 α 4)f 2 + (x 0 α 3 x 1 α 4)f 3 0, which is a syzygy of f 1, f 2, f 3, by the last part, it is generated by µ 1 and µ 2. As a result, the module of syzygies of f 1, f 2, f 3, f 4 is generated by µ 1, µ 2, µ 3, and ν 3,4. Now, let s start the proof of the general case, which follows the similar idea. Moreover, to make the proof more accessible, we will precede in the n-tuple language. Proof. Let F n be a rank n free module with basis B {b 1,..., b n }, and define R n map φ : F n I n such that φ(b i ) f i. The first syzygy module of f 1,..., f n is ker φ and since we have a basis for F n, we can write element of F n as n-tuples. For any arbitrary element α M n F n, write α as α (α 1, α 2,..., α n ), then for each α i, we can rewrite it as x 0 α i + α i where x 0 doesn t divide any term of α i for all i, and by taking out proper number of ν 1,i, we can have x 0 α i for all i 2. By mapping the syzygy into I n, we get (x 0 α 1 + α 1)f 1 + (x 0 α 2 + α 2)f 2 + i1 (x 0 α i + α i )f i 0 (x 0 α i + α i )(2x 0 x i 1 + S(f i 2 )) 0 and that implies n α i S(f i 2) 0 because any other terms are divisible by x 0. Then let because of Lemma 1, we have α β i+1 S(µ i 2 ) + α (0, 0, α 3,..., α n), 3 j,k n,j k β j,k S(ν j 2,k 2 ).

10 10 AUTHOR: YUZHE BAI SUPERVISOR: DR. EUGENE GORSKY For any j, k 3 such that j k. Because S(ν j 2,k 2 ) S(f k 2 )e j + S(f j 2 )e k (f k 2x 0 x k 1 )e j + (f j 2x 0 x j 1 )e k ν j,k + 2x 0 x k 1 e j 2x 0 x j 1 e k, we can take out all the S(ν j 2,k 2 ) as ν j,k and what remains will go to the x 0 α i parts. Let α α β j,k ν j,k, then α M n. Write α as 3 j,k n,j k α (x 0 α 1 + α 1, x α α 2,..., x 0α n + α n). Let α (0, 0, α 3,..., α n), notice that α n (n 3)x 1 β n and we have α β i+1 S(µ i 2 ). Now, we consider the image of S(µ i 2 ), for i 3,.., n 1. Therefore, we have φ(s(µ i 2 )) φ( α) (x 0 α 1 + α 1 )f 1 + (x 0 α i (i + 1 3k)x k f i+2 k + (2i 1)x i f 2 k1 (i + 1)x 0 S(f i ) + (2i 1)x i 1 f 2 α 2 + (2i 1)β i+1 x i 1 )f 2 + By removing the terms divisible by x 2 0, we get ( α (2i 1)β i+1 x i 1 )f x 0 α i f i (i + 1)β i+1 x 0 S(f i ) x 0 α i S(f i 2 ) (i + 1)β i+1 x 0 S(f i ) 0. Since f 2 2x 0 x 1, we can cancel a x 0 in each of the terms and simplify the equation as 2( α (2i 1)β i+1 x i 1 )x 1 + α i S(f i 2) (i + 1)β i+1 S(f i ) Because S(f i 2 ) 2x 1 x i 2 + S 2 (f i 4 ) for i 5, we can write α i x 1δ i + δ i and β i x 1 γ i + γ i where δ i and γ i contain no term divisible by x 1. Therefore, we notice the terms that are not divisible by x 1 will sum up to be 0. That is δ i S 2 (f i 4 ) (i + 1)γ i+1s 2 (f i 2 ) 0 n+1 δ i S 2 (f i 4 ) (i 1)γ i 1S 2 (f i 4 ) 0 (δ i (i 1)γ i 1)S 2 (f i 4 ) + nγ ns 2 (f n 3 ) 0. Because the equation is a relation of S 2 (f i ), where i 1,..., n 3, by Lemma?? and induction hypothesis, we know γ n x 2 η + η i S 2 (f i 4 ).

11 RECURSIVE RELATIONS FOR THE HILBERT SERIES FOR CERTAIN QUADRATIC IDEALS 11 Then, so that As a result, let β n x 1 γ n + x 2 η + η i (f i 2x 0 x i 1 2x 1 x i 2 ) x 1 γ n + x 2 η + η i (f i 2x 0 x i 1 ) + S(f n 2 ), α n (n 3)[x 2 1γ n + x 1 x 2 η + η i x 1 (f i 2x 0 x i 1 ) + x 1 S(f n 2 )] (n 3)[(f 3 2x 0 x 2 ) γ n (f 4 2x 0 x 3 )η + η i x 1 (f i 2x 0 x i 1 ) 1 n 3 (n 3 3i)x i+1 S(f n 2 i ) n 3 i1 (n 3)[(f 3 2x 0 x 2 ) γ n (f 4 2x 0 x 3 )η + η i x 1 (f i 2x 0 x i 1 ) 1 n 3 (n 3 3i)x i+1 (f n i 2x 0 x n i 1 ) n 3 i1 θ i f i x 0 θ α α θ i ν i,n 1 n 1 ( α n θ)µ, then α M n and the last entry is 0. Then by induction hypothesis, it is a combination of µ i and ν j,k for i 1, 2,..., n 2 and j, k 1, 2,..., n 1 such that j k. In conclusion, M n is generated by µ i and ν j,k for i 1, 2,..., n 1 and j, k 1, 2,..., n such that j k. In fact, this result of the structure of the first syzygy module will induce a recursive formula for the minimal free resolution of the quotient ring, and it can further imply an explicit formula of the Hilbert Series, which, as we stated in the introduction, can be found in [7]. References [1] C. Bruschek, H. Mourtada, J. Schepers. Arc spaces and the Rogers Ramanujan identities. Ramanujan J. 30 (2013), no. 1, [2] B. Feigin. Abelianization of the BGG Resolution of Representations of the Virasoro Algebra. Funct. Anal. Appl. 45:4 (2011), [3] B. Feigin, A. Stoyanovsky. Functional models of the representations of current algebras, and semi-infinite Schubert cells. Funct. Anal. Appl. 28:1 (1994), [4] C. Calinescu, J. Lepowsky, A. Milas. Vertex-algebraic structure of the principal subspaces of certain A (1) 1 -modules. I. Level one case. Internat. J. Math. 19 (2008), no. 1, [5] S. Capparelli, J. Lepowsky, A. Milas. The Rogers-Ramanujan recursion and intertwining operators. Commun. Contemp. Math. 5 (2003), no. 6, [6] E. Gorsky, A. Oblomkov, J. Rasmussen. On stable Khovanov homology of torus knots. Exp. Math. 22 (2013), no. 3, [7] Y. Bai, E. Gorsky, O. Kivinen. Quadratic ideals and Rogers-Ramanujan recursions. arxiv: v1 [math.ag].