QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS

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1 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Abstract. We give an exlicit recursive descrition of the Hilbert series and Gröbner bases for the family of uadratic ideals defining the jet schemes of a double oint. We relate these recursions to the Rogers-Ramanujan identity and rove a conjecture of the second author, Oblomkov and Rasmussen.. Introduction In this aer, we study a family of uadratic ideals defining the jet schemes for the double oint D = Sec k[x]/x 2. Here k is a field of characteristic zero. Recall that the (n -jet scheme of X is defined as the sace of formal mas Sec k[t]/t n X []. In the case of the double oint, such a formal ma is defined by a olynomial x(t = x 0 + x t + + x n t n, such that x(t 2 0 mod t n. By exanding this euation, we get a system of euations n f = x 2 0, f 2 = 2x 0 x,..., f n = x i x n i. We denote the defining ideal of Jet n D A n by i=0 I n := f,..., f n R n := k[x 0,..., x n ]. The ring R n is Z 2 0-graded by assigning the grading (i, to x i. It is then clear that the ideal I n is bihomogeneous. Let H n (, t = i,j 0 dim k (R n /I n i,j i t j Z[[, t]] denote the bigraded Hilbert series for R n /I n. Our first main result is the following. Theorem.. The series H n (, t satisfies the recursion relation with initial conditions H n (, t = H n 2(, t + th n 3 (, 2 t n t H 0 (, t =, H (, t = + t, H 2 (, t = t + t. Using this recursion relation, we obtain exlicit combinatorial formulas for H n (, t: Theorem.2. The Hilbert series H n (, t is given by the following exlicit formula: ( h(n,+ ( t H n (, t = ( n h(n, t ( n t, =0 where h(n, = n 2.

2 2 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN In the limit n, we rerove the theorem of Bruschek, Mourtada and Scheers [4], which relates the Hilbert series of the arc sace for the double oint to the Rogers- Ramanujan identity. In fact, we refine their result by considering an additional grading, see euation (7.. Similar results for n = were obtained by Feigin-Stoyanovsky [9, 0], Leowsky et al. [5, 6], and the second author, Oblomkov and Rasmussen in [8]. Although our aroach to the comutation of the Hilbert series is insired by [4], it is uite different. The key result in [4] shows that for n = the olynomials f k form a Gröbner basis of the ideal I. As we will see below, the Gröbner basis of the ideal I n for finite n is larger and has a very subtle recursive structure. We comletely describe such a basis in Theorems 4.2 and 4.6. In articular, we rove the following. Theorem.3. Let k > 2. Then the reduced Gröbner basis for I n contains ( n k+ 2 k 2 olynomials of degree k. Our roof of Theorem. does not use Gröbner bases at all. First, by an exlicit inductive argument in Theorem 2.2 we give a comlete descrition of the first syzygy module for f i. Then, we define a shift oerator S : R n R n+, which sends x i to x i+, and identify I n x 0 R n and I n /(I n x 0 R n with the images of I n 3 and I n 2 under aroriate owers of S. This imlies the recursion relation in Theorem.. We also observe a recursive structure in the minimal free resolution of R n /I n. In articular, we rove the following: Theorem.4. Let b(i, n denote the rank of the i-th term in the minimal free resolution for R n /I n, in other words the i-th Betti number. Then b(i, n = b(i, n + b(i, n 3 + b(i 2, n 3. As a conseuence, we can comute the rojective dimension of R n /I n. Corollary.5. The rojective dimension of R n /I n euals 2n 3. Remark.6. It is easy to see that the reduced scheme (Jet n D red is a linear subsace given by the euations x 0 =... = x n = 0 and has dimension 2 n n dim Jet n D = n =. 2 2 A more careful analysis of the gradings in Theorem.4 imlies another formula for the series H n (, t which was first conjectured in [8]. Theorem.7. The Hilbert series of R n /I n has the following form: H n (, t = n i=0 ( ( ( k t i t =0 k=0 ( ( ( 52 3 n t n 2 2 t 2+2. The aer is organized as follows. In Section 2 we introduce the shift oerator S, describe its roerties and rove Theorem 2.2 which exlicitly describes all syzygies between the f i. In Section 3, we use the shift oerator to find a recursive relation for the Hilbert series and to rove Theorem.. In Section 4, we use the recursive structure to describe a Gröbner basis for I n. In Section 5, we give a recursive descrition of the minimal free resolution of R n /I n and rove Theorem.4. In Section 6, we solve both of the above recursions exlicitly (with the given initial conditions and give two exlicit combinatorial formulas for H n (, t. Finally, in Section 7 we briefly discuss the limit of all these techniues at n and the connection to the Rogers-Ramanujan identity.

3 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 3 Acknowledgments E. G. would like to thank Boris Feigin, Mikhail Bershtein, James Leowsky, Kirill Paramonov and Anne Schilling for useful discussions, and Russian Academic Excellence Project 5-00 for its suort. O. K. thanks Eric Babson and Jésus de Loera for discussions in the initial stages of the roject. The work of E. G. and O. K. was suorted by the NSF grants DMS and DMS The work of E. G. in section 6 was suorted by the RSF grant O.K. was also suorted by the Ville, Kalle and Yrjö Väisälä foundation of the Finnish Academy of Science and Letters. 2. Ideals and syzygies 2.. Ideals. Let R n = k[x 0,..., x n ] and f k = k i=0 x ix k i. Define I n R n to be the ideal generated by f,..., f n. Let F n be the free R n -module with the basis e,..., e n. Consider the ma φ n : F n R n given by the euation φ n (α,..., α n = f α f n α n. The R n -module Ker(φ n is called the first syzygy module of I n. Lemma 2.. One has (2. Proof. Indeed, (n 3ix i f n+ i = 0. i=0 (n 3ix i f n+ i = i=0 The coefficient at each monomial x i x k x l euals i+k+l=n (n 3ix i x k x l. (n 3i + (n 3k + (n 3l = 3n 3(i + k + l = 3n 3n = 0. For 0 < k < n, define µ k := ( 2kx k, ( 2k + 3x k,..., kx 0, 0,..., 0 F n. By (2., we have φ n (µ k = 0. Denote also ν ij = f i e j f j e i (for i j. It is clear that φ n (ν ij = 0. The main result of this section is the following. Theorem 2.2. The first syzygy module Ker(φ n is generated by µ k and ν i,j over R n. We rove Theorem 2.2 in Section The shift oerator. We define a ring homomorhism S : R n R n+ by the euation S(x i = x i+. Note that S is injective and we can uniuely write any olynomial in R n in the form f = x 0 f + S(f, f R n, f R n. The following euation is clear from the definition and will be very useful below: (2.2 f n = 2x 0 x n + S(f n 2. By abuse of notation, denote also S : F n F n+2 the ma which is given by (2.3 S(α,..., α n = (0, 0, S(α,..., S(α n. Lemma 2.3. Let α F n. Then φ n+2 (S(α is divisible by x 0 if and only if φ n (α = 0.

4 4 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Proof. By (2.2 we have φ n+2 (S(α = ( S(α i f i+2 S α i f i i= i= mod x 0. Therefore φ n+2 (S(α is divisible by x 0 if and only if S( α i f i is divisible by x 0. But since no shift contains x 0, this haens if and only if ( S αi f i = 0 α i f i = φ n (α = 0. Since φ n (µ k = φ n (ν ij = 0, by Lemma 2.3 the images of S(µ k and S(ν ij under φ n+2 are divisible by x 0. The following lemma describes these images exlicitly. Lemma 2.4. One has φ n+2 (S(µ k = (2k + 6x k+3 f + (2k + 3x k+2 f 2 (k + 3x 0 f k+4, φ n+2 (S(ν ij = 2x 0 x j+ f i+2 2x 0 x i+ f j+2. Proof. By definition, so S(µ k = (0, 0, 2kx k+, ( 2k + 3x k,..., kx, 0,..., 0 = µ k+3 + (2k + 6x k+3 e + (2k + 3x k+2 e 2 (k + 3x 0 e k+4, φ n+2 (S(µ k = (2k + 6x k+3 f + (2k + 3x k+2 f 2 (k + 3x 0 f k+4. Also, S(ν ij = S(f i e j+2 S(f j e i+2, so φ n+2 (S(ν ij = S(f i f j+2 S(f j f i+2 = (f i+2 2x 0 x i+ f j+2 (f j+2 2x 0 x j+ f i+2 = Corollary 2.5. One has Proof. 2x 0 x j+ f i+2 2x 0 x i+ f j+2. φ n+2 (S(µ k = (2k + 3x k+2 f 2 (k + 3x 0 S(f k+2 = kx k+2 f 2 (k + 3x 0 S 2 (f k. φ n+2 (S(µ k = (2k + 6x k+3 f + (2k + 3x k+2 f 2 (k + 3x 0 f k+4 = (2k + 6x k+3 f + (2k + 3x k+2 f 2 (k + 3(2x 2 0x k+3 + 2x 0 x x k+2 + x 0 S 2 (f k = (2k + 3x k+2 f 2 (k + 3x 0 S(f k+2 = kx k+2 f 2 (k + 3x 0 S 2 (f k. Examle 2.6. µ = ( 2x, x 0, so S(µ = (0, 0, 2x 2, x, and φ 4 (S(µ = 2x 2 (2x 0 x 2 + x 2 + x (2x 0 x 3 + 2x x 2 = 2x 3 x 0 x 4x 0 x 2 2 = x 3 f 2 4x 0 S 2 (x 2 0. Lemma 2.7. The olynomial x S(f n 2 can be exressed via f,..., f n modulo x 0. Proof. We have (n 3x 0 f n 2 + (n 6x f n (n 3x n 2 f 0 = 0, so (n 3x S(f n 2 + (n 6x 2 S(f n (n 3x n S(f 0 = 0. It remains to notice that S(f i f i+2 mod x 0. Lemma 2.8. Assume that Ker(φ n 2 is generated by µ k and ν i,j and suose that φ n (α is divisible by x 0. Then α n = Ax 0 + Bx + n γ if i for some A, B and γ i.

5 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 5 Proof. As above, we can write α i = x 0 α i + S(α i 2 for i 3. Since f and f 2 are divisible by x 0, we get φ n (S(α = S(α i 2f i α i f i 0 mod x 0. By Lemma 2.3 we get φ n 2 (α = 0. By the assumtion, we can write α = β k µ k + γ i,j ν ij. Therefore and k<n 2 i= i<j n 2 α n 2 = β n x 0 + j n 3 α n = x 0 α n + S(α n 2 = x 0 α n + S(β n x + γ j,n 2 f j, j n 3 S(γ j,n 2 (f j+2 2x 0 x j Examles. Before roving Theorem 2.2, we would like to resent the roof for n 4. Examle 2.9. For n = 2 we have f = x 2 0 and f 2 = 2x 0 x, so the module of syzygies is clearly generated by ( 2x, x 0 = µ. Examle 2.0. Let n = 3, suose that α f +α 2 f 2 +α 3 f 3 = 0. We can write α 3 = α 3x 0 + α 3, where α 3 does not contain x 0. Since f and f 2 are divisible by x 0 and f 3 = 2x 0 x 2 + x 2, we get x 2 α 3 = 0, so α 3 = 0. Now α = 2 α 3µ 2 + γ, where γ is a syzygy between f i with γ 3 = 0. By the revious examle, γ is a multile of µ, so the module of syzygies is actually generated by µ and µ 2. Examle 2.. Let n = 4, suose that α is a syzygy. We can write α 3 = α 3x 0 + α 3 and α 4 = α 4x 0 + α 4 where α i do not contain x 0. Similarly to the revious case, we obtain (2.4 α 3x 2 + α 4 2x x 2 = 0. This means that there exists some β such that α 3 = 2x 2 β and α 4 = x β. Now α x α 2 2x 0 x + (α 3x 0 2x 2 β(2x 0 x 2 + x 2 + (α 4x 0 + x β(2x 0 x 3 + 2x x 2 = 0. The terms without x 0 cancel, and the linear terms in x 0 are the following: x 0 (2α 2 x + α 3x 2 4x 2 2β + 2α 4x x 2 + 2βx x 3 = 0. Note that all terms but 4x 2 2β are divisible by x, so β is divisible by x, β = mx. Then α 4 = α 4x 0 + mx 2 = (α 4 2x 2 mx 0 + mf 3. By subtracting mν 3,4 + 3 (α 4 2x 2 mµ 3 from α, we obtain a syzygy between f, f 2, f 3 and reduce to the revious case.

6 6 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN 2.4. Syzygies. In this section, we rove Theorem 2.2 by induction on n. The base cases were covered in Section 2.3. Suose that α = (α,..., α n Ker(φ n, i. e. is a linear relation between f,..., f n. As above, write α i = α ix 0 + S(α i 2 for i 3. Without loss of generality, we can assume that α i do not contain x 0 (otherwise we can subtract a multile of ν,i. Since f i = 2x 0 x i + S(f i 2, by collecting terms without x 0 we get n S(α i 2S(f i 2 = 0. This means that φ n 2 (α = 0 and by the induction assumtion we may then write Because n α = β i+ µ i j<k n,j k β j,k ν j 2,k 2. S(ν j 2,k 2 = S(f k 2 e j + S(f j 2 e k = ν j,k + 2x 0 x k e j 2x 0 x j e k, without loss of generality we can assume α = S( n β i+µ i 2. By Corollary 2.5 we get φ n (S(µ i 2 = (i + x 0 S(f i + (2i x i f 2, hence n φ n (α = α f +(α 2 + (2i S(β i+ x i f 2 + n x 0 α if i (i+s(β i+ x 0 S(f i = 0. By collecting the terms linear in x 0, we get n n (α 2 + (2i S(β i+ x i 2x + α is(f i 2 (i + S(β i+ S(f i = 0, so is divisible by x, and n α is(f i 2 (i + S(β i+ S(f i n α i f i 2 (i + β i+ f i is divisible by x 0, where α i = S(α i. By Lemma 2.8, this imlies n 2 β n = Bx 0 + Cx + γ i f i for some constants B, C. Now we can rewrite n 3 α n = α nx 0 + S(β n x 0 = α nx 0 + Bx 2 + Cx x 2 + γ i x (f i+2 2x 0 x n + γ n 2 x S(f n 2. Observe that x 2 = f 3 2x 0 x 2, x x 2 = 2 (f 3 2x 0 x 3 and by Lemma 2.7 x S(f n 2 can be exressed via f,..., f n modulo x 0. In other words, n α n = δx 0 + δ i f i for some coefficients δ i. Then α δµ n n n δ iν i,j is a syzygy between f,..., f n, so by the induction assumtion it can be exressed as an R n -linear combination of the µ i and ν i,j.

7 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 7 Remark 2.2. The above roof shows that the syzygies ν,k and ν 2,k are not necessary, and can be exressed as linear combinations of other syzygies. Indeed, since the coefficients at e k are divisible by x 0, one can subtract an aroriate multile of µ k and get a syzygy involving e,..., e k only. 3. Hilbert series In this section, we rove Theorem 3.5 by studying the relation between the ideals I n and x 0 R n. Lemma 3.. One has R n /(x 0 R n + I n S(R n 2 /I n 2 [x n ] as R n -modules, the module structure on the right coming from S : R n R n. Proof. We have x 0 R n + I n = x 0, f,..., f n = x 0, S(f,..., S(f n 2, so R n /(x 0 R n + I n = R n / x 0, S(f,..., S(f n 2 = S(R n 2 /I n 2 [x n ]. Lemma 3.2. The subsace x 0 S 2 (I n 3 [x n ] does not intersect the ideal f, f 2 in R n. Furthermore, x 0 S 2 (I n 3 [x n ]+ f, f 2 is an ideal in R n which is contained in I n x 0 R n. Proof. Given a nonzero olynomial g I n 3, the iterated shift S 2 (g does not contain x 0 or x, so that x 0 S 2 (g is not contained in f, f 2. Furthermore, I n 3 is stable under multilication by x 0,..., x n 4, so S 2 (I n 3 is stable under multilication by x 2,..., x n 2, and x 0 S 2 (I n 3 [x n ] is stable under multilication by x 2,..., x n. Multilication by x 0 or x sends the latter subsace to f, f 2, so x 0 S 2 (I n 3 [x n ] + f, f 2 is an ideal in R n. Finally, to rove that this ideal is contained in I n, it is sufficient to rove that x 0 S 2 (f k I n for k n 3. On the other hand, by Corollary 2.5: Lemma 3.3. One has x 0 S 2 (f k = k + 3 φ n(s(µ k mod f, f 2. I n x 0 R n = x 0 S 2 (I n 3 [x n ] + f, f 2. Proof. By Lemma 3.2, the right hand side is a submodule of the left hand side, so it remains to rove the reverse inclusion. We have f i = 2x 0 x i + S(f i 2 = 2x 0 x i + 2x x i 2 + S 2 (f i 4. Suose that n i= α if i I n x 0 R n. Then by Lemma 2.8, α n = Ax 0 + Bx + j γ j f j = A x 0 + B x + j γ j S 2 (f j 4. Now by (2. and Corollary 2.5, x 0 f n and x f n can be exressed as R n -linear combinations of f,..., f n and elements of x 0 S 2 (I n 3 [x n ] + f, f 2, so n i= α if i can be exressed as such a combination as well. Induction on n finishes the roof. Corollary 3.4. One has x 0 R n /(I n x 0 R n = x 0 S 2 (R n 3 /I n 3 [x n ].

8 8 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Proof. We have Therefore x 0 R n / f, f 2 = x 0 R n /(x 2 0, x 0 x = x 0 k[x 2,..., x n ] = x 0 S 2 (R n 3 [x n ] x 0 R n /(I n x 0 R n = x 0 R n /(x 0 S 2 (I n 3 [x n ] + f, f 2 = x 0 S 2 (R n 3 /I n 3 [x n ]. Theorem 3.5. Let H n (, t denote the bigraded Hilbert series of the uotient R n /I n. Then one has the following recursion relation (3. H n (, t = H n 2(, t + th n 3 (, 2 t n t with initial conditions H 0 (, t =, H (, t = + t, H 2 (, t = t + t. Remark 3.6. This recursion is similar, but not identical to the various recursions considered by Andrews [, 2, 3] in his roofs of the Rogers-Ramanujan identity. It is also similar to the recursions recently considered by Paramonov [2] in a different context. Proof. We have an exact seuence 0 x 0 R n /(x 0 R n I n R n /I n R n /(x 0 R n + I n 0. By Lemma 3., the Hilbert series of R n /(x 0 R n + I n euals H n 2(,t, and by Corollary n t 3.4 the Hilbert series of x 0 R n /(x 0 R n I n euals th n 3(, 2 t. n t 4. Gröbner bases We will now comute Gröbner bases for the ideals I n. Recall that a Gröbner basis for an ideal I is a subset G = {g,..., g s } I such that, for a chosen monomial ordering <, LT < (g,..., LT < (g s = LT < (I, where LT < denotes leading term. Let us order the monomials in R n in grevlex order, that is x α < x β if α < β or α = β and the rightmost entry of α β is negative. Remark 4.. In fact, any order refining the reverse lexicograhic order will work, but for definiteness and its oularity in comuter algebra systems we shall fix grevlex order throughout. Theorem 4.2. Let G = {f } R, G 2 = {f, f 2 } R 2 and recursively define the sets G n, n 3 as follows: G n = x 0 S 2 (G n 3 {f, f 2 } S(G n 2, where S is a modified shift oerator as exlained below. Then G n is a Gröbner basis for I n.

9 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 9 Remark 4.3. The notation reuires exlanation. Note that any G m is naturally a subset of R n, n m so we can and will identify G m inside a larger olynomial ring without exlicit mention. Furthermore, we denote by x 0 S 2 (G n 3 the image of G n 3 under S 2 : R n 2 R n multilied by x 0. The oerator S is defined on elements I n 2 as follows: write = n i= ϕ if i, and let S( = S(ϕ i f i+2. i= Note that by (2.2, we have S( = S( + n i= x 0x i+2 S(ϕ i I n+2. In articular, if 0 and is homogeneous then LT < ( S( = S(LT < (. Therefore the construction of S( reuires a choice if ϕ i, but the leading term of the result does not deend on this choice. Proof. We will roceed by induction. The base cases n =, 2 are clear because the ideals are monomial. Consider now the ideal LT < (I n generated by all the leading terms of elements of I n. It is clear by Lemma 3. and the fact that S resects the reverse lexicograhic order that if g I n is not divisible by x 0, its leading term is the image of a leading term in I n 2 under S. Since we assumed G n 2 to be a Gröbner basis, we must have LT < (g divisible by some monomial in S(LT < (G n 2. Similarly, if g is divisible by x 0, we know by Lemma 3.2 and order reservation that its leading term is the image under x 0 S 2 of a leading term in I n 3 or divisible by f, f 2. By the induction assumtion LT < (g is then divisible by an element of x 0 S 2 (LT < (G n 3 {f, f 2 }. In articular, LT < (I n LT < (G n. But the reverse inclusion is clear, so we have LT < (I n = LT < (G n as desired, and G n is a Gröbner basis for I n. Examle 4.4. We have G 3 = {f, f 2, f 3 } G 4 = {f, f 2, f 3, f 4, x 0 x 2 2} G 5 = {f, f 2, f 3, f 4, f 5, x 0 x 2 x 3 } G 6 = {f,..., f 6, x 0 x x 0 x 2 x 4, 2x x x 0 x 3 x 4 x 0 x 2 x 5 }. Note that the last olynomial in G 6 can be identified with S(x 0 x 2 2 S(G 4. Indeed, so 4x 0 x 2 2 = 2x 2 (2x 0 x 2 + x 2 x (2x 0 x 3 + 2x x 2 + x 3 (2x 0 x = 2x 2 f 3 x f 4 + x 3 f 2, S(4x 0 x 2 2 = 2x 3 f 5 x 2 f 6 + x 4 f 4 = 2x 3 (2x 0 x 4 + 2x x 3 + x 2 2 x 2 (2x 0 x 5 + 2x x 4 + 2x 2 x 3 + x 4 (2x 0 x 3 + 2x x 2 = 4x x x 0 x 3 x 4 2x 0 x 2 x 5. Remark 4.5. The Gröbner basis constructed in Theorem 4.2 is far from being reduced. The following theorem describes the reduced basis imlicitly. Since all G n contain {f,..., f n } and none of their leading terms divides one another, we can throw away other olynomials in G n in a controlled manner to obtain a minimal Gröbner basis. That is to say, if the leading terms of G n \{g} still generate the leading ideal we are in business. Therefore after aroriate reduction [7, Proosition 6 on. 92] we get a reduced Gröbner basis with the same leading terms.

10 0 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Let us call a monomial x a i i divisible by x 2 i or by x i x i+. admissible if a i + a i+ for all i, that is, it is not Theorem 4.6. Fix k > 2. The leading terms of (t-degree k in a reduced Gröbner basis for I n have the form m(x LT < (f n+k 2 where m(x is an admissible monomial of degree k 2 in variables x 0,..., x n+k 7. The number of degree k olynomials in the reduced 2 Gröbner basis euals ( n k+ 2. k 2 Remark 4.7. It is easy to see that there are no linear olynomials in the Gröbner basis (or in the ideal I n, and f,..., f n are the only uadratic olynomials in the reduced Gröbner basis. Proof. We rove the statement by induction in n. Suose that it is true for G n 2 and G n 3. By Theorem 4.2, the leading monomials in the degree k art of G n consist of shifted degree k monomials in G n 2, and twice shifted degree (k monomials in G n 3, multilied by x 0. Consider first the case k = 3. We will rove that the leading terms in the reduced Gröbner basis have the form x j LT < (f n+ for j n 4. Indeed, in the first case we 2 get S(x j LT < (f (n 2+ = x j+ LT < (f n+. In the second case we have to consider the olynomials x 0 S 2 (f i for all i n 3. Observe that for i n 4 we get LT < (x 0 S 2 (f i = x 0 LT < (f i+4 and hence divisible by the leading term of f i+4 and can be eliminated. For i = n 3 we get LT < (x 0 S 2 (f n 3 = x 0 LT < (f n+. Assume now that k > 3. In the first case we get S(m(x LT < (f (n 2+k 2 = S(m(x LT < (f n+k 2. If m(x is an admissible monomial in x j, 0 j (n 2+k 7 then S(m(x is an admissible monomial in x j, j (n 2+k 7 + = n+k In the second case we get x 0 S 2 (m(x LT < (f (n 3+(k 2 = x 0 S 2 (m(x LT < (f n+k 2. Now S 2 (m(x is an admissible monomial in x j, 2 j (n 3+(k 7 +2 = n+k 7, so 2 2 x 0 S 2 (m(x is also an admissible in a correct set of variables. In fact, all such monomials not divisible by x 0 aear from the first case, and the ones divisible by x 0 aear from the second case. It is easy to see that none of these leading monomials are divisible by each other. Therefore after aroriate reduction [7] we get a reduced Gröbner basis with the same leading terms. Finally, we can count monomials of given degree k. The number of admissible monomials of degree l in s variables euals ( s l+, so the number of olynomials in Gn of degree k euals l ( ( + n+k 7 (k n k+ = 2. k 2 k 2 Examle 4.8. Let n = 2. The reduced Gröbner basis for I 2 contains uadratic olynomials f,..., f 2. It also contains 5 cubic olynomials with leading terms 6 uartic olynomials with leading terms x 0 x 2 6, x x 2 6, x 2 x 2 6, x 3 x 2 6, x 4 x 2 6, x 0 x 2 x 6 x 7, x 0 x 3 x 6 x 7, x 0 x 4 x 6 x 7, x x 3 x 6 x 7, x x 4 x 6 x 7, x 2 x 4 x 6 x 7

11 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS and 4 uintic olynomials with leading terms x 0 x 2 x 4 x 2 7, x 0 x 2 x 5 x 2 7, x 0 x 3 x 5 x 2 7, x x 3 x 5 x 2 7. Observe that LT < (f 3 = x 2 6, LT < (f 4 = x 6 x 7 and LT < (f 5 = x Minimal resolution In this section we describe the bigraded minimal free resolutions of I n and R n /I n. We write them as follows: and 0 I n F (, n F (2, n F (3, n 0 R n /I n R n = F (0, n F (, n F (2, n F (3, n Theorem 5.. Let F (i, n be the i-th term in the minimal free resolution for I n. Then there is an injection F (i, n F (i, n, and F (i, n/f (i, n S(F (i, n 3 x 0 S(F (i 2, n 3 as R n -modules, and the shift of a free R n -module is as in (2.3. Note that the gradings in the right hand side are shifted by the bidegree of f n (which euals n t 2. Proof. Observe that the ideal generated by f,..., f n in R n is isomorhic to I n [x n ], so its minimal resolution over R n is identical to the one for I n over R n tensored over R n. Moreover, since I n = f,..., f n, the minimal free R n -resolution of I n [x n ] is naturally a subcomlex of the minimal free resolution for I n. In other words, F (i, n Rn R n can be identified with a subsace in F (i, n, which we will by abuse of notation also denote F (i, n. We have a short exact seuence 0 F (i, n F (i, n F (i, n/f (i, n 0. From the long exact seuence in cohomology, it is easy to see that F (i, n/f (i, n is acyclic in ositive degrees. Now I n = f,... f n, so F (, n/f (, n = R n is generated by a single vector corresonding to f n. Furthermore, by Theorem 2.2 F (2, n has generators corresonding to µ,..., µ n and ν i,j for 3 i < j n, so F (2, n/f (2, n = Rn n 2 is sanned by the basis elements corresonding to µ n and ν i,n for 3 i n. The differential d : F (2, n F (, n descends to d : F (2, n/f (2, n F (, n/f (, n, sending µ n to x 0 f n and ν i,n to f i f n. Therefore, the uotient comlex with terms F (i, n/f (i, n is isomorhic to the minimal resolution of R n / x 0, f 3,..., f n = R n / x 0, S(f,..., S(f n 3. The latter is nothing but the (shifted minimal resolution for I n 3 tensored with the two-term comlex x R 0 n Rn. Corollary 5.2. Let b(i, n denote the rank of F (i, n. Then (5. b(i, n = b(i, n + b(i, n 3 + b(i 2, n 3. Corollary 5.3. Let H n (, t denote the Hilbert series for R n /I n, and let Hn (, t = H n (, t n i=0 ( i t. Then H n (, t satisfies the following recursion relation: (5.2 Hn (, t = H n (, t n t 2 ( t 2 H n 3 (, t. Corollary 5.4. The rojective dimension of I n euals 2n. The rojective dimension 3 of R n /I n euals 2n. 3

12 2 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Proof. By definition, the rojective dimension d(i n is eual to the length of the minimal free (or rojective resolution. By (5. we have d(i n = d(i n The minimal free resolutions for I, I 2 and I 3 are easy to comute: ( f I ( f f 2 R I 2 R2 2 R 2 2x 0 4x 2 ( x x f f 2 f 3 0 2x I 3 R3 3 0 R3. 2 The minimal resolution of R n /I n is one ste longer than the one for I n. 2x x 0 6. Combinatorial identities We define ( a ( ( a = b ( ( b ( ( a b. If a < b, we set ( a = 0. The following lemma is well known. b Lemma 6.. The following identities holds: ( ( ( a a a + + b+ = = a b( a b b + b + b Proof. One has ( a = ( ( a b a, b + ( b+ b hence ( ( a a + b+ = b b + ( a ( a+ b ( b+ = + ( ( a + b+ ( a b b ( b+ ( a + b +. ( a. b + = Theorem 6.2. The Hilbert series H n (, t is given by the following exlicit formula: ( h(n,+ ( t (6. H n (, t = ( n h(n, t ( n t, where h(n, = n 2. =0 Proof. By Theorem 3.5 it is sufficient to rove that the right hand side of (6. satisfies the recursion relation (3.. Let us denote the -th term in (6. by H n, (, t so that H n (, t = H n,(, t. We have h(n 2, = h(n 3, = h(n,, so ( h(n, ( t H n 2, (, t = ( n h(n, t ( n 2 t, ( h(n, H n 3, (, 2 ( ( 2 t 2 2 t = ( n h(n, t ( n 2 t,

13 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 3 therefore (6.2 H n 2, (, t + th n 3, (, 2 t = ( t ( n h(n, t ( n 2 t ( t ( n h(n, t ( n 2 t [ ( ( ] h(n, h(n, + = ( h(n, + = ( n th n, (, t. This roves (3., and the initial conditions are easy to check. The free resolution of I n gives another formula for the Hilbert series of R n /I n. Proosition 6.3. Let b(i, n, as above, denote the rank of i-th module in the free resolution of R n /I n. Then b(i, n = [( ( ( ( ] n 2 + n 2 + i i Remark 6.4. The terms in the first sum are nonzero if (n + /3 and i/2 i. The terms in the second sum are nonzero if (n /3 and (i /2 (i. Proof. Let ( ( ( ( n 2 + n 2 A(n,, i =, B(n,, i =. i i Then ( n 2 A(n,, i + A(n 3,, i + A(n 3,, i 2 = ( n 2 ( i ( i + + ( n 2 ( i ( ( n 2 = i + ( ( n 2 = i ( n 2 + ( i = A(n,, i. Similarly, B(n,, i + B(n 3,, i + B(n 3,, i 2 = B(n,, i, so the right hand side satisfies the recursion relation (5.. It remains to check the base cases: ( n f(0, n = =, 0 ( ( n n 3 f(, n = n = +, 0 ( ( ( ( n 2 n n 3 n 3 f(2, n = (n + = By Corollary 5.4 b(i, n = 0 for i > 2 and n 3. We have the following (, t-analogue of Proosition 6.3.

14 4 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Proosition 6.5. Let b(i, n denote the bigraded Hilbert olynomial for the generating set in F (i, n. Then ( ( (6.3 b(i, n = 52 3+(i (i n t 2+(i + >0 i ( ( (i (i n 2 2 t 2+2+(i i Proof. The roof is comletely analogous to the roof of Proosition 6.3, but we include it here for comleteness. By Theorem 5. we have a recursion relation (6.4 b(i, n = b(i, n + n t 2 b(i, n 3(, t + n t 3 b(i 2, n 3(, t. We need to rove that the right hand side of (6.3 satisfies (6.4. Let ( ( Â(n,, i = 52 3+(i (i n t 2+(i i Then ( ( Â(n 3,, i (, t = (i (i + n 2 2 t 2 2+(i, i ( ( Â(n 3,, i 2(, t = (i (i n 2 2 t 2 2+(i i so Â(n 3,, i (, t + tâ(n 3,, i 2(, t = ( ( (i (i n 2 2 t 2 2+(i i Now Â(n,, i + n t 2 Â(n 3,, i (, t + n t 3 Â(n 3,, i 2(, t = [ (n ( ( ( ] 52 3+(i (i 2 n 2 2 t 2+(i + n 3+ = i i ( ( 52 3+(i (i n t 2+(i = Â(n,, i. i A similar recursion holds for B(n,, i. It remains to check the initial conditions: b(0, n =, ( ( n n 3 b(, n = (t 2 + t n t 2 = t 2 + t 2, 0 ( ( ( ( n 2 n n 3 n 3 b(2, n = t 3 [n ] + 5 t 4 = t t t The following result was conjectured by the second author, Oblomkov and Rasmussen in [8, Conjecture 4.]...,

15 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 5 Theorem 6.6. The Hilbert series of R n /I n has the following form: (6.5 H n (, t = ( n 52 3 ( i=0 ( i t =0 ( n t 2 ( k t k=0 ( n 2 2 t 2+2. Proof. It is clear that H n (, t = n i=0 ( i t i=0 ( i b(i, n. The latter can be comuted by (6.3, and it remains to use the identity ( ( k t = ( j j(j /2 t j. j k=0 j=0 7. Limit at n In the limit n both formulas for the Hilbert series simlify. Indeed, for fixed we have ( n lim = n ( (, so we can take the limit of all the above results. Proosition 7.. The limit of the Hilbert series H n (, t has the following form: ( t (7. H (, t = ( ( 2 (. =0 Proosition 7.2. The limit of the bigraded rank of the i-th syzygy module F (i, n euals ( (7.2 b(i, = ( 52 3+(i (i 2 t 2+(i i >0 ( ( + ( (i (i 2 t 2+2+(i i ( ( Proosition 7.3. The limit of the Hilbert series H n (, t has the following form: (7.3 H n (, t = i=0 ( i t =0 ( k=0 ( k t 52 3 k+ 2 t t 2+2. The euality between the right hand sides of (7.3 and (7. was roved in [0, Theorem 3.3.2(b]. At t = and t = one recovers more familiar Rogers-Ramanujan identities. The following roosition concerning Gröbner bases in the limit was roved first in [4], but we give an alternative roof here. In fact, [4] use a slightly different basis of Bell olynomials. Yet another roof can be obtained by taking the limit in Theorem 4.6. Proosition 7.4. For n the olynomials f i form a Gröbner basis for the ideal I.

16 6 YUZHE BAI, EUGENE GORSKY, AND OSCAR KIVINEN Before embarking on the roof, we record the following lemmas concerning Gröbner bases here for the convenience of the reader. Lemma 7.5 ([7] Proosition 8 on. 06. Given (g,..., g s F s, the S-airs (7.4 S ij := lcm(lt <(g i, LT < (g j LT < (g i e i lcm(lt <(g i, LT < (g j e j LT < (g j form a homogeneous basis for the syzygies on {LT < (g,..., LT < (g s }. Lemma 7.6 ([7] Proosition 9 on. 07. Let I = g,..., g s. Then G = {g,..., g s } is a Gröbner basis for I if and only if every element of a homogeneous basis for the syzygies on LT < (G reduces to zero modulo G. Lemma 7.7 ([7] Proosition 4 on.03. G = {g,..., g s } R n, and suose g i, g j G have relatively rime leading monomials. Then the S-olynomial (7.5 S(g i, g j := φ n (S ij = lcm(lt <(g i, LT < (g j LT < (g i reduces to zero modulo G. g j lcm(lt <(g i, LT < (g j g j LT < (g j Proof of Proosition 7.4. Consider S(f i, f j. By Lemma 7.7 gcd(lt < (f i, LT < (f j = imlies that S(f i, f j reduces to zero modulo {f k } k=. Write i = 2 + r, where r = 0,. Then LT < (f i = x 2 if i is even and LT < (f i = 2x x + if i is odd. So the only case we need to consider is j = i +. In this case, we have { 2x 2 lcm(lt < (f i, LT < (f i+ = x +, i even 2x x 2 +, i odd. Additionally S(f i, f i+ = { 2x + f i x f i+, x f i 2x + f i+, i even i odd. But from (2. it follows that these S-airs aear in the relations φ n (µ n = 0 for n 0. Since n =, we always have these relations in I. Additionally, moving the S-air to the right-hand side we reduce S(f i, f i+ 0 modulo {f k } k=. In articular, Lemma 7.6 imlies that {f k } k= is a Gröbner basis for I. References [] G. E. Andrews and R. J. Baxter. A motivated roof of the Rogers-Ramanujan identities. Amer. Math. Monthly, 96 (989, no. 5, [2] G. E. Andrews. On the roofs of the Rogers-Ramanujan identities. In -series and artitions (Minneaolis, MN, 988, volume 8 of IMA Vol. Math. Al., ages 4. Sringer, New York, 989. [3] G. E. Andrews. The theory of artitions. Cambridge Mathematical Library. Cambridge University Press, Cambridge, 998. Rerint of the 976 original. [4] C. Bruschek, H. Mourtada, J. Scheers. Arc saces and the Rogers Ramanujan identities. Ramanujan J. 30 (203, no., [5] C. Calinescu, J. Leowsky, A. Milas. Vertex-algebraic structure of the rincial subsaces of certain -modules. I. Level one case. Internat. J. Math. 9 (2008, no., [6] S. Caarelli, J. Leowsky, A. Milas. The Rogers-Ramanujan recursion and intertwining oerators. Commun. Contem. Math. 5 (2003, no. 6, [7] D. Cox, J. Little, and D. O Shea. Ideals, varieties, and algorithms. Undergraduate Texts in Mathematics. Sringer Verlag, New York, third edition, An introduction to comutational algebraic geometry and commutative algebra. [8] E. Gorsky, A. Oblomkov, J. Rasmussen. On stable Khovanov homology of torus knots. Ex. Math. 22 (203, no. 3, A (

17 QUADRATIC IDEALS AND ROGERS-RAMANUJAN RECURSIONS 7 [9] B. Feigin. Abelianization of the BGG Resolution of Reresentations of the Virasoro Algebra. Funct. Anal. Al. 45 (20, no. 4, [0] B. Feigin, A. Stoyanovsky. Functional models of the reresentations of current algebras, and semiinfinite Schubert cells. Funct. Anal. Al. 28 (994, no., [] S. Ishii. Jet schemes, arc saces and the Nash roblem. C. R. Math. Acad. Sci. Soc. R. Can. 29 (2007, no., 2. [2] K. Paramonov. Cores with distinct arts and bigraded Fibonacci numbers. Discrete Math. 34 (208, no. 4, Deartment of Mathematics, University of California, Davis, One Shields Avenue, 9566 Davis, CA address: yzbai@ucdavis.edu Deartment of Mathematics, University of California, Davis, One Shields Avenue, 9566 Davis, CA International Laboratory of Reresentation Theory and Mathematical Physics, National Research University Higher School of Economics, Moscow, Russia address: egorskiy@math.ucdavis.edu Deartment of Mathematics, University of California, Davis, One Shields Avenue, 9566 Davis, CA address: kivineo@math.ucdavis.edu