About categorification of cyclotomic integers and tensored N-complexes
aa r X i v : . [ m a t h . K T ] J u l ABOUT CATEGORIFICATION OF CYCLOTOMIC INTEGERS ANDTENSORED N -COMPLEXES DJALAL MIRMOHADESA
BSTRACT . We prove that the ideal used in recent works to categorify the cyclo-tomic integers is generated by a cyclotomic polynomial. Moreover, we publish aproof by T. Ekedahl that the q -binomial relations used in the tensor product of N -complexes makes it necessary for the category to be enriched over the cyclotomicintegers.
1. A
CKNOWLEDGEMENTS
It was during my master’s thesis that I started to investigate the question posedas Theorem 3.1. I could not prove it, nor find any counterexamples. Finally, theproblem reached Ekedahl in October 2010 who swiftly replied by mail with theproof published here.About four years later at a conference in Montreal, Khovanov asked if someonecould work on categorification of the ring of cyclotomic integers.Motivated by Ekedahls proof, I started to look for a suitable ideal in a someproduct category of N -complexes so that I could apply the methods in the proof byEkedahl to prove that the decategorified ideal is generated by the cyclotomic poly-nomial. The resulting paper [Mir] however only dealt with two distinct primes,and a simple calculation was sufficient to show that the ideal is principal.However, without Ekedahls solution, I would most likely not have worked on[Mir], and I would certainly not have written this paper. So I would like to ac-knowledge my gratitude to Torsten Ekedahl.2. A BOUT CATEGORIFICATION OF CYCLOTOMIC INTEGERS
Recently, Laugwitz & Qi in [LQ] constructed a monoidal category toghetherwith a thick ideal such that in the ring Z [ q, q − ] , this ideal is generated by certainelements of the form ( q m − / ( q m/p k − , p k | m . Here, we show that this idealis generated by Φ m ( q ) . This result generalizes the case in [Mir, Last step of mainTheorem] where m is equal to a product of two primes. Note that we can dothe calculations in Z [ q ] instead of Z [ q, q − ] because in the quotient ring we have q m = 1 , hence q − = q m − .We refer to [LQ, Mir] and the references within them for a more detailed intro-duction to the subject. Lemma 2.1.
Any maximal ideal of Z [ x ] contains a prime number.Proof (unknown origin). Let I be an ideal of Z [ x ] . If I ∩ Z = (0) , we need to showthat I cannot be maximal. Let I ′ be the ideal of Q [ x ] generated by I . We have I ′ =( f ( x )) for a polynomial f coming from Z [ x ] of content . The polynomial f hasdegree > , because if could be written as a linear combination of polynomialsfrom I in Q [ x ] , then I ∩ Z = (0) . But any h ∈ I is of the form gf for some g ∈ Z [ x ] .It follows from Gauss lemma that g ∈ Z [ x ] . This shows that I ⊆ ( f ( x )) in Z [ x ] .To see that ( f ( x )) is not maximal in Z [ x ] , pick an m ∈ Z such that f ( m ) = 0 , ± . Evaluation at m then induces a map Z [ x ] −→ Z / ( f ( m )) which has kernel strictly between ( f ( x )) and Z [ x ] . Hence, any maximal ideal I of Z [ x ] has I ∩ Z = ( n ) for some integer n > . But since n is the characteristic of thefield Z [ x ] /I , it must be prime. (cid:3) Lemma 2.2.
An ideal I of Z [ x ] is the unit ideal if and only if, for every prime p , thecanonical homomorphism Z [ x ] → Z / ( p )[ x ] maps I to the unit ideal in Z / ( p )[ x ] .Proof. The canonical homomorphisms Z [ x ] → Z / ( p )[ x ] maps the unit ideal to aunit ideal. So assume I is not the unit ideal of Z [ x ] , then by Zorn’s lemma I liesin a maximal ideal M , and by Lemma 2.1, there is a prime p ∈ M . But then, theimage of M in Z / ( p )[ x ] cannot be the unit ideal. (cid:3) Lemma 2.3.
For a prime p , and distinct positive integers n and m not divisible by p ,we have gcd { Φ n ( q ) , Φ m ( q ) } = 1 in the principal ideal domain Z / ( p )[ q ] .Proof. Let ϕ denote Euler’s totient function and k denote the least common multi-ple of n and m . The integer p ϕ ( k ) − is then divisible by k , n and m . Let GF p ϕ ( k ) denote a splitting field of q p ϕ ( k ) − q over Z / ( p ) . Since the polynomial Φ n ( q )Φ m ( q ) isa divisor of ( q p ϕ ( k ) − − q in Z [ q ] , it has only simple roots in GF p ϕ ( k ) . This provesthe lemma. (cid:3) Theorem 2.4.
Let n = p p · · · p t , where p k are distinct primes. In the ring Z [ q ] , wehave the following equality of ideals (cid:18) [ n ] q [ n/p ] q (cid:19) + (cid:18) [ n ] q [ n/p ] q (cid:19) + · · · + (cid:18) [ n ] q [ n/p t ] q (cid:19) = (cid:0) Φ n ( q ) (cid:1) where [ m ] q = ( q m − / ( q − and Φ n denotes the n :th cyclotomic polynomial.Proof. Recall that q n − Q d | n Φ d ( q ) , hence [ n ] q (cid:14) [ n/p k ] q = ( q n − (cid:14) ( q n/p k −
1) = Y d | n,p k | d Φ d ( q ) . Since Φ n ( q ) divides every generator [ n ] q (cid:14) [ n/p k ] q , it is equivalent to show that thepolynomials [ n ] q (cid:14) [ n/p k ] q Φ n ( q ) , ≤ k ≤ t generate the unit ideal in Z [ q ] . We useLemma 2.2, so we need to show this in Z / ( p )[ q ] for an arbitrary prime p . Since Z / ( p )[ q ] is a PID, it is enough to show that gcd([ n ] q (cid:14) [ n/p ] q , · · · , [ n ] q (cid:14) [ n/p t ] q ) =Φ n ( q ) .In the case p n , the minimum multiplicity of Φ m ( q ) in [ n ] q (cid:14) [ n/p k ] q over ≤ k ≤ t is equal to if m = n and equal to otherwise. It then follows from Lemma2.3 that the gcd of the generators is equal to Φ n ( q ) .In the case p | n , we again use Lemma 2.3. We only need to count the multiplic-ities of Φ m ( q ) where p m , because by [Nag, p. 160] (assuming p m ) and the factthat we work over characteristic p , we have Φ pm ( q ) = Φ m ( q p )Φ m ( q ) = Φ m ( q ) p − . Moreover, when p k = p [ n ] q (cid:14) [ n/p k ] q = ( q n − (cid:14) ( q n/p −
1) = ( q n/p − p − = Y d | n/p Φ d ( q ) p − BOUT CATEGORIFICATION OF CYCLOTOMIC INTEGERS AND TENSORED N -COMPLEXES 3 and when p k = p [ n ] q (cid:14) [ n/p k ] q = ( q n/p − p (cid:14) ( q n/p k p − p = Y d | n/p,p k | d Φ d ( q ) p . The multiplicity of Φ n/p ( q ) in [ n ] q (cid:14) [ n/p k ] q is then equal to p − when p k = p and equal to p otherwise. Hence the minimum multiplicity over ≤ k ≤ t is equalto p − , which is equal to the multiplicity of Φ n/p ( q ) in Φ n ( q ) . The minimummultiplicity of Φ m ( q ) , where m = n/p and p m is equal to for the same reasonas before. (cid:3)
3. A
BOUT TENSORED N - COMPLEXES
To define a tensor product for N -complexes, Kapranov [Kap] uses q -commuta-tivity in the construction of the total complex. Assuming that q is a primitive N :throot of unity (that is Φ N ( q ) = 0 ) implies that (cid:0) Ni (cid:1) q = 0 for < i < N . This turnsthe total complex into an N -complex. See [Kap, Prop. 1.8–1.10] for details.The following theorem shows that the assumption that Φ N ( q ) = 0 is not onlysufficient in the above construction, but also necessary. Theorem 3.1.
In the ring Z [ q ] , we have the following equality of ideals (cid:18) n (cid:19) q ! + (cid:18) n (cid:19) q ! + · · · + (cid:18) nn − (cid:19) q ! = (cid:0) Φ n ( q ) (cid:1) where (cid:0) ni (cid:1) q denote q -binomial coefficients and Φ n denotes the n :th cyclotomic polynomial.Proof (T. Ekedahl). The claim is that the ideal in Z [ q ] generated by (cid:0) ni (cid:1) q , < i < n ,is equal to Φ n ( q ) , the n’th cyclotomic polynomial (all (cid:0) ni (cid:1) q are clearly divisible by Φ n ( q ) as the factor Φ n ( q ) in q n − appearing in the numerator doesn’t cancel fromthe denominator). Hence an equivalent formulation is that the ideal I generatedby (cid:0) ni (cid:1) q / Φ n ( q ) is equal to the unit ideal. If not it is contained in a maximal idealand any maximal ideal of Z [ q ] contains a prime number p . Hence we may replace Z [ q ] by Z /p [ q ] . As the latter ring is a PID, what we need to show is that the GCDof the (cid:0) ni (cid:1) q is equal to Φ n ( q ) . Now we recall that q m − Q d | m Φ d ( q ) and if m = p k m ′ we have Φ m = Φ m ′ ( q ) p k (everything computed in Z /p [ q ] ). Finally, if p m, m ′ , then Φ m ( q ) and Φ m ′ ( q ) are relatively prime. Hence, for p d we havethat the multiplicity with which Φ d divides q m − is equal to if d m and equalto ψ ( k ) := p k + p k − + · · · + 1 if d | m and k is the largest power of p dividing m .To show the result it is enough to show that for every d with p d the largestpower of Φ d ( q ) which divides all (cid:0) ni (cid:1) q is if n = p k d and p k if n = p k d . Assumetherefore that p d . Applied to i = 1 this gives d | n and we write n = p k m with p m . Assume first that d = m and consider (cid:18) np k d (cid:19) q = ( q n − q n − − · · · ( q n − p k d +1 )( q − q − · · · ( q p k d − Now, the multiplicity with which Φ d ( q ) divides q p k d − j − , for ≤ j < p k d is equalto the same multiplicity for q p k m − j − and hence the Φ d ( q ) -factors in the numer-ator and denominator cancel exactly. If instead n = p k d , then we use (cid:0) np k − d (cid:1) q andthe argument is the same except that we get an extra contribution of multiplicity p k in q n − . (cid:3) DJALAL MIRMOHADES
Note that for a prime p we have, by [Nag, p. 160] Φ np ( q ) = Φ n ( q p ) if p | n , Φ n ( q p )Φ n ( q ) if p n .So in characteristic p we get Φ np k ( q ) = Φ n ( q p k ) / Φ n ( q p k − ) = Φ n ( q ) ( p − p k − if p n .R EFERENCES[Kap] M. M. K
APRANOV . On the q –analog of homological algebra , arXiv:q-alg/9611005.[LQ] R. L AUGWITZ , Y. Q I . A categorification of cyclotomic rings , arXiv:1804.01478.[Mir] D. M
IRMOHADES . Categorification of the ring of cyclotomic integers for products of two primes ,arXiv:1506.08755.[Nag] T. N