We show that every semigroup which is a finite disjoint union of copies of
the free monogenic semigroup (natural numbers under addition) has linear
growth. This implies that the the corresponding semigroup algebra is a PI
algebra.Comment: 7 page

Abughazalah, Nabilah

Etingof, Pavel

Archiv der Mathematik

English

arXiv

We show that every semigroup which is a finite disjoint union of copies of the free monogenic semigroup (natural numbers under addition) has linear growth. This implies that the corresponding semigroup algebra is a PI algebra.National Science Foundation (U.S.) (Grant DMS-1000113)Saudi Aramco (Ibn Khaldun Fellowship

Etingof, Pavel I.

DSpace@MIT

Arch. Math. 105 (2015), 223–228c© 2015 Springer Basel0003-889X/15/030223-6published online August 20, 2015DOI 10.1007/s00013-015-0801-2 Archiv der MathematikLinear growth for semigroups which are disjoint unionsof finitely many copies of the free monogenic semigroupNabilah Abughazalah and Pavel EtingofAbstract. We show that every semigroup which is a ﬁnite disjoint union ofcopies of the free monogenic semigroup (natural numbers under addition)has linear growth. This implies that the corresponding semigroup algebrais a PI algebra.Mathematics Subject Classification. 20M05.Keywords. Monogenic semigroup, Unions, Linear growth, PI algebra.1. Introduction. It is well known that a semigroup may sometimes be decom-posed into a disjoint union of subsemigroups which is unlike the structuresof classical algebra such as groups and rings. For instance, the Rees theoremstates that every completely simple semigroup is a Rees matrix semigroup overa group G, and is thus a disjoint union of copies of G, see [6, Theorem 3.3.1];every Cliﬀord semigroup is a strong semilattice of groups and as such it is adisjoint union of its maximal subgroups, see [6, Theorem 3.3.1]; every commu-tative semigroup is a semilattice of archimedean semigroups, see [5, Theorem3.3.1].If S is a semigroup which can be decomposed into a disjoint union of sub-semigroups, then it is natural to ask how the properties of S depend on thesesubsemigroups. For example, if the subsemigroups are ﬁnitely generated, thenso is S. Arajo et al. [3] consider the ﬁnite presentability of semigroups whichare disjoint unions of ﬁnitely presented subsemigroups; Golubov [4] showedthat a semigroup which is a disjoint union of residually ﬁnite subsemigroupsis residually ﬁnite.For semigroups S which are disjoint unions of ﬁnitely many copies of thefree monogenic semigroup, the authors in [1] proved that S is ﬁnitely presentedand residually ﬁnite; in [2] the authors proved that, up to isomorphism andanti-isomorphism, there are only two types of semigroups which are unions oftwo copies of the free monogenic semigroup. Similarly, they showed that there224 N. Abughazalah and P. Etingof Arch. Math.are only nine types of semigroups which are unions of three copies of the freemonogenic semigroup and provided ﬁnite presentations for semigroups of eachof these types.In this paper we continue investigating semigroups which are disjoint unionsof ﬁnitely many copies of the free monogenic semigroup. Speciﬁcally, our mainresult is that every semigroup which is a ﬁnite disjoint union of copies of the freemonogenic semigroup has linear growth. This implies that the correspondingsemigroup algebra is a PI algebra.2. The linear growth theorem. Let S be a semigroup which is a disjoint unionof ﬁnitely many copies of the free monogenic semigroup:S =⋃a∈ANa,where A is a ﬁnite set and Na = 〈a〉 for a ∈ A. Deﬁne the functionsf : S × A × Z>0 → A and n : S × A × Z>0 → Z>0 by the formulaxyi = f(x, y, i)n(x,y,i).For y, z ∈ A, x ∈ S, let E(x, y, z) be the set of i ∈ Z>0 such that f(x, y, i) = z(i.e., xyi is a power of z).Definition 2.1. We say that z is (x, y)-persistent if E(x, y, z) is inﬁnite.Lemma 2.2. (i) Let z be (x, y)-persistent. Then E(x, y, z) = i0 + E0(x, y, z),where i0 ∈ Z>0 and E0(x, y, z) is a nonzero submonoid of Z≥0 (an addi-tively closed subset containing 0). Moreover, there exists a rational num-ber Mx(y, z) ≥ 0, such that for any i ∈ E(x, y, z),n(x, y, i) = n0 + Mx(y, z)(i − i0).(ii) Mx(y, z) is independent on x.(iii) The function f(x, y, i) is periodic in i starting from some place.Proof. First note that if ztyq = zs for some t, q, s > 0, then for N ≥ t we havezNyq = zN+s−t, and hencezNyqm = zN+(s−t)m (1)for all m > 0 and N ≥ mt.Now let i0 be the smallest number in E(x, y, z), and xyi0 = zn0 . By ourassumption, there exists r > 0 such that xyi0+r = zn; let us take the smallestsuch r. Thus zn0yr = zn. Let M = Mx(y, z) := n−n0r . Then by (1), for anyp > 0 and N ≥ pn0, we havezNypr = zN+Mpr. (2)Suppose now that ztyq = zs for some t, q, s > 0. Then setting m = r in (1),p = q in (2), and comparing these equations, we gets − t = Mq. (3)Now, let E0(x, y, z) = E(x, y, z)− i0. By (3), for all j ∈ E0(x, y, z), we havezn0yj = zMj+n0 . (4)Vol. 105 (2015) Linear growth for semigroups 225Since this holds for inﬁnitely many j, we have Mj+n0 > 0 for inﬁnitely manyj, hence M ≥ 0.Let us show that E0(x, y, z) is closed under addition. Suppose j, k ∈ E0(x,y, z). Multiplying (4) by yk, we getzn0yj+k = zMj+n0yk = zMjzMk+n0 = zM(j+k)+n0 .Thus, j + k ∈ E0(x, y, z). This proves (i).To prove (ii), let x′ ∈ S be another element such that z is (x′, y)-persistent.Then by (2), we havezNypr = zN+Mpr, zNypr′= zN+M′pr′ (5)for all p > 0 and suﬃciently large N , where M ′ = Mx′(y, z). Thus,zNyrr′= zN+Mrr′= zN+M′rr′for large enough N , implying M = M ′, as desired.Finally, note that it follows from (i) that if f(x, y, i) = z for some (x, y)-persistent z, and if j ∈ E0(x, y, z), then f(x, y, i + j) = z. So statement (iii)follows from the fact that for suﬃciently large j, xyj = zm for an (x, y)-persistent z. Namely, the period T for f(x, y, j) can be taken to be any positiveelement of the intersection⋂z is (x,y)−persistentE0(x, y, z).Remark 2.3. Lemma 2.2 in [1] states that if xiyj = xk holds in S for somex, y ∈ A and i, j, k ∈ N then i ≤ k. By this lemma and Lemma 2.2, if theset E(x, y, z) contains more than one element then it contains an arithmeticprogression, and hence is inﬁnite.In view of Lemma 2.2(ii), we will say that z is y-persistent if there is x ∈ Ssuch that z is (x, y)-persistent, and will denote Mx(y, z) simply by M(y, z).Notice that y is obviously y-persistent, with M(y, y) = 1.Lemma 2.4. Let x, y, z ∈ A. If z is y-persistent and y is x-persistent then z isx-persistent and M(x, z) = M(x, y)M(y, z).Proof. Suppose z is (v, y)-persistent, and y is (u, x)-persistent. Then, byLemma 2.2,uxi0+rm = yn0+M(x,y)rm, vyi′0+r′m′ = zn′0+M(y,z)r′m′for appropriate i0, i′0, n0, n′0, r, r′ ∈ Z>0 and any m,m′ ∈ Z≥0. Let m = sr′with s ∈ Z≥0, and  ≥ 0 be such that +n0 − i′0 = pr′ for some p ∈ Z≥0. Then(vyu)xi0+rm = vy+n0+M(x,y)rm = vyi′0+r′(p+M(x,y)rs)= zn′0+M(y,z)r′(p+M(x,y)rs).Thus,(vyu)xi0+rr′s = zn′0+M(y,z)pr′+M(x,y)M(y,z)rr′s.226 N. Abughazalah and P. Etingof Arch. Math.By Lemma 2.2, this means that z is (vyu, x)-persistent, and M(x, z) =M(x, y)M(y, z), as desired. Lemma 2.5. Let A be a finite set, P ⊂ A×A be a reflexive transitive relation,and M : P → Z≥0 be a nonnegative integer function such that M(y, y) = 1,and M(x, z) = M(x, y)M(y, z) whenever (x, y), (y, z) ∈ P . Then there existsa positive integer function d on A such thatd(y) ≥ d(z)M(y, z)whenever (y, z) ∈ P .Proof. We say that z is reachable from y (denote by z ≥ y) if (y, z) ∈ P . Wedeﬁne a relation ρ on A as follows:ρ = {(y, z) ⊆ A×A such that y is reachable from z and z is reachable from y}.It is clear that the relation ρ is reﬂexive, symmetric, and transitive, so it is anequivalence relation, splitting A into equivalence classes A1, A2, . . . , An. Nowwe show that d exists if n = 1, which means that A is a single equivalence class.In this case, M(x, y) is deﬁned for all x, y, and M(x, y)M(y, x) = M(y, y) = 1,so M(x, y) > 0. Let us pick a ∈ A and deﬁne d as follows:d(b) = M(b, a) ∀b ∈ A.We can multiply this function by an integer to make it integer-valued. By ourassumption, this function satisﬁes the required condition [in fact we have astronger condition d(y) = d(z)M(y, z)].Now consider the case n > 1. We say that an equivalence class C is a sinkif each element reachable from C is in C. We claim that there is a sink classAi among the equivalence classes A1, A2, . . . , An. To prove this, assume thecontrary, i.e., that there is no sink class. Pick an arbitrary equivalence classAi1 . It is not a sink, so we can reach some Ai2 from it with i2 = i1, and wecan reach Ai3 from Ai2 , with i2 = i3, and so on. For some k < l, we musthave ik = il+1, which is a contradiction, since Aik = Aik+1 , and yet they arereachable from each other.Now we continue the proof of the lemma by induction on |A| = N . It isclear that the statement is true when |A| = 1 . Suppose that the statement istrue when |A| < N , and let us prove it for |A| = N . Let C be a sink class,and consider A′ = A\C. By the induction assumption, d exists on A′ and C.Since C is a sink, if y ∈ A′ and z ∈ C then y is not reachable from z. So wemay multiply the function d on A′ by a large integer (without changing d onC) so that d(y) ≥ d(z)M(y, z) for all y ∈ A′, z ∈ C such that z is reachablefrom y. Then d satisﬁes the required condition. Corollary 2.6. For the semigroup S, there exists a positive integer function don A such thatd(y) ≥ d(z)M(y, z) (6)whenever z is y-persistent.Vol. 105 (2015) Linear growth for semigroups 227Proof. This follows from Lemmas 2.4 and 2.5, applying the latter to the set Pof pairs (y, z) such that z is y-persistent (and A, M as above). Now pick a function d as in Corollary 2.6. We extend this function from Ato S by setting d(xi) = d(x)i, i ∈ Z>0.Lemma 2.7. There exists a constant K > 0 such that d(x) ≤ K and d(xu) ≤K + d(u) for every u ∈ S and x ∈ A.Proof. Let y ∈ A. It suﬃces to prove that there exists Ky > 0 such thatd(xyi) ≤ Ky + d(yi) = Ky + d(y)ifor all i. Then we can take K ′ = maxyKy, and then pick K ≥ K ′ so thatd(x) ≤ K for all x ∈ A. Further, it is enough to show that for each (x, y)-persistent z, there exists Ky,z > 0 such that d(xyi) ≤ Ky,z + d(y)i wheneverf(x, y, i) = z. Indeed, taking K ′y := maxzKy,z, we see that the inequalityd(xyi) ≤ K ′y +d(y)i holds for almost all i [namely, for all i such that f(x, y, i)is (x, y)-persistent], so there exists Ky ≥ K ′y such that d(xyi) ≤ Ky + d(y)ifor all i.By Lemma 2.2, if f(x, y, i) = z then xyi = zn0+M(y,z)(i−i0), so by (6),d(xyi) = d(zn0+M(y,z)(i−i0)) = d(z)(n0 + M(y, z)(i − i0))≤ d(z)n0 + d(y)(i − i0) = d(z)n0 − d(y)i0 + d(y)i.So we may take Ky,z to be any positive number such that Ky,z ≥ d(z)n0 −d(y)i0. Corollary 2.8. If u is a word of length n, then d(u) ≤ Kn.Proof. This clearly follows from Lemma 2.7 by induction in n. Proposition 2.9. For r > 0, let I(r) = {w ∈ S such that d(w) ≤ r}. Thenthere exists L > 0 such that |I(r)| ≤ Lr for all r.Proof. Let a ∈ A and suppose am ∈ I(r). Since d(am) = d(a)m, we get thatm ≤ rd(a) . So the number of elements of I(r) of the form am is[rd(a)]. Hencewe may take L =∑a∈A1d(a) . Theorem 2.10. The semigroup S has linear growth.Proof. Let J(m) be the set of w ∈ S which can be represented by a wordof length ≤ m. Then by Corollary 2.8 J(m) ⊂ I(Km). Hence |J(m)| ≤|I(Km)| ≤ LKm by Proposition 2.9. This implies the theorem. Corollary 2.11. Let S := 1 ∪ S be the monoid obtained by adding a unit to S,and A = FS be the algebra spanned by S over any field F. Then A is a PIalgebra.Proof. By a theorem of Small et al. [7] any ﬁnitely generated algebra of lineargrowth (i.e., GK dimension 1) is PI. Thus the result follows from Theorem2.10. 228 N. Abughazalah and P. Etingof Arch. Math.Acknowledgements. The work of P.E. was partially supported by the NSFGrant DMS-1000113. Both authors gratefully acknowledge the support of IbnKhaldun Fellowship (Aramco). The ﬁrst author is deeply grateful for the gen-erous hospitality and excellent working conditions with Prof. Pavel Etingofat MIT. The ﬁrst author wants to thank Princess Nourah bint AbdulrahmanUniversity (PNU) for supporting the fellowship.References[1] N. Abughazalah and N. Ruskuc, On disjoint unions of ﬁnitely many copiesof the free monogenic semigroup, Semigroup Forum. Vol. 87. No. 1. Springer US,2013.[2] N. Abughazalah et al., A classiﬁcation of disjoint unions of two or three freemonogenic semigroup, Semigroup Forum, Online ﬁrst December (2014).[3] I.M. Arau´jo et al., On generators and relations for unions of semigroups,Semigroup Forum 63 (2001), 49–62.[4] E.A. Golubov, Finitely approximable regular semigroups (Russian), Mat. Za-metki 17 (1975), 423–432. (English translation: Math. Notes 17 (1975), 247–251.)[5] P.A. Grillet, Semigroups, Marcel Dekker, New York, 1995.[6] J.M. Howie, Fundamentals of Semigroup Theory, Clarendon Press, Oxford, 1995.[7] L. Small, J. Stafford, and R. Warfield, Aﬃne algebras of Gelfand-Kirillovdimension one are PI, Math. Proc. Cambridge Philos. Soc., 97 (1984), pp. 407174.Nabilah AbughazalahMathematical Sciences Department,Princess Nourah Bint Abdulrahman University,P.O. Box 84428,Riyadh 11671,Saudi Arabiae-mail: nhabughazala@pnu.edu.saPavel EtingofDepartment of Mathematics,Massachusetts Institute of Technology,Cambridge, MA 02139,USAe-mail: etingof@math.mit.eduReceived: 8 May 2015

Linear growth for semigroups which are disjoint unions of finitely many copies of the free monogenic semigroup

Nabilah Abughazalah

Pavel Etingof

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Linear growth for semigroups which are disjoint unions of finitely many copies of the free monogenic semigroup

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