WHEN ARE REES CONGRUENCES PRINCIPAL?

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1 proceedings of the american mathematical society Volume 106, Number 3, July 1989 WHEN ARE REES CONGRUENCES PRINCIPAL? C. M. REIS (Communicated by Thomas H. Brylawski) Abstract. Let p be the Rees congruence modulo the ideal / of the free monoid X*. In this short note we give necessary and sufficient conditions, in terms of the partial order induced by division on the complement of /, for p to be principal. In particular, we prove that if / is principal, so is p. Introduction Let X be a finite alphabet with at least two letters and X* the free monoid on X. Given a language L on X, the principal congruence determined by L, and denoted PL, is defined by: u = v(pl) if the following is valid: for all x, y G X*, xuy G L if and only if xvy g L. Recall that L is a union of PL-classes and that PL is the coarsest congruence saturating L. A congruence p on X* is said to be principal if p = PL for some L ç X*. If / is an ideal of X*, the Rees congruence p, is defined by: u s v(p ) if either u,v G I or u = v when u, v I. In this short note we find necessary and sufficient conditions for the Rees congruence p. to be principal and deduce that the Rees congruence modulo a prime ideal is principal as is the Rees congruence modulo a principal ideal. As for notation, the length of an element w g X* (called a word) will be denoted by tzz and the complement of a language L over X will be denoted by I. As a general reference we recommend G. Lallement's book [2]. Main results Definition 1. (a) Define a partial order < on X* by u < v if v = puq for some p,q G X*. (b) A set of incomparable elements relative to the partial order in (a) is called an infix code [3]. Notation. If L is a language over X, the sets {u G X*\u < I for some / G L} and {u G X*\l < u for some / G L} will be denoted by (L] and [L), respectively. Received by the editors July 30, 1988 and, in revised form, October 12, Mathematics Subject Classification (1985 Revision). Primary 20M05; Secondary 20M American Mathematical Society /89 $1.00+ $.25 per page

2 594 C. M. REIS Lemma 2. Let I be an ideal of X* and L a language over X. If L ç I, then [L) C I while if LC I, the complement of I, then (L] c 7. Proof. Clear from the definitions. Q Definition 3. A language W over A" is a chain if any two words of W are comparable. A chain 9f is said to be maximal in T, where T is a language over X, if W is contained in T and is not properly contained in any chain contained in T. A simple application of Zorn's lemma proves that, given a language T over X and ugt, there exists a chain W maximal in T and containing u. Lemma 4. Let I be an ideal of X*. Then every chain W maximal in 7 has the property that, for any x G W and any natural number n < \x\, there exists y G'W with \y\ = n. Proof. Let m = \x\. If m = 1, the lemma is valid. Assume the lemma is true for a word of length zzz - 1 belonging to W and let \x\ = m > 1. Let xf and x be the words obtained by deleting the first and last letters of x, respectively. Clearly (jc] = {x} U (x A U (x \, whence, since (x] ç 7 by Lemma 2 and W is maximal, it follows that either x, or x is in W. But \xf\ = \x \ = m- I and thus by the induction hypothesis, given a natural number zz < m - 1, there is a word in W of length n. The result now follows by induction. G The following is the main result. Theorem 5. Let I be an ideal of X*. Then p{ is principal iff 7 contains at most one word which is maximal in 7. Proof. Suppose first that 7 contains no maximal words. If 7 = 0, then I = X* and p, is clearly principal. Thus assume / # 0 and let 7 = {xx,x2,...}. Since 7 has no maximal words, each x; belongs to an infinite chain C maximal in 7. By Lemma 4, given any natural number zz, there exists a word w G C( with \w\ = zz. Let px < p2 < p3-- be the sequence of primes. For each Ct, choose a subchain Z)( = {z{xl) < z(2l) < } such that {z^l = 2P'> and let L = \JjDi- Let u,v G L and let «= 2P', \v\ = 2P'<. If \u\ = \v\, then p's = p\ wnence Ps P, and i = j Therefore s = t and i = j whence u = v. Thus no two words of L can have the same length. Let u, v g L, u ^ v. Then jw = 2s, \v\ = 2' with s > t, say. Now u G D; for some z, whence there exist x,y G X* such that xy ^ 1 and xuy G Di ç L. Let \xuy\ = 2. Then \x\ + \y\ = 2l -2s,1 > s> t. But \xvy\ = 2l -2s + 2l = 2t(2i~t -2s~l +1) and hence, since 2 "' - 2S~' + 1 is odd, it follows that xvy L. Therefore u ^ v(pl) and hence every PL-class contained in L is a singleton class. Let w g7. Then w G C for some z and thus there exist p, q G X* such that pwq G L. But, denoting the P -class of a word x by [x]l, we have p[w]lq ç [pwq]l = {pwq} whence \[w]l\ = 1. Moreover, pj saturates L and so /

3 WHEN ARE REES CONGRUENCES PRINCIPAL? 345 is contained in a PL-class since PL is the coarsest congruence saturating L. Since the P -classes in 7 are singletons, it follows that / is a P -class and hence PL = p,. Assume now that / contains exactly one maximal word, say w. If / = (w], then clearly p = P.,. Thus assume that (w] ^ 7. Each word x* of 7\(w] = {xx,x2,...} belongs to an infinite chain Ci maximal in /. Choose the subchains D of Ci as above and let L = (IJ, D )u{w}. Let u, v G L\{w}, u ^ v. Then \u\ = 2s, \v\ = 2', s > t and there exist x,y G X* such that \x\ + \y\ > \w\ and xuy G L\{w}. Arguing as before and since x + j> > \w\, it follows that xvy L. Therefore u ^ v(pl). Let u G L\{w}. Then u^w(pl) since the only context of w is (1,1). Thus the PL-classes contained in L are singleton classes. As above, since (X* ux* ) n L ^ 0 for all u G 1, it follows that Pj = PL. Conversely, suppose 7 contains more than one maximal word, say u, v with u t v. Suppose there exists L with p, = PL. Since PL saturates L, it follows that either I nl = 0 or I ç L. Assume, therefore, without loss of generality, that LCI. Since u and v are maximal in /, xuy G I or xvy G I implies xy = 1. If u L (resp. v $ L), then u = wpl (resp. v = wpl ) for all w G I, a contradiction. Hence u,v G L and u = vpl. This contradiction shows that p, cannot be principal. D We now obtain a necessary and sufficient condition for a Rees congruence not to be principal in terms of the minimal generating set of the ideal. Definition 6 [1]. Let / be an ideal of X*. Then the set of minimal words of / is called the infix root of / and is denoted by r(i). Clearly / = [r(i)) = X*r(I)X*. generating set of /. Definition 7. The word zz is a prefix (resp. (resp. w = xu ) for some x G X*. It is also obvious that r(i) is the minimal suffix) of w if w = ux Notation. For a given language L, let y(l) = {u\u / 1 and «is a prefix of some word in L}, y (L) = {u\u ± 1 and «is a suffix of some word in L}. Corollary 8. Let I be an ideal of X*. Then p is not principal iff there exist distinct words u and v in 7 such that \<?(Xu)nr(I)\ = \SS(uX)nr(I)\ = \&>(Xv)nr(I)\ = \^(vx)nr(i)\ = \x\. Proof. If Pj is not principal, by Theorem 5 7 contains two distinct words u and v which are maximal in 7. Clearly (ux) u (Xu) U (vx) u (Xv) ç /. Let a G X. Then there exists w G r(i) such that w < au, whence w = au where uu = u.

4 596 C. M. REIS Similar statements apply to ua, av, and va and the necessity of the result follows. Conversely, if the condition holds, u and v are maximal in / and thus by Theorem 4, p is not principal. Recall that an ideal / of a semigroup S is prime, if for all a,b G S, asb ç / implies a G I or b G I. We are now able to prove the following. Theorem 9. / is either a prime ideal or \r(i)\ < \X\, then p, is principal. Hence, if I is principal, p is principal. Proof. If / is prime, then given u,v I, there exists x G X* such that uxv ^ /. Hence 7 contains no maximal word whence, by Theorem 5, p} is principal. If r(/) < \X\, \3 (Xu) n r(i)\ < \X\ for all ugx*. Therefore, by Corollary 8, p is principal. D Remark. It can easily be proved that if A > 3 all principal ideals are prime. However, if \X\ = 2, X*uX is prime iff u ab+ U a+b U ba+ U b+a where X = {a,b}. Definition 10. An infix code L is said to be maximal if Lli{w} code for any w G X*\L. is not an infix Theorem 11. Let X be an alphabet with more than one letter and let I be an ideal of X* such that r(i) is a finite maximal infix code. Then p, is not principal provided r(i) ^ X. Proof. Since r(i) is finite and maximal, 7 is also finite and hence contains words which are maximal in 7. Thus, by Theorem 5 7 contains exactly one maximal word w if p is principal. Clearly 7 = (w] and 7 ^ {1} since r(i) ^ X. By the maximality of r(i), there exists t G r(i) such that w < t. Letting tf and t denote the words obtained by deleting the first and last letters of t, respectively, we have t{ < w < t and t < w < t. Hence t, = w = t whence w = as, s > 0 for some a G X and t = as+. It follows that each word of r(i) is a power of the letter a. Hence r(i) = {a }, contradicting the maximality of the infix code r(i). Hence p is not principal, o The above theorem is not valid if \r(i)\ = oo as the following example shows. Example 12. Let X = {a,b} and let w = a"'a22 a"s ± 1 where zz(. > 0 and ai ^ ai+] for all i. Then the skeleton of w, denoted sk(w), is the word axa2--as. Let L m {ab'ajb\i,j > 0} U {ba'tfa^j > 0} u a2bja2\j > 0} U {b2ajb2\j > 0}. L is clearly an infix code. That it is maximal is a consequence of the following: first, if sk(tzj) < 2, clearly w G (L] ; if sk(tü) = 3, w = ab1 a or w = b'ajb. In either case, if i,k > 2, w g [L), while if one of i and zc is 1, w G (L]. If sk(tfj) > 4,w g[l) thus proving that L is indeed maximal infix.

5 It is easy to check that WHEN ARE REES CONGRUENCES PRINCIPAL? 597 X*LX* = {w\ sk(uz) < 2} U {ab'ct,bajbk\j,k > 0} \j{aibja,biajb\i,j >0}o{l} and that no word of X*LX* is maximal in X*LX*. By Theorem 5, it follows that the Rees congruence modulo the ideal X* LX* is principal. Theorem 13. Let I be an ideal of X* such that p, is principal and let L ç 7 be a language with p = PL. Let F be a finite subset of L not containing the maximal word of 7 (if such exists). Then PL<F = p[. Proof. Let T = L\F and let u,v G 7. Assume first that / contains no maximal word. Then u belongs to an infinite chain in 7. Hence there exist p,qgx* suchthat \puq\ and \pvq\ are greater than max{\w\\w G F} and puq G 7. Let u = v(pt). Then puq s pvq(pt). Let xpuqy G L. Since \xpuqy\ > max{\w\\w g F}, it follows that xpuqy G T, whence xpvqy G TGL. Similarly, xpvqy G L implies xpuqy G L, thus proving that puq = pvq(pj). Hence puq = pvq and u = v. Therefore p, = PT. Suppose now that 7 contains a maximal word y and let u, v G 7. If one of u and v belongs to an infinite chain of 7, the same argument as above shows that u = v(pt) implies u = v. Suppose therefore that neither word belongs to an infinite chain of 7. Then u < y, v < y and thus there exist p, q G X* such that puq = y. Since y F, it follows that puq G T. Hence, if u = v(pt), then pvq G T. If pvq = y, then u = v. Suppose if possible that pvq ^ y. If pvq < y, there exist s,t G X*, \st\ > 1 such that spvqt = y G T. Hence spuqt = syt G T ç 7, a contradiction since y is maximal in 7. Thus y and pvq are incomparable. Therefore pvq belongs to an infinite chain of 7. An argument similar to the one given above proves puq = pvq = y, a contradiction. Thus PT is equality on 7. Clearly / is a P^-class, whence PT = p,. D We conclude by giving an example to show that the above theorem is not valid if 7 contains a maximal word y and y G F. Example 14. Let X = {a,b}, 7 = a* U (a b2]. Then p = pl where L = {a2"\n > l}u(tf2z32]. However PL\{a2b2} # P,- In fact ab2 = a2b(pl^{a2b2}) and ab2,a2b g7. References 1. Y. Q. Guo, H. J. Shyr and G. Theirrin, Qf-disjunctive languages, Papers on automata and language VII, Department of Mathematics, Karl Marx University of Economics, Budapest, 1985, pp G. Lallement, Semigroups and combinatorial applications, Wiley-Interscience, New York, C. M. Reis, Infix congruences on a free monoid, Trans. Amer. Math. Soc. vol. 310, Number 2, Dec Mathematics Department, University of Western Ontario, London, Ontario N6A 5B7, Canada

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