Cohen Macaulay Rings Associated with Digraphs

Transcription

1 JOURNAL OF ALGEBRA 206, ARTICLE NO. JA CohenMacaulay Rng Aocated wth Dgaph Kazufum Eto Depatment of Mathematc, Nppon Inttute of Technology, Satama, Japan Communcated by Cag Hunee Receved June 10, 1997 Let G be a dgaph, that, a pa of et contng of a et of vetce and a et of dected edge Ž fo a moe pece defnton, ee Secton 3.. It an nteetng poblem to now how to count the Hamlton cycle of G, that, cycle contanng all vetce of G. In th pape, we wll gve the uppe bound of the numbe of them by ung the theoy of commutatve ng. It natual to aume that G tongly connected, othewe thee no Hamlton cycle. The way to gve t to aocate G wth a CohenMacaulay ng whch potvely gaded of dmenon 1 and to compute t Macaulay type. In fact, the aocated ng a monod ng, whch defned fo a ubmodule of Z whee Z the ng of ntege Žfo the defnton, ee Secton 1.. Fo th eaon, we wll compute the Macaulay type of CohenMacaulay monod ng n a pecal cae n Secton 2. Let A X,..., X Ž a feld. 1. In Secton 1, we defne the polynomal F fo Z and the deal IV of A fo a ubmodule V of Z Ž fo the defnton, ee Secton 1.. And we call the ng AIŽ V. a monod ng. Thee defnton and a few popete of th ng ae found n 5. The monod ng alway a gaded ng. If t potvely gaded, we can ue the theoy of Gobne bae and how a bac lemma about geneaton of element n t defnng deal IV Ž ee Popoton In Secton 2, we only teat a pecal cae: V a ubmodule geneated by,..., n Z wth Ž. 0, Ž. 1 j 0 fo each j and 1 0, whee denote the th enty. Futhe aume that the monod ng AIŽ V. potvely gaded of dmenon 1. We wll ee n the next ecton that thee condton ae atfed f V aocated wth a tongly connected dgaph. Thu to nvetgate th cae mae ene. Unde thee condton AIŽ V. CohenMacaulay and we wll compute t Macaulay type $25.00 Copyght 1998 by Academc Pe All ght of epoducton n any fom eeved.

2 542 KAZUFUMI ETO In Secton 3, we wll aocate a dgaph G wth a ubmodule VG of Z, whee the numbe of vetce of G. We can pove that the monod ng AIŽVG. potvely gaded of dmenon 1 f and only f G tongly connected. And we wll pove the man theoem: THEOREM. Let G be a tongly connected dgaph. Then the numbe of Hamlton cycle of G le than o equal to the Macaulay type of aocated monod ng AIŽVG.. We alo gve a uffcent condton that they ae equal. 1. PRELIMINARIES Thoughout th pape, let be a feld, 1 an ntege, A X,..., X a polynomal ng ove, and Z the ng of ntege. 1 DEFINITION. Let V be a ubmodule of Z of an. Fo V, we put the polynomal Ž bnomal. FŽ. of A: Ł Ł F F F X X, Ž. 0 Ž. 0 whee Ž. denote the th enty of. And we put IV to be the deal of A geneated by all polynomal FŽ. fo V. We call the ng AIŽ V. a monod ng. Ft note ht IV an V 5, Pop If V contaned n Ke n 1,...,n, the enel of the map fom Z to Z whch natually nduced by ntege n,...,n, the deal IŽ V. 1 Ž weghted. homogeneou by the degee wth deg X n fo each. And we ay that V potve f thee ae potve ntege n 1,...,n wth V KeŽ n,...,n. Žcf.. 6. If V potve, the ng AIŽ V. 1 a potvely gaded ng. LEMMA 1.1. Fo,..., Z, thee ae monomal M,...,M of A 1 1 uch that Ž. 1 GCDŽ M,...,M Ž. 2 M F Ž. M F Ž. 1 1 fo 1,...,1. Ž 3. If M,..., M ae monomal atfyng M F Ž. 1 M F Ž. 1 1 fo 1,..., 1, thee a monomal G wth M GM fo each. Ž. Ž 4 Thee a monomal H wth M F HF. 1 1.

3 COHENMACAULAY RINGS 543 Poof. We wll pove the aeton by nducton on. If 2, put M F Ž. GCDŽF Ž., F Ž.. and M F Ž. GCDŽF Ž , F Ž... Then GCDŽ M, M. 1 and MF Ž. M F Ž If M1 and M ae monomal atfyng MF Ž. M F Ž , we have M1 GM1 and M GM wth G GCDŽ M, M.. And MF M FŽ a bnomal, hence of the fom HF whee H a monomal and Z. And mut be 1 2. Aume 2. By the nducton hypothe, thee ae monomal N,...,N uch that 1 1 Ž. 1 GCDŽ N,...,N Ž. 2 NF Ž. N F Ž. 1 1 fo 1,..., 2. Ž 3. If N,..., N ae monomal atfyng N F Ž. 1 1 N F Ž. 1 1 fo 1,..., 2, thee a monomal G wth N GN fo each. Ž. 1 Ž 1. 4 NF HF whee H a monomal Put w, M F Ž. GCDŽHF Ž w., F Ž.. 1, M MN fo 1,..., 1, and M HF Ž w. GCDŽHF Ž w., F Ž... Snce GCDŽ N,...,N. GCDŽ M, M. 1, we have GCDŽ M,...,M And M F Ž. M F Ž. fo 1,..., 2 and M F Ž MHF w MF. Futhe, we have M FŽ. HFw Ž 1. Ž HF. 1 whee H a monomal. If M,...,M ae monomal atfyng M F Ž. M F Ž fo 1,...,1, thee a monomal G wth M G N fo 1,..., 1 1. And M FŽ. GHFŽ w.. Snce GHF Ž w. M F Ž MF Ž., thee a monomal G wth G GM and M GM. Hence M GM fo 1,..., 1. Q.E.D. LEMMA 1.2. Let 1, 1,..., Z, and M 1,...,M monomal of A atfyng M F Ž. M F Ž. fo 1,..., 1. If Ž fo 1, then deg M Ž. X 1 2. Poof. If Ž. 0 and Ž. 0, deg M Ž. 1 X 1 deg M. The aeton follow fom th. Q.E.D. X 1 Let V be a potve ubmodule of Z. If 1, 2 V and G 1, G2 ae monomal of A wth GF Ž. G F Ž., then GF G FŽ the polynomal of the fom GFŽ. 1 2 fo ome monomal G. Moe geneally, let J ŽFŽ.,...,FŽ.. 1 be an deal of A whee 1,..., V. Then we can get a Gobne bae of J fom FŽ.,...,FŽ. by Buch- 1

4 544 KAZUFUMI ETO bege algothm. And t eay to ee that th Gobne bae cont of the polynomal of the fom t j 1 GF w G F d, whee G and G ae monomal, d 1, 1 j fo each, w d GF Ž w. GF Ž d., G F Ž d. G F Ž d. j 1 1 j j 1 1 j fo 1 1 1,...,t1, and GF Ž w. GF Ž d. t t j t. Futhe, PROPOSITION 1.3 Žcf. 4, Theoem Let V be a pote ubmodule of Z and let J ŽFŽ.,...,FŽ.. 1 be an deal of A wth 1,..., V. Fo Z and G a monomal, f GFŽ. J, thee an expeon t j 1 GF G F d, whee d 1, 1 j, G a monomal fo each, d, F Ž. GF Ž d., G F Ž d. G F Ž d. j 1 1 j j 1 1 j fo 1 1 1,...,t1, and F Ž. GF Ž d. t t j t. The followng lemma ueful. LEMMA 1.4 4, Pop Let, Z. If thee no pa, wth 1 2 0, Ž 1. Ž 1. 0, 0, Ž 2. Ž 2. 0, thee ae monomal M, M wth FŽ. MF M FŽ In patcula, FŽ. contaned n the deal ŽFŽ., FŽ MACAULAY TYPE OF MONOID RINGS In th ecton, we wll ue the followng notaton. Let,..., Z wth Ž. 0, Ž. 0 fo j uch that 1 j Ž Ž. ² : 2 V 1,...,, a ubmodule of Z panned by 1,...,, potve of an 1. Fom thee condton, we note that the ng AIŽ V. a potvely gaded CohenMacaulay ng of dmenon 1 by 5, Pop. 1.1 and 5, Lemma 1.2.

5 COHENMACAULAY RINGS 545 Let S be the th ymmetc goup. By Lemma 1.1, fo each S wth Ž. 1 1, thee ae monomal M Ž.,...,M Ž. uch that 1 Ž. 1 GCDŽM Ž.,...,M Ž Ž 2. M Ž. F Ž. M Ž. F Ž. fo 1,..., 1. Ž. Ž. Ž1. Ž1. Futhe, nce 0, we have M Ž. FŽ. 1 1 Ž. Ž. 0, hence M Ž. F Ž. M Ž. F Ž. 1 1 Ž. Ž.. Let J be an deal of A geneated by M Ž. fo all S 1 atfyng Ž. 1 1, I ŽFŽ.,...,FŽ.. an deal of A, and J J I I. Then THEOREM 2.1. The ng AIŽ V. a potely gaded CohenMacaulay ng of type Ž J., whee denote the leat numbe of geneato. To pove th theoem, we need eveal lemma. LEMMA 2.2. Fo S wth 1 1, 1 Ž. 1 deg M fo 1. X 1 Ž. 2 1 Poof. Let M M fo 1,..., and Ž. 0. A n the poof of Lemma 1.2, nce Ž. 0 fo, we have Ž deg M deg M Ž. X 1 X Ž 0. Ž. 1 and deg X M Ž. deg X M Ž. fo 1,..., 0. If deg X M Ž. 0, the 0 0 aeton poved. So we pove th. If deg X M Ž. 0, we have deg X M Ž. 0 fo 1,..., 0. Snce 0 MF Ž. M F Ž. and Ž. 0, we have 1 1 Ž. Ž. Ž. If l, we have 0 deg M deg M F Ž. 0. X Ž. X Ž. Ž. deg M deg M, X Žl. Ž. X Ž. l1 nce Ž. Žl. 0f l 0. Then X dvde each M Ž. fo 1,...,, a contadcton. Hence deg X M Ž. 0. Q.E.D. Note. We have deg M Ž. 0 fo any. X 1 LEMMA 2.3. The deal IŽ V. geneated by all polynomal of the fom FŽ. whee 1 and 1 j j j. j1 j2 j 1 2 Poof. Let V. Then we can wte d wth d 0 and we j1 j j j may aume that thee j wth d 0, nce 0. Let I j 1 2

6 546 KAZUFUMI ETO be the deal geneated by FŽ.,...,FŽ. 1 and all polynomal of the fom FŽ. j j j whee 1 and 1 j1 j. We wll 1 2 pove FŽ. I by nducton on d max j d j. If d 1, the aeton clea. Suppoe d 1. Let w Ž d 1. 1 d 0 j j j and w2 d 0 j. We have w1 w 2. By the nducton hypothe, j Fw and Fw ae contaned n I. And f Ž w , we have d 0 and Ž w. 0. Hence FŽ. contaned n the deal ŽFw, Fw by Lemma 1.4. And FŽ. I. Th complete the poof. Q.E.D. LEMMA 2.4. Let N, N be monomal wth N F Ž. NF Ž j 2 j fo j j. Then thee ae monomal N, N, and G wth N F Ž j NF Ž., N N GFŽ., and N N GFŽ.. 2 j 1 2 j 2 1 j j Ž j. j Ž j. Poof. We have F X and F X j j j j. By the aumpton, thee a monomal G atfyng N Ž 1 GF j. and N2 GF Ž.. Then N GF Ž. and N GF Ž. j 1 j 2 j atfy the condton of the lemma. Q.E.D. Ž. PROPOSITION 2.5. I : IV J I. 0 0 Poof. Step 1. We wll how ŽI : FŽ J I 0. By the defnton of J, we Ž. Ž Ž.. have I 0: F 1 J ecall I0 F 2,...,F. Let : A A be a homomophm wth Ž e. FŽ. j j fo j 1,..., whee e 1,...,e fom the canoncal bae of A. Let R be a ubmodule of A geneated by all element of the fom M Ž. e fo S atfyng Ž and let R be a ubmodule of A geneated by all element of the fom FŽ. j ej FŽ. e j j fo j j. Then both R and R ae contaned n Ke. We clam Ke R R. To ee th, n ecton 1, thee a Gobne bae of the deal ŽFŽ. 1,....,F contng of bnomal of the fom G FŽ d. 1 j whee G a monomal, d 1, and 1 j fo each wth G F Ž d. j G F Ž d. 1 1 j 1 fo 1,...,1. Hence Ke geneated by element of the fom 1 dg ej whch atfe the condton ½ G a monomal, d 1, 1 j fo each, and G FŽ d j. G1FŽ d1 j. fo 1,..., 1. Futhe, nce t contaned n Ke, we have GF Ž d. GF Ž d. 1 1 j1 j and 1 d j 0. Now we how that z 1 dg ej contaned n R R. Let Ž z. j d fo 1,...,. Snce an V 1 and 1 0, we have Ž z. Ž z. Ž z We may aume z 0 nce dg e R R f and only f 1 1 j 1 1

7 COHENMACAULAY RINGS 547 d G e R R 1 1 j 1. We wll pove the aeton by nducton on Ž z. 1. Aume that thee 0 wth d 1 and d 1 1. If j j 1, we have G G nce G F Ž. G F Ž. 1 j 1 j. And we al o have d G e d G e 0 and d G F Ž. j 1 1 j 1 1 j d G F Ž. 2 2 j. Hence z dg ej 1 dg ej at fe condton and contaned n Ke. Aume j j 1. Then, by Lemma 2.4, thee ae monomal N 1, N 2, G 0 0 wth NF Ž. NF Ž., N G GFŽ. 1 j 2 j 1 1 j, and N2 G GFŽ.. Put Then and j 0 1 w dg e Ne Ne dg e. j 1 j 1 2 j j z w N G e N G e 1 j 1 j j ž Ž j. j 1 j 1 j / G F e F e R, G F d NF, j 1 1 j 0 01 NF G F d. 2 j j 0 02 Thu w atfe condton and contaned n Ke. And we have z R R f and only f w R R. Hence, by condeng w, we can change the ode of j and j f d 1 and d Theefoe, n any cae, thee ext z d G 1 ej Ke atfy- ng condton uch that thee t wth d1f t and d 1f t and that z z R. By eplacng z wth z, we may aume that thee t wth d1f t and d 1f t. t1 Then w G e Ž 1. 1 t1 j t Gt1ej atfe t1 t1 condton n th ode and w z. Fo each t, thee t wth e e, nce Ž z. j j j 0 fo each j. Agan by the above agument, thee ext w dg e atfyng condton 1 j wth d 1 fo any and z w R. In the cae of Ž z. 1 0, th mple that z contaned n R. Suppoe Ž z Then we eplace z wth w. Hence we may aume z 1 Ge j. Note that we may aume j1 1. Fo, thee 0 wth j 1 and the element 1 01 G 1ej G e atfe condton Ž j 01 Fo each l, let be the numbe wth e e and e e fo l j l j l. And let S wth. Then Ž By Lemma Ž1. Ž2. Ž.

8 548 KAZUFUMI ETO 1.2, deg G deg M Ž. fo each, thu M Ž. X 1 X 1 1 dvde G 1. Put H GMŽ. 1. Then 1 1 w H M e G e j2 Ž 2j. Ž 2j. j 2 atfe condton and z w R. Futhe, nce Ž w. Ž z , we have w R R by the nducton hypothe. Hence z R R. Thu the clam poved. Step 2. By Step 1, Ž I : IV. Ž I : FŽ J I 0. By Lemma 2.3, IV geneated by FŽ.,...,FŽ. and FŽ. 1 1 j j fo and 2 j1 j, nce Hence we clam that, fo any S wth Ž 1. 1, M F I 1 1 j1 j 0 fo 1 1 and 2 j j. 1 By Lemma 2.2, 1 Ž. deg M fo 1. X 1 Ž. 2 1 Ž. 1 Let j,..., j be the dtnct numbe wth j Ž j. 1 t 1 t and j,..., j 4 2,...,4 j,..., j 4. Fom the expeon 1 t 1 ž 1 j/ j1 jt 1 we have 0, ž 1 j/ j1 j1 jt jt 1 GF G F G F 0, wth GCDŽ G, G,...,G. j j 1, by Lemma 1.1. Then, a n the poof of 1 t Lemma 2.2, deg G 0 fo m 1,...,t, X jm m1 1 Ž j m. 1 deg G X j j j Ž. fo m 1,...,. j m m l m l1 Hence G dvde M. Theefoe ž 1 j/ 0 1 M F I. Q.E.D.

9 COHENMACAULAY RINGS 549 Poof of Theoem 2.1. By the defnton of 1,..., and V, an V 1 and AIŽ V. a potvely gaded CohenMacaulay ng of dmenon 1 by 5, Pop. 1.1 and 5, Lemma 1.2. By Popoton 2.5, we have whee K fom th. A IŽV. K I : IŽ V. I J I I, A IŽV the canoncal module of AIŽ V.. The aeton follow Q.E.D. 3. MAIN RESULT In th ecton, we tudy the elaton between dgaph and monod ng. And we wll pove that the numbe of Hamlton cycle of a connected dgaph le than o equal to the Macaulay type of monod ng aocated wth t, f the aocated monod ng potvely gaded of dmenon 1. Futhe, we wll gve the uffcent condton that they ae equal. DEFINITION. A dgaph G cont of VeŽ G. a fnte et of vetce and a et of dected edge whch ae odeed pa of dtnct vetce Žwe denote the edge fom to j by j.. It poble that G ha edge j and j. A path alway mean a dected path; that, f thee a path fom to j, thee ae n 0, dtnct vetce 1,..., n, edge l l1 fo l 1,...,n1, and an edge n j. A cycle a path whch ha the ame ft and lat vetex. A Hamlton cycle of G a cycle whch contan all vetce of G. Let G be a dgaph wth VeŽ G. 1,..., 4. Fo G, we defne the -matx Ž a. j wth the ente n Z a follow: If thee an edge j, we et aj a negatve ntege. If j and thee no edge j, et aj 0. Put aja j. Let j be the jth column vecto of the matx Ž aj. fo j 1,..., and ² : VG 1,..., be a ubmodule of Z. We ft note 1 0 and Ž. 0. And Ž. 0 f and only f thee no edge whch tat at. PROPOSITION 3.1. Let G be a dgaph. The followng ae equalent: Ž 1. VŽ G. a pote and an VG 1. Ž. 2 Fo any pope et S of VeŽ G., thee an edge fom ome etex S Žep. S. to ome etex S Žep. S Ž. 3 G tongly connected; that, fo any dtnct etce 0, 1, thee ae a path fom 0 to 1 and a path fom 1 to 0. Ž. 4 Fo each edge, thee a cycle contanng t.

10 550 KAZUFUMI ETO Poof. Ž. 1 Ž. 2 Suppoe that thee a pope ubet S of VeŽ G. uch that thee no edge fom any 0 S to any 1 S. Put w S. Note w 0 nce an V 1. Then Ž w. 0 fo each S, by aumpton. And Ž w. 0 fo each S. Hence VG not potve. Ž. 2 Ž. 1 Let S be a pope ubet of VeŽ G. and w j S j.we clam that thee ae and wth Ž w. 0 and Ž w. 0. By condton Ž. 2, thee an edge fom ome vetex S Žep. S. 0 0 to ome vetex S Žep. S.. Then Ž w. 0 and Ž w Hence the clam 0 0 poved. Snce IVG Ž. geneated by all polynomal of the fom Fw whee S a pope et of VeŽ G. by Lemma 2.3, VG potve by 6, Lemma 1.2. Suppoe an VG 1. Then thee a nontval elaton dj j 0 wth dj 0. We may aume that thee j wth dj 0, and futhe aume that thee wth dj 0 fo j 1,...,, by enumbe- ng. Then we have Ž. j 0 fo 1,..., and j 1,...,. Put w. Then Ž w. 0 fo 1,..., and Ž w. 1 0 fo 1,...,. Snce VG potve, Ž w. 0 fo any and Ž. j 0 fo 1,..., and j 1,...,. Th mple that G dconnected, a contadcton. Hence an VG 1. Ž. 2 Ž. 3 Suppoe that thee ae, VeŽ G. 0 1 uch that thee no path fom to. Let S be the ubet of VeŽ G. 0 1 contng of the vetce uch that thee a path fom 0 to. Then S a pope et of VeŽ G.. And thee no edge fom the vetex of S to the vetex of the complement of S. Ž. 3 Ž. 4 Let be an edge n G. Thee a path fom to by Ž. 3. Hence thee a cycle contanng Ž. 4 Ž. 2 Suppoe that thee a pope ubet S of VeŽ G. uch that thee no edge fom the vetex of S to the vetex of the complement of S. Snce G connected, thee an edge fom the vetex of the complement of S to the vetex of S. And thee no cycle whch contan t. Q.E.D. Fom now on, we teat a tongly connected gaph. By Popoton 3.1, the aocated module VG ²,..., : 1 potve of an 1, hence we can ue the eult n Secton 2. And we alo ue the notaton n Secton 2, uch a S, M Ž., I, J, and J. 0 THEOREM 3.2. Let G be a tongly connected gaph. The numbe of Hamlton cycle of G le than o equal to the Macaulay type of CohenMacaulay ng AIŽVG..

11 COHENMACAULAY RINGS 551 Poof. Step 1. Fo a Hamlton cycle 1 m m 1, let S 2 wth Ž. m. Then we can defne the monomal N Ž. 1. And we clam that the canoncal mage of M Ž. 1 n J a pat of a mnmal geneatng ytem of J. If the clam poved, the numbe of Hamlton cycle le than o equal to the leat numbe of geneato of J, whch equal to the Macaulay type of AIŽ V. by Theoem 2.1, hence the aeton poved. Step 2. Let be an element n S wth Ž. 1 1 uch that M Ž. 1 a pat of a mnmal geneatng ytem and M Ž. ŽM Ž I 0. Then we wll how. Ft note Ž. 0 fo 1,..., 1 and Ž. Ž. Ž1. Ž Snce thee a monomal H wth M Ž. HM Ž. I, thee an expeon Ž j. 1 M HM G F d, whee G a monomal, d 1, 2 j fo each, and G F Ž d. F Ž d. fo 1,..., 1, by Popoton 1.3. A j 1 j1 n the poof of Popoton 2.5, we may aume that thee 0 wth d 1f 0 and d 1f 0. By Lemma 2.2, deg X M1 Ž. deg F Ž. fo each. Hence d 1 fo any. X 1 Ž. 1 2 Let fo 1,...,. We wll how 1. If the clam poved, we have fo each and, hence M Ž. 1 a pat of a mnmal geneatng ytem of J. 1 Let l 1. Suppoe and Ž j. 1 2 l l fo each. Fo 1 each, nce Ž j. l and d 1, we have Ž d. Žl. j 0. Thu deg M Ž. deg M Ž. X 1 X 1 by Lemma 1.2. On the othe hand, nce Žl. Žl. 1 2 l, deg M X 1 Žl. Ž j. Žl. Ž j. deg X M1, Žl. Žl. l1 jl by Lemma 2.2. Hence deg M Ž. deg M Ž. X 1 X 1. Then l1 l, Žl. Žl. Ž. 1 othewe deg M deg M. Futhe, Ž j. X Žl. 1 X Žl. 1 Žl. Žl1. l 1 fo each, nce j1 X 1 X 1 Žl. Ž j. Žl. Žl. 1 deg M deg M and Ž. 0. The clam follow fom nducton on l. Q.E.D. Žl. Žl1. Fnally, we wll gve a uffcent condton that two nvaant ae equal.

12 552 KAZUFUMI ETO PROPOSITION 3.3. condton Suppoe that G tongly connected and atfe the If thee ae edge j and j, thee an edge between j and j. Then the numbe of Hamlton cycle of G equal to the Macaulay type of Ž. AI VG. Poof. Step 1. By the aumpton, f Ž. 0 and Ž. j j 0, we have Ž. 0oŽ. 0. Let be an element n S wth Ž 1. j j j j 1 uch that M Ž. 1 a pat of a mnmal geneatng ytem of J. We clam Ž. 0 fo 1,..., 1 and Ž. Ž. Ž1. Ž If the clam poved, thee a Hamlton cycle 1 Ž. 2 Ž. 1 and aocated wth t. It follow that thee a one-to-one coepondence between element n a mnmal geneatng ytem of J and Hamlton cycle of G. Hence the aeton poved. Step 2. Let S wth Ž 1. 1 and let Ž j j 1. S be a tanpoton fo j 1. Put Ž jj1.. Then, by Lemma 2.2, Ž j. Ž Ž j1.. Ž j., deg M Ž. deg M Ž. X 1 X 1 Ž j1. Ž Ž j.., Ž j 1., 0, othewe. Hence, f Ž. 0, M Ž. dvde M Ž.. Futhe f Ž. Ž j. Ž j Ž j1. Ž j. 0, M Ž. 1 not pat of a mnmal geneatng ytem of J. 1 Let 1 1 2,..., S and Ž 1. Ž 1 2,...,. 1. Then Ž By Lemma 2.2, ½ Ž., Ž 2. Ž2. Ž2., deg X M1Ž. deg X M1Ž. Ž Ž2.., 1, Ž 2.. Hence, f Ž. 0, we have M Ž. M Ž. HFŽ. 1 Ž Ž2. fo ome monomal H, that, M Ž. M Ž. mod I Step 3. Let S wth Ž 1. 1 uch that M Ž. 1 a pat of a mnmal geneatng ytem of J. If Ž. 1 Ž2. 0, we can eplace wth Ž 1. Ž 12,...,. 1, nce M Ž. M Ž. 1 1 mod I0 by Step 2. Hence we may aume Ž. 0 Žnote that thee j wth Ž. 1 Ž2. 1 j 0 nce G tongly connected.. Next, f Ž. Ž. 0, we eplace wth Ž 1.Ž 1 Ž3. Ž2. Ž3. 1, 2,...,. 1 Ž 23.. Thu we may aume Ž. 0o Ž. 1 Ž3. Ž2. Ž3. 0. If Ž. 0, we have Ž. 0 and Ž. 0, by aumpton. Ž2. Ž3. 1 Ž3. Ž3. Ž2.

13 COHENMACAULAY RINGS 553 Let Ž By Step 2, M Ž. dvded by M Ž. 1 1 and t not a pat of a mnmal geneatng ytem of J, a contadcton. Thu Ž. Ž2. Ž3. 0. In geneal, uppoe Ž. Ž. Ž1. 0 fo 1,...,l 1. Snce G tongly connected, thee ae j l and l wth Ž. Ž. Ž j. 0. Hence we may aume Ž. Ž. Žl1. 0, f neceay, by eplacng wth a utable element n S. Let max l 1 Ž. 04 Ž. Žl1.. Aume l. Snce Ž. 0 and Ž. Ž. Ž 1. Ž 1. Žl1. 0, we have Ž. Ž.Ž 0. Let Ž ll1.by. Žl1. Ž 1. Step 2, M Ž. dvded by M Ž. 1 1 and t not a pat of a mnmal geneatng ytem of J, a contadcton. Thu l and Ž. Žl. Žl1. 0. Hence we conclude Ž. Ž. Ž1. 0 fo 1,..., 1. Fnally, we how Ž. 0. Suppoe Ž. Ž. 1 Ž Let max Ž. 04. Then. Snce Ž. 0 and Ž. Ž. 1 Ž. Ž 1. Ž , we have Ž Let 12,..., 1,. Then Ž. 1 Ž , Ž. fo 2,...,, and Ž. 1 fo 1,...,. Snce Ž. 0 fo 1,...,, M Ž. dvded by M Ž. Ž up to modulo I0 by the ame agument of Step 2. Futhe, nce Ž. 1 0, M M not 1 and M Ž. 1 Ž not a pat of a mnmal geneatng ytem of J, a contadcton. Hence Ž. Ž ŽNote that we pove that, f the mage of M Ž. 1 a pat of a mnmal geneatng ytem of J, thee S wth Ž 1. 1, aocated wth a Hamlton cycle, uch that M Ž. M Ž. 1 1 I 0. By the poof of Theoem 3.2, th mple, hence telf aocated wth a Hamlton cycle.. Q.E.D. EXAMPLE 1. and the et of edge of G Let G be a dgaph wth VeŽ G. 1,2,..., 4 wth 2 j j 4. Then t atfe the condton of Popoton 3.3 and the numbe of t Hamlton cycle Ž 1.!. Hence the Macaulay type of aocated monod ng alo Ž 1!.. EXAMPLE 2. and all edge ae Let G be a dgaph wth VeŽ G. 1,2,..., 4 wth 3 1 fo 1,..., 1, 2 fo 1,..., 2,

14 554 KAZUFUMI ETO and 1, Ž 1. 1, 2. Then the numbe of Hamlton cycle 2 f odd, 1 f even. Hence the Macaulay type of aocated monod ng 2 f odd and 1 f even. Futhe, f odd, we can pove that the defnng deal IVG Ž. geneated by FŽ.,...,FŽ. 1, that, an almot complete nteecton by ung Lemma 1.4. Th cae eentally nvetgated by Gatnge 7, Satz 6.5. If 4, th eult nown a the unque example whch defne a noncomplete nteecton, Goenten monod ng n affne 4-pace 1, 6. ACKNOWLEDGMENTS I am gateful to T. Endo fo ueful conveaton and T. Kanzo fo nd advce. REFERENCES 1. H. Beny, Symmetc emgoup of ntege geneated by 4 element, Manucpta Math. 17 Ž 1975., W. Bun and J. Hezog, CohenMacaulay Rng, Cambdge Stud. Adv. Math., Vol. 39, Cambdge Unv. Pe, D. Eenbud, Commutatve Algeba wth a Vew Towad Algebac Geomety, Gad. Text n Math., Vol. 150, Spnge-Velag, New YoBeln, K. Eto, Almot complete nteecton monomal cuve n A, Gaujutu Kenyu Ž Academc Stude. Math. 43 Ž 1994., K. Eto, Defnng deal of complete nteecton monod ng, Toyo J. Math. 18 Ž 1995., K. Eto, Defnng deal of emgoup ng whch ae Goenten, Comm. Algeba 24 Ž 1996., W. Gatnge, Ube de Vechwndungdeale monomale Kuven, Ph.D. the, Unvety of Regenbug, Landhut, F. Haay, Gaph Theoy, Addon-Weley, Readng, MA, 1972.