Zero Divisor Graph on Modules

Transcription

1 J. Appl. Envron. Bol. Sc., 5S)7-3, 05 05, Textod Publcton ISSN: Journl of Appled Envronmentl nd Bologcl Scences Zero Dvsor Grph on odules Shbn Sedgh, rym Yzdn, Yhy Shbnpour 3. Deprtment of themtcs, Fculty ember, Qemshhr Brnch, Islmc Azd Unversty, Qemshhr, Irn.. Deprtment of themtcs, Fculty ember, Qemshhr Brnch, Islmc Azd Unversty, Qemshhr, Irn. 3. Deprtment of themtcs, Tbr Unversty of Bbol, Irn. eceved: y 4, 05 Accepted: August 7, 05 ABSTACT Suppose s dsplcement nd unt rng nd s module. In ths rtcle, the grph depends on, we show wth Γ, so tht, then Γ s clssc zero dvsor grph. We show tht the Γ grph wth dm Γ ) 3 s coected grph nd n Lesser module wth condton Z /{ O}, hve gr Γ ), f nd only f Γ s str grph. KEYWODS: odule, Zero dvsor grph of modulus, ound, Dmeter, Complete bprtte grph. INTODUCTION AND PELIINAIES The frst tme n 988, Beck [0] stted the concept of zero dvsor grph for commuttve rng. Beck ws consdered ll members of the dsplcement nd unt rng s vertex of the grph nd hs mn tsk ws to fnd the necessry nd suffcent condtons for the fnteness of chromtc number of grph. Also, ccordng to defnton of0 nd,two vertex of x nd y Were djcent, f nd only f xy0. evews relted to grph colorng contnued by Anderson nd Nsr [4]. But from ths grph s not obtned nterestng results. And n ddton to were obvous propertes. For exmple, ll ts vertces were djcent to zero. ecently, the zero dvsor grph from dsplcement rng hs been extended to the grph mker bsurd del from dsplcement rng. Two dels I nd J re djcent, whenever IJ O)). In [8], the clssc zero dvsor grph hs been extended to modules on dsplcement rngs. Accordng to [], m, n re djcent, f nd only f m: nm: o tht s drect extenson of clsscl zero dvsor grph. In [4] nd [3],the uthors presented two dfferent grphs to module ccordng to the frst dulty Hom, ). Although they necessrly s not generlztons of the clsscl dvsor grph, but there re some deep mutul reltons between these two grphs nd type of ts clssc. We frst nlyzed the expresson of severl bsc defntons. Defnton -): Suppose s Abeln collectve group nd s dsplcement rng, n ths cse the clls rght module, whenever sclr multplcton of elements defned n the followng wy. : m, r) mr. So tht for ech r, r, r nd m, m, m hve:. m + m) mr+ mr. m r + r) mr + mr 3. m r r) mr) r And f s unt ndm. m, then cll s untry module. Defnton -): In the G V, E) cycle wth n length s seres ofx V dstnct vertces, s x x x... x grph V s represents the verte nd E s represents the grph edge) x x 3 nsuch n. Correspondng uthor: Shbn Sedgh, Deprtment of themtcs, Fculty ember, Qemshhr Brnch, Islmc Azd Unversty, Qemshhr, Irn. E-ml: sedgh_gh@yhoo.com 7

2 Sedgh et l., 05 Defnton -3): In the G V, E) grph the shortest pth length between two u nd v vertex show wth du, v) nd the dmeter of the grph re defned s follows: dm G) SUP{ d u, v) u, v V} Defnton -4): The shortest dstnce n grph clled bck grph tht show wth grg) symbol. For exmple, the bck cube to length 4. Defnton -5): The G V, E) grph clled bprtte whenever cn the set of v vertces prttoned nto two sets, so tht between the vertces of ny set, there s no edge. In ddton, the G bprtte grph cll complete f ny two vertces n set re not prttonng be djcent to ech other. Complete grph wth n of vertex showed wth n Defnton -6): Suppose s unt dsplcement rng nd s module. In the dependent grph on mens Γ, we sym, n re djcent f nd only f m: n: o.. n esults In ths chpter, we look to rtculte defntons, theorems, concepts nd the mn results n conjuncton wth the complete nd bprtte grph from zero dvsor grph. In the followng fgure, we show zero dvsor grph the some of the Zmodules. Exmple -) K Fgure. Z s complete grph. Accordng to the bove exmple, we know tht zero dvsor grph from 3 Therefore, t s possble for the P frst number cn be result tht the Z module Z p Z s complete grph wth Z P. Theorem -): Suppose S nd S re the two Z \{ nd Γ s complete grph. 3 Z module of smple dentcl nd p S S then Proof: Suppose tht nd re two dentcl modules. Then Γ Γ ).Wth ths fct s enough to prove the theorem. Web show tht~s S. For echo x S hve x, o) : S). Also for ech, b) hve, b) o) : o nd lso o) s djcent to ech element of, b) nonzero. Ths result s cheved to o). Now suppose tht x nd y s two nonzero element from S. It s esy to show tht x) S) s del mxml from. It s obvous tht : s ncludng S). On the other hnd f : then we hve x, o) nd lso for ech r, we hve x, o) r. Then yr 0 be result tht r nd lso x xr 0, whch s nconsstency. Therefore, : S) nd )~ s djcent ny nonzero element from. 8

3 J. Appl. Envron. Bol. Sc., 5S)7-3, 05 Theorem -): Suppose s commuttve rng. In ths cse, s feld f nd only f Γ modulus, the s complete grph. ) for ech Proof: Suppose tht s feld. If dm ) then Γ φnd therefore, Γ s complete grph. If dm ) for echo m, we hve m:, becuse there s o r m:, tht result s r m nd lso mr mtht s nconsstency. Therefore, for ech m, n element from, we hven m: o. ) The No ssumpton s del mxml from. Put. Then for ech No x / N o ndo r, we hve o) o, r) : ) o. So for ech dstnct element of s, r pposng No zero n, we hve o, r) o, s) : ) o. Then for ech o, S, we hve o, )No S o. No Becuse NoS o,s): ). And ths mples tht for ech s o,, we hve N o S o) nd N N o s) ond therefore, N o o) nd so o s feld. Theorem -4): Suppose s modulus of sem-smple fnte genertor, whch homogeneous components s smple, then y s djcent member of \{, f nd only fx I y o. Proof: Suppose tht I S, whch Sre non-dentcl smple below modules from. suppose tht y Z re djcent. We need to show thtx I y o. SupposexI y o, therefore, there s α IthtS xiy, snce y nd x re below modules from, there re subsdres of A nd B from I α tht x S ) then: nd y S ). Lemm t the source [7]). Suppose x y: o) A B y: y: y BS) BT) I B S) And x ~ I\ AS s result: x) I\ AS) I I\ A S).However, snce x y: o, therefore, y: nd lso we hve S I S ). However, snce for ech, j I S) Sj) re prelmnry, so for ech r I\ A, we hve I S ) S ) I S ) S ). However, forr I\ A, there s j r B, whch B B I\ A r S jr) Sr) S jr) Sr nd so we hve ). So S jr ~ Sr nd ccordng to our ssumpton S jr S r. Becuse, there s α I tht Sα xiy. Snce S x~ α I\ AS, therefore, there s I\ A, whch Sα ~S. However, the bove equtons, there s j j Btht S α S Sj, whch n result, we hve Sα yi BS) o), whch s n nconsstency. Proof the second sde, s resultng usng the followng lemm. Lemm -3): Suppose s module nd m nd n re n nonzero element from. ) If m nd n re djcent, then for echr,s tht mr ond ns o, we hve mr ns o. ) If m I n o, then m nd n re djcent. Proof ): Supposem n: o, esly t cn be shown tht for echr, mr n: o.on the other hnd for echs, we hve ns: n:. So mr ns: mr n: o. 3) Snce n: nim s resulted m n: otht cn be concludedm n o B I\A. 9

4 Sedgh et l., 05 esult -4): Suppose tht nd re two non-dentcl smple sub grph from. Then Γ s complete bprtte grph. Proof proof of ):Suppose We re showng tht x nd y Z re djcent, ccordng to the before theorem x y o) y j, whch non-dentcl smple sub modulus From, therefore, I. jnd, j {, }. It s esy to show tht x nd y re two x nd Second proof):for echx nd y, we hve y for ech x I y j j. o. Therefore, from lemm -5), x nd re not djcent. If y \{ so tht x y: o, : nd therefore, y re djcent. Show tht ny two elements from then x y: : o. On the other ) )) otht s resultng ) ). Becuse ) s del mxml from, so ) ), then ~, whch s contrdcton. As result, we show ) ) tht foro x nd o y, x+ re not djcent of ny element from for, Z, we hve: z : : ) therefore, x z: o x+ z: x+ ) y nd for ech + s resulted tht o )) nd so ) y )) ). Becuse ) s del mxml from. Therefore ), ths s contrdcton. If z x+ : o, then by lemm -9) n the source [7] one of the followng concluson: Cse ):If x+, then ~ ~, n the other words ~ ~, whch s not smple module, becuse: x+ I I, I, whch result x) ~, whch s contrdcton. Therefore, I I nd therefore < s contrdcton. Cse ): If x+ then smlr cse, contrdcton occurs. Cse 3): Suppose x+ therefore, x + : :, so z x+ : o s resultng tht z o tht ths s contrdcton. Smlrly, by replcng nsted, lso contrdcton occurs. Fnlly, bout the x + x + y ) : o s resultng tht n y x + y ) : o, x x + y ) : o, whch s mpossble. Suppose for n 3tht ts homogeneous components re smple. Cn be predcted n ths cse the Γ s complete n-prt grph.in theorem the source 4- [6]) hs been shown to reducton the dsplcement rng, we hve tht the non-empty Γ), n s wth gr Γ )), f nd only f Γ ) K for ech n, then, we generlze ths result for Γ. Defnton -7): The module of clled reducton, whenever for ech nd m tht we hve. m othen m o. Lemm -8): Suppose s reducton module wth Z \{. If Γ s bprtte grph by sector of V nd V then V U{ for ech, s sub module from. V Proof: Supposer, xx V. We must showv + V V nd rx V. If then V Now supposerx o. From ssumpton, xs djcent of n element from V n the nme ofy. If y rx o rx. rx then 0

5 J. Appl. Envron. Bol. Sc., 5S)7-3, 05 y y : o r y: o, o from lemm -5), tht s resultng for ech m, mr, snce the s reducton module, so mr o, whch s resultng m s djcent ytht s contrdcton. So rx y nd from lemm -5, rxs djcent y, snce y V, therefore rx V. If xor xre equl to zero, then x+ x V, so t cn be ssumed tht none of thex or xs not equl to zero. Snce tht x V, therefore, there s y, y V, whch x re djcent y, for ech,. From lemm -5, we hve y y o) w y y, scence tht for ech,, we hve x I, therefore, o I w: x w VI V o) V nd f x + x o, becuse V s belongng to s sub module of. Lemm -9): If I, we hve x + x )w: o), now f x + x o, tht w therefore, x+ x V, smlrly, cn be shown thtv m Z, then m s fundmentl sub module from. m I K o, from m Z, whch s contrdcton, therefore, Proof: If m s not fundmentl, then there s non-zero sub module of K from, whch lemm -5, the m s djcent ech nonzero element from K, so m s fundmentl sub module from. Theorem -0): Suppose s reducton module wth \{ then followng tems concluson: ) Γ s bprtte grph. ). d U n Z. If Γ s bprtte grph, Proof ): SupposeZ V UV V IV nd no element ofv re not djcent. From lemm -8, we hve: V V U{ nd V V U{ re sub modules of. For ech Z Vnd y V, we hve Z I y V IV o) Proof ): Snce V) z nd tht φ nd from lemm -5, z nd y re djcent. zv ) re empty, of lemm -9, ech submodule of V nd V re fundmentl. So, V nd V re unform sub module from. Now we showv Vn s fundmentl. Suppose K s sub module from, whch K IV V ) o) nd o) y K. Then, for ech o z V nd o z V, we hve zi y o) ziw. Therefore, z s djcent of y nd w, so tht z VI V o), whch s contrdcton, therefore, V V n s fundmentl. Lemm -): Suppose s reducton module wth Z \{. Then gr Γ ) f nd only f Γ s str grph. Proof ):It s obvous. ) Suppose Γ s not ncludng cycle, Γ s tree nd therefore s bprtte grph. Now from theorem-0, Γ s complete bprtte grph. SupposeV nd V re prt of Γ grph. Snce the Γ s not ny cycle, we hve V or V, whch concluded tht Γ s str grph. Defne -): A hlf group grph s bprtte zero dvsor, f nd only f s not ncludng ny trngle. [4]) Lemm -3): Suppose s module. Γ s ncludng cycle of odd length, then Γ s trngle. Proof: By nducton, we show for ech cycle the odd length, n + 5, there s cycle of length of k+ for k< n. Supposex x x3... x n x n+ x s cycle of odd length of n +. If two non-consecutve vote of xnd x jre djcent, then proof s complete. Otherwse element of o) z x Ix3 o) from lemm -5, z for ech n +. So gn z s djcent to both elements of xndx n+. Therefore, we x x + z x x... hve the cycle of n 4 5 n+, tht ths s desred sme dstnce. x

6 Sedgh et l., 05 Theorem -4): Suppose s module. If s gr Γ ) 4, then Γ s bprtte grph wth prts of V nd V, so tht V, V. Conversely, t s true f the s reducton module wth Z \{. Proof: Supposegr Γ ) 4. Usng lemm -3, we show tht the length of ech cycle of Γ s even. Snce Γ hs cycle of length 4, ths hypothess s confrmed. On the contrry, lso theorem -0cn s proved. In the followng, we expln the between generlzton reltonshp from the defnton of zero dvsor grphs n [], t wll show byγ b ) nd wht, we hve shown n ths pper. Frst, t s worth notng tht Γ s sub grph from Γb tht f m, n Z re two djcent vertces n Γ, Or such s equvlence n m: o or m n: othen n: m: o. However, the reverse s not true s the followng exmple. Exmple -5): Suppose Z Z4s Z module. Then Γ b Kv, however Γ dfferent fromk v, whch we re showng n fgure.. Fgure. However, when the s multplctve module As for ech sub modules of N from, there s del of I from, whch N I), then Γ Γb. Suppose m: n: o. Therefore, n : nnd m : m, so both of m n: ondn m: o. EFEENCES [] G. Alpour, S. Akbr,. Behbood,. Nkndsh, J Nkmehr, nd F. Shves, The clssfcton of the hltng-del grphs of commuttve rngs, Algebr Colloquum, to pper). [] S. Akbr nd A. ohmmdn, on zero-dvsor grphs of fnte rngs, J. Algebr ), no., [3] DF Anderson, C Axtell, nd JA Stckles, Jr., Zero-dvsor grphs n commuttve rngs, n Commuttve Algebr, Noethern nd Non-Noethern Perspectves. Fontn, S.-E. Kbbj, B. Olberdng, I. Swnson, Eds.), 3-45, Sprnger-Verlg, New York, 00. [4] DF Anderson, A. Frzer, A. Luvem nd PS Lvngston, The zero-dvsor grph of commuttve rng, II, n: Lecture Notes n Pure nd Appl. th., Vol. 0, pp. 6-7, Dekker, New York, 00. [5] DF Anderson, nd PS Lvngston, The zero-dvsor grph of commuttve rng, J. Algebr 7 999), no., [6] DF Anderson, nd PS uly, On the dmeter nd grth of zero-dvsor grph, J. Pure Appl. Algerbr 0 007), no., [7] FW Anderson nd K Fuller, ngs nd Ctegores of odules, Sprnger-Verlg, New York, 99. [8]. Bzr, E. omthn, nd S. Sfeeyn, A zero-dvsor grph for modules wth respect to ther frst) dul, J. Algebr Appl. 03), no., 5-36, pp. [9] A zero-dvsor grph for modules wth respect to elements of ther frst) dul, submtted to Bull. Of IS. [0] I. Beck, Colorng of commuttve rngs, J. Algebr 6 988), no., []. Behbood, zero dvsor grphs for modules over commuttve rngs, J. Commut, Algebr 4 0), no.,

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