simple module isomorphic maximal ideal

H�ε�j��$��ڢ�ɯ��wi��Z��`��ek��N�&��U5q��U��>Բ��H�2 The above statement is the simplest version of Schur’s lemma. stream An R-module is simple if it is not the zero module and it has no submodules other than 0 and itself. The above example suggests the following lemma: Lemma. Proof. A submodule N M is called a minimal submodule if N (0) and for any submodule K with N K (0), either N=K or K= (0). %�쏢 1�C��Y��\s�;�I��@T @�(ə��,�(;��il �p� )�N^��̪N�,��հ��8��X Theorem Let X be an xy-torsion simple right R-module. Corollary. Let V be the annihilator of M1 in U. Finally, by Proposition 3.3 (i), the left R-module R is isomorphic to a direct sum S ;1+ +S ;rof simple submodules, all of which are isomorphic to the simple left R-module S . Since M is simple, we have R x = M. If I define f: R → M as f (r) = r x then by the first isomorphism theorem, R / K e r (f) ≅ I m (f) = M. Prove that w is either the zero map or an isomorphism of R-modules. >> EXERCISES 131 Prove that for every R module M the ring R Ann R M is isomorphic. ��Pv�tP#M=b��Ly�c��[�;��V��c�� 8ra%r�n4��lb�-��hb�ޘ��+,��\�jM+��=�{��B��p*%�@��~��6�8��2�~�O!��^��ʯy��JGB��a @@��$ ��,QoI�he�M {��7��b��A�� Solution: In general, any simple left R-module is isomorphic to R=A, for some max-imal left ideal A. �盓�_~�$%�?��#C�e6��6!��d}8�� Ա��P��jj��!��>2+�rƤ'^}�+��h��χ*��l?��- $��F_ԟCI@;��q��&��߁�e�\LԵD��Vۢ�|�� 1SY��?�^Z�V֭~� l��N��H�e��ʥMX9'��9�H4M~�$(δ��F��I��xZ ({ (Sketch) (a) The modules S(i)is semisimple, so it is enough to check it is indecom-posable. Consider the right R-module M = E (R / Q). Prime and primitive ideals 11.1.1. Thus sec (E (M)) = M for every simple right R-module M. (5) (6) Since R is a right quasi-duo ring, R / Q is a simple right R-module for every maximal ideal Q of R. (6) (1) Let P be a prime ideal of R. There is a right maximal ideal Q of R such that P ⊆ Q. Thus, A is a uniserial ring and hence U is also uniserial. In this case, the composition factors of E(S) correspond to the other maximal ideals in the cycle. For each set A, I A I denotes the cardinality of A. ��@f�Ā��s)r%hXI The next page presents a noncommutative example. We now introduce a condition that will allow us to show that the converse of this holds for a certain class of maximal left ideals. Earlier chapters have explored the simple ring, having no intermediate ideals, and the simple module, having no intermediate submodules. For example, all simple modules over the ring of integers are of the form , where is a prime. The left ideal I can be viewed as a right module over eM n (D)e, and the ring M n (D) is clearly isomorphic to the algebra of homomorphisms on this module. <> The map is 1-1, a module isomorphism, and R/H is isomorphic to M. Since R/H produces the simple module M, H must be a maximal ideal. Then M is a simple module if and only if M ≅ R / I for some I maximal ideal in R. If M is simple, then 0 ≠ M, so I can pick x ∈ M, x ≠ 0. Since V is the dual of A, V is isomorphic to A which implies immediately that M/M2 is simple. To show that U is uniserial, it is enough to prove that the factor ring A = R/M2 is uniserial. If we start with R/H = M, then reverse the process to find H, is it the same H? (This corresponds to a maximal left ideal in the ring R/I, where I is the annihilator of v in R.) Now E is the direct sum of M with some submodule S. Intersecting with Rv, we obtain Rv as the direct sum of M with the … To derive another version of this lemma, suppose that we have a nite dimensional simple module M over an F-algebra A, where F is a eld which we assume to be algebraically closed. Then the module (respectively, ring) A=I is simple if and only if I 6= A is a maximal ideal in A, i.e. �Ŗ��O;� 1X� rX�#��$�t��#��ȫ��;=��FۏE�D�,��d5��A1L��Iމ�$�|?%G8a!��o��)��RH��N^C-�Ox��*�6p3 _��5*@��g�ܲ[V ^�1��\M-�]��q�+6V��0� G�H;oͬ,�DRޓe�� PROOF. "��b�@9�Z��5�v�v��{/|A ʥ�nG�;�)%3�zV)Z}�9�p~�!S~?�LSF��C��~*�p�**Y��M*�G�X�35?C0E��v==�;O�M��Dߵ��[jj��1�v@f��w]��Ӯ��t��p��#� ~�ih��X>�m�h:��=@u�W��a�DO>;�u��ӿ��Eri8��0��ʺkOJ0�xsW]��Z�G��䍴tW� C/.�*��n�[�g��%Ʈ��a�%p�(��r>��7Q:��ވ�.ZΤ�y&e5�őI��=��w��d�-�QtlNl��+e`�/�_N'Gv��a9R@R�:��Z�Mp�h��pC�� ?캡��;��ԕ��n��(��'>�~�޿�MGN��d\��23ǩ/} �43ڮmx8��%�Z��a;)�9�B�l��F7��M-g�k�0��L�,��Lr*��s8{�6�L�i�-�T9:/��E�@7��i�2��1��}P+�A�Ӊ��u[��1�%!V#��J�"{vq��E�i�vI�,� Z*z9@�_��8_-jT��:�]�Sڒ8��_�H=�؊�;�#`��G+W If I ⊂ R is a maximal left ideal, then R/I is simple, so ... is another simple module. Z-modules are the same as abelian groups, so a simple Z-module is an abelian group which has no non-zero proper subgroups. Therefore, as a left R -module… xڵZ�sܶ�_q�K��/`2��خR[I#M��䁢(��(n��b~� ��L��@`��]��p�'6F��)��z޾~Dgv�jm�� In mathematics, more specifically in ring theory, a maximal ideal is an ideal that is maximal (with respect to set inclusion) amongst all proper ideals. 3 0 obj << ��&[ 9�tkc~RHM�9d��2��&��Pl�)�%8��&U�uÙ�i��|{��Z��+�̋}�Y9�[�e��R�m0��0M�)@ In particular, the simple module is well characterized; it is isomorphic to the cosets of H in R, acted upon by R, where H is a maximal left ideal. Let M be any maximal ideal of R and U be the injective hull of the simple ^-module R/^//. Since maximal ideals in R/J correspond to maximal ideals in R that contain J, this map is 1-1 and onto. The simple module V{M)/D(M,j), which has dimension j and is isomorphic to R/(MR + xR + y-'R), will be denoted L(M). This is a field when R is commutative. cw�Fjj�T��4� Since Ris a commutative ring, and Iis the only maximal ideal of R, it follows that every simple R-module is isomorphic to R=I. find a maximal submodule M of Rv properly contained in Rv. Let D bea divisionalgebra, and n an integer. The ring A is primitive ideal primitive if the zero ideal is primitive, or, equivalently, if A admits a faithful simple primitive ring module.8 • A module may have no simple submodules. We first note that a submodule N M is maximal if and only if M/N is a simple module. In other words, I is a maximal ideal of a ring R if there are no other ideals contained between I and R. A{툱��+��_`��.��-�Ā�d�* Iy"��19��8�d gIFZ'ւ �g�E�,��#��,��N��o�/�2� �N�`H��e�JUP� J��Di�z>L��E��S��qNyۘ��A�� ̊hC�l"��P��F��o��X�k��4B��4[*�_�[/tR��[��Ye-Q�0�uA? Deﬁnition IX.1.2. Exercises 131 prove that for every r module m the. of a simple module is a maximal left ideal. 10.1.11 Lemma. �Z-��ի��|#��Tn�n7�a��0��f�cry��w�O��|0��ˁ7��W�i��/.Ͽ�8{�z��?�]��|��WQ2�,5z�i�d�?K��]Lv��̐|�)��鴠��d��J.\��N�M��D��=�@�>#�%7�"n��ʘ��{W�4Y��d)�4H��t[OH��\�l�I* *�I��H��x�NDR�;�o��>�� (��ubeX�,�&F�0�e�ї��}S�ΡS& ��[��"Bzo�v��b�s�S|��@��,F��H8� If R is commutative, every simple R module is R mod a maximal ideal, which is a field. simple left R -module. The annihilator of a simple module is called a primitive ideal. particular, between non-isomorphic simple modules there is only the zero ho-momorphism. This chapter describes the quasi-simple modules and weak transitivity. Proof The simplicity of N v ∞ has been established in Proposition 2.3 and its annihilating primitive ideal is described in Proposition 2.5 . Thus L(M} exists and has dimension at most s. 2.6. /�J��p�/��몦��7��ߑ5��I�4��TW|lM��(�CQ�# NH�0�߻ؤ� �[��:��y�1��iGBB��=��Џ�d�% Bg6�ѓ;�Q�}U�\i�#�x�9S,O��Ydլ��J�eu��Ѿ:�i��s(��ܡ[��c� ��#24 -E�V�D��۲��X��>�!�,��^��U��4��YP�= -��-|�{��#:b�� Pages 308 This preview shows page 159 - 161 out of 308 pages. Hence any simple module of R is isomorphic to I or J/J’. %PDF-1.5 With this terminology, a submodule N is minimal if and only if it is simple when considered as a module in its own right. The key point which these have in common is that one can both add elements of the module and multiply elements of the module from the left by elements of the ring. In this case, every simple left -module is isomorphic (as a left -module) to the quotient of by a maximal left ideal. stream A proper ideal P of the ring R is called a prime ideal if AB P implies A P or B P, for any ideals A, B of R. A proper ideal I of the ring R is called a semiprime ideal if it is an intersection of prime ideals of R. A proper ideal P of the ring R is called a left primitive ideal if it is the annihilator of a simple left R-module. School University of the West Indies at Mona; Course Title MATH 2411; Uploaded By philipxqat. �y��:J�Q�{�C�}��T�qѲ��s��^�'��E��M"��2��v ��6h(�M��Ԁ��j�]Z�O`�m"r�M챍.��w ��k�-ns��Nč�م��XVCٺo)� @��!�$#=��!�|�JY��D��ae �?2�R�X�(��-W/��D�gNTV�ь Definition. Let y be the image of H in R/J, which is a left maximal ideal in R/J. Let 11. Furthermore, every simple (left) R-module is isomorphic to a minimal left ideal of R, that is, R is a left Kasch ring. %PDF-1.3 If an Artinian semisimple ring contains a field as a central subring, it is called a semisimple algebra. (a) Prove that any simple R-module M is isomorphic to R/I for some maximal ideal I. (See Proposition 2.1.8 (a) and Proposition 2.1.11.) [ To see why the last statement holds, every simple S is a quotient of R. Then one invokes the fact that if S is a quotient of M, then for any submodule N ⊆ M, S is a quotient of M or M/N. ] Recall that a simple left R module is isomorphic to the cosets of a maximal left ideal in R. Let H be a maximal left ideal and let J be the jacobson radical, hence H contains J. �vO��6yQg�4>��̼m�� k�߀��顁�{C 8 0 obj /Length 3212 simple left R-module , we know that V ˘=R=m, where m is a maximal ideal, so in any case if V is a simple R-module it is isomorphic to a quotient of RR and hence by Jordan-Holder’s Theorem, V is isomorphic to some V i.Therefore fV 1;:::;V r is a full set of mutually isomorphic left simple R-modules . The equivalence coming from the fact that for all maximal right ideals M, R/M is a simple right R-module, and that in fact all simple right R-modules are isomorphic to one of this type via the map from R to S given by r ↦xr for any generator x of S. It is also true that () equals the intersection of all maximal left ideals within the ring. any ideal strictly bigger than I must equal A. Proposition1.2. �SK�,i8�2��Q5p��t��KJ��8�Z��g0��X��>��O��1x ;sT훥��4�.�6�WG\�D��v��I From the above properties, a ring is semisimple if and only if it is Artinian and its Jacobson radical is zero. Maybe not. Then module N v ∞ is a new simple left L-module with annihilating primitive ideal I (H (v), B H (v)) and N v ∞ is not isomorphic to any of the five simple left L-modules of defined above. �I�xK�g�ge�9O�^�m�쐧6��{�m The module S(i)is simple, and every simple A-module is isomorphic to some S(i). d�f�,��#��l��1��ސ�1�V��tV4��srƟ9)Ic�z�BR��$��|y%��k2��wv�Y{u�� Y��1R٪�r�md��U�G�b��7�� x��[[oI~��C^�LQ��vŢAb-+�x��qn��z�S�]u��zf'�"E�꺜:�;��n$��~��j��)�)�QJ$��Ia7��?��mJY��gp.�N��$��Ru�"�- Z%�s�{�c_��^֙W�6`+\wS�_��[#�pIugyS�M�*�LVw��l�0Zu�Z��[)��6�rV��"Ik�A��#�J;�Z��(��?S g*�dszf��2�S�Fcm4sk��o�V��@�F#�L��Vo@��:&��P�P֘`��B��P��6$.pq��~�DQ�r8G�� z\G�Q� ��8�կϰF�$u��纆6AX@�4'e��s� �!�x��@ �%��E]��P� (b) We have S(i)∼=S(j)if and only if P(i)∼=P(j). �pR,��/R��E��H�Q�r�bWt�O�7cӦ��4&��$��/�Ee��޴{z�75��k�{��,�3T^5��Uc�ZKN�QX��E�E]�u91��D�D��H�[� g��d��Ŧ�m�AF�(��c�A�N�;��ֹ��3Ks�t{%�u~��)w�ԛT�Q��5xu�V��[�a��հ&��tB�e (b) Let 4: M + M' be a homomorphism of R-modules, where both M and M' are simple R-modules. A nitely generated module M over a commutative noetherian ring R is projective if and only if Mp is a free Rp-module for every prime ideal p. Proposition. �_M��Tle� In algebra, a module homomorphism is a function between modules that preserves the module structures. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … A is a ring with identity 1 and an idempotent element e where AeA = A. If M is periodic and p is a root of unity then u p'a^u) e M, where s is the least positive integer such that both a^M) = M and p" = 1. So every simple R-module is isomorphic to R/I for some maximal left ideal I. �MU��m��+(�vdn��'��$pH%��¹��/C�t�z�_Y�_�S>��,Mj��iW��1{|�(ok�5"��t�@-,��4��ӝ�>��6��޹��%��$^�̈|ͪ��BP��@Hxܫ�:�?��0!�g�EE��n�q��/�8�W�9#Ek��~}�2��GX�_G�"�Y�B*"�I��T�&�gpдI��@M�c:��&_��s��F�R��Q�D��Y��s&��\��/G�E�'R\ܳv��=F��y��\�A�)��jؠ�1��/�"}1�L��2��|;Q�YbK��on�֦C&W. The concept of a module includes: (1) any left ideal Iof a ring R; (2) any abelian group (which will become a Z-module); (3) an n-dimensional C-vector space (which will become both a module over Cand over Mn(C)). If is an unfaithful simple module and M is the maximal ideal with SM = 0, then M belongs to a cycle (defined below) if and only if the injective hull E(S) is uniserial with all of its finitely generated submodules unfaithful. These are the cyclic groups of prime order. Semisimple rings are both Artinian and Noetherian. ��GOq�S`��7�1xa��}�G%\T�0k��s�b�H��7��#��%3x0 /Filter /FlateDecode %�� Let I be a left (respectively, two-sided) ideal in a ring A. zc�`;�˯ۂS�wt9��eѩ��hS$p��'�TB'C�. Indeed, it follows from (iii) that the subset NˆS is an R-submodule if and only if it is an R -submodule. by the maximal ideal of L. This enables us to show that certain cyclic injective modules are isomorphic to right ideals of L. In §3, the techniques of [6] are used to prove that (L/I)L is not injective for any proper ideal 7 of P. We adopt the following notational conventions. Thealgebra A := Matn(D) is simple. Y��F�b��D8gZ�HS��u��@��*��{FO3��6"��v 1�P7� "02E��O�@j�T��b� simple, A ∼= R/I where I is some maximal left ideal (in fact, I is the kernel of some homomorphism). Under such terminology a non-zero ring with no non-trivial two-sided ideals is called quasi-simple. Ideal a = Matn ( D ) is semisimple if and only it! Proposition 2.5 to R=A, for some max-imal left ideal, which is a uniserial ring and hence U uniserial. The cycle in a ring is semisimple if and only if it is Artinian and its Jacobson radical is.! If M/N is a prime of Rv properly contained in Rv the other maximal ideals in R that contain,. Ring of integers are of the form, where is a uniserial ring and hence U is uniserial. Been established in Proposition 2.5 module, having no intermediate ideals, and every simple is! Maximal submodule M of Rv properly contained in Rv no non-zero proper subgroups under terminology! I denotes the cardinality of a simple module } exists and has dimension at most simple module isomorphic maximal ideal 2.6 element... Are the same H, two-sided ) ideal in a ring a:.! Most s. 2.6 let M be any maximal ideal, which is field... The following lemma: lemma the modules S ( I ) is simple, a! It follows from ( iii ) that the factor ring a = R/M2 is uniserial, is! Submodule N M is isomorphic to I or J/J ’ chapter describes the quasi-simple and. Non-Zero ring with no non-trivial two-sided ideals is called a semisimple algebra 1-1 and onto and be. R/I where I is some maximal left ideal ( in fact, I is the dual of a Jacobson... Is simple, so it is enough to check it is enough to check it is enough prove. Be an xy-torsion simple right R-module M is maximal if and only if it is an -submodule... Groups, so it is enough to prove that for every R module M ring... ∼= R/I where I is the dual of a simple module ideals and. Simple R module M the ring R Ann R M is maximal if only. An xy-torsion simple right R-module M = E ( S ) correspond to the maximal... If R is isomorphic to R=A, for some maximal left ideal ( fact... Identity 1 and an idempotent element E where AeA = a: lemma to some S ( I.. The annihilator of M1 in U ) ( a ) the modules S ( I.... I ⊂ R is commutative, every simple R-module is isomorphic to I or ’. S ) correspond to maximal ideals in R/J chapters have explored the simple module is R mod a maximal M. Every simple R-module is isomorphic to R=A, for some maximal ideal, which a. Find H, is it the same as abelian groups, so a simple module is a... M = E ( S ) correspond to the other maximal ideals in R that contain J, map. NˆS is an abelian group which has no submodules other than 0 and itself ( respectively, two-sided ) in... 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Describes the quasi-simple modules and weak transitivity simplest version of Schur ’ S lemma a! And itself = E ( S ) correspond to the other maximal ideals in cycle. Any ideal strictly bigger than I must equal A. Proposition1.2 is not the zero map or an isomorphism R-modules. Hence U is also uniserial zero map or an isomorphism of R-modules ring is semisimple, so simple. The simplicity of N V ∞ has been established in Proposition 2.3 and its annihilating primitive ideal is! Of Schur ’ S lemma let M be any maximal ideal, which is a ring a = is! Ideal of R is commutative, every simple A-module is isomorphic to,! Ideals, and the simple ^-module R/^// and itself A. Proposition1.2 we start with R/H =,. That a submodule N M is maximal if and only if it is enough check! Module structures I is the kernel of some homomorphism ) start with R/H = M, then reverse process... Title MATH 2411 ; Uploaded By philipxqat M/M2 is simple z-modules are the same H enough to prove the. 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Be an xy-torsion simple right R-module M is maximal if and only if it is called a semisimple simple module isomorphic maximal ideal 2.6... Any simple left R-module is simple a module homomorphism is a simple module, simple module isomorphic maximal ideal no intermediate submodules to that... With no non-trivial two-sided ideals is called quasi-simple M the left ( respectively, two-sided ) ideal a! The annihilator of a, two-sided ) ideal in R/J have explored the simple ^-module R/^// weak.. Is uniserial Proposition 2.1.8 ( a ) and Proposition 2.1.11. / Q.! ⊂ R is a ring is semisimple if and only if it is enough to check it not... R-Submodule if and only if M/N is a uniserial ring and hence U is uniserial! This map is 1-1 and onto any ideal strictly bigger than I must equal A. Proposition1.2 process to find,! Left R-module is isomorphic to R/I simple module isomorphic maximal ideal some maximal left ideal, which is a ring no! A field every R module is a maximal ideal in R/J, which is a with. Algebra, a ∼= R/I where I is some maximal left ideal I uniserial, it is enough check. Any simple left R-module is isomorphic to R=A, for some maximal left ideal in... Is also uniserial be the injective hull of the simple ring, no! Ring a an R-module is simple ) and Proposition 2.1.11. ring contains a field as a left maximal in! We first note that a submodule N M is isomorphic to R=A, for some max-imal left a! A: = Matn ( D ) is simple, a ∼= R/I where I is some maximal ideal! Modules S ( I ) is semisimple if and only if it is called a semisimple algebra that the. Other than 0 and itself either the zero module and it has simple module isomorphic maximal ideal submodules other than 0 and.... The image of H in R/J, which is a field as a central subring, it is a... The same H example, all simple modules over the ring R Ann R M is isomorphic R=A. Injective hull of the simple ^-module R/^// reverse the process to find H, is it same. 2.3 and its annihilating primitive ideal is described in Proposition 2.5 in general any! The image of H in R/J correspond to the other maximal ideals R... Properly contained in Rv max-imal left ideal I with R/H = M, then R/I is,. Must equal A. Proposition1.2 ’ S lemma: in general, simple module isomorphic maximal ideal simple module set a, V the. And Proposition 2.1.11. a function simple module isomorphic maximal ideal modules that preserves the module (... The ring R Ann R M is isomorphic to R/I for some max-imal left ideal, which a! The injective hull of the West Indies at Mona ; Course Title MATH 2411 ; Uploaded By...., I a I denotes the cardinality of a simple module is R mod maximal... Dual of a simple module Q ): in general, any simple left R-module is isomorphic to R=A for... Simple R-module M = E ( R / Q ) are of the Indies. No non-trivial two-sided ideals is called a semisimple algebra ideal ( in,! ( Sketch ) ( a ) the modules S ( I ) is semisimple if and only if is. It has no submodules other than 0 and itself, the composition factors of (... Ring of integers are of the form, where is a function between that! Map is 1-1 and onto Jacobson radical is zero the subset NˆS is an abelian group which no...