Jukl, Marek

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Institute of Mathematics AS CR, v. v. i.

Acta Universitatis Palackianae Olomucensis. Facultas RerumNaturalium. MathematicaMarek JuklLinear forms on free modules over certain local ringActa Universitatis Palackianae Olomucensis. Facultas Rerum Naturalium. Mathematica, Vol. 32 (1993), No.1, 49--62Persistent URL: http://dml.cz/dmlcz/120298Terms of use:© Palacký University Olomouc, Faculty of Science, 1993Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document must containthese Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech Digital MathematicsLibrary http://project.dml.czACTA UNIVERSITATIS PALACKIANAE OLOMUCENSIS FACULTAS RERUM NATURЛLШM 1993 Mathematica XXXII Vol. 110 LINEAR FORMS O N F R E E M O D U L E S OVER C E R T A I N LOCAL R I N G MAREK JUKL (Received December 12, 1992) A b s t r a c t A real linear algebra A having a E-basis < 1, r/,.. ., r/m~] > with r/m = 0 will be called the plural algebra. The linear forms on a free finite-dimensional module M — especially their kernel — are investigated. K e y words : Linear algebra, free module, linear form. MS Classification: 13C99 The problem solved in the article may be formulated as follows: It is known that the kernel of linear form on a vector space is a n - 1-dimensional subspace. Can this be suitably generalizated in the case the real vector space be replaced by a free finite-dimensional module over a certain local ring? In the article, it will be shown that an analogic relation between linear forms on this module and its (certain) hyperplanes can be found. 1 Real plural algebra of finite order. Definition 1.1 Real plural algebra of order m is every linear algebra A o n 1 having as a vector space over 31 a basis {1, r/, r/2,..., r/ m _ 1 } , where r/m = 0. Definition 1.2 By a system of projections A—>E it is meant a system of map­pings pk : A onto E, defined for k = 0, . . . , m — 1, as follows: m— 1 V/?€A, /?= £ 6 , V ; pfc(/i)dJ:V £=0 Proposit ion 1.3 An element e ~A is a unit if and only if p0(e) ^ 0. Proof 1) Let e EA be a unit and let p0(e) = 0, p0(e) = 0 => 3/i G A; e = 77/1. Then 1 = ee~l = (rjp)s~l = rj(jie~l). Multi-plying the equality rj(jie~l) = 1 by rjm~l we get 0 = rjm~l, which contradicts D. l . l . 2) Let p0(e) 7̂ 0. Let £ = ]T)i=o Cj/]1. Then e~l = _^™0 / - ^ e x i s ^ s if an (* o n ty if ^n e following system of equations (expressing just the fact _^i_o eirf ' __^=o /*r7* = 1) is solvable. (*) eQfk + eiffc.i + ... + ekf0 = 60k, 0 < k < m - 1. It is solvable if and only if e0 = p0(e) ^ 0. Proposit ion 1.4 Let a unit a ~ A be given. Then there exists a f3 G A wtM /?2 = a if and only if p0(a) > 0. Proof Let a = YlT^ ^rjk. Let us take /?,/? = X^™"1 6,V. Then m— 1 /?2= £ W+f *+i=o Thus m—1 m—1 a = /?2 <* a = £ a*-,* = £ W+'', fc-O i + J = 0 which is equivalent to the system of equations: (0) a0 = b20 (1) a, = 2606i (2) a2 = 26062 + b\ (m - 1) am_i = 2606m_i + 6x6m_2 + . . . + 6m_26i With respect to the condition p0((3) = 60 7- 0 (P.L3) it is solvable if and only if a0 = p0(a) > 0. Proposit ion 1.5 A is a local ring with the maximal ideal rjA. The ideals if A, 1 < j < rri, are the all ideals in A. 5:0 Proof 1) 77A is the only maximal ideal in A 77A is evidently an ideal. According to P. 1.3 A\r)A consists just of units of A. From this follows (see the consequence 1.6.(1) of theorem 1.3. in [1]) that A is a local ring and 77A the maximal ideal of one. 2) rf'A, 1 < j < m are the only ideals in A Let J, J / A, is an ideal in A and let us suppose that Vj, l < j < r a ; J^rfA. For such ideal certainly 3k, 1 < k < m; J C r)k A A J <JL r]k + 1 A. Let a £ J, rn-1 ex g rjk+lA => ex = V^ ajrf , a0 = ' . . . = a^ — ] = 0 , a^ / 0. i=o Thus e = ^TL"*.1 cijrf~k is a unit, a = r)ke. If £ E 7?fc A then: 3 /?GA; e = ^ / ^ = ( ^ _ 1 ) « - > ^ J ^ J = 77*A which is a contradiction. Proposit ion 1.6 The ring A is isomorphic to the factor ring of polynoms R[x]/(xm). P r o o f Let us consider the mapping / : R[x] -> A, f(h(x)) = /I(T/), V/I(X) E R[x]. Then / is clearly an epimorphismus with the kernel (x**1). Therefore following diagram commutes: j [x ] / ( í f f l ) and the mapping F is an isomorphism. Proposit ion 1.7 The ring A is isomorphic to the linear algebra of matrix 3Vtmm(IR) of the form: bo bi . . . bm_i \ 0 b0 . . . 6m_2 0 0 60 / 51 P r o o f Let us define g : A —> M m m ( E ) in the following way: m— 1 a = \Y1 ajrf => g(a) = (a,-,-) <£> [(j at;- = 0) A (; > t => atJ- = cy _,-)]. i=o Considered mapping is evidently the founded isomorphism A—KM. 2 Free finite-dimensional modules over the algebra A A g r e e m e n t 2.1 In the following text we denote by A the E-algebra introduced in section 1. We will have a deal with the free finite-dimensional modules over the algebra A *. The capital M denotes always such module. P r o p o s i t i o n 2.2 Let { F i , . . . K n } be some system of generators of a module M . If U_\) - - - ylLk are Nearly independent elements from M then: (1) k<n (2) by a suitable renumbering of elements E±y..., Enr {IL\ >•••> iZt > i£fc+i > • • • i En } WM be a set of generators of M . P r o o f (by induction) (a) k = 1 (1): evidently fulfiled n (2): let U_x be linearly independent, U_x = Y^^tiS* (*) «=i We will show that there exists at least one unit among £ i , . . . £n . In the opposite case multiplying (*) by nm~~l we have: r)m~~lU_x = oA/ /m _ 1 / 0 => Hx is linearly dependent — contradiction. Let for example £i be a unit. Then from (*) it follows: J =2 Consequently \U_l, E^,..., E„\ = M . (b) Let P.2.2 be fulfiled for k - 1. As A is a local ring, that M is an A-space in the sence of [2]. 52 As U_l}..., f/j. are linearly independent, then U_x,. . . , U_k-\ a r e linearly indepen-dent as well. By the induction supposition we have by a suitable renumbering ofE,-: \U1,...,Uk_1,Ek,...,En)=M. Now k — 1 n Ut € M => ___ = Y_ tiHi + Y. ZiEj (**) 1 = 1 j=fc Let us derive that there exists at least one unit among fo,..., £n • Otherwise after multiplying (**) by rjm~l we would obtain: (vm-1Zi)Ui + --- + (vm-1tk-i)Uk-1-r,m-1Uk=oAT,m-1?0 which contradicts to linear independence of Ux,. . . , U_k. Let for example £j. be a unit. Then from (**) we have: Ek = (-er'eoiZi + • • •+(-^16-1)iz*_1+ +^]/z* + (-Vfc+o.a.+i + • • • + (-4"^n)^. It follows from this that: [£/-_,. .. ,U_kiMLk+ii • • • ?£n] —M, i.e. (2) is true. From the induction supposition we get that k — 1 < n. From (**) it follows that k — 1 = n implies the linear dependence of U1, .. ., U_k) which is not possible, i.e. (1). Consequence 2.3 If the module M has one basis consisting of n elements then any its basis consists of the same number n elements. Any linear independent system of n elements of M forms a basis of M. The number n is called the dimension (more precisely A-dimension) of the M. Moreover it follows from the proof of P.2.2 that a linear independence of the system {K1 ; . . . , E_n} implies the linear independence of the system {U^ , . . . , £ / * , Kjb+i, • • •, __W } • Proposition 2.4 Let M be a free n-d^mens^onal module on A. Then M is an mn-dimensional vector-space oi>er IR (m denotes —as usually— the order of A). P r o o f Let £ = < E_x,. . ., E_n > be a basis .of A-module M. Let U E M, & G A, n m—\ n rn—l u-.£&.&, & = Y_*»>v. i fl = EE-••;(,r!-s.)-2 = 1 jzzO izzl jzzO I.e. M is evidently a vector-space over E. It remains to prove that the system of generators 'B=<E1,...,E_l,r,E1,...,VEn,...,rim-1E1,...,r)m-lEn> 53 is (over E) linearly independent. Let us suppose that n m—1 3e<;€K; £ £ e ^ ) = o i = l j=o It follows from this n m — 1 m — 1 _C( _C e^)^i = fi =* _C e*jV = °> 1=1 j = 0 j = 0 Vi, 1 Vi, 1 < t < n, Vj, 0 < j < m - 1; e{j = 0. Therefore _3 is a basis of M as a vector-space on E, thus card _3= dimM= mn. R Proposit ion 2.5 Let £ = < Fi, •. •, ___„ > 6e a bas.s of A-module M. Le. ws define a system of vector-spaces P o , . . . , P m - i over E ; P i = [v ,__i, . . . ,7 , '__.] , 0 < . < m - l , Considering M as an M-vector space, then the following statements are valid: (i) M = e £ - )1 P . (2) V__GM 3! (__ D , . . . ,__ n _ 1 )GPgl ; __ = __£ _ V _ _ , -Proof 1) As £ is a basis of A-module M, then according to the proof of P.2.4 'B=<E1,...,En,T,E1,...,VEn,...,r,m-1E1,...,T]m-1En> is a basis of a vector-space M over M, from which we have (1). 2) Let X EM. n m ~ l Then X = _^&J__,-, £ = _T) «oV", *ij 6 l , 1 < i < n, 0 < j < m. t = i j=o n m—1 Then X = J2 E ««^--. = E ^ ' E *«--<)• 1 = 1 j=0 j i n m— 1 Let us put 5^&_E«- = 2C;. Then X = ] T i^Xj i 2Cj £ p o , 0 < j < m. t=i i=o As _3 is a basis of the vector-space M, we get from this that the system of elements Xij E l , 1 < i < n, 0 < j < m, and thus also vectors Kj arc unique i.e. (1). Notat ion 2.6 The system of vector-spaces P _ , . . . , P m - i is determined by a given basis of A-module M. Therefore "unique" in 2.5 (2) means unique up to selection of a basis of M. 54 3 Linear forms on modules over the algebra A Proposition 3.1 Let <j> be a linear form on M (A.2.1). Then there exists ex-actly one system of linear forms <f>o,. . . , (/>m-i M into E such that: m — l <t>= 53 trf j=0 Proof m— 1 u e M => <j>(u) = 5 3 M JV =» Pj(HLD) = t/j, 0 < j < m. i=o Denoting <^ = 0 o pj, 0 < j < m, we cleary obtain a system of mappings <̂ o, • • •, 0m-i satisfying the equality <j> = X ^ o ^jV- Exactly one such system exists for arbitrary linear form <j>. [if {<f)j}, \ipj} are two such systems then m —1 m — l m — l j=0 j=0 j=0 Due to D.L2 it follows that {<j>j} is a system of linear forms M into JR. Proposition 3.2 If (fro, • • •, ^m-i are linear forms then the mapping m~ 1 </>= $3 trf 3=0 is a linear form M into A if and only if VX E M ; MvK) = o, ) 1 < k < m- 1 > (*) MiX) = 4>k-i(20 J Proof 1) Let <f> = J2T=0 't'irf ^ e a ' i n e a r f ° r m As <̂> is a linear form M into A, then VX € M; <j>(r)X) = rf<j>(X). Thus m - 1 m - 2 m—l E MvKW = _P MK)rik+1 = E ^-iQO^. j = 0 Jfc=0 j = l we get from this !>;(20 G i ] : 0o(i?2O = o. 0i(»»2Q = ^ - i (2D, ] (tjX) = r\4(X)\ m— 1 *fo.2Q = £ *ifa2Q^ [(*)] = i=o m - 1 m - 2 m - 1 = E ^-iQO^ = •?(£ <M*K) = ? ( £ W W ) = ^(20 i = l i=0 j=0 (c) we prove: VX G M, Vo G A, a = ^ J L " 1 a j V ! <Ha20 = ar.^QQ: m—1 m—1 *(*20 = *(E(*JV)20 t(a)] = E ^KV*) = i=o i=o m—lm—1 m—1 m—1 = £ £ M'rfxtf = £ « ; £ f̂c(̂ 2C)r?* = j fc=0i=0 i = 0 fc=0 m —1 m—1 = £ a^rf )0[(b)] = ("T a,V)«K2D = M(2Q i=o i=o It follows that ^ is a linear form. P r o p o s i t i o n 3.3 Lel <j>o,... ><t>m-\ : M ™> E 6e a system of linear forms such that m—1 i=o 25 the linear form M into A. Then m - 1 fc VXGM, 2£=£»r\£j.2£..GPo; ^-(20 = £^*-i(2Ci). 0 < j < m - l . ; = 0 j = 0 Proof Let X = Y^=o Tf2Ly Then **(20 = M2La + f)X1+-+ r)kxk + ••• + rr-'iLsn-x) = = <J>k(Xjo + v2Ll + r)k2Lk + ••• + +nk(ri(2Lk+i + ••• + nm-k-2xm-i))) [P-3.2] = = MX0) + ^-i(2Ci) + • • • + Mxk) + +4>o(ri(Xk+l + --- + r,m-k-22Lm-i)) = = Mx0) + *k-i(Xi) + ••• + Mxk) + o, 0 < k < m - 1. 56 Proposition 3.4 If (j) : M —> A is a linear form then there exists exactly one system of linear forms / 0 , . . . , fm_x : P 0 -> E such that m-\ VK€M, X=^2^Xji XJGPQ ; = 0 the following relation is valid: k M £ ) = £ /* - i (2g) , 0 < * < m - l . (*) i=o where m - l ] T <j>jrf = (j) i=o Proof Putting fj = <£j/Po, 0 < jf < m - 1, we get (due to P.3.3) the system of linear forms Po —• 3R fulfiling (*), i.e. M*) = £/*-;(2Q), o<*<m-i . i=o We prove the unicity of this system: {//}, {#j} being two systems fulfiling (*) and determining systems of linear forms M into A {<!>]}, {ipj} consecutively. From the equality <j> = J ^ 1 far? = EJL"^ ^ V it follows (due to P.3.1): <j>j = i/jj, 0 < j < m — 1. From this we arrive in equalities 3.4 (*) in the form as follows yX_, X_ = Y2T=o rf 2L.: (0) * = 0 : (7o(X0) = V>o(20 = M20 = /o(2Co) => fo = 9o (1) * = 1 : ^i(2Co) + ^o(Xi) = ^ i ( Z ) = ^ i ( ^ ) = /i(Z.o) + / o ( i i ) , due to (0) => #i = / i (m — 1) k = m — 1 : flfm-l(2Co) + !7m-2(2Cl)+ " * + -0p(2Cm-2) = ^m-l(2Q = = *m-lQ0 = /m-l(io) + /m-2(2£l) + ' ' ' + fo(X*n-l), due to ( 0 ) , ( l ) , . . . , ( m - 1) => gm-X = / m _ i . Thus fj = gj, 0 < j <m — I and the unicity of the system is proved. Proposit ion 3.5 If {/y}™-"̂ 1 is a system of linear forms P0 into A and let {(j)k}^0 be the system of linear forms M into E defined as follows: 57 VX6M, X = J2T=oWxj; k MX) = ][>-;(*;), 0<k<m-l (**) i=o then the mapping </> = J2fc=o f̂c7?* *5 ^ e ^near form M —• A determined uniquely by the system {/?}. P r o o f If /o, • • •, /m- i : P —• IB- are linear forms then the system of mappings {(/>k} defined by (**) is evidently the system of linear forms M into E. Hands the supposition is correct. It is necessary to show that the mapping <j> is the linear form M —* A. According to P.3.2 it is sufficient to show linear forms defined by (**) have the property 3.2.(*) i.e.: V K E M ; (1) MvX) = 0, (2) ^(i720 = ^ - i ( 2 0 , l < * < m - l . LetX = E7="01V^r Then obviously (rjX)j = Xj_i, 1 < j < m and (i7K)o = 0. So we have: (1) Mv20[(**)] = M(riX)o) = /o(fi) = 0 (2) MnXM**)] = Ef=o/*-y(('?2C)i) f/*(e) = o] = E?=i/*-y((»?2C)i) = = £ ) _ . /*-/G2Q_i) =- [j - 1 = h] = E*Io /(*-i)-i.(2C*) = ^ - i (20 -From this <j> = £]JL0 ^ji?7, {<j>j} defined by (**) is the linear form. Due to P.3.1 the unicity is evident. Definition 3.6 A linear form <j> M into A is called a linear form of order k (0 < ib < m) if: (1) V X G M ; <KX)er?*A, (2) 3 Y G M ; </>(Y) $ *r*+1A. In the special case k = 0 the linear form is called the epiform. Proposit ion 3.7 If <j> is a linear form of order k then there exists at least one epiform \ su^h that <t> = l k X -58 Proof Let <j> be a linear form of order k. We get clearly from this: 0o = 4>i = • • • = ^ - i = 0 A 3Y G M; ^ ( Y ) + 0. Let us denote <j>* = <j>k + • • • + nrn~k~~l(j)rn„i. Then (j) = nk<j>*, though 0* is not a linear form from M to A generally. According to P.3.4 there is the system {,/}} of linear forms Po into M. fulfilling 3.4 (*) for the linear form (j). Since 0 is the form of order k from 3.4 (*) we have: /o = / i = . . . = /*.-! = 0 . Let us define the system {hj}JtTQl of linear forms P 0 into E as follows: ho = fki h\ = /jfc + l , . . . , / l m - f c - l = frn-l- (*) and linear forms bm-jt, • • •, Am- i a r e arbitrary. According to P.3.5 to the {hj} we can construct the system {xj} Dv means of 3.5 (**) for which x = YlT=o Xj7/"7 1S ^n e n n e a r form. And due to (*) we get: <t>k(x) = fk(xQ) + fk-i(Xi) + • • • + /o(2£*) = fk(x0) = fc0(2Co) = xo(20 **+i(20 = A+i(2£o) + h(X_) + 0 = /n(Xo) + A0(2Ci) = xi(20 <t>m-l(X) = /m- l (No) + • • • + / f c ( I m - ^ l ) + 0 = = Am-Jb-l(2Co) + r- Ao(Km-ifc-l) = Xm-A;-l(N)-Thus n^x = ^ ar-d since 3Y_ G M; (^(Y) = XoQQ / 0, x is t n e epiform. 4 Kernels of linear forms Definition 4.1 Let M be a n-dimensional A-module (by C.2.1). A free (n — l)-dimensional submodule of M is called a hyperplane of the M, Theorem 4.2 If (j) is an epiform then there exists exactly one hyperplane Ji of the M such that 3V = A'er (j). Proof Let £= {E_X). . . , Fn} be a basis of the A-module M. X_ = YA=I £--Ei *s a vector from M. Let us put ^(Ej) = c*i, 1 (2Q = J2i=\£iCti. As (j) is an epiform there exists an c\j, 1 < j < n, being a unit. We may suppose that an is a unit. We will construct vectors Vi, • •, Y-n-i a s follows: Vj, 1 < j < n\ Vj = anE- - cxjEn. 59 Evidently each of them turns the form <j> to zero. Let us prove their linear independence over A: Let us suppose that 3/3j E A, 1 < j < n — 1; n—1 n—1 53 & Zi = ^ => 53 /?iK£i - a ^ ) = o => i = i i = i n - l => ^1(53(an(^^i) --Sn(ft«i))) = 2=> i = i n —1 n —1 => E ^ i i + (-««l(E/%«-i)).e« = e=> i=i i=i =>& = ... = /?„_!=(). Let us put N = [ F _ i , . . . , ] L . - 1 ] We show 3N = Ker <j>. 1) Let X G X => X = Er="/ & £ => *(2Q = 0 => X 6 Ker <£ 2) Let X_ G Ker <£. The X has the expression X_ = 53*=l 6I?; • We get ФШ = 5]č; tat Л ф(X) = 0 Л (V6 ЗA{; & = anA*) => ť=i n n —1, n —1 => o = 53 ţiai = an 53 Лt*a* + a ^ n ^ 5 3 Л i a i + ^ n = °5 t = i t = i t = i n - l i.e. £n = - y ^ A,at. t = i Then of course: n n—1 n—1 2L = 5 > £ . = ]Ţ>«A,të. - ( £ A<a,)l„ = ť = l ť = l ť = l n - 1 n - 1 = 53 A.(a„Яť - or.ЯJ = 5 3 AiZ* => X Є K. t=i І = I Theorem 4.3 IfTi is an hyperplane of module M then there exists an epiform 4> such that Ker <j> = N. 60 Proof Let {Vu . . . , Vn_x} be a basis of 5V. Then (by P.2.2) there is Vn _ M such that {V l 5 . . • , V n _ i , Vn} is a basis of M. Take X G M, X = £ " = 1 &&• Let us define a mapping 0 : M —> A by the relation 0(K) = £n • Then 0 evidently is the epiform. [0(Vn) = 1] with Ker 0 = Tsf. Consequence 4.4 Let 3NfC M. JN is a hyperplane of M if and only if there exists the epiform 0 such that JN" = Ker 0 Theorem 4.5 Let 0, i/> 6e epiforms. Then Ker 0 = A'er xp if and onli/ zf there exists a unit e G A sucA Ma/ 0 = eij). Proof 1) Let 0 = £"0 where e is a unit, then obviously Ker 0 =Ker -0. 2) Let Ker 0 = Ker ip =jN" where X is (by T.4.2) the hyperplane of M. Let {Vi, • • - ,V n _ i} be a basis of JN. Then there exists a E_n G M such that \V,,...,V^_1,En] = M (by P.2.2). Then VX € M: n _=_3^Y i=>^(_)=en^(_n) and WX)=tmKEn)-1 = 1 As 0,-0 are epiforms 4>(En)} ifr(En) are units and therefore we can find £ for which 4>(E_n) — £-^(E-n) a n d thus 0 = ex/;. Theorem 4.6 If x : M —• A is a linear form of order k then there exists a hyperplane Jf OfM such that KerX = { N G M ; ^XE'N Proof If x is a form of order k then there is an epiform 0 such that x — r]k<t> (by P.3.7). Then by T.4.2 there exists J\"CM, ^'=Ker 0. 1) ^ K G X ^ x ( i ) = ^0(K) = 0(^X) = O^N 6 Kerx 2) Let X G Ker x => x(2Q = 0 => ^ 0 ( X ) = 0 => (j>(rjkX) = 0 => 17* X G JN". 61 References [1] Ativan, M.F. and MacDonald, I.G.: Introducion to commutative algebra (Russian), Mir, Moscow, 1972. [2] McDonald, B.R.: Geometric algebra over local rings, Pure and applied mathematics, New York, 1976. Author's address: Department of Algebra and Geometry Faculty of Science Palacký University Tomkova 38, Hejčín 779 00 Olomouc Czech Republic 62

Linear forms on free modules over certain local ring

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Linear forms on free modules over certain local ring

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Institute of Mathematics AS CR, v. v. i.