aa r X i v : . [ c s . I T ] D ec WHEN THE EXTENSION PROPERTY DOES NOT HOLD
SERHII DYSHKO
Abstract.
A complete extension theorem for linear codes over a mod-ule alphabet and the symmetrized weight composition is proved. It isshown that an extension property with respect to arbitrary weight func-tion does not hold for module alphabets with a noncyclic socle. Introduction
The famous MacWilliams Extension Theorem states that each linearHamming isometry of a linear code extends to a monomial map. The resultwas originally proved in the MacWilliams’ Ph.D. thesis, see [14], and waslater generalized for linear codes over module alphabets.Starting from the work [16] of Ward and Wood, where they used thecharacter theory to get an easy proof of the classical MacWilliams ExtensionTheorem, several generalizations of the result appeared in works of Dinh,L´opez-Permouth, Greferath, Wood, and others, see [4, 5, 10, 18]. For finiterings and the Hamming weight, it was proved that the extension theoremholds for linear codes over a module alphabet if and only if the alphabet ispseudo-injective and has a cyclic socle, see [18].Regarding the symmetrized weight composition, the extension theoremfor the case of classical linear codes was proved by Goldberg in [8]. There isa recent result in [7] where the authors proved that if an alphabet has a cyclicsocle, then an analogue of the extension theorem holds for the symmetrizedweight composition built on arbitrary group. The result was improved in [1]where the author showed, in some additional assumptions, the maximalityof the cyclic socle condition for the symmetrized weight composition built onthe full automorphism group of an alphabet. There the author showed therelation between extension properties with respect to the Hamming weightand the symmetrized weight composition.There exist various results on the extension property for arbitrary weightfunctions, and particularly for homogeneous weights and the Lee weight, asfor example in [2], [9] and [13].In this paper we give the complete proof of the maximality of cyclic soclecondition for the extension theorem in the context of codes over a modulealphabet and symmetrized weight composition built on arbitrary group, seeCorollary 1. The result is used to show that for a noncyclic socle alphabetand arbitrary weight function the extension property does not hold, seeCorollary 2. Preliminaries
Let R be a ring with identity and let A be a finite left R -module. Considera group Aut R ( A ) of all R -linear automorphisms of A . Let G be a subgroupof Aut R ( A ). Consider the action of G on A and denote by A/G the set oforbits.Let F(
X, Y ) denote the set of all maps from the set X to the set Y .Let n be a positive integer and consider a module A n . Define a mapswc G : A n → F( A/G, Q ), called the symmetrized weight composition builton the group G . For each a ∈ A n , O ∈ A/G ,swc G ( a )( O ) = |{ i ∈ { , . . . , n } | a i ∈ O }| . The
Hamming weight wt : A n → { , . . . , n } is a function that counts thenumber of nonzero coordinates. There is always a zero orbit { } in A/G .For each a ∈ A n , swc G ( a )( { } ) = n − wt( a ).Consider a linear code C ⊆ A n and a map f ∈ Hom R ( C, A n ). The map f is called an swc G -isometry if f preserves swc G . We call f a Hammingisometry if f preserves the Hamming weight.A closure of a subgroup G ≤ Aut R ( A ), denoted ¯ G , is defined as,¯ G = { g ∈ Aut R ( A ) | ∀ O ∈ A/G, g ( O ) = O } . Obviously, G is a subgroup of ¯ G . If G = ¯ G , then the group is called closed . Also, swc G = swc ¯ G , since both groups have the same orbits. Moreon a closure of a group and its properties see in [17].A map h : A → A is called G -monomial if there exist a permutation π ∈ S n and automorphisms g , g , . . . , g n ∈ ¯ G such that for any a ∈ A n h (( a , a , . . . , a n )) = (cid:0) g ( a π (1) ) , g ( a π (2) ) . . . , g n ( a π ( n ) ) (cid:1) It is not difficult to show that a map f ∈ Hom R ( A n , A n ) is an swc G -isometry if and only it is a G -monomial map.We say that A has an extension property with respect to swc G if forany code C ⊆ A n , each swc G -isometry f ∈ Hom R ( C, A n ) extends to a G -monomial map. Characters and the Fourier transform.
Let A be a left R -module. Themodule A can be seen as an abelian group, i.e., A is a Z -module. Considera multiplicative Z -module C ∗ . Denote by ˆ A = Hom Z ( A, C ∗ ) the set ofcharacters of A . The set ˆ A has a natural structure of a right R -module, see[10] (Section 2.2).Let W be a left R -module. The Fourier transform of a map f : W → C is a map F ( f ) : ˆ W → C , defined as F ( f )( χ ) = X w ∈ W f ( w ) χ ( w ) . Recall the indicator function of a subset Y of a set X is a map Y : X →{ , } , such that Y ( x ) = 1 if x ∈ Y and Y ( x ) = 0 otherwise. For a HEN THE EXTENSION PROPERTY DOES NOT HOLD 3 submodule V ⊆ M , F ( V ) = | V | V ⊥ , where the dual module V ⊥ ⊆ ˆ M is defined as V ⊥ = { χ ∈ ˆ M | ∀ v ∈ V, χ ( v ) = 1 } . Note that the Fourier transform is invertible, V ⊥⊥ ∼ = V , see [10] (Section2.2), and it is true that for any R -submodules V, U ⊆ W , ( V ∩ U ) ⊥ = V ⊥ + U ⊥ . 3. Extension criterium
Let W be a left R -module isomorphic to C . Let λ ∈ Hom R ( W, A n ) be amap such that λ ( W ) = C . Present the map λ in the form λ = ( λ , . . . , λ n ),where λ i ∈ Hom R ( W, A ) is a projection on the i th coordinate, for i ∈{ , . . . , n } . Let f : C → A n be a homomorphism of left R -modules. Define µ = f λ ∈ Hom R ( W, A n ).An R -module A is called G -pseudo-injective , if for any submodule B ⊆ A ,each injective map f ∈ Hom R ( B, A ), such that for any O ∈ A/G , f ( O ∩ B ) ⊆ O , extends to an element of ¯ G . Proposition 1.
The map f ∈ Hom R ( C, A n ) is an swc G -isometry if andonly if for any O ∈ A/G , the following equality holds, (1) n X i =1 λ − i ( O ) = n X i =1 µ − i ( O ) . If f extends to a G -monomial map, then there exists a permutation π ∈ S n such that for each O ∈ A/G the equality holds, (2) λ − i ( O ) = µ − π ( i ) ( O ) . If A is G -pseudo-injective and eq. (2) holds, then f extends to a G -monomialmap.Proof. For any w ∈ W , O ∈ A/G ,swc( λ ( w ))( O ) = n X i =1 O ( λ i ( w )) = n X i =1 λ − i ( O ) ( w ) . Therefore, the map f is an swc G -isometry if and only if eq. (1) holds.If f is extendable to a G -monomial map with a permutation π ∈ S n andautomorphisms g , . . . , g n ∈ ¯ G , then for all i ∈ { , . . . , n } , µ π ( i ) = g i λ i .Hence, for all O ∈ A/G , µ − π ( i ) ( O ) = λ − i ( g − i ( O )) = λ − i ( O ).Prove the last part. Fix i ∈ { , . . . , n } . From eq. (2) calculated in theorbit { } , Ker λ i = Ker µ π ( i ) = N ⊆ W . Consider the injective maps¯ λ i , ¯ µ π ( i ) : W/N → A such that ¯ λ i ( ¯ w ) = λ i ( w ) and ¯ µ π ( i ) ( ¯ w ) = µ π ( i ) ( w ) forall w ∈ W , where ¯ w = w + N .One can verify that for all O ∈ A/G , ¯ λ − i ( O ) = ¯ µ − π ( i ) ( O ). Then, itis true that for the injective map h i = ¯ µ π ( i ) ¯ λ − i ∈ Hom R ( A, A ), h ( O ∩ SERHII DYSHKO ¯ λ i ( W/N )) ⊆ O , for all O ∈ A/G . Since A is G -pseudo-injective, there existsa G -monomial map g i such that g i = h i on ¯ λ i ( W/N ) = λ i ( W ) ⊆ A . It iseasy to check that λ i = g i µ π ( i ) . Hence f extends to a G -monomial map. (cid:3) Proposition 2.
A module A is G -pseudo-injective if and only if for any code C ⊂ A , each swc G -isometry f ∈ Hom R ( C, A ) extends to a G -monomialmap.Proof. Prove the contrapositive. By definition, A is not G -pseudo-injective,if there exists a module C ⊆ A and an injective map f ∈ Hom R ( C, A ),such that for each O ∈ A/G , f ( O ∩ C ) ⊆ O , but f does not extend to anautomorphism g ∈ ¯ G . Equivalently, swc G ( x ) = swc G ( f ( x )), for all x ∈ C ,yet f does not extend to a G -monomial map. (cid:3) Remark 1.
A module A is called pseudo-injective , if for any module B ⊆ A , each injective map f ∈ Hom R ( B, A ) extends to an endomorphism inHom R ( A, A ). In [4] and [18] the authors used the property of pseudo-injectivity to describe the extension property for the Hamming weight. Theyshowed that an alphabet is not pseudo-injective if and only if there exists alinear code C ⊂ A with an unextendable Hamming isometry. Remark 2.
Not all the modules are G -pseudo-injective. It is even truethat not all pseudo-injective modules are G -pseudo-injective, for some G ≤ Aut R ( A ). In our future paper we will give a description of G -pseudo-injectivity of finite vector spaces. Apparently, despite the fact that vectorspaces are pseudo-injective, almost all vector spaces, except a few families,are not G -pseudo-injective for some G .4. Matrix module alphabet
Let R = M k × k ( F q ) be the ring of k × k matrices over the finite field F q ,where k is a positive integer and q is a prime power. It is proved in [12,p. 656] that each left(right) module R -module U is isomorphic to M k × t ( F q )(M t × k ( F q )), for some nonnegative integer t . Call the integer t a dimension of U and denote dim U = t .Let m be a positive integer, m > k . Let M be an m -dimensional left(right) R -module. Let L ( M ) be the set of all R -submodules in M . Consider a poset( L ( M ) , ⊆ ) and define a map E : F( L ( M ) , Q ) → F( M, Q ) , η X U ∈L ( M ) η ( U ) U . The set F( L ( M ) , Q ) has a structure of an |L ( M ) | -dimensional vector spaceover the field Q . In the same way, F( M, Q ) is an | M | -dimensional Q -linearvector space. The map E is a Q -linear homomorphism. Similar notions of a multiplicity function and the “ W function ” were observed in [17] and [18].Let V be a submodule of M . Consider a map in F( L ( M ) , Q ), η V ( U ) = ( ( − dim V − dim U q ( dim V − dim U ) , if U ⊆ V ;0 , otherwise . HEN THE EXTENSION PROPERTY DOES NOT HOLD 5
In fact, η V ( U ) = µ ( U, V ), where µ is the M¨obius function of the poset( L ( M ) , ⊆ ), see [18] (Remark 4.1). Lemma 1.
For any V ⊆ M , if dim V > k , then E ( η V ) = 0 .Proof. First, note that for a submodule U ⊆ M and an element x ∈ M theinclusion x ∈ U holds if and only if for the cyclic module xR the inclusion xR ⊆ U holds. Calculate, for any x ∈ M , E ( η V )( x ) = X U ∈L ( M ) η V ( U ) U ( x ) = X U ⊆ V µ ( U, V ) U ( x ) = X xR ⊆ U ⊆ V µ ( U, V ) . From the duality of the M¨obius function, see [15] (Proposition 3), the lastsum is equal to P V ⊇ U ⊇ xR µ ∗ ( V, U ), where µ ∗ is the M¨obius function of theposet ( L ( M ) , ⊇ ). Since dim xR ≤ dim R R = k < dim V , V ⊃ xR . From thedefinition of the M¨obius function the resulting sum equals 0. (cid:3) Let ℓ be a positive integer, m ≥ ℓ > k . Fix a submodule X in M ofdimension m − l . Define two subsets of L ( M ), S = ℓ = { V ∈ L ( M ) | dim V = ℓ, V ∩ X = { }} ,S <ℓ = { V ∈ L ( M ) | dim V < ℓ, V ∩ X = { }} . Consider a map, E ′ : F( S = ℓ , Q ) → F( S <ℓ , Q ) , ξ X V ∈ S = ℓ ξ ( V ) η V . The map E ′ is a Q -linear homomorphism of Q -linear vector spaces. Lemma 2.
Let A be an a -dimensional F q -linear vector space and let B beits b -dimension subspace. Then, |{ C ⊆ A | C ∩ B = { } , dim F q C = c }| = q bc (cid:18) a − bc (cid:19) q . Proof.
See [3] (Lemma 9.3.2). (cid:3)
Lemma 3.
For any positive integer t there exists an integer x > t such that t − X i =0 (cid:18) xi (cid:19) q < q t ( x − t ) . Proof. If t = 1 the inequality holds for any x >
1. Let t > x ≥ t . Then, t − X i =0 (cid:18) xi (cid:19) q < t (cid:18) xt − (cid:19) q = t ( q x − . . . ( q x − t +2 − q t − − . . . ( q − < c ( t ) q x q x − . . . q x − t +2 = c ( t ) q ( t − x − t +2)2 = q ( t − x + c ′ ( t ) , for some constants c ( t ) , c ′ ( t ) that depend on t . The inequality q ( t − x + c ′ ( t ) ≤ q t ( x − t ) holds for any x ≥ t + c ′ ( t ). Thus we can take x large enough to begreater than 2 t and to satisfy the inequality. (cid:3) SERHII DYSHKO
Lemma 4.
There exists m such that Ker E ′ = { } .Proof. For any positive integer m there exists an isomorphism between theposet of subspaces of an m -dimensional vector spaces and the poset of sub-modules of an m -dimensional R -module, see [19]. Therefore, we can useLemma 2 for R -modules.Calculate the cardinalities of sets. From Lemma 2, | S = ℓ | = q l ( m − l ) (cid:18) ll (cid:19) q = q l ( m − l ) . Since S <ℓ ⊆ { U ∈ L ( M ) | dim U < ℓ } , | S <ℓ | ≤ ℓ − X i =0 (cid:18) mi (cid:19) q . From Lemma 3, there exists m such that | S = ℓ | > | S <ℓ | . Thereforedim Q Ker E ′ ≥ dim Q F( S = ℓ , Q ) − dim Q F( S <ℓ , Q ) = | S = ℓ | − | S <ℓ | > . (cid:3) From now we assume that dim M = m is such that Ker E ′ = { } . It ispossible due to Lemma 4. Let 0 = ξ ∈ Ker E ′ ⊆ F( S = ℓ , Q ). Define a map, η ( V ) = (cid:26) ξ ( V ) , if V ∈ S = ℓ ;0 , otherwise . Lemma 5.
The equality E ( η ) = 0 holds.Proof. From Lemma 1, for any V ∈ L ( M ) of dimension ℓ > k ,0 = E ( η V ) = X U ∈L ( M ) η V ( U ) U = V + X U ∈ S <ℓ η V ( U ) U . Then, E ( η ) = X V ∈L ( M ) η ( V ) V = X V ∈ S = ℓ ξ ( V ) V = − X V ∈ S = ℓ ξ ( V ) X U ∈ S <ℓ η V ( U ) U = − X U ∈ S <ℓ X V ∈ S = ℓ ξ ( V ) η V ( U ) U = − X U ∈ S <ℓ E ′ ( ξ )( U ) U = 0 . (cid:3) One can see that we can choose ξ ∈ Ker E ′ to have integer values, bymultiplying by the proper scalar λ ∈ Q , so we assume η also has integervalues.The module of characters W = ˆ M is a right(left) R -module of dimension m . The poset ( L ( M ) , ⊆ ) is isomorphic to the poset ( L ( W ) , ⊆ ). For anymodule V ∈ L ( W ) the dual module V ⊥ is in L ( M ) and vice versa. Definea dual map η ⊥ ∈ F( L ( W ) , Q ) as follows, for any V ∈ L ( W ), η ⊥ ( V ) = η ( V ⊥ ) . HEN THE EXTENSION PROPERTY DOES NOT HOLD 7
Note that the map η ⊥ has only integer values. Lemma 6.
The equality E ( η ⊥ ) = 0 holds.Proof. The Fourier transform is a Q -linear map. Note that since for all V ∈ S = ℓ , dim V = ℓ , | V | = q kℓ and | V ⊥ | = q ( m − ℓ ) k . Calculate, F ( E ( η ⊥ )) = X V ∈L ( W ) η ⊥ ( V ) F ( V ) = X V ⊥ ∈ S = ℓ η ( V ⊥ ) | V | V ⊥ = X U ∈ S = ℓ | U ⊥ | η ( U ) U = q ( m − ℓ ) k X U ∈ S = ℓ η ( U ) U = q ( m − ℓ ) k X U ∈L ( M ) η ( U ) U = q ( m − ℓ ) k E ( η ) . From Lemma 5, E ( η ) = 0, so F ( E ( η ⊥ )) = 0. The Fourier transform isinvertible, thus E ( η ⊥ ) = F − (0) = 0. (cid:3) Define S ⊥ = { V ∈ L ( W ) | V ⊥ ∈ S = ℓ } . For each V ∈ S ⊥ , by duality, V + X ⊥ = W . Recall dim V = m − ℓ and dim X ⊥ = m − ( m − ℓ ) = ℓ , andtherefore V ∩ X ⊥ = { } . Alternatively, S ⊥ can be defined as, S ⊥ = { V ∈ L ( W ) | V ∩ X ⊥ = { } , dim V = m − ℓ } . Note that X V ∈ S ⊥ η ⊥ ( V ) = X V ∈L ( W ) η ⊥ ( V ) V (0) = E ( η ⊥ )(0) = 0 . Proposition 3.
Let R = M k × k ( F q ) and let A be an ℓ -dimensional R -module,where ℓ > k . Then there exist a positive integer n and an R -linear code C ⊂ A n with an unextendable swc G -isometry f ∈ Hom R ( C, A n ) , for any G ≤ Aut R ( A ) .Proof. Use the notation of this section. Since dim X ⊥ = dim A there is amodule isomorphism ψ : X ⊥ → A . Define the length of the code n as thesum of the positive values η ⊥ ( U ), U ∈ S ⊥ . From the calculations above, itis equal to the sum of the negative values η ⊥ ( U ), U ∈ S ⊥ .For any i ∈ { , . . . , n } , let us define λ i ∈ Hom R ( W, A ). Choose a sub-module V i ⊆ W , such that η ⊥ ( V i ) >
0, (do it for the submodule V = V i exactly η ⊥ ( V ) times). Since V i ∈ S ⊥ , V i ∩ X ⊥ = { } , dim V i = m − ℓ anddim X ⊥ = ℓ , and therefore W = V i ⊕ X ⊥ . Define λ i : W = V i ⊕ X ⊥ → A, ( v, x ) ψ ( x ) . Then Ker λ i = V i ⊆ W and for any a ∈ A , λ − i ( a ) = ψ − ( a ) + V i . In thesame way, for any i ∈ { , . . . , n } , define maps µ i : W → A for the modules U i , such that η ⊥ ( U i ) < { e } < Aut R ( A ). The trivialgroup has one-point orbits in A . Calculate, using Lemma 6, for any a ∈ A , SERHII DYSHKO for any w ∈ W , n X i =1 ( λ − i ( { a } ) − µ − i ( { a } ) )( w ) = X U ∈ S ⊥ η ⊥ ( U ) ψ − ( a )+ U ( w )= X U ∈ S ⊥ η ⊥ ( U ) U ( w − ψ − ( a )) = E ( η ⊥ )( w − ψ − ( a )) = 0 . Also, it is easy to see that for any i, j ∈ { , . . . , n } , Ker λ i = Ker µ j .Since eq. (1) holds for the orbit { } , it is true that Ker λ = T ni =1 Ker λ i = T ni =1 Ker µ i = Ker µ = N ⊂ W . Let ¯ λ, ¯ µ be two canonical injective maps¯ λ, ¯ µ ∈ Hom R ( W/N, A n ) such that ¯ λ ( ¯ w ) = λ ( w ) and ¯ µ ( ¯ w ) = µ ( w ) for all w ∈ W , where ¯ w = w + N .Use the notation of Section 3. Define a code C ⊂ A n as the image λ ( W ).Define a map f ∈ Hom R ( C, A n ) as f = ¯ µ ¯ λ − . It is true that f λ = µ . FromProposition 1, f is an swc { e } -isometry. Therefore, f is an swc G -isometry.However, f does not extend even to an Aut R ( A )-monomial map, since eq. (2)does not hold for the orbit { } . (cid:3) Main result
Let R be a finite ring with identity. Recall several results that appearin [18] in order to generalize Proposition 3 for the case of arbitrary modulealphabet. For the finite ring R there is an isomorphism, R/ rad( R ) ∼ = M r × r ( F q ) ⊕ · · · ⊕ M r n × r n ( F q n ) , for nonnegative integers n, r , . . . , r n and prime powers q , . . . , q n , see [18]and [11] (Theorem 3.5 and Theorem 13.1).Consider a finite left R -module A . Since soc( A ) is a sum of simple R -modules, there exist nonnegative integers s , . . . , s n such that,soc( A ) ∼ = s T ⊕ · · · ⊕ s n T n , where T i ∼ = M r i × ( F q i ) is a simple M r i × r i ( F q i )-module, i ∈ { , . . . , n } . Proposition 4 (see [18] (Proposition 5.2)) . The socle soc( A ) is cyclic ifand only if s i ≤ r i , for all i ∈ { , . . . , n } . In [7] (Theorem 3) the authors proved the following.
Theorem 1.
Let R be a finite ring with identity. Let A be a finite R -module with a cyclic socle. The alphabet A has an extension property withrespect to the symmetrized weight composition swc G , built on any subgroup G ≤ Aut R ( A ) . We prove the complementary part.
Theorem 2.
Let R be a finite ring with identity. Let A be a finite R -module with a noncyclic socle. The alphabet A does not have an extensionproperty with respect to the symmetrized weight composition swc G , built onany subgroup G ≤ Aut R ( A ) . HEN THE EXTENSION PROPERTY DOES NOT HOLD 9
Proof.
Our proof repeats the idea of [18] (Theorem 6.4). From Proposition 4,soc( A ) is not cyclic, then there exists an index i with s i > r i . Of course, s i T i ⊆ soc( A ) ⊆ A . Recall that s i T i is the pullback to R of the M r i × r i ( F q )-module B = M r i × s i ( F q ). Denote the ring M r i × r i ( F q ) = R ′ .Because r i < s i , Proposition 3 implies the existence of R ′ -linear code C ⊂ B n and an swc { e } -isometry f ∈ Hom R ′ ( C, B n ) that does not extend toa Aut R ′ ( B )-monomial map.Recall the notation of Section 3. Denote V = Ker λ . Define a subcode C ′ = λ ( V ) ⊆ λ ( W ) = C . The first column of C ′ is a zero-column. Assumethat the code f ( C ′ ) has a zero-column. Then there exists i ∈ { , . . . , n } such that V ⊆ Ker µ i . The code C is constructed from the map η ⊥ , definedin Section 4, so dim V = dim A = dim Ker µ i for all i ∈ { , . . . , n } . Also,Ker λ i = Ker µ j for all i, j ∈ { , . . . , n } . Therefore it is impossible to have V ⊆ Ker µ i for some i ∈ { , . . . , n } and thus f ( C ′ ) does not have a zero-column.The projection mappings R → R/ rad( R ) → R ′ allows us to consider C ′ and f as an R -module and an R -linear homomorphism correspondingly.We have C ′ ⊂ ( s i T i ) n ⊆ soc( A ) n ⊂ A n as R -modules. The map f is thusan swc { e } -isometry of an R -linear code over A . Since { e } ≤ G , obviously, f is an swc G -isometry. The codes C ′ and f ( C ′ ) have different number of zerocolumns and hence f does not extend to an Aut R ( A )-monomial map. (cid:3) Corollary 1.
Let R be a finite ring with identity. Let A be a finite R -module. Let G be a subgroup of Aut R ( A ) . The alphabet A has an extensionproperty with respect to swc G if and only if soc( A ) is cyclic.Proof. See Theorem 1 and Theorem 2. (cid:3)
Let ω : A → C be a function. For a positive integer n define a weightfunction ω : A n → C , ω ( a ) = P ni =1 ω ( a i ). Consider a code C ⊆ A n . We saythat a map f ∈ Hom R ( C, A n ) is an ω -preserving function if for any a ∈ A n , ω ( a ) = ω ( f ( a )).Let U ( ω ) = { g ∈ Aut R ( A ) | ∀ a ∈ A, ω ( g ( a )) = ω ( a ) } be a symmetrygroup of the weight. An alphabet A is said to have an extension propertywith respect to the weight function ω if for any linear code C ⊆ A n , any ω -preserving map f ∈ Hom R ( C, A n ) extends to an U ( ω )-monomial map. Corollary 2.
Let R be a finite ring with identity. Let A be a finite R -modulewith a noncyclic socle. Let ω : A → C be arbitrary weight function. Thealphabet A does not have an extension property with respect to the weight ω .Proof. Let C ⊆ A n be a code and let f ∈ Hom R ( C, A n ) be an unextendableswc U ( ω ) -isometry that exist due to Theorem 2. The map f is then an ω -preserving map and it does not extend to a U ( ω )-monomial map. (cid:3) Remark 3.
The length of the code with an unextendable isometries that isobserved in Theorem 2 and later used in Corollary 2 can be large. For thecode observed in Proposition 3 we can give a lower bound n ≥ Q ki =1 (1 + q i ), which is proved in [6]. In a future paper we will show that for the spacialcase k = 1 we can give an explicit construction for the code observed inProposition 3. The length of the resulting code is q + 1. Moreover theresulting unextendable swc G -isometry can be an automorphism. Remark 4.
Unlike the case of the Hamming weight, for arbitrary weightfunction it is not true that the cyclic socle and pseudo-injectivity conditionslead to the extension property. For example, it is still an open question ifthe extension property holds for the Lee weight and a cyclic group alphabet,see [2]. In a recent work of Langevin, see [13], the extension property for theLee weight with an alphabet that is a cyclic group of prime order is proved.
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