Finite Frobenius Rings and the MacWilliams Identities

Transcription

1 Finite Frobenius Rings and the MacWilliams Identities Jay A. Wood Department of Mathematics Western Michigan University jwood/ Algebra and Communications Seminar University College Dublin November 14, 2011

2 Finite Frobenius Rings (from last week) Finite ring R with 1. The (Jacobson) radical Rad(R) of R is the intersection of all maximal left ideals of R; Rad(R) is a two-sided ideal of R. The (left/right) socle Soc(R) of R is the ideal of R generated by all the simple left/right ideals of R. R is Frobenius if R/ Rad(R) = Soc(R) as one-sided modules (both left and right). JW (WMU) MacWilliams Identities November 14, / 26

3 Two Useful Theorems About Finite Frobenius Rings (also, last week) (Honold, 2001) R/ Rad(R) = Soc( R R) as left modules iff R/ Rad(R) = Soc(R R ) as right modules. R is Frobenius iff R = R as left modules iff R = R as right modules (Hirano, 1997; indep. 1999). Corollary: R is Frobenius iff there exists a character π of R such that ker π contains no nonzero left (right) ideal of R. This π is a generating character. JW (WMU) MacWilliams Identities November 14, / 26

4 Characters on a Finite Ring Let R be a finite ring, with 1. A character of R is a homomorphism π : (R, +) (C, ). The set R of all characters of R is an (R, R)-bimodule with scalar multiplications for π R, r, x R. ( r π)(x) = π(xr), (π r )(x) = π(rx), JW (WMU) MacWilliams Identities November 14, / 26

5 Generating Characters For any character π R, there are two homomorphisms R R: r r π, r π r. The first is left linear; the second is right linear. A character π is a left (right) generating character if the first (second) map is surjective. JW (WMU) MacWilliams Identities November 14, / 26

6 Equivalent Conditions Remember R = R. A map R R is surjective iff it is injective iff it is bijective. The first map r r π is injective iff ker π contains no nonzero left ideal of R. The second map r π r is injective iff ker π contains no nonzero right ideal of R. JW (WMU) MacWilliams Identities November 14, / 26

7 Left-Right Symmetry A character χ is left generating iff it is right generating. Suppose χ is right generating and that Rx ker χ. χ(rx) = 1 for all r R; χ r (x) = 1 for all r R. Thus x is annihilated by every character of R, implying x = 0. JW (WMU) MacWilliams Identities November 14, / 26

8 Generating Characters and Frobenius Rings Theorem. R is Frobenius iff R admits a generating character. All isomorphisms below are as one-sided modules. Fact: (R/ Rad(R)) = Soc( R). Matrix fact: (R/ Rad(R)) = R/ Rad(R). If R = R (existence of generating character), then R/ Rad(R) = Soc(R), and R is Frobenius. JW (WMU) MacWilliams Identities November 14, / 26

9 Producing a Generating Character The converse remains: if R is Frobenius, how to produce a generating character? Start with Soc(R) = R/ Rad(R), which is a sum of square matrix modules. Suppose θ is a character of Soc(R) such that ker θ contains no nonzero left submodule of Soc(R). Take any extension of θ to a character χ of R. Theorem: χ is a generating character of R. JW (WMU) MacWilliams Identities November 14, / 26

10 Some Details Why does an extension exist? 0 Soc(R) R R/ Soc(R) 0 induces 0 (R/ Soc(R)) R (Soc(R)) 0. Suppose I is a left ideal with I ker χ. Then Soc(I ) Soc(R) ker χ = ker θ, because χ = θ on Soc(R). By hypothesis on θ, Soc(I ) = 0. Thus I = 0. JW (WMU) MacWilliams Identities November 14, / 26

11 Characters on Matrix Modules Let R = M m (F q ) and M = M m k (F q ); q = p e. Theorem. M admits a character θ such that ker θ contains no nonzero left R-submodule iff m k. R itself has generating character: χ(p) = exp(2πi Tr q,p (Tr P)/p), where Tr is the matrix trace and Tr q,p : F q F p is the field trace, Tr q,p (x) = x + x p + x p2 + + x pe 1 for x F q. For m k, embed M into R and restrict χ to M. Failure when m < k is an exercise in linear algebra. JW (WMU) MacWilliams Identities November 14, / 26

12 Final Argument If R is Frobenius, then Soc(R) = R/ Rad(R) is a sum of square matrix modules. Each of these matrix modules admits a character with no nonzero left submodules in its kernel. The product of these characters is a character of Soc(R) with no nonzero submodules in its kernel. Extend this character (in any way) to a character of R, and the extension is a generating character of R JW (WMU) MacWilliams Identities November 14, / 26

13 Examples Finite fields F q : χ(x) = exp(2πi Tr q,p (x)/p). Z/nZ: χ(x) = exp(2πix/n). Galois rings (Galois extensions of Z/p m Z). Soc(R) = F p m; use χ for F p m and extend. Finite chain rings (all ideals form a chain). Extend from Soc(R) = F q. Products of Frobenius rings. Use product of the generating characters. Matrix rings over a Frobenius ring: M n (R). χ = χ R Tr. Finite group rings over a Frobenius ring: R[G]. χ( a g g) = χ R (a e ), e is identity element. JW (WMU) MacWilliams Identities November 14, / 26

14 MacWilliams Identities The MacWilliams identities relate the weight enumerator of a linear code to that of its dual code, via a linear change of variables. The identities date from 1962 in the doctoral dissertation of MacWilliams. The proof given here is modeled on a proof due to Gleason over finite fields. The proof uses the Poisson summation formula for the Fourier transform. JW (WMU) MacWilliams Identities November 14, / 26

15 Definitions Let R be a finite Frobenius ring. A linear code over R of length n is a left R-submodule C R n. The Hamming weight wt(x) = {i : x i 0} for x = (x 1, x 2,..., x n ) R n. The Hamming weight enumerator of C R n is W C (X, Y ) = x C X n wt(x) Y wt(x) = n A i X n i Y i, i=0 where A i is the number of x C of weight i. JW (WMU) MacWilliams Identities November 14, / 26

16 Annihilators The dot product on R n is x y = n x i y i R, i=1 for x = (x 1,..., x n ), y = (y 1,..., y n ) R n. Define the right annihilator of C R n by r(c) = {y R n : C y = 0}. JW (WMU) MacWilliams Identities November 14, / 26

17 MacWilliams Identities over Finite Frobenius Rings Theorem (1999) Let R be a finite Frobenius ring. If C R n is a left linear code, then the MacWilliams identities hold: W C (X, Y ) = 1 r(c) W r(c)(x + ( R 1)Y, X Y ). JW (WMU) MacWilliams Identities November 14, / 26

18 Characters of Finite Abelian Groups Let (G, +) be a finite abelian group. A character π of G is a group homomorphism π : (G, +) (C, ), where (C, ) is the multiplicative group of nonzero complex numbers. The set Ĝ of all characters of G is itself a finite abelian group called the character group. Ĝ = G. JW (WMU) MacWilliams Identities November 14, / 26

19 Two Useful Formulas π(x) = x G π(x) = π Ĝ { G, π = 1, 0, π 1. { G, x = 0, 0, x 0. JW (WMU) MacWilliams Identities November 14, / 26

20 Fourier Transform Given a function f : G V, with V a complex vector space, its Fourier transform is a function ˆf : Ĝ V defined by ˆf (π) = x G π(x)f (x), π Ĝ. Fourier inversion: f (x) = 1 G π( x)ˆf (π), x G. π Ĝ JW (WMU) MacWilliams Identities November 14, / 26

21 Poisson Summation Formula For a subgroup H G, define its annihilator (Ĝ : H) = {π Ĝ : π(h) = 1}. (Ĝ : H) = G / H. For a subgroup H G and any a G, 1 f (a + h) = h H (Ĝ : H) π( a)ˆf (π). π (Ĝ:H) In particular, for a subgroup H G, 1 f (h) = ˆf h H (Ĝ : H) (π). π (Ĝ:H) JW (WMU) MacWilliams Identities November 14, / 26

22 Proof of the MacWilliams Identities (a) The proof follows a proof due to Gleason (1970). Let R be Frobenius with generating character χ. Let G = R n, an abelian group under addition. Let H = C, a left linear code. Let V = C[X, Y ], a complex vector space. Let f : G V be f (x) = X n wt(x) Y wt(x). JW (WMU) MacWilliams Identities November 14, / 26

23 Proof of the MacWilliams Identities (b) By Frobenius hypothesis, every character of G = R n has the form π a, for some a R n, with π a (x) = χ(x a), x R n. π a (Ĝ : H) if and only if a r(c). (Ĝ : H) = r(c). JW (WMU) MacWilliams Identities November 14, / 26

24 Proof of the MacWilliams Identities (c) For f (x) = X n wt(x) Y wt(x), ˆf (π a ) = (X + ( R 1)Y ) n wt(a) (X Y ) wt(a). This requires some manipulations and use of π(x) formulas. (Next slide.) Recognize ˆf (π a ) as summand of W r(c) (X + ( R 1)Y, X Y ). JW (WMU) MacWilliams Identities November 14, / 26

25 Idea of Manipulation Let n = 1, f (x) = X 1 wt(x) Y wt(x). ˆf (π a ) = x R π a (x)x 1 wt(x) Y wt(x) = X + x 0 π a (x)y = { X + ( R 1)Y, a = 0, X Y, a 0, = (X + ( R 1)Y ) 1 wt(a) (X Y ) wt(a) JW (WMU) MacWilliams Identities November 14, / 26

26 References These slides and other papers are available on the web: http : //homepages.wmich.edu/ jwood Many references and generalizations (complete weight enumerators, additive codes, etc.) in the paper Foundations of Linear Codes... JW (WMU) MacWilliams Identities November 14, / 26