Characters and finite Frobenius ringshomepages.wmich.edu/~jwood/eprints/ASReCoM-3.pdfQuasi-Frobenius rings I Let R be a nite ring with 1. I R is quasi-Frobenius (QF) if R is an injective

Characters and finite Frobenius rings

Jay A. Wood

Department of MathematicsWestern Michigan University

http://homepages.wmich.edu/∼jwood/

Algebra for Secure and Reliable CommunicationsModeling

Morelia, Michoacan, MexicoOctober 9, 2012

Motivation

I Recall from my last lecture that the MacWilliamsidentities for the Hamming weight hold for any finitering R that satisfies

R ∼= R

as one-sided R-modules.

I The extension theorem for Hamming weight willalso hold for such rings (later).

JW (WMU) Frobenius rings October 9, 2012 2 / 28

Recall definitions

I Let R be a finite associative ring with 1.

I Recall: the Jacobson radical J is the intersection ofall the maximal left ideals of R ; J is a two-sidedideal.

I Recall that the left socle Soc(RR) is the left idealgenerated by all the simple left ideals of R .

I Similarly, for the right socle Soc(RR).

I Each Soc(R) is a two-sided ideal.


Finite Frobenius rings

I A finite ring R is Frobenius if R(R/J) ∼= Soc(RR)and (R/J)R

∼= Soc(RR).

I It is a theorem of Honold, 2001, that each of theseisomorphisms implies the other.

I From my first lecture: Fq and Z/mZ are Frobenius.Klemm’s example F2[X ,Y ]/(X 2,XY ,Y 2) is notFrobenius.


Character modules

I Suppose M is a finite left R-module.

I The character group M admits the structure of aright R-module via

(πr)(m) := π(rm), r ∈ R ,m ∈ M , π ∈ M .

I Similarly, if N is a right R-module, then N is a leftR-module.


Finite fields

I Consider a finite field Fq.

I Fq is an Fq-vector space.

I Since |Fq| = |Fq|, Fq has dimension 1, and Fq∼= Fq

as Fq-vector spaces.

I The character θq = θp ◦ Trq/p is a basis.


Matrix modules

I R = Mn(Fq) is the ring of n × n matrices over Fq.

I Let M = Mn×k(Fq) and N = Mk×n(Fq); M is a leftR-module, and N is a right R-module.

I Define a character on R : ρ = θq ◦ Tr, where Tr isthe matrix trace.

I M ∼= N via P 7→ (Q 7→ ρ(PQ)).

I N ∼= M via Q 7→ (P 7→ ρ(PQ)).

I In particular, R ∼= R as left and as right modules.


Main theorem

TheoremLet R be a finite ring with 1. The following areequivalent:

1. R is Frobenius;

2. R ∼= R as left R-modules;

3. R ∼= R as right R-modules.

I Due independently to Hirano, 1997, and Wood,1999.


Short exact sequence

I is an exact contravariant functor on R-modules.

I A short exact sequence of left R-modules

0→ M1 → M2 → M3 → 0

induces a short exact sequence of right R-modules

0→ (M2 : M1)→ M2 → M1 → 0.

I Similarly with left-right reversed.


Socles of character modules

TheoremLet M be a finite left R-module. Then

Soc(M) ∼= (M/JM) .I J annihilates simple modules, so that

Soc(M) = (M : JM).

I Use (M/JM) ∼= (M : JM).


One direction

I Suppose R ∼= R as right R-modules.

I R/J is a sum of matrix rings, so

(R/J)R∼= (R(R/J)) .

I Use M = R (as left R-module) from Theorem.

I Then (R(R/J)) ∼= Soc(RR) ∼= Soc(RR).

I Repeat on other side, or use Honold’s theorem.


Generating characters (a)

I Let M be a finite left R-module.

I A character ρ of M is a (left) generating character ifker ρ contains no nonzero left submodules of M .

I Similarly for right modules.


Generating characters (b)

TheoremM has a left generating character iff M injects into R.

I If f : M ↪→ R , set ρ(m) = f (m)(1R).

I If ρ is a generating character, definef (m) = (r 7→ ρ(rm)).

I If m ∈ ker f , then Rm ⊂ ker ρ.

I Because |R | = |R |, R ∼= R as left modules iff R hasa left generating character. Same for right.


Left generating iff right generating

TheoremLet ρ be a character of R. Then ρ is left generating iff ρis right generating.

I If ρ is right generating, then RR∼= RR , so every

π ∈ R has the form ρr for some r ∈ R .

I If Ra ⊂ ker ρ, then for all r ∈ R ,1 = ρ(ra) = (ρr)(a). Thus π(a) = 1 for all π ∈ R .This implies a = 0, and ρ is left generating.


Simple R-modules

I For any finite ring R , R/J is a sum of matrix rings:

R/J ∼=k⊕

i=1

Mµi(Fqi

).

I Let Ti = Mµi×1(Fqi). Ti is a simple left

Mµi(Fqi

)-module and a simple left R-module.

I Fact: the Ti are the only simple left R-modules, upto isomorphism.


Structure of R/J and Soc(R)

I As a left R-module, R(R/J) ∼= ⊕ki=1µiTi .

I Because Soc(RR) is generated by simple modules,Soc(RR) ∼= ⊕k

i=1siTi , for some si , nonnegativeintegers.

I Thus, R is Frobenius iff µi = si for all i = 1, . . . , k .

I In general, Soc(RR) is a sum of matrix modulesMµi×si (Fqi

).


Generating characters for Soc(R)

I If R is Frobenius, then Soc(RR) ∼= ⊕Mµi(Fqi

).

I We saw earlier that Mµi(Fqi

) admits a leftgenerating character θi .

I The product of the θi is a left generating characterof Soc(RR).


Extending generating characters (a)

TheoremLet M be a finite left R-module. If Soc(M) admits a leftgenerating character θ, then θ extends to a leftgenerating character of M.

I 0→ Soc(M)→ M → M/ Soc(M)→ 0, induces

0→ (M : Soc(M))→ M → Soc(M) → 0.

I Let ρ be any extension of θ.


Extending generating characters (b)

I Claim: ρ is a left generating character of M .

I Suppose I is a left submodule of ker ρ. Then

Soc(I ) ⊂ Soc(M) ∩ ker ρ = Soc(M) ∩ ker θ.

I Since θ is a left generating character, Soc(I ) = 0.

I Thus, I = 0.


Other direction

I Suppose R is Frobenius.

I Soc(RR) ∼= ⊕Mµi(Fqi

) admits a left generatingcharacter θ.

I Any extension ρ of θ is a left generating characterof R

I Thus R ∼= R as left R-modules.

I Any left generating character is also rightgenerating, so R ∼= R as right R-modules.


Examples of Frobenius rings

I Fq, ρ = θq.

I Z/mZ, ρ(a) = exp(2πia/m).I Chain rings: all the left ideals form a chain under

inclusion. Soc(R) ∼= Fq. Extend θq on Soc(R) to ρon R . Examples of chain rings:

I Any finite commutative local ring with principal maximalideal.

I Z/pkZ.I Galois rings: Galois extensions of Z/pkZ.I Fq[X ]/(X k).I Certain quotients of skew polynomial rings.


More examples

I Mn(Fq), ρ = θq ◦ Tr.

I Mn(R), where R is Frobenius. ρ = ρR ◦ Tr.

I R[G ], the group ring of a finite group G withcoefficients in a Frobenius ring R . Every element ofR[G ] is of the form a =

∑g∈G agg , with ag ∈ R .

ρ(a) = ρR(ae), where e is the identity of G .

I (Algebraic topology) Certain finite subalgebras ofthe Steenrod algebra.


Commutative case

I Every finite commutative ring R splits as a sum oflocal rings (Ri ,mi), where mi is the unique maximalideal of Ri . R is Frobenius iff each Ri is Frobenius.ρ =

∏ρi .

I A local commutative ring (Ri ,mi) is Frobenius iffSoc(Ri) = ann(mi) has dimension 1 overFq∼= Ri/mi . Extend θq on Soc(Ri) to ρi on Ri .


Quasi-Frobenius rings

I Let R be a finite ring with 1.

I R is quasi-Frobenius (QF) if R is an injective left(or right) R-module.

I That is, for every short exact sequence of R-modules

0→ A→ B → C → 0,

we have a short exact sequence

0→ HomR(C ,R)→ HomR(B ,R)→ HomR(A,R)→ 0.


Frobenius implies QF

I In general, M = HomZ(M ,C×) ∼= HomR(M , R), via

π ∈ M 7→ (m 7→ (r 7→ π(rm))).

I is an exact functor represented by R , so R isalways injective.

I If R is Frobenius, then R ∼= R . Thus R is injective,hence QF.


Benson’s example

I Let R be a ring consisting of all matrices over F2 ofthe following form:

a1 0 a2 0 0 00 a1 0 a2 a3 0

a4 0 a5 0 0 00 a4 0 a5 a6 00 0 0 0 a9 0

a7 0 a8 0 0 a9

.

I This R is QF but not Frobenius.


Role of QF rings in duality

I Recall the dot product on Rn: a · b =∑

aibi .

I For RC ⊂ Rn and DR ⊂ Rn, recall the annihilators

l(D) := {b ∈ Rn : b · d = 0, d ∈ D},r(C ) := {b ∈ Rn : a · b = 0, a ∈ C}.

I For all C ,D, l(r(C )) = C and r(l(D)) = D iff R isQF.


Role of Frobenius rings in duality

I The MacWilliams identities are true over Frobeniusrings:

WC (X ,Y ) =1

|r(C )|Wr(C )(X + (|R | − 1)Y ,X − Y ).

I Setting X = Y = 1, yields |C ||r(C )| = |R |n, for RFrobenius.

I If R is QF but not Frobenius, there exists a leftideal I ⊂ R with |I ||r(I )| < |R |.

I The MacWilliams identities (in standard form)cannot hold over a non-Frobenius ring.


Documents

Characters and finite Frobenius ringshomepages.wmich.edu/~jwood/eprints/ASReCoM-3.pdfQuasi-Frobenius rings I Let R be a nite ring with 1. I R is quasi-Frobenius (QF) if R is an injective