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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.
JW (WMU) Frobenius rings October 9, 2012 3 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 4 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 5 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 6 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 7 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 8 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 9 / 28
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).
JW (WMU) Frobenius rings October 9, 2012 10 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 11 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 12 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 13 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 14 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 15 / 28
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
).
JW (WMU) Frobenius rings October 9, 2012 16 / 28
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).
JW (WMU) Frobenius rings October 9, 2012 17 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 18 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 19 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 20 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 21 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 22 / 28
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 .
JW (WMU) Frobenius rings October 9, 2012 23 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 24 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 25 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 26 / 28
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.
JW (WMU) Frobenius rings October 9, 2012 27 / 28
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.
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