Finite Frobenius Rings and
Their Applications
Xiang-dong Hou
Department of Mathematics
University of South Florida
May 16, 2007
Lecture 1
Quasi-Frobenius Rings and Frobenius Rings
• Quasi-Frobenius Rings
• Frobenius Rings
• The Character Module
• The Extension Property
1
Quasi-Frobenius Rings
Quasi-Frobenius rings. A left noetherian ring
R is called quasi-Frobenius (QF) if R R is an injective module. Recall that an R-module R A is
called injective if for every injection i : A → B
and R-map f : A → E, there exists an R-map
g : B → E such that
i
0 ....................... A ....................... B
...
..
.
..
.
..g
.
f ....
.
... ....
. ..
E
commutes.
2
Left and right annihilators.
Let R be a ring and S ⊂ R. The left and right
annihilators of S are respectively
l(S) = {x ∈ R : xs = 0 ∀s ∈ S},
r(S) = {x ∈ R : sx = 0 ∀s ∈ S}.
Theorem 1 R is QF ⇔ R is left noetherian
and for every left ideal A and right ideal B of
R, l(r(A)) = A and r(l(B)) = B.
Facts.
(i) The definition of QF rings and the statement of Theorem 1 are left-right symmetric.
(ii) A QF ring is artinian (left and right).
Theorem 2 A ring R is QF iff projective Rmodules are precisely injective R-modules.
3
Proposition 3 Let R be a right artinian ring
and J = rad R. Let 1R = e1 + · · · + en be a
decomposition of 1R into orthogonal primitive
idempotents.
(i) Every non nilpotent right ideal I of R contains an idempotent.
(ii) Every right ideal properly contained in eiR
is nilpotent.
(iii) eiR/eiJ is a simple R-module.
(iv) Let M be a simple right R-module. Then
∼ e R/e J.
M ei 6= 0 ⇔ M =
i
i
4
Proposition 4 let R be a QF ring and let e be
a primitive idempotent of R. Let J = rad R.
Then l(J)e is the unique simple submodule of
Re and Re is the injective hull of l(J)e.
Proof. 1◦ l(J)e is simple. We have
l(J)e = l(eJ) ∩ Re = l(eJ) ∩ l((1 − e)R)
= l(eJ + (1 − e)R).
Since
R
eR ⊕ (1 − e)R ∼
=
= eR/eJ
eJ + (1 − e)R
eJ ⊕ (1 − e)R
is simple, eJ + (1 − e)R is a maximal right ideal
of R. Since R is QF, l(eJ + (1 − e)R) is a
minimal left ideal of R.
2◦ Re is the injective hull of l(J)e. Since Re is
projective and R is QF, Re is injective. Thus
Re contains an injective hull H of l(J)e. Since
H is injective, Re = H ⊕ H 0 for some submodules H 0 ⊂ Re. Since Re is indecomposable,
Re = H.
5
3◦ l(J)e is the unique simple submodule of Re.
It suffices to show that Re is essential over
l(J)e, i.e., l(J)e has a nonzero intersection with
any nonzero submodule of Re. Since Re is
injective, Re contains a maximal essential extension M of l(J)e ([Lam98, Proof of Lemma
3.29]). It suffices to show M = Re. By Zorn’s
lemma, ∃ submodules A ⊂ Re maximal with respect to the property that A ∩ M = 0. Then in
Re/A, any nonzero submodule B/A intersects
A ⊕ M/A nontrivially. So Re/A is essential over
∼ M . By the maximality of M , we
A ⊕ M/A =
have Re/A = A ⊕ M/A, i.e., Re = A ⊕ M . Thus
M is injective. By 2◦, M = Re.
Frobenius Rings
Nakayama permutation and Frobenius rings.
Let R be a QF ring and let
1R = e11 + · · · + e1m1 + · · · + et1 + · · · + etmt (1)
be a decomposition of 1R into orthogonal prim∼ e iff i = k. (If e
itive idempotents, where eij =
kl
∼ f means
and f are idempotents of a ring R, e =
∼ f R, which is equivalent to Re =
∼ Rf .
that eR =
Cf. [Lam91, Prop. 21.20].) Put ei = ei1,
1 ≤ i ≤ t. By Proposition 3 (iv) and Proposition 4, ∃ a function π : {1, . . . , t} → {1, . . . , t}
such that
∼ Re
l(J)ei =
π(i) /Jeπ(i) ,
1 ≤ ni ≤ t.
∼ l(J)e , hence
If π(i) = π(k), then l(J)ei =
k
∼
Rei = E(l(J)ei) = E(l(J)ek ) = Rek . So i = k.
Thus π is a permutation of {1, . . . , t}; it is
6
called the (left) Nakayama permutation of R.
We have
eiR/eiJ ,→ (eiR/eiJ)∗∗
∗
∼
(note below)
= l(J)ei
∼ (Re
/Je
)∗ (note below)
=
π(i)
∼e
=
π(i) r(J).
π(i)
So,
∼e
eiR/eiJ =
π(i) r(J).
(2)
Therefore, the right Nakayama permutation of
R is π −1. R is called Frobenius if mi = mπ(i),
1 ≤ i ≤ t.
Note. A∗ = HomR (A, R) is the dual module of
A. If R is QF and e is a primitive idempotent
of R, then
∼ e r(J),
(Re/Je)∗ =
∼ l(J)e.
(eR/eJ)∗ =
Theorem 5 Let R be an artinian ring and let
R̄ = R/rad R. Then the following statements
are equivalent
(i) R is Frobenius.
∼ R̄.
(ii) R is QF and soc(R R) =
R
∼ R̄ .
(iii) R is QF and soc(RR ) =
R
∼ R̄ and soc(R ) =
∼ R̄ .
(iv) soc(R R) =
R
R
R
∼ R̄
Note. If |R| < ∞, then (iv) ⇔ soc(R R) =
R
∼ R̄ , see [Hon01].
or soc(RR ) =
R
The proof uses the following
Proposition 6 Let R be an artinian ring. Then
R is QF ⇔ the dual of every simple left or right
R-module is simple.
7
The Character Module
The character module. Let R A be a left Rmodule and let  = HomZ(R AZ, C×
Z ). Then Â
is a right R-module. (∀f ∈ Â and r ∈ R, f r ∈ Â
is defined by (f r)(x) = f (rx) ∀x ∈ R.) Â is
called the character module of A.
If φ ∈ HomR (R A, R B), define
φ̂ : B̂ −→
Â
f 7−→ f ◦ φ.
Then φ̂ ∈ HomR (B̂R , ÂR ). Moreover, (ˆ) : R M →
MR is a contravariant functor.
Similarly, if AR is a right R-modules, the character module of A is defined to be  =
HomZ(ZAR , ZC×), which is a left R-module. If
φ ∈ HomR (AR , BR ), also define φ̂ : B̂ → Â by
φ(f ) = f ◦ φ. Then φ̂ ∈ HomR (R B̂, R Â) and
(ˆ) : MR → R M is a contravariant functor.
8
The character group. Let A be an abelian
group (Z-module). Then  = HomZ(A, C×)
is the character group of A. When |A| < ∞,
∼ Â (non canonically).
A=
The character module of a finite module.
If R A is a finite left R-module, then for each
f ∈ HomZ(R AZ, C×
Z ), ∃!g ∈ HomZ(R AZ, (Q/Z)Z)
such that
f = α ◦ g,
where α : Q/Z → C×, r + Z 7→ er2πi. Thus the
character module of R A can also be defined
as A[ = HomZ(R AZ, (Q/Z)Z). (If AR is a finite
right R-module, then A[ = HomZ(ZAR , Z(Q/Z)).)
Let R M0 (M0
R ) denote the category of finite
left (right) R-modules. Then ( )[ : R M0 → M0
R
[
0
0
and ( ) : MR → R M are contravariant functors.
9
Fact. ( )[[ : R M0 → R M0 is naturally equivalent to the identity functor of R M0. The same
0.
is true for ( )[[ : M0
→
M
R
R
Proposition 7 Let R be a finite ring, J =
rad R, and e a primitive idempotent of R. Then
∼ eR/eJ.
(Re/Je)[ =
(3)
Generating characters. Let R be a finite
ring. The character group
(R, +)[ = HomZ(R, Q/Z)
is an (R, R)-bimodule since R is an (R, R)bimodule. A character χ ∈ (R, +)[ is called
a generating character of R if (R, +)[ = Rχ.
10
Proposition 8 Let R be a finite ring and χ ∈
(R, +)[. The following statements are equivalent.
(i) χ is a generating character of R.
(ii) The map (R, +) → (R, +)[, a 7→ χ(· a) = aχ
is a left R-module isomorphism.
(iii) ker χ does not contain any nonzero left
ideal.
Proposition 9 Let R be a finite ring and χ a
generating character of R. Then there is an
automorphism (the Nakayama automorphism)
η of R such that
χ(yx) = χ xη(y)
∀x, y ∈ R.
(4)
11
Corollary 10 The definition of generating characters and the statements in Proposition 8 are
left-right symmetric.
Theorem 11 (J. Wood 99) Let R be a finite ring. Then the following statements are
equivalent.
(i) R is Frobenius.
(ii) R has a generating character χ.
∼R .
(iii) (R R)[ =
R
∼ R.
(iv) (RR )[ =
R
12
Proof. 1◦ (iii) or (iv) ⇒ (i). Obvious since R R
and RR are cyclic R-modules.
2◦ (ii) ⇒ (iii) and (iv). Define
φ : R R −→ (RR )[
a 7−→ χ(· a),
ψ : RR −→ (R R)[
a 7−→ χ(a ·).
Then φ and ψ are isomorphisms.
3◦ (ii) ⇒ R is QF. Let L be a left ideal of
R. Since χ is a generating character of R,
the map (R, +) → (R, +)[, a 7→ χ(· a) is onto.
Since the restriction map (R, +)[ → L[ is also
onto, f : R → L[, a 7→ χ(· a)|L, is an onto homomorphism of abelian groups. Since ker f =
∼ L[. In particular,
r(L), we have R/r(L) =
|R|
|R/r(L)| = |L[| = |L|, i.e., |r(L)| = |L| . In
|R|
the same way, |l(J)| = |J| for every right ideal
|R|
J of R. Thus |r(l(J))| = |l(J)| = |J|. Since
J ⊂ r(l(J)), we have r(l(J)) = J. In the same
way, l(r(L)) = L. Therefore R is QF.
13
4◦ (i) ⇔ (iii). (By symmetry, we have (i) ⇔ (iv).) By
1◦ and 3◦ , we may assume that R is already QF. Let
1R = e11 + · · · + e1m1 + · · · + et1 + · · · + etmt
be a decomposition of 1R into orthogonal primitive idem∼
potents, where
L eij = ekl iff i = k. Put ei = ei1, 1 ≤ i ≤ t.
Then R R = i mi · Rei, so
M
[ ∼
mi · (Rei)[ .
(R R) =
i
Since Rei is indecomposable, so is (Rei)[ . Since Q/Z
is an injective Z-module, (R R)[ = HomZ(R RZ, (Q/Z)Z)
is an injective R-module. So (Rei)[ is injective. From
Rei Rei/Jei, we have
∼ e R/e J,
(Rei)[ (Rei/Jei)[ =
i
i
where the isomorphism is from Proposition 7. Thus
∼ the injective hull of e R/e J. By (2), e R/e J =
∼
(Rei)[ =
i
i
i
i
eπ(i) r(J), where π is the Nakayama permutation of R,
and by Proposition 4, eπ(i) R is the injective hull of eπ(i) r(J).
So
∼ e R.
(Rei)[ =
π(i)
Thus
M
M
M
[ ∼
[ ∼
( R) =
m · (Re ) =
m ·e R=
m −1 · e R.
i
R
i
i
i
Meanwhile, RR =
i
L
i mi
π(i)
π (i)
i
· eiR. So
∼ R ⇔ m −1 = m ∀i ⇔ R is Frobenuius.
(R R)[ =
i
R
π (i)
14
i
The Extension Property
Codes over modules. Let R be a finite ring
and R A a finite R-module. For a = (a1, . . . , an) ∈
An, the Hamming weight of a is defined to be
|a| = |{i : 1 ≤ i ≤ n, ai 6= 0}|. An R-submodule
of An is called a linear code of length n over
A. A monomial transformation of An is an Rautomorphism An → An of the form
f (a1, . . . , an) = φ1(aρ(1)), . . . , φn(aρ(n)) ,
(a1, . . . , an) ∈ An,
where ρ is a permutation of {1, . . . , n} and
φ1, . . . , φn ∈ AutR (A). Monomial transformations of An are precisely the automorphism
which preserves the Hamming weight; they are
the isometries of An with respect to the Hamming distance.
15
More generally, let C1, C2 ⊂ An be two linear codes. An R-isomorphism g : C1 → C2
which preserves the Hamming weight is called
an isometry from C1 to C2. If g can be extended to an isometry of An, C1 and C2 are
called equivalent. A module R A is said to have
the extension property if every isometry between two codes in An can be extended to an
isometry of An. A finite ring R is said to have
the extension property if R R has the property.
The MacWilliams extension theorem in coding
theory states that finite fields have the extension property.
16
Lemma 12 (Bass) Let R be a left or right
artinian ring.
(i) If A is a left ideal of R and r ∈ R such that
A + Rr = R, then A + r contains a unit of
R.
(ii) If B is a right ideal of R and s ∈ R such
that A+sR = R, then B +s contains a unit
of R.
A partial order. Let R be a ring and MR a
right R-module. For x, y ∈ M , define x ≤ y
if xR ⊂ yR; define x ≈ y if x = yu for some
u ∈ R×. Then ≤ is a transitive relation and ≈
is an equivalence relation on M . Moreover, ≤
induces a transitive relation, also denoted by
≤, on M/ ≈.
Corollary 13 If R is left or right artinian, then
the above relation ≤ on M/ ≈ is a partial order.
17
Cauchy binomial theorem.
n
Y
(1 + xq j ) =
j=0
n h i
X
n
j=0 j q
j
(
q 2) xj .
Theorem 14 (J. Wood 07) Let R be a finite ring. Then R is Frobenius ⇔ R has the
extension property.
proof (⇒) Let C ⊂ Rn be a linear code and let f =
(f1 , . . . , fn) : C → Rn be a weight preserving R-map. We
want to show that f can be extended to an isometry of
Rn. Let πi : Rn → R be the ith projection.
Let χ be a generating character of R. ∀a ∈ R, we have
X
|R| if a = 0,
e2πiχ(ar) =
0
if a =
6 0.
r∈R
So, ∀x ∈ C,
n X
X
j=1 r∈R
e2πiχ(fj (x)r) = |R| |f (x)| = |R| |x| =
n X
X
e2πiχ(πj (x)r) .
j=1 r∈R
(5)
e2πiχ(fj ( )r)
e2πiχ(πj ( )r)
For each 1 ≤ j ≤ n and r ∈ R,
and
are characters of C. By the orthogonality relation of
characters, the terms in the sum at the sum at the LHS
18
of (5), after a suitable ordering, are the same as the
terms in the sum at the RHS of (5).
The dual module C ∗ as a right R-module has a transitive
relation ≤. Among f1 , . . . , fn, π1 , · · · , πn ∈ C ∗ , choose one
say f1 maximal with respect to ≤. From the above, we
have e2πiχ(f1 ( )) = e2πiχ(πj1 ( )r1 ) for some 1 ≤ j1 ≤ n and
r1 ∈ R. Then χ(f1 ( )) = χ(πj1 ( )r1 ). Thus im(f1 −
πj1 r1 ) ⊂ ker χ. Since χ is a generating character, we
have f1 = πj1 r1 . So f1 ≤ πj1 . The maximality of f1
implies that πj1 ≤ f1 . By Corollary 13, f1 = πj1 u1 for
some u1 ∈ R× . Since
n
X
e2πiχ(f1 (x)r) =
n
X
e2πiχ(πj1 (x)u1 r) =
e2πiχ(πj1 (x)r) ,
r∈R
r∈R
r∈R
n
X
Equation (5) is reduced to
n
X X
2≤j≤n r∈R
e2πiχ(fj (x)r) =
n
X X
e2πiχ(πj (x)r) .
(6)
1≤j≤n r∈R
j6=j1
By induction, ∃ a permutation ρ of {1, . . . , n} and u1 , . . . , un ∈
R× such that fi = πρ(i) ui, 1 ≤ i ≤ n. Now (πρ(1) u1 , . . . , πρ(n) un)
is an isometry of Rn and an extension of f .
(⇐) Assume to that R is not Frobenius. We want to
show that R does not have the extension property.
1◦ Write
R/J = Mn1 (Fq1 ) × · · · × Mnk (Fqk ).
Let Si = Mni ×1 (Fqi ) viewed as a left Mni (Fqi )-module; Si
is the unique simple Mni (Fqi )-module up to isomorphism.
By an observation of Dinh and López-Permouth, we may
assume that (n1 + 1)S1 ,→ R R. Put n = n1 and q = q1 .
∼M
Then (n + 1)S1 =
n×(n+1) (Fq ) as Mn (Fq )-module.
2◦ Let
N =
n
Y
i=1
i
(1+q ) =
X
0≤i≤n+1
i even
hn + 1 i
i
i
q ( 2) =
q
X
0≤i≤n+1
i odd
hn + 1 i
i
i
q (2) .
q
Define two N -tuples v+ , v− ∈ Mn+1 (Fq )N as follows: The
entries of v+ are reduced column echelon forms of even
rank in Mn+1 (Fq ); each reduced column echelon form of
2i
rank 2i appears q ( 2 ) times in v+ . The entries of v− are
reduced column echelon forms of odd rank in Mn+1 (Fq );
each reduced column echelon form of rank 2i+1 appears
2i+1
q ( 2 ) times in v− .
Define
C+ = {Xv+ : X ∈ Mn×(n+1) (Fq )} ⊂ Mn×(n+1) (Fq )N ,
C− = {Xv− : X ∈ Mn×(n+1) (Fq )} ⊂ Mn×(n+1) (Fq )N ,
19
where Xv+ is the result of multiplying X to each component of v+ . Then obviously,
f :
C+ −→ C−
Xv+ 7−→ Xv− , X ∈ Mn×(n+1) (Fq ),
(7)
is a well defined Mn(Fq )-isomorphism. We claim that f
preserves the Hamming weight. Let X ∈ Mn×(n+1) (Fq )
with rank X = r. Then the number of reduced column
echelon forms Y ∈ Mn+1 (Fq ) of rank i such that XY 6= 0
n+1−r
is n+1
−
. So
i
i
q
q
|Xv+ |−|Xv− | =
n+1
X
i=0
i
2
(−1) q ( )
i
hn + 1i
i
−
q
hn + 1 − r i i
= 0.
q
3◦ C+ and C− are not equivalent. C+ has a component X · 0 which is always 0; C− does not have such a
component.
4◦ Using the embedding Mn×(n+1) (Fq ) ,→ R R, C+ and C−
can be viewed as R-submodules of (R R)N . The map f
in (7) is an R-isomorphism which preserves the Hamming weight of (R R)N . We claim that 6 ∃ monomial
transformation h : (R R)N → (R R)N such that h|C + = f .
Otherwise, h|Mn×(n+1) (Fq )N : Mn×(n+1) (Fq )N → Mn×(n+1) (Fq )N
would be a monomial transformation whose restriction
to C+ is f , a contradiction to 3◦ . Therefore R does not
have the extension property.
Observation [Dinh and López-Permouth] Let
R be a left (or right) artinian ring and let J =
rad R and R̄ = R/J. Then R̄ is semisimple, so
∼M
R̄ =
n−1 (D1 ) × · · · × Mnk (Dk ),
where ni > 0 and Di is a division ring. Let
0

ij = 

...
1
0
...
j

∈ Mni (Di).
0 ni ×ni
P
Then 1R̄ = i,j ij is a decomposition of 1R̄
∼
into primitive orthogonal idempotents with ij =
i0j 0 iff i = i0. Let i = i1. Then R̄1, . . . , R̄k
is the list of all nonisomorphic simple R̄ modules. Note that simple R̄-modules are precisely
Lk
simple R-modules. We have R R̄ = i=1 ni ·
∼ Lk m · R̄ . Then
R̄i. Write soc(R R) =
i
i=1 i
m1 + · · · + mk ≥ n1 + · · · + nk . (This can be
seen by lifting the idempotents ij to R.) Thus
∼ R̄ ⇔ m ≤ n for all i. If |R| < ∞,
soc(R R) =
i
i
R
by the note after Theorem 5, R is Frobenius
⇔ mi ≤ ni for all i. So, if a finite ring R is not
Frobenius, then ∃i such that (ni +1)·R̄i ,→ R R.
20
Lecture 2. Frobenius Local Rings
Finite Frobenius Local Rings
Theorem 15 Let R be a finite ring.
statements are equivalent.
The following
(i) R is Frobenius and local.
(ii) R is QF and local.
(iii) R has a unique minimal left ideal.
(iv) R has a unique minimal right ideal.
Proof. (i) ⇒ (ii). Obvious.
(ii) ⇒ (iii) and (iv). Since R is local, R has a unique
maximal right ideal M . Since R is QF, l(M ) is the unique
minimal left ideal of R. So we have (iii). (iv) follows by
symmetry.
(iii) ⇒ (iv) and (i). By symmetry, we may assume (iii).
Let L be the unique minimal left ideal of R. Choose
a character χ of (R, +) such that χ is nontrivial on L.
Then ker χ does not contain any nonzero left ideal of R,
so χ is a generating character of R. By Theorem 11, R
is Frobenius. Since R is QF, r(L) is the unique maximal
right ideal of R. Thus R is local.
21
Example. Consider
R = Zpm [X]/(X n, pa X n−1 − pb),
where p is a prime and n ≥ 2, 0 ≤ a, b ≤ m are integers.
If b ≤ a, then in R, pb = pa X n−1 = pa−bX n−1 pb
∼ Z b [X]/(X n), which is
= pa−bX n−1 pa X n−1 = 0. Thus R =
p
not interesting. (R has a unique minimal ideal pb−1 X n−1 R.)
So we assume b > a. Since in R, p2b−a = pb−a pb =
∼
pb−a pa X n−1 = pbX n−1 = pa X n−1 X n−1 = 0, we have R =
Zpmin{m,2b−a} [X]/(X n, paX n−1 − X b). Thus we may assume
m ≤ 2b − a.
If a = 0 and n = 2, then X 2 = (X + pb)(X − pb) in Zpm [X]
∼
since m ≤ 2b − a ≤ 2b. So R = Zpm [X]/(x2 , X − pb) =
∼ Zpm .
Zpm [X]/(X − pb) =
If a = 0 and b = m, then R = Zpm [X]/(X n−1 ), which is
also uninteresting.
Therefore, we assume
0 ≤ a < b ≤ m, 2b−a ≥ m, (a, n) 6= (0, 2), (a, b) 6= (0, m).
We have the following facts.
∼ Zpm × Zn−2
(i) (R, +) =
× Zpa .
pb
(ii) R is Frobenius and local ⇔ 2b − a = m. When
2b − a = m, pm−1 R is the unique minimal ideal of R.
22
Lemma 16 (A variation of Hensel’s lemma)
Let R be a commutative ring, m a nilpotent
ideal of R, and k = R/m. Let ( ) : R → k
be the natural homomorphism. Let f, g ∈ k[x]
such that f is monic and f, g are coprime. Let
H ∈ R[x] such that H̄ = f g. Put
A = {F ∈ R[x] : F is monic and F̄ = f },
B = {L ∈ m[x] : deg L < deg f }
and define
φ : A −→ B
F 7−→ L,
where L is given by the Euclidean algorithm
H = GF + L,
G, L ∈ R[x], deg L < deg F.
Then φ is a bijection. In particular, ∃!F, G ∈
R[x] such that F is monic, F̄ = f , Ḡ = g and
H = F G.
23
Basic polynomials. Let p be a prime and n ∈ Z+ . The
homomorphism ( ) : Zpn → Zp induces a homomorphism
Zpn [x] → Zp[x]. A monic polynomial f ∈ Zpn [x] is called
basic if its image f¯ ∈ Zp [x] is irreducible.
Lemma 17 If f1 , f2 ∈ Zpn [x] are two basic polynomials
of the same degree, then Zpn [x]/(f1 ) ' Zpn [x]/(f2 ).
Proof. Let R = Zpn [x]/(f1 ) and k = R/pR. Then k = Fpt ,
where t = deg f1 . Since f¯2 ∈ Zp [x] has degree t, f¯2 has
a zero α ∈ k, so f¯2 = (x − α)g for some g ∈ k[x]. By
Lemma 16, f2 = (x − a)G for some a ∈ R with ā = α
and some G ∈ R[x].
We claim that R = Zpn [a]. For each b ∈ R, since b̄ ∈ k =
Zp[α], we have b̄ = h0(α) for some h0 ∈ Zp[x]. Choose
H0 ∈ Zpn [x] such that H̄0 = h0 . Then b̄ = H0 (a), so
b = H0 (a) + pb1 for some b1 ∈ R. By induction and the
fact that pn = 0 in R, we have b = H0 (a)+pH1 (a)+· · ·+
pn−1 Hn−1 (a) for some H0 , . . . , Hn−1 ∈ Zpn [x]. So b ∈ Zpn [a]
and the claim is proved.
Now the ring homomorphism
φ : Zpn [x] −→
R
f
7−→ f (a)
induces an onto homomorphism φ̄ : Zpn [x]/(f2 ) → R.
Since |Zpn [x]/(f2 )| = pnt = |R|, φ̄ is an isomorphism.
24
Galois rings. Let f ∈ Zpn [x] be a basic polynomial of degree t. The ring Zpn [x]/(f ), which
depends only on p, n, t, is called the Galois ring
and is denoted by GR(pn, t).
• (GR(pn, t), +) = Ztpn .
• GR(pn, t) is a commutative local ring with
maximal ideal (p) and residue field
GR(pn, t)/(p) = Fpt .
• All ideals of GR(pn, t) are GR(pn, t) = (p0) ⊃
(p1) ⊃ · · · ⊃ (pn) = 0. Thus GR(pn, t) is
a finite commutative Frobenius local ring
with a unique minimal ideal (pn−1).
25
Teichmüller set. Let f ∈ Zp[x] be a monic
primitive irreducible polynomial of degree t. Then
t
f | xp −1 − 1 in Zp[x]. By Lemma 16, ∃ basic
polynomial F ∈ Zpn [x] such that F̄ = f and
t
F | xp −1 − 1 in Zpn [x]. We have
GR(pn, t) = Zpn [x]/(F ) = Zpn [ω],
where ω is the image of x in Zpn [x]/(F ). ω̄ ∈
GR(pn, t)/(p) = Fpt is a root of f . Hence
the multiplicative order o(ω̄) = pt − 1. Thus
t −2
t
p
o(ω) = p − 1. The set T = {0, 1, ω, . . . , ω
}
is called the Teichmüller set of GR(pn, t). Since
t
0̄, 1̄, ω̄, . . . , ω̄ p −2 are all distinct in Fpt , T is a set
of representatives of cosets of (p) in GR(pn, t).
The elements
ξ0 + pξ1 + · · · + pn−1ξn−1,
ξi ∈ T
(8)
are all distinct for different (ξ0, . . . , ξn−1). Thus
every element in GR(pn, t) can be uniquely expressed in the form (8), which is called the
p-adic expansion of the element.
26
Multiplicative group. Using the p-adic expansion (8),
we see that the multiplicative group of GR(pn, t) is
∼ T × × (1 + pGR(pn, t)),
GR(pn, t)× = T × (1 + pGR(pn, t)) =
(9)
∼
n
×
where
= T \ {0} = hωi = Zpt −1 . Since |GR(p , t) | =
(pt−1)p(n−1)t, T × is the only cyclic subgroup of GR(pn, t)×
of order pt − 1. It follows that the Teichmüller set of
GR(pn, t) is unique. In (9), it can be shown that


Ztpn−1
if p is odd or



p = 2 and n ≤ 2,
n
∼
1 + pGR(p , t) =



Z × Z n−2 × Zt−1 if p = 2 and n ≥ 3.
2
2
2n−1
T×
Automorphism group.
Proposition 18 Define
σ : GR(pn, t) −→ GR(pn, t)
n−1
n−1
X
X p
i
ξi p
7−→
ξi pi, ξi ∈ T.
i=0
(10)
i=0
Then σ ∈ Aut(GR(pn, t)), o(σ) = t and Aut(GR(pn, t)) =
hσi. σ is called the Frobenius map of GR(pn, t).
27
Theorem 19 Let R = GR(pn, r) and let σ be
the Frobenius automorphism of R. Let R MR
be an (R, R)-bimodule. Then
R MR = M1 ⊕ · · · ⊕ Mr
(as (R, R)-bimodue),
where
xa = σ i(a)x
∀x ∈ Mi, a ∈ R.
28
Finite Chain Rings
A finite chain ring is a finite ring whose left
ideals form a chain under inclusion. This definition is left-right symmetric. (Since a finite
chain ring R is QF, A 7→ r(A) is an inclusion
reversing bijection from the set of all left ideals
of R to the set of all right ideals of R.) The
structure of finite chain rings was determined
by Clark and Drake (1973). Let R be a finite
chain ring with maximal ideal M . The following steps lead us to a concrete description of
R.
1◦ M = Rπ for some π ∈ R. Write M = Ra1 + · · · + Ran,
ai ∈ M . Since Ra1 , . . . , Ran form a chain, M = Rai for
some i.
2◦ Since M is the unique maximal left ideal of R, M =
rad R. So M is an ideal of R. Moreover, π is nilpotent.
Let s be the nilpotency of π. We claim that for 0 ≤ i ≤
s − 1, π i ∈
/ Rπ i+1 . Otherwise, π = rπ i+1 for some r ∈ R,
i.e., π i(1 − rπ) = 0. Since 1 − rπ is invertible, π i = 0,
which is a contradiction. In the same way, π i ∈
/ π i+1 R.
∼ R/Rπ.
3◦ For each 0 ≤ i ≤ s − 1, Rπ i/Rπ i+1 =
29
Define f : R/Rπ → Rπ i/Rπ i+1 , r + Rπ 7→ rπ i + Rπ i+1 .
Then f is an onto R-map. By 2◦ , Rπ i/Rπ i+1 6= 0. Since
R/Rπ is simple, f is onto.
4◦ Since R/Rπ is a simple R-module, by 3◦ ,
R = Rπ 0 ⊃ Rπ 1 ⊃ · · · ⊃ Rπ s = 0
are all left ideals of R. Let R/Rπ = Fpr . Then |Rπ i/Rπ i+1 | =
pr , 0 ≤ i ≤ s − 1, so |Rπ i| = p(s−i)r .
5◦ Since Rπ is a two-sided ideal of R, all Rπ i are twosided ideal of R. By 2◦ , R = π 0 R ) π 1 R ) · · · ) π s R = 0.
Since R is QF, it has precisely s + 1 right ideals. So we
must have π iR = Rπ i, 0 ≤ i ≤ s.
6◦ Let char R = pn, then R contains a copy of GR(pn, r).
We first show that ∃ω ∈ R× such that o(ω) = pr − 1.
Choose ω1 ∈ R such that its image ω̄1 ∈ R/Rπ = Fpr
t
has order pt − 1. Then ω1p −1 = 1 + rπ for
some r ∈ R.
u
Choose u ∈ Z+ large enough such that pi ≡ 0 (mod pn)
t
u
for all 1 ≤ i ≤ s. Then ω1(p −1)p = (1 + rπ)p = 1 +
Ppu pu
pu
i
i=1 i (rπ) = 1. Let ω = ω1 .
u
Zpn [ω] is a subring of R. So Zpn [ω] is a commutative
local ring. (In general, a subring of a finite local ring R
is local. ∀a ∈ S \ S ∩ (rad R), a ∈ R× . Since |R× | < ∞,
am = 1 for some m > 1. Then a ∈ S × .) Let m be the
maximal ideal of Zpn [ω]. Let ω̄ ∈ Zpn [ω]/m be the image of
ω. Let f ∈ Zp [x] be the minimal polynomial of ω̄ over Zp .
t
Since f | xp −1 − 1, by Lemma 16, ∃ a basic polynomial
t
F ∈ Zpn [x] such that F̄ = f and xp −1 − 1 = F G for some
G ∈ Zpn [x]. Since F̄ (ω̄) = 0, we have Ḡ(ω̄) 6= 0 (since
t
xp −1 − 1 is separable.) Thus G(ω) is invertible in Zpn [ω].
t
Since 0 = ω p −1 − 1 = F (ω)G(ω), we have F (ω) = 0. Let
φ : GR(pn, r) = Zpn [x]/(F ) Zpn [ω]
be the ring homomorphism mapping x + (F ) to ω. φ induces an isomorphism GR(pn, r)/(pm ) → Zpn [ω] for some
1 ≤ m ≤ n. (Note that every ideal of GR(pn, r) is
of the form (pm ).) Since pm = char GR(pn, r)/(pm ) =
∼
char Zpn [ω] = pn, we must have m = n. Thus GR(pn, r) =
Zpn [ω].
7◦ Let S ⊂ R be an isomorphic copy of GR(pn, r). πadic expansion. The embedding S/pS ,→ R/Rπ is an
isomorphism since both fields are Fpr . Let T be the
Teichmüller set of S. Then T is a set of representatives
of cosets of Rπ in R. By induction, every element of R
can be uniquely written as
ξ0 + ξ1 π + · · · + ξs−1 π s−1 ,
ξi ∈ T.
(11)
8◦ Since M is an (S, S)-bimodule, by the above theorem,
S MS
= M1 ⊕ · · · ⊕ Mr
(as (S, S)-bimodule),
where
xa = σ i(a)x
∀x ∈ Mi, a ∈ S.
∃Mi 6⊂ M 2 . So we may choose π in 1◦ such that π ∈ Mi.
Then
πa = σ i(a)π
∀a ∈ S.
9◦ Let e ≤ s be the largest integer such that p ∈ Rπ e .
Then p = uπ e for some u ∈ R× . Thus π ne = u−npn = 0
but π (n−1)e = u−(n−1) pn−1 6= 0. So (n − 1)e < s ≤ ne.
Write
s = (n − 1)e + t,
1 ≤ t ≤ e.
Note that if n = 1, then s = t and by definition, e = s =
t. We claim that
SR
= S ⊕ Sπ ⊕ · · · ⊕ Sπ k−1
(12)
and
S
i ∼
S (Sπ ) =
∼ Sp
S/pn−1 S =
if 0 ≤ i ≤ t − 1,
if t ≤ i ≤ k − 1.
(13)
Define f : S → Sπ i, a 7→ aπ i. Then clearly,
0
if 0 ≤ i ≤ t − 1,
ker f =
pn−1 S if t ≤ i ≤ e − 1.
So we have the isomorphism in (13).
By (12) we have
SR
= S + Sπ + · · · + Sπ s−1
= S + Sπ + · · · + Sπ e−1 + Rπ e
= S + Sπ + · · · + Sπ e−1 + Rp.
(14)
By Nakayama’s lemma,
SR
= S + Sπ + · · · + Sπ s−1 .
(15)
By (13),
|S| = |Sπ| · · · |Sπ s−1 | = |S|t|Sπ|e−t = pprtp(n−1)r(e−t)
= pr[(n−1)e+t] = prs = |R|.
So the sum in (15) is direct.
10◦ Since π e ∈ pR, by 9◦ ,
π e = p(ae−1 π e−1 + · · · + a1 π + a0 ),
ai ∈ S.
(16)
If k < s, we must have a0 ∈ S × since p = uπ e , u ∈ R× . If
e = s, then π e = 0 = p and we can choose a0 = 1 ∈ S.
Thus in (16), we can always choose a0 ∈ S × .
Let S[X; σ i] be the skew polynomial ring. (In S[X; σ i],
aX = Xσ i(a) ∀a ∈ S.)
Define
f :
S[X; σ i]
−→
R
m
a0 + · · · + am x
7−→ a0 + · · · + am π m .
Clearly, f is an onto homomorphism. We claim that
ker f = (g, pn−1 X t),
where g = X e −p(ae−1 X e−1 +· · ·+a0 ). Clearly, (g, pn−1 X t) ⊂
ker f . Suppose h ∈ ker f . We may assume deg h ≤ e − 1
because of g. Write h = α0 + · · · + αe−1 X e−1 , αi ∈ S.
Then α0 + · · · + αe−1 π e−1 = 0. By (12), αiπ i = 0 for all
0 ≤ i ≤ e, which implies that α0 = · · · = αt−1 = 0 and
αt, . . . , αe−1 ∈ pn−1 S. So h ∈ (g, pn−1 X t).
Now we arrive at the conclusion that
∼ S[X; σ i]/(g, pn−1 X t).
R=
Facts about finite chain rings. Here is a summary of
properties of finite chain rings.
• Every finite chain ring R is of the form
∼ S[X; α]/(g, pn−1 X t),
R=
(17)
where p is a prime, n ≥ 1, S = GR(pn, r), α ∈
Aut(S), g = X e − p(ae−1 X e−1 + · · · + a0 ) ∈ S[X; α],
ai ∈ S, a0 ∈ S × , t = e if n = 1 and 1 ≤ t ≤ e if n > 1.
Let π be the image of X in R. Then rad R = Rπ,
R/Rπ = Fpr , Rp = Rπ k , and the nilpotency of π is
(n − 1)e + t. We call R in (17) a finite chain ring of
type (p, n, r, k, t).
∼ Zrtn × Zr(e−t)
• (R, +) =
p
pn−1 .
• Let T × be the cyclic subgroup of S × . (Note that
S ,→ R.) Then
R× = T × (1 + Rπ) = T × n (1 + Rπ).
The structure of R× is not known in general.
• Aut(R) =?.
30
Finite Commutative Chain Rings
Let p be a prime and let n, r, e, t be positive
integers such that t = e if n = 1 and 1 ≤ t ≤ e
if n > 1. It follows from (17) that every finite
commutative chain ring of type (p, n, r, e, t) is
of the form
R = S[X]/(g, pn−1X t),
(18)
where S = GR(pn, r), g = X e − p(ae−1X e−1 +
· · · + a0) ∈ S[X], ai ∈ S, a0 ∈ S ×.
Multiplicative group. We have
R× = T ×(1 + Rπ) = T × × (1 + Rπ).
T × is the unique cyclic subgroup of R× of
order pr − 1. 1 + Rπ is a finite abelian pgroup. In most cases, the structure of 1+Rπ is
completely determined by the type (p, n, r, e, t)
(Theorems 20,T2.8). There is a remaining
case where the structure of 1 + Rπ depends
not only on the type (p, n, r, e, t) but also on
the Eisenstein polynomial g. The result in this
case is less explicit (Theorem 22).
31
Theorem 20 (H-L-M 03) Let R be a finite
commutative chain ring of type (p, 1, r, e, t), i.e.,
R = Fpr [X]/(X e). Let I = {i ∈ Z : 1 ≤ i < e, p i}. Then
M
∼
1 + Rπ =
Zr
p .
i∈I
dlogp i e
p
Theorem 21 (H-L-M 03) Let n > 1 and let R
be a finite commutative chain ring of type (p, n, r, e, t)
given by (18). Assume that either (p−1) - e or (p−1) | e
but the image of −a0 in R/Rπ = Fpr is not a (p − 1)st
power. Let s = (n − 1)e + t and
n
pe o
I = i ∈ Z : 1 ≤ i ≤ min s,
p−1
, p - i . (19)
For each i > 0, let
k
ι(i) = max 0, dlogp
e ,
(p − 1)i
s − ipι(i)
α(i) = ι(i) + d
e.
e
Then
M
∼
1 + Rπ =
Zrpα(i) .
i∈I
32
Theorem 22 (H-L-M 03) Let n > 1 and let
R be a finite commutative chain ring of type
(p, n, r, e, t) given by (18). Assume that e =
c(p − 1)pu, p - c, u ≥ 0 and that the image of
−a0 in R/Rπ = Fpr is a (p − 1)st power. Let
u
u+1
ψ : Fpr → Fpr , z 7→ z p + ā0z p
. Choose −
1, . . . , r ∈ R such that their images ¯
1, . . . , ¯
r ∈
R/Rπ = Fpr form a basis of Fpr over Fp and
ker ψ = h¯
r i. Choose δ ∈ R such that ψ(¯
1), . . . ,
ψ(¯
r−1), δ̄ form a basis of Fpr over Fp. Then
∃γ ∈ Z and γ(i, j) ∈ Z for each (i, j) ∈ Ω such
that
u+1
u+1
(1+π cr )p
= (1+π cp
δ)γ
Y
(1+π ij )γ(i,j).
(i,j)∈Ω
(20)
1 + Rπ is the abelian p-group with generators
xij (i ∈ I, 1 ≤ j ≤ r) and x subject to the
relations
 α(i)
p


x
= 1,
(i, j) ∈ Ω,


 ij u+1
α(p
)
p
= 1,
x




γ(i,j)
xpu+1 = xγ Q
x
.
cr
(i,j)∈Ω ij
Finite commutative chain rings and p-adic fields.
Let Qp be the field of p-adic numbers and let Z
– p denote
the ring of p-adic integers. (Since we use Zn for Z/nZ,
the notation for the ring of p-adic integers needs to be
a little different from Zp .)
Let K/Qp be a finite extension, with residue degree f and
ramification index e. Let oK be the ring of integers of
K, πK a prime of K, and K̄ = oK /πK oK the residue field
of K. Let k/Qp be the maximal unramified subextension
of K/Qp . Let s, t, n be positive integers such that s =
(n − 1)e + t, where 1 ≤ t ≤ e.
Choose a ∈ ok be such that k̄ = Zp [ā], where ā is
the image of a in k̄. Let Φ ∈ –
Zp[X] be the minimal
polynomial of a over Qp . Then the image Φ̃ of Φ in
(–
Zp/pn–
Zp)[X] = Zpn [X] is a basic of degree r and
∼ Zpn [X]/(Φ̃) =
∼ GR(pn, r).
o /pno =
k
k
The minimal polynomial of πK over k is an Eisenstein
polynomial Ψ ∈ ok [X] of degree e such that
∼ (o /pno )[X]/(Ψ̃, pn−1 X t)
o /π s o =
K
K K
n
k
k
∼ GR(p , f )[X]/(Ψ̃, pn−1 X t),
=
∼ GR(pn, r)[X].
where Ψ̃ is the image of Ψ in (ok /pnok )[X] =
Ψ̃ is an Eisenstein polynomial over GR(pn, r). Thus
s o
oK /πK
K is a finite commutative chain ring with invariants
(p, 1, r, t, t) if n = 1,
(21)
(p, n, r, e, t) if n > 1.
33
Enumeration of isomorphism classes. Let
C(p, n, r, e, t) be the number of isomorphism classes
of finite commutative chain rings type (p, n, r, e, t)
and let I(Qp, f, e) denote the number of isomorphism classes of finite extensions K/Qp with
residue degree r and ramification index e.
C(p, n, r, e, t) = I(Qp, f, e)
when
p
+ νp(e) e.
(n − 1)e + t >
p−1
The number C(p, n, f, e, t) is difficult to compute when νp(e) > 0. In certain cases, explicit
formulas for C(p, n, r, e, t) are known; these formulas are given in the following theorems. (When
n = 1, obviously, C(p, n, r, e, t) = 1, So we assume n > 1.)
34
Theorem 23 (Clark and Liang 73 and Hou 01)
Let p be a prime and let n, f, e, t be positive integers such that n ≥ 2, 1 ≤ t ≤ e, and p - e.
Then
f −1
φ(c)
1 X (i,f )
C(p, n, f, e, t) =
=
p
−1, e ,
τ (c)
f i=0
c|(e,pf −1)
(22)
where φ is the Euler function, (a, b) is the greatest common divisor of a and b, and τ (c) is the
smallest positive integer m such that pm ≡ 1
(mod c).
X
35
Theorem 24 (Hou-Keating 04) Assume p k e and n >
1
3 + p−1
− et . Let e0 = e/p. Then
C(p, n, r, e, t) =
1
re
eX
0 −1
h=0
−
p2 br
+
ch
b−1
X
r
ch
−1
X
e0 f
!
(p2 + ωhij )p o(h,i,j) ,
i=0 j=0
where
(i) b = (e0 , pr − 1),
(ii) ch is the smallest positive integer such that
b | (pch − 1)h
(ii)
(
t(h, i, j) =
plcm(r,ch j) −1
i
pch j −1
+
uhch j
(r,ch j)
i
if j > 0,
if j = 0,
where u ∈ Z satisfies e0 u ≡ b (mod pr − 1),
(iv) o(h, i, j) =
r
(r,ch j)
·
b
,
(b,t(h,i,j))
(v)
ωhij =

p2








p
pr −1
(mod (p − 1)), and
p−1
pch j −1
e0 u−b
≡
(mod (p − 1)),
pr −1
p−1
if (p − 1) | e, −h ≡
e0 i
b
+
(pch j −1)h
b
·
otherwise.
36
Theorem 25 (Hou-Keating 04) Assume p > 2, p2 k
1
e, n > 4 + p−1
− et , and (pf − 1, e) = 1. Let e0 = e/p2 .
Write r = pir0 with p - r0 . Then
C(p, n, r, e, t)
"
r0 i
p e0
X
τ
1
1
(p
− 1)2
2
= i 0
φ(τ ) (p − 1, τ )
pr
2
p−1
0
τ |r
r0
r0
i
r0
i
i
p τ p e0 (p τ p e0 − 1)
1 2
(p τ 2p e0 − 1)
− (p − 1, τ )
+ (p − 1, τ )
p−1
2
p−1
0
r0
τ
1+ rτ (p+1)pi e0
pi e0 n
2
+ (p + 1)(p
−p
) − (p − 1)p
r0 pi−j e0
r0 i−j
i
p e0 −1
X
τ
τ
−
1)(p
− 1)
(p
j−1
+
p
p+1
j=1
r0
i−j+1
+ (p − 1)(−p2+ τ p
r0
e0
i−j
+ (p2 − 1)(p1+ τ (p+1)p
+
r0
i−j+1
+ pτ p
e0
r0
e0
i−j
− pτ p
r0
i−j
+ p τ 2p
e0
e0
r0
τ
pi+1 e0
+p
r0
τ
2pi e0
)
)
r0
r0 i−j
1
1
i−j
(p − 1, τ )2 (p τ p e0 − 1)2 + (p2 − 1, τ )(p τ 2p e0 − 1)
2
2
#
.
Lecture 3. Partial Difference Sets
An Introduction to Partial Difference Sets
Let G be a finite group. A partial difference
set (PDS) in G is a subset D ⊂ G such that
the differences d1d−1
2 (d1 , d2 ∈ D, d1 6= d2 ) represent each element in D \ {e} exactly λ times
and each element in G \ (D ∪ {e}) exactly µ
times. The integers (v = |G|, k = |D|, λ, µ)
are called the parameters of the PDS. When
λ = µ, the PDS is called a difference set.
Example 26 (Paley PDS) Let q be a prime
power such that q ≡ 1 (mod 4). Let D be
the set of nonzero squares in Fq . Then D is a
1 (q − 1), 1 (q − 5), 1 (q − 1)) PDS.
(q, 2
4
4
37
Example 27 (PCP) Let G be a group of order n2 . An
(n, r) partial congruence partition (PCP) of G is a set P
of r subgroups of order n of G such that if H1 , H2 ∈ P
with H1 6= H2 , then H1 ∩ H2 = {e}.
S
Claim. If P is an (n, r) PCP of G, then D = H∈P (H \
{e}) is a (n2 , r(n − 1), n + r2 − 3r, r 2 − r) PDS in G.
Proof. Let P = {H1 , . . . , Hr }. Note that if i 6= j, then
Hi × Hj → G, (a, b) 7→ ab−1 is a bijection. Let x ∈ G \ {e}.
If x ∈
/ D, then for each pair i 6= j, ∃!(a, b) ∈ (Hi \ {e}) ×
(Hj \ {e}) such that ab−1 = x. So ab−1 = x has r(r − 1)
solutions (a, b) ∈ D × D. Assume x ∈ D, say x ∈ H1 .
Then for each pair i 6= j with i, j 6= 1, ∃!(a, b) ∈ (Hi \
{e})×(Hj \{e}) such that ab−1 = x. If exactly one of i, j is
1, Then 6 ∃(a, b) ∈ (Hi \{e})×(Hj \{e}) such that ab−1 = x.
Also ∃n − 2 pairs (a, b) ∈ (H1 \ {e}) × (H1 \ {e}) such that
ab−1 = x. So ab−1 = x has (r − 1)(r − 2) + n − 2 =
n + r 2 − 3r solutions (a, b) ∈ D × D.
An example of PCP. Let G = F2q . Any collection of
1-dimensional Fq -subspaces of F2q is a PCP.
38
PDS and strongly regular graphs. A graph Γ with v
vertices is called a (v, k, λ, µ) strongly regular graph if it
has the following properties.
(i) Every vertex belongs to k edges.
(ii) If two vertices are adjacent, there are λ additional
vertices adjacent to both of them; if two vertices
are not adjacent, there are µ additional vertices adjacent to both of them.
Strongly regular graphs are precisely association schemes
with two classes.
Let G be a finite group and D ⊂ G\{e} such that D(−1) =
D. The Cayley graph Γ(G, D) of D has vertex set G such
that a, b ∈ G are adjacent iff ab−1 ∈ D. Γ(G, D) admits
G as a regular automorphism group.
Fact. The Cayley graph of D ⊂ G is a (v, k, λ, µ) strongly
regular graph iff D is a (v, k, λ, µ) PDS with D(−1) = D
and e ∈
/ D. Moreover, every strongly regular graph admitting a regular automorphism group can be obtained
this way.
39
PDS and two weight codes.
Theorem 28 Let A = [a1, . . . , an] ∈ Ms×n(Fq )
such that rank A = s and a1, . . . , an are pairwise linearly independent. Then A generates
an (n, s) code with two nonzero weights w1
and w2 iff
D = {tai : 1 ≤ i ≤ n, t ∈ F×
q}
is a PDS in Fsq with parameters


v = q s,




k = n(q − 1),

λ = k2 + 3k − q(k + 1)(w1 + w2) + q 2w1w2,




µ = k 2 + k − qk(w + w ) + q 2w w .
1
2
1 2
40
Equation in group ring. LetPG be a finite group.
Identify a subset D ⊂ G with
g∈D g ∈ Z[G]. Define
P
D(−1) = y∈D g −1 . Then D is a (v = |G|, k = |D|, λ, µ)
PDS ⇔
DD(−1) = µG + (λ − µ)D + γe,
(23)
where
γ=
k−µ
k−λ
if e ∈
/ D,
if e ∈ D.
Call a PDS reversible if D = D(−1) . All PDS with λ 6= µ
are reversible [Ma94].
Proposition 29 Let G be a finite abelian group and
D ⊂ G. Then D is a reversible PDS iff ∃α, β ∈ R such
that
χ(D) = α or β
∀1 6= χ ∈ Ĝ.
(24)
When (24) is satisfied, the parameters of the PDS D
are


v = |G|,



k = |D|,

λ = k + αβ + δ(α + β),


µ = λ − (α + β),
where
δ=
0
1
if e ∈ D,
if e ∈
/ D.
41
Proof of Example 26 Let q = ps. Then every nontrivial character χ of Fq is of the form
TrFq /Fp (a·)
ζp
, a∈
F×
q . We have
1 X TrFq /Fp (ax2)
χ(D) =
ζp
2
×
x∈Fq
X
TrFq /Fp (ax2 )
1
=
ζp
−1
2 x∈F
q
i
1h 1
= ±q 2 − 1
(Gauss quadratic summ).
2
1
1
1
1
2
So α = 2 (q − 1), β = 2 (−q 2 − 1), λ = k +
1 (q − 1) − 1 (q − 1) − 1 = 1 (q − 5),
αβ + α + β = 2
4
4
1
µ = 4 (q − 1).
42
Latin square type PDS. Let G be a finite abelian group
with |G| = n2 . Let D ⊂ G \ {e} be a reversible PDS such
that
χ(D) = −r or − r + n
∀1 6= χ ∈ Ĝ,
where r ∈ R. By (23),
DD = µG + (λ − µ)D + (k − µ)e
= (|D| − r(−r + n))G + (−2r + n)D + r(−r + n)e.
(25)
Counting the sum of coefficients at both sides of (25),
we have
|D|2 − |D|(n2 − 2r + n) + r(−r + n)(n2 − 1) = 0,
which gives two solutions for |D|:
|D| = r(n − 1) or (n − r)(n + 1).
If |D| = r(n − 1), D is called a Latin square type PDS; if
|D| = (n − r)(n + 1), D is called a negative Latin square
type PDS.
43
Constructing Latin Square Type PDS by Latin Shells
Latin shells. Let G ⊃ G1 ⊃ G2 be finite abelian groups
such that |G| = |G1 ||G2 | and let r be a real number. A
subset D ⊂ G \ G1 is called a (G, G1 , G2 ; r)-Latin shell if
for each character χ of G,
χ(D) =

1
1 
r|G| 2 1 −
,



|G
|
2


 |G| 21
−r
,
|G
|

2



0,


1

−r or − r + |G| 2 ,
if χ is principal,
if χ is principal on G1 but not on G,
if χ is principal on G2 but not on G1 ,
if χ is not principal on G2 .
The group G2 is called the thickness of the Latin shell
D.
Note that when G1 = G2 , a (G, G1 , G1 ; r)-Latin shell
is a Latin square type1 PDS with non principal character
values −r and −r+|G| 2 . The following proposition shows
that one can nest one Latin shell inside another to form
a new Latin shell with greater thickness.
44
Proposition 30 Let G ⊃ G1 ⊃ G2 be finite
abelian groups such that |G| = |G1||G2| and
let D be a (G, G1, G2; r)-Latin shell. Let G1 ⊃
H1 ⊃ H2 ⊃ G2 be subgroups such that |G1||G2| =
|H1||H2| and let E be a
(G1/G2, H1/G2, H2/G2; r/|G2|)-Latin shell. Let
Ē ⊂ G1 \H1 be the pre image of E. Then D ∪ Ē
is a (G, H1, H2; r)-Latin shell.
45
Proposition 31 (Hou03) Let R be a finite Frobenius
local ring and I a proper ideal of R. Let rR (I) be the
right annihilator of I. Let σ : rR (I) → {0, 1} be such
that
X
σ(a + x) = e for every a ∈ rR (I),
x∈soc(RR )
where e is any fixed integer with 0 ≤ e ≤ |soc(RR )| =
|R/rad(R)|. Then
D = (x, ȳ) ∈ rR (I)×(R/I) \ rR (I)×(rad(R)/I) : σ(y −1 x) = 1
is an rR (I) × (R/I), rR (I) × (rad(R)/I), soc(RR ) × {0};
e|rad(R)|/|I| -Latin shell where
∼ (R/I) × (R/I)
r (I) × (R/I) =
R
and
∼
rad(R)/I × rad(R)/I
rR (I)×(rad(R)/I)
soc(RR )×{0} =
as abelian groups.
46
A Construction of Finite Frobenius Rings
The Latin shell in Proposition 31 construction
applies to R × R whenever R is a homomorphic
image of a finite Frobenius local ring. So what
kind of rings are homomorphic images of finite
Frobenius local rings? We show that
• every finite ring is a homomorphic image
of a finite Frobenius ring and
• every finite local ring is a homomorphic image of a finite Frobenius local ring.
47
Proposition 32 Let R be a ring and R MR an
(R, R)-bimodule. Define R n M = R × M , and
for each (r1, m1), (r2, m2) ∈ R n M , define
(r1, m1) + (r2, m2) = (r1 + r2, m1 + m2),
(r1, m1) · (r2, m2) = (r1r2, r1m2 + m1r2).
Then R n M is a ring with identity (1, 0). R × 0
∼ R. 0 × M is
is a subring of R n M and R × 0 =
an ideal of R n M and
R −→ (R n M )/(0 × M )
(r, 0)
r 7−→
is a ring isomorphism.
The construction in Proposition 32 is known
as the “trivial extension”.
48
Let R be a finite ring and let R[ be the
character[ group
[
of (R, +), i.e., R = Hom ZR, Z(Q/Z) . Then R is an
(R, R)-bimodule: For each ω ∈ R[ and a ∈ R,
ωa : R −→ Q/Z
is defined by (ωa)(x) = ω(ax) and
aω : R −→ Q/Z
is defined by (aω)(x) = ω(xa).
Theorem 33 (Hou-Nechaev 07) Let R be any finite
ring. Then R n R[ is a finite Frobenius ring. In fact
χ : R n R[ −→ Q/Z
(r, ω) 7−→ ω(1)
is a generating character of R n R[ .
Proof. Assume to the contrary that ker χ contains a
nonzero left ideal I of R n R[ . First we claim that there
is an element (r, ω) ∈ I such that ω 6= 0. (Otherwise,
choose 0 6= (r, 0) ∈ I. There is an ω ∈ R[ such that
ω(r) 6= 0. Then we have (0, ω) · (r, 0) = (0, ωr) ∈ I,
where ωr(1) = ω(r) 6= 0. Hence ωr 6= 0.) Since ω 6= 0,
ω(s) 6= 0 for some s ∈ R. Thus (s, 0)·(r, ω) = (sr, sω) ∈ I
but
χ(sr, sω) = sω(1) = ω(s) 6= 0,
which is a contradiction.
Corollary 34 A finite ring is local if and only if it is a
homomorphic image of a finite Frobenius local ring.
49
Proof. Let R be a finite local ring. By Proposition 32
∼ (R n R[ )/(0 × R[ ), where R n R[ is
and Theorem 33, R =
Frobenius. Since (0 × R[ ) is a nilpotent ideal of R n R[ ,
R n R[ is also local.
Theorem 35 (i) Let R be a finite local ring with residue
field R/rad(R) = GF(pr ). Then
∼ Zr1n × Zr2n × · · · × Zrkn
(R, +) =
(26)
p
p
1
p
2
k
where n1 > n2 > · · · > nk ≥ 1, ri > 0, and r|ri (1 ≤ i ≤ k).
Moreover,
∼ Zr1n−r × Zr n −1 × Zr2n × · · · × Zrkn
(rad(R), +) =
(27)
p
p
1
p
1
p
2
k
(ii) For each prime p and positive integers ni, ri (1 ≤ i ≤
k) and r such that n1 > n2 > · · · > nk ≥ 1 and r|ri, there
is a finite local ring R such that R/rad(R) = GF(pr ) and
∼ Zr1n × Zr2n × · · · × Zrkn
(28)
(R, +) =
p
1
p
2
p
k
∼ Zr1n−r × Zr n −1 × Zr2n × · · · × Zrkn
(rad(R), +) =
pk
p1
p2
p1
(29)
Theorem 36 Let p be a prime and let ni, ri (1 ≤ i ≤ k)
and r be positive integers such that n1 > n2 > · · · > nk
2r2
2rk
1
and r | ri. Let G = Z2r
pn1 × Zpn2 × · · · × Zpnk and let 0 ≤ e ≤
pr . Then there exists a (G, G1 , G2 ; epn1 r1 +···+nk rk −r )-Latin
shell with |G2 | = pr and
2r2
2rk
1 −r)
∼ Z2(r
G1 /G2 =
× Z2r
pn1
pn1 −1 × Zpn2 × · · · × Zpnk
Proof. By Theorem 35 (ii), there is a finite local ring R
such that R/rad(R) = GF(pr ) and
∼ Zr1n × Zr2n × · · · × Zrkn
(R, +) =
p1
p2
pk
∼ Zr1n−r × Zr n −1 × Zr2n × · · · × Zrkn
(rad(R), +) =
p
1
p
1
p
p
2
k
By Corollary 34, there is a finite Frobenius local ring
∼ T /I. Note that
T and an ideal I C T such that R =
∼ rad(R) and |rad(T )|/|I| = |rad(R)| = pn1 r1 +···+nk rk −r .
rad(T )/I =
By Proposition 31, there is an (R×R, G1 , G2 ; epn1 r1 +···+nk rk −r )∼ G, |G | = |soc(R ) × {0}| = pr
Latin shell with R × R =
2
R
and
1 −r)
∼ rad(R)×rad(R) =
∼ Z2(r
G /G =
×Z2rn −1 ×Z2rn 2 ×· · ·×Z2rn k .
n
1
2
p
1
p
1
p
2
p
k
Corollary 37 In the notation of Theorem 36, there is a
Latin square 1type PDS in1 G with1 non principal character
values −e|G| 2 /pr and |G| 2 − e|G| 2 /pr .
Proof. Simply use Theorem 36 and Proposition 30 repeatedly.
References
W. E. Clark and D. Drake, Finite chain rings, Abhandlungen Math. Sem. Univ. Hamburg, 39 (1973), 147 –
153.
W. E. Clark and J. J. Liang, Enumeration of finite commutative chain rings, J. Algebra 27 (1973), 445 – 453.
C. W. Curtis and I. Reiner, Representation Theory of
Finite Groups and Associative Algebras, John Wiley &
Sons, Inc., New York, 1988.
H. Q. Dinh and S. R. Ló pez-Permouth, On the equivalence of codes over rings and modules, Finite Fields
Appl. 10 (2004), 615 – 625.
T. Honold, Characterization of finite Frobenius rings,
Arch. Math. 76 (2001), 406 – 415.
X. Hou, Finite commutative chain rings, Finite Fields
Appl. 7 (2001), 382 – 396.
X. Hou, Rings and constructions of partial difference
sets, Discrete Math. 270 (2003), 149 – 176.
X. Hou and K. Keating, Enumeration of isomorphism
classes of extensions of p-adic fields, J. Number Theory,
104 (2004) 14 – 61.
X. Hou, K. H. Leung, S. L. Ma, On the group of units
of finite commutative chain rings, Finite Fields Appl. 9
(2003), 20 – 38.
50
X. Hou and A. Nechaev, A construction of finite Frobenius rings and its application to partial difference sets,
J. Algebra 309 (2007), 1 – 9.
T. W. Hungerford, Algebra, Springer-Verlag, New YorkBerlin, 1980.
T. Y.Lam, A First Course in Noncommutative Rings,
Springer-Verlag, New York, 1991.
T. Y. Lam Lectures on Modules and Rings Springer,
New York, 1998.
S. L. Ma, A survey of partial difference sets, Des. Codes
and Cryptogr. 4 (1994), 221 – 261.
B. R. McDonald, Finite Rings with Identity, Marcel
Dekker, Inc., New York, 1974.
J. Wood Duality for modules over finite rings and applications to coding theory, Amer. J. Math. 121 (1999),
555–575.
J. Wood Code equivalence characterizes finite Frobenius
rings, Proc. Amer. Math. Soc. in press.