Associative rings and their adjoint groups

36 Pages·2005·0.287 MB·English

by  Ya. Sysak

Associative rings and their adjoint groups.
Ya. Sysak
One of purposes of these lectures is to state some results and methods (in-
cluded some recent of them) concerning investigations of relationships between
thestructureofanassociativering,inparticularitsLie-structure,andthestruc-
ture of the adjoint group and semigroup of this ring.
1 Preliminaries. Irreducible modules and
the density theorem of Jacobson.
Let R bearing, notnecessarilywithaunity, and M anabeliangroup. Recall
that M isaright R-moduleifamapping M×R→M whoseimageisusually
denoted by {mr |m∈M, r ∈R} is determined and for all m,n∈M and all
r,s∈R the following conditions hold:
1) (m+n)r =mr+nr,
2) m(r+s)=mr+ms, and
3) m(rs)=(mr)s.
If R has a unity 1 and m·1 = m for all m, then M is called unital.
Obviously every abelian group can be viewed as a unital Z-module.
As usually, if V is a non-empty subset of M , then VR denotes the sub-
group of M generated by the set {vr | v ∈ V, r ∈ R}. A subgroup N of
M is a submodule, if NR ⊆ N . The factor group M/N with the mapping
(m+N)r = mr+N for all m ∈ M and r ∈ R is a right R-module called
the factor module of M modulo N . A module M is a direct sum of modules
M with i from a set I if M , regarded as a group, is the direct sum of the
i
subgroups M .
i
Example. 1. If mr = 0 for all m ∈ M and r ∈ R, then M is an
R-module called trivial. Every subgroup of M is a submodule in M .
2. If M =R+ and and the mapping M ×R→M is the multiplication in
R, then R can be considered as a right R-module whose submodules coincide
with the right ideals of R.
A module M is simple if 0 and M are the only submodules of M . A
simplemodule M isirreducibleif mR=M forsome(andsoforany)non-zero
m ∈ M . In particular, every simple module is either irreducible or trivial. A
module M is semisimple if it is a direct sum of irreducible submodules.
1
2
Lemma 1.1. Let M be a semisimple R-module. Then every submodule N
of M is complemented in M , i.e. there exists a submodule V of M such
that M = N ⊕V (which means M = N +V and N ∩V = 0). Moreover, if
m∈M , then m∈mR.
Proof. By Zorn’s lemma, there exists a submodule V which is a direct sum
of irreducible submodules of M and is maximal with respect to the condition
N ∩V =0. If N +V (cid:54)=M , then there exists an irreducible submodule W of
V such that (N +V)∩W =0 and so N ∩(V +W)=0 which contradicts to
the choice of V . Thus M =N ⊕V .
Let 0(cid:54)=m∈M and N =mR+mZ. Then N is a submodule of M such
that NR⊆mR. By proved above, N =mR⊕W for a submodule W of N .
Since WR⊆mR∩W =0, this implies W =0 and so m∈mR.
For two right R-modules M and M , a group homomorphism α:M →
1 2 1
M is called an R-homomorphism or a module homomorphism from M into
2 1
M if (mr)α = mαr for all m ∈ M and r ∈ R. Clearly the kernel Kerα
2 1
and the image Imα of the R-homomorphism α are submodules of M and
1
M , respectively, and the factor module M /Kerα is isomorphic to Imα.
2 1
If M isaright R-module,thentheset End M ofall R-homomorphisms
R
from M into M (called R-endomorpisms of M ) forms a ring if the addition
and the multiplication of two elements α and β of End M are defined by
R
the rules
mα+β =mα+mβ and mαβ =(mα)β
for every m∈M . The identity mapping on M is a unity of this ring and M
can be viewed as a unital right End M -module if we put mα = mα for all
R
m∈M and α∈End M .
R
For a non-empty subset V of M , the set Ann (V) = {r ∈ R | vr =
R
0 for all v ∈ V} is a right ideal of R which is called the annihilator of V
in R. Moreover, the annihilator Ann (M) of M in R is an ideal of R. The
R
module M is faithful if Ann (M)=0.
R
Lemma 1.2. Let M be an irreducible right R-module and P =Ann (m) for
R
a non-zero element m∈M . Then the following statements hold:
1) P is a maximal right ideal of R such that the modules M and R/P are
isomorphic,
2) there exists an element e∈R\P such that r−er ∈P for every r ∈R,
and
3) the ideal I of R which is maximal with respect to the condition I ⊆ P
coincides with the annihilator Ann M .
R
Conversely, if for a maximal right ideal P of R there exists an element
e∈R\P such that r−er ∈P for every r ∈R, then the factor module R/P
is irreducible.
3
Proof. Since the mapping r (cid:55)→ mr with r ∈ R is a module homomorphism
from R onto M , its kernel P is a right ideal of R and the image mR is
a submodule of M . As M is irreducible and mR (cid:54)= 0, we have mR = M .
Therefore M is isomorphic to the factor module R/P and so P is a maximal
right ideal of R.
As m = me for some e ∈ R, we have mr = mer for every r ∈ R which
implies m(r−er)=0 and so r−er ∈P . Clearly e(cid:54)=P because m(cid:54)=0.
Finally, from Ann (M)⊆P itfollows Ann (M)⊆I. Ontheother hand,
R R
MI =(mR)I ⊆mI =0, so that Ann M =I
R
Conversely, if P is a maximal right ideal of R and r −er ∈ P for all
r ∈ R and some e ∈ R\P , then (e+P)R = R and so the R-module R/P
is irreducible.
Corollary 1.3. If a ring R, regarded as a right R-module, is irreducible, then
R is a division ring.
Proof. By Lemma 1.2, R has a left unity e. Therefore re = re2 and hence
(r−re)e=0 for every r ∈R. Since Ann (e)=0, it follows that r =re, so
R
that e is a unity of R. Finally, every non-zero element r ∈R is left invertible
in R because rR=R. Thus, r mustbeinvertiblein R andso R isadivision
ring.
Let R be a ring, M a right R-module and K = End M . Then M is
R
a unital right K-module, so that we can consider the ring L=End M . For
K
each r ∈ R, we define the mapping rˆ : M → M by the rule mrˆ = mr for
every m∈M . Then rˆ is a K-endomorphism of M because
(mα)rˆ=(mα)r =(mr)α =mrˆα
for all m ∈M and r ∈R. The mapping r (cid:55)→ rˆ determines a ring homomor-
phismˆ:R→L whose kernel coincides with Ann M and whose image Rˆ is
R
a subring of L. In particular, Rˆ is isomorphic to R provided that M is a
faithful R-module.
Let E = EndM be the ring of all group endomorphisms of M . Clearly
all rings K, L and Rˆ are subrings of E. Moreover, K = C (Rˆ) and L =
E
C (K), so that Rˆ ⊆ L = C (C (Rˆ)). It turns out that in many important
E E E
cases Rˆ coincides with L. A precise description of this situation is given by
the density theorem of Jacobson.
Theorem 1.4. Let M be a semisimple R-module and K = End M . Then
R
for each f ∈End M and any elements m ,...,m of M there exists r ∈R
K 1 k
such that mf =m r for all 1≤i≤k.
i i
Proof. Let k =1. Wewanttofindanelement r ∈R suchthat mf =m r. It
1 1
followsfromLemma1.1that M =m R⊕N forsomesubmodule N of M and
1
m∈mR forevery m∈M . Let π :M →m R betheprojectionfrom M onto
1
m R, i.e. m−mπ belongs to N for every m ∈ M . Obviously π is an R-
1
endomorphism of M , so that π ∈K. Therefore mf =(mπ)f =(mf)π =m r
1 1 1 1
for some r ∈R.
4
Considernowthegeneralcase. Let Mk =M⊕...⊕M with k summands.
Then Mk is a semisimple R-module. The mapping fk : Mk → Mk with
(x ,...,x )fk = (xf,...,xf) for any x ,...,x ∈ M is a K-endomorphism
1 k 1 k 1 k
of Mk. Hence, by proved above, there exists r ∈R such that
(m ,...,m )fk =(m ,...,m )r.
1 k 1 k
Thus mf =m r for every 1≤i≤k, as required.
i i
Corollary 1.5. Let M be a faithful irreducible R-module and K =End M .
R
Then K is a division ring, so that M is a vector space over K. If dim M <
K
∞, then R is isomorphic to End M .
K
Proof. Let m ,...,m be a basis of M over K and f ∈ End M . Then
1 k K
there exists r ∈ R such that mf = m r for every 1 ≤ i ≤ k by the density
i i
theorem 1.4. This implies mf = mr for every m ∈ M and so f = rˆ. Thus
Rˆ =End M .
K
Corollary 1.6. Let M be a finite-dimensional vector space over an algebraic
closed field F and let R be a subring of End M(∼= Mat (F)). If M is an
F k
irreducible R-module, then R is isomorphic to End M .
F
Proof. Since M is irreducible, then the ring K = End M is a division ring
R
by Schur’s lemma. Clearly every element of K is algebraic over F because
K =C (R) and every matrix over F is a root of its characteristic poly-
EndFM
nomial. Thus, for each α ∈ K, there exists a minimal polynomial p(x) over
F suchthat p(α)=0. Since F isalgebraicclosedandthepolynomial p(x) is
irreducible over F , we have degp(x) = 1 and so α ∈ F . Therefore K = F .
Let m ,...,m be a basis of M over F and f ∈End M =End M . Then
1 k K F
there exists r ∈R such that mf =m r for all 1≤i≤k and hence mf =mr
i i
for every m∈M . Thus f =r and so R=End M .
F
Aring R iscalled(right)primitiveifthereexistsafaithfulirreducible(right)
R-module. It is known that right primitivity does not imply left primitivity.
An example of such a ring was given by G. Bergman (1964).
Example. 1. If R isaringand M isanirreducibleright R-module, then
thefactorring R/Ann M isprimitivebecause M canbeviewedasafaithful
R
irreducible R/Ann M -module.
R
2. Every division ring is primitive.
3. The ring Z and any ring with zero multiplication are not primitive.
Proposition 1.7. If a commutative ring R is primitive, then R is a field.
Proof. If M is a faithful irreducible R-module, then Ann (m) = 0 for each
R
non-zero m∈M by Lemma 1.2.3) and so R is a field by Corollary 1.3.
Proposition 1.8. Let R be a primitive ring. Then either R is isomorphic
to the ring Mat (D) of all (n×n)-matrices over a division ring D for some
n
integer n ≥ 1 or for each n ≥ 1 there exists a subring S of R such that
n
Mat (D) is a homomorphic image of S .
n n
5
Proof. Let M beafaithfulirreducible R-moduleand D =End M . Then M
R
isavectorspaceover D. If M isfinite-dimensionalover D and dim M =n
D
for some n≥1, then R is isomorphic to End M by Corollary 1.5 and so to
D
Mat (D).
n
Let M be infinite-dimensional and V its subspace of dimension n ≥ 1.
Take a basis v ,...,v of V and denote by W a vector subspace of M such
1 n
that M =V ⊕W . Every non-zero element f ∈End V can be considered as
D
an element of End M if we put Wf =0. Thus, by the density theorem 1.4,
D
there exists r ∈ R such that vf = v r for all 1 ≤ i ≤ n. Clearly Vr ⊆ V .
i i
Put S = {s ∈ R | Vs ⊆ V}. Then S is a subring of R and the factor ring
n
S /Ann V is isomorphic to End V and so to Mat (D).
n Sn D n
The following lemma gives an internal description of the division ring D
from Proposition 1.8 by means of the structure of R itself. If S is a subring
of R, then the set
Id S ={r ∈R|rS ⊆S and Sr ⊆S}
R
is a subring of R in which S is an ideal. The subring Id S is called the
R
idealizer of S in R.
Lemma 1.9. Let M be a faithful irreducible right R-module and P the an-
nihilator of a non-zero element of M in R. If S = Id P , then the factor
R
ring S/P is a division ring whose multiplicative group is isomorphic to the
multiplicative group of End M .
R
Proof. By Lemma 1.2, P is a maximal right ideal of R such that there exists
anelement e∈R\P with r−er ∈P forevery r ∈R. Inparticular, e−e2 ∈P
and eP ⊆P , so that e∈S. Show first that s−se∈P for all s∈S. Since
(s−se)e=s(e−e2)∈P , it suffices to prove that the inclusion se∈P implies
s∈P if s∈S.
Indeed,inthiscase ser ∈P and sr−ser =s(r−er)∈P forevery r ∈R,
so that sR⊆P . As eR(cid:34)P and the set {r ∈R|rR⊆P} is a right ideal of
R containing P , we have P ={r ∈R|rR⊆P} and so s∈P , as desired.
Thus, e is a unity of S modulo P . Next, for each s ∈ S \P , we have
sR+P =R andso e=sr+x forsome r ∈R and x∈P . Therefore e−sr ∈P
and, in particular, sr ∈S. This implies r ∈S because otherwise R=rP +P
and so sR = srP +sP ⊆ P , contrary to the above. Clearly r ∈/ P , so that
there exists t ∈ S with e−rt ∈ P . Since the elements (e−sr)t, s(e−rt),
t−et and s−se areallcontainedin P ,wehave s−t∈P andso e−rs∈P .
Hence the factor ring D = S/P is a division ring. To prove that the
multiplicative groups of D and End M are isomorphic it suffices to assume
R
that M =R/P becausetheright R-modules M and R/P areisomorphicby
Lemma 1.2. For each d = s+P of D and every m = r+P of M , we put
dm=sr+P and show that M is a left vectorspace over D and the mapping
dˆ:m(cid:55)→dm is an R-endomorphism of M . .
Indeed, if d = s +P and m = r +P for some s ∈ S and r ∈ R,
1 1 1 1
then both elements s − s and r − r belong to P , so that sr − s r =
1 1 1 1
6
(s−s )r+s (r−r ) ∈ P and therefore s r +P = sr+P . Furthermore,
1 1 1 1 1
for all d,d ∈ D and m,m ∈ M , the relations d(m+m ) = dm+dm ,
1 1 1 1
(d+d )m=dm+d m and (dd )m=d(d m) areimmediatelyverified. Finally,
1 1 1 1
for every t ∈ R, we have (mt)dˆ= d(mt) = srt+P = (dm)t = (mdˆ)t, so that
dˆ∈End M .
R
Itiseasilyseenthatthemapping d−1 (cid:55)→dˆ determinesagroupisomorphism
(cid:98) from D\{0} into the multiplicative group of End M . Now, if α is a non-
R
zero element of End M and v = e+P , then vα = s+P = vs for some
R
s ∈ R\P . As P = Ann (v) and v(sP) = (vα)P = (vP)α = 0, it follows
R
that sP ⊆P and so s∈S\P . Take the element d=s+P ∈D\{0}. Since
s+P =se+P and m=r+P =vr,wehave mα =(vr)α)=(vα)r =sr+P =
dm and so α = dˆ. Thus the group homomorphism(cid:98) is surjective and so the
multiplicative groups of D and End M are isomorphic.
R
2 The Jacobson radical and its properties.
Foraring R,theintersectionoftheannihilatorsofallsimpleright R-modules
isanideal J(R) of R calledtheJacobsonradicalof R. Asweshallseefurther,
J(R) coincidesalsowiththeintersectionoftheannihilatorsofallsimpleleft R-
modules. But before we give a characterization of J(R) by internal properties
of the ring R.
Consideron R the”circlemultiplication” r◦s=r+s+rs forall r,s∈R
under which the set of all elements of R forms a semigroup Rad with neutral
element 0∈R. The group of all invertible elements of this semigroup is called
the adjoint group of R and denoted by R◦. The following lemma describes a
relationship between J(R) and R◦.
Lemma 2.1. Let R be a ring and J a right ideal of R. Then J ⊆ J(R) if
and only if J ⊆R◦.
Proof. Let J ⊆ J(R) and assume that J (cid:34) R◦. Then there exists an
element a ∈ J such that the set P = {r+ar | r ∈ R} is a proper right ideal
of R.
Indeed, otherwise for every a ∈ J there exist elements b,c ∈ R such that
b+ab = −a and c+bc = −b. This means that a◦b = b◦c = 0 and so
a = a◦b◦c = c. Therefore a◦b = b◦a = 0 and hence a ∈ R◦, contrary to
the assumption.
Clearly R =aR+P and thus a does not belong to P , so that by Zorn’s
lemma there exists a maximal right ideal S of R containing P . Since the
element −a is a left unity modulo S, the right R-module M = R/S is
irreducible by Lemma 1.2 and Ma = M (cid:54)= 0, contrary to the inclusion a ∈
J ⊆J(R). Thus J ⊆R◦.
Conversely, let J ⊆ R◦ and J (cid:34) J(R). Then there exists an irreducible
right R-module M such that MJ (cid:54)= 0 and so mJ = M for some m ∈ M .
Therefore ma = −m for some a ∈ J. Since a◦b = 0 for some b ∈ R, we
7
have −m = ma+(m+ma)b = m(a◦b) = 0 and therefore M = 0, contrary
to the choice of M . Thus J ⊆J(R).
ThefollowingcharacterizationoftheJacobsonradicalof R isanimmediate
consequence of Lemma 2.1.
Corollary 2.2. For every ring R, the Jacobson radical of R is the largest
ideal of R which is contained in the adjoint group R◦ of R.
By symmetry, Corollary 2.2 implies that J(R) coincides with the intersec-
tion of the annihilators of all left simple R-modules.
The second important property of the Jacobson radical of R concerns the
adjoint group of a factor ring of R.
Corollary 2.3. If R is a ring and J is an ideal of R contained in J(R),
then J◦ is a normal subgroup of R◦ and (R/J)◦ =R◦/J ∼=R◦/J◦.
Proof. Since J ⊆R◦ by Lemma 2.1, this implies that J =J◦ and so r+J ∈
(R/J)◦ for some r ∈R if and only if r ∈R◦.
Notethatif I isanidealof R,theningeneraltheadjointgroup (R/I)◦ of
the factor ring R/I need not be a factor group of R◦. For instance, (Z/5Z)◦
hasorder 4 and Z◦ hasorder 2. Ontheotherhand, itisclearthattheimage
of R◦ in R/I is contained in (R/I)◦, so that (J(R)+I)/I lies in J(R/I).
Finally, the third property of J(R) allows us to divide the class of all rings
intotwoimportantsubclasses: theclassofringsoverwhichirreduciblemodules
exist and the class of rings whose all simple modules are trivial.
Corollary 2.4. If R is a ring, then J(R/J(R))=0.
Proof. Let J be the full preimage of J(R/J(R)) in R. Since J/J(R) ⊆
(R/J(R))◦ =R◦/J(R) byLemma2.1andbyCorollary2.3,thisimplies J ⊆R◦
and so J =J(R).
We shall say that a ring R is semiprimitive if there exists a collection of
ideals of R modulo each of which the ring R is (right) primitive and the
intersectionofallidealsofthiscollectioniszero. Thefollowingstatementshows
that in contrast to primitivity the definition of semiprimitivity is symmetrical.
Lemma 2.5. A ring R is semiprimitive if and only if J(R)=0.
Proof. If I is an ideal of R modulo which R is primitive, then J(R) ⊆ I
because J(R)+I/I ⊆ J(R/I) = 0. Thus, J(R) = 0 provided that R is
semiprimitive. Conversely, the annihilator of every irreducible right R-module
is an ideal of R modulo which R is primitive. Since the intersection of the
annihilators of all irreducible right R-modules is J(R), this means that R is
semiprimitive provided that J(R)=0.
The following theorem may be considered as one of the main tools of inves-
tigation of the structure of an arbitrary associative ring.
8
Theorem 2.6. Every ring is semiprimitive modulo its Jacobson radical.
Proof. It follows immediately from Corollary 2.4 and Lemma 2.5.
3 Artinian rings.
Let R be a ring. A right R-module M is called artinian if M satisfies
the minimal condition for the submodules of M . In other words, every set
of submodules of M has a least element with respect to the set-theoretical
inclusion.
Obviously submodules and factor modules of artinian modules are artinian.
It is also clear that each direct sum of finitely many simple R-modules is an
artinian R-module and that every semisimple artinian R-module has only
finitely many irreducible direct summands. In particular, vector spaces over a
division ring are artinian if and only if they are finite-dimensional.
A ring R is called (right) artinian if R, regarded as a right R-module, is
artinian. Thefollowingdescriptionofartinianprimitiveringsisinfactobtained
as an application of the density theorem of Jacobson.
Proposition3.1. Aprimitivering R isartinianifandonlyif R isisomorphic
to the ring Mat (D) of all (n×n)-matrices over a division ring D for some
n
n≥1.
Proof. If R is isomorphic to Mat (D), then R can be considered as an n2-
n
dimensionalvectorspaceover D,sothat R satisfiestheminimalconditionfor
the subspaces of R. Since every right ideal of R is simultaneously a subspace,
the ring R is artinian.
Conversely, let R be an artinian ring, M a faithful irreducible R-module
and D = End M . Then D is a division ring and M can be viewed as a
R
vector space over D. Show that M is finite-dimensional over D.
Suppose the contrary and let {m ,...,m ,...} be a basis of M over D.
(cid:84) 1 s
Foreach s≥1,let P = s Ann (m ) andlet f denotethelinearoperator
s i=1 R i s
of End M defined by the rule mfs =0 for i≤s and mfs =m for i>s.
D i i i
Then P ⊆ P and, by the density theorem, there exists some r ∈ R such
s+1 s
that mfs = m r for all 1 ≤ i ≤ s+1. Therefore m r = o for 1 ≤ i ≤ s and
i i i
m r = m , so that r ∈ P \P . But then P ⊃ ... ⊃ P ⊃ P ⊃ ...
s+1 s+1 s s+1 1 s s+1
is an infinite strongly descending chain of right ideals of R, contrary to the
artinianity of R .
Hence dim M =n forsome n≥1 andthus R isisomorphicto End M
D D
and so Mat (D) by Corollary 1.5.
n
A ring R is called simple if R has no non-trivial ideals. Clearly every
division ring is simple.
Lemma 3.2. For each n ≥ 1, the matrix ring Mat (D) over a division ring
n
D is simple.
9
Proof. For each pair (i,j) with 1 ≤ i,j ≤ n, let E denote the (n×n)-
ij
matrix whose elements, excepting the unity 1 in the position (i,j), are zero.
Define the element δ of D by the rule δ = 0 if i (cid:54)= j and δ = 1 if
ij ij ij
i = j. Then E E = δ E for any pairs (i,j) and (k,l). Clearly the set
ij kl jk il
{E |1≤i,j ≤n} is a basis of Mat (D), regarded as a vector space over D.
ij n (cid:80)
Thus a non-zero ideal I of Mat (D) contains a matrix A= a E with
n i,j ij ij
a ∈ D and a (cid:54)= 0 for some pair (p,q). A routine calculation shows that
ij pq
E =a−1E AE ∈I for each pair (k,l) and so I =Mat (D).
kl pq kp ql n
As an application of Lemma 3.1 and Lemma 3.2, we can now obtain a com-
plete description of the structure of semiprimitive artinian rings.
Theorem 3.3. (Wedderburn-Artin) A semiprimitive ring R is artinian if and
only if R is isomorphic to a direct sum of finitely many matrix rings over
division rings.
Proof. Sinceeachdirectsumofprimitiveringsisasemiprimitivering,itfollows
from Proposition 3.1 that R is semiprimitive and artinian. To prove the con-
verse, takeaminimalideal V of R. Clearlythereexistsanideal I of R such
that I∩V =0 andthefactorring R/I isprimitive. ItfollowsfromProposition
3.1 that R/I is isomorphic to a matrix ring Mat (D) for some division ring
n
D and n≥1. As (V +I)/I is a non-zero ideal of R/I, we have R=I +V
by Lemma 3.2 and so V , as a ring, is isomorphic to Mat (D). Furthermore,
n
VI =IV ⊆I∩V =0, so that every ideal of I is an ideal of R. In particular,
the Jacobson radical J(I) of I is an ideal of R. Since J(I) ⊆ I◦ ⊆ R◦,
this implies J(I) ⊆ J(R) = 0 by Lemma 2.1 and so I, regarded as a ring, is
semiprimitive and artinian. Taking a minimal ideal of I which is simultane-
ously a minimal ideal of R and repeating the same arguments, we obtain after
finitelymanystepsthat R isadirectsumoffinitelymanyidealseachofwhich,
regarded as a ring, is isomorphic to a matrix ring over a division ring.
If S and T are subrings of a ring R, then ST denotes the additive sub-
group of R generated by the set {st | s ∈ S,t ∈ T}. It is easily verified that
(ST)U =S(TU) foreverysubring U of R. Furthermore, ST isevena(right)
ideal of R provided that both S and T are (right) ideals of R. Finally, for
each positive integer n we put S1 = S and Sn = Sn−1S. A subring S is
called nilpotent if Sn =0 for some n≥1.
Theorem 3.4. The Jacobson radical of an artinian ring is nilpotent.
Proof. Let R be an artinian ring and J the Jacobson radical of R. Then
the descending chain J ⊇ ... ⊇ Jn ⊇ Jn+1 ⊇ ... of ideals of R cannot be
infinite, so that Jn = Jn+1 for some n ≥ 1. Assume that Jn (cid:54)= 0 and put
I ={r ∈R|rJn =0}. Then I isanidealof R and J (cid:34)I becauseotherwise
JJn =Jn+1 =0. In particular, R(cid:54)=I and so the factor ring R/I contains a
minimal non-zero right ideal S/I. Since S/I is a simple right R-module, it
must be annihilated by J, so that SJ ⊆I. Therefore (SJ)Jn =0 and hence
10
SJn+1 = SJn = 0. This implies S ⊆ I, contrary to the choice of S. Thus
Jn =0.
As it follows from Wedderburn-Artin theorem 3.3, every semiprimitive ring
is right artinian if and only if it is left artinian. The following example shows
that a corresponding statementis not true in general. In other words, there are
rings satisfying the minimal condition for the right but not the left ideals.
Example. (Small)Let F beafieldand F(x) thefieldofrationalfunctions
(cid:181) (cid:182)
F F(x)
over F . Consider the ring R of (2×2)-matrices of the form
0 F(x)
(cid:181) (cid:182)
0 F(x)
over F(x). Then J(R) = , so that J(R)2 = 0 and the factor
0 0
ring R/J(R) is isomorphic to the direct sum of two fields F and F(x). It is
easily seen that the right ideals of R contained in the Jacobson radical J(R)
are vector spaces over F(x) and, on the other hand, the left ideals of that are
vectorspacesover F . Since J(R),regardedasavectorspace,is 1-dimensional
over F(x) and infinite-dimensional over F , it is a minimal right ideal of R
and contains infinite strongly descending chains of left ideals of R. Thus the
ring R is right artinian but not left artinian.
4 Division rings. Wedderburn’s theorem on fi-
nite division rings.
4.1 Cyclotomic polynomials.
Let n be a positive integer and let (cid:178) ,...,(cid:178) be all roots of degree n from
0 n−1
1 in the field C of complex numbers. We know that (cid:178) =cos2πm +isin2πm
m n n
for every 0 ≤ m ≤ n−1. The root (cid:178) is primitive if (m,n) = 1. The
m
number of all primitive roots of degree n is equal to ϕ(n), Euler’s function
of n. Let δ ,...,δ be the primitive roots of degree n. Put Φ (x) =
1 ϕ(n) n
(x−δ )...(x−δ ). Then Φ (x) is the n-th cyclotomic polynomial on x.
1 ϕ(n) n
It is easy to see that
(cid:81)
1) Φ (x)= Φ (x), where d runs all distinct divisors of n, and
n d|n d
2) (Φ (x),Φ (x))=1 provided that m(cid:54)=n.
n m
Proposition 4.1. For every positive integer n, the polynomial Φ (x) has
n
integer coefficients, and if d is a proper divisor of n, then Φ (x) is a divisor
n
of xn−1 .
xd−1
Proof. If n=p isaprime,then Φ (x)=xp−1+...+x+1 andtheproposition
p
is proved. By induction on n, for every proper divisor d of n the polynomial
(cid:81)
Φ (x) hasdesiredproperties. Thereforethepolynomial f(x)= Φ (x)
d d|n,d(cid:54)=n d
has integer coefficients and f(x) is a divisor of xn−1 by 2). Since xn−1=
Φ (x)f(x) by 1), Φ (x) has the same property and clearly is coprime with
n n
xd−1 for every d|n,d(cid:54)=n. Hence Φ (x) is a divisor of xn−1 .
n xd−1