Table Of ContentAssociative 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
