, "
PROPERTIES OF MATRIX RINGS
S.M~' Kaye ()
SOME PROPERTIES OF A RING WHICH ARE INHERITED
BY THE RING OF nxn MATRICES OVER THAT RING
She ilaM. Kaye
Q V
Submitted to the Faculty of Graduate Studies and Research in
pa:rtial~- fulfilment of the requirements for the degree of Master of
Science
Department of Mathematics,
McGil1 University,
Montreal, P.Q. August, 1966
@) Shie1aM. Kaye 1967
- .. '".". o Preface
As one might expect, many theorems on properties of a ring which
are inherited by the matrix ring appear in papers concerned with a
particular property of the ring. Basically, the aim of this thesis
is to collect such results. There are two advantages in making such
a collection. The first is that by suitable organization, the task
of proving these theorems may be simplified considerably. This is
seen in Sections 1 and 3. Secondly, the question arises: "Are any
other properties of a ring inherited by the matrix ring?" Theorems
3.6 and 3.7 provide a partial answer to this questiono
l would like to thank Professor l. Connell for suggesting this
topic, and for the great deal of help and encouragement he has given () me.
(
1 ~) Contents
page
Conventions 1
1. Simple, Primitive, and Prime Rings 3
Simple Rings 4
Prime Rings 4
Primitive Rings 5
Semi-simple, Semi-primitive, and Semi-prime Rings 8
2. Chain Conditions 9
Artinian Rings 9
Noetherian Rings 9 ()
3. A One-to-one Correspondence between Left R-modules and
Left R(n)-modules 11 Self-injective Rings 18
Hereditary and semi-hereditary Rings 19
Perfect and Semi-perfect Rings 20
Regular Rings 22
Homological Dimension 25
Appendix: Alternative Proof for Theorem 3.10 30
4. Coherent Rings 32 o Summary 38
Bibliography 40 1
1. Every ring R has a unit e1ement 1 ~ 0 which acts as the identity
operator on every R-modu1e.
2. The ring of aU n x n matrices over a ring R is denoted by R(n).
If S is a subset of R, S (n) represents the set of aU n x.'.n matrices
with entries in S. If 1 ~ i ~ n and 1 ~ j ~ n, e.. € R( ) is defined ~J n
t o b e th e ma t r~x· wh ose on1 y non-zero ent ry ~s . 1·~n t h e J . th pos~t~on.. 0 f .th t h e ~ row. Every matrix ri ri be written unique1y in the form ~~r ..e ..• I!:.I J" ~J ~J • • • rr~n) may . . . () nn
3. n remains a fixed integer greater than 1. The resu1ts are true
for n = 1, but the proofs are not necessari1y va1id in this case.
4. By "idea1" is meant a 2-sided ideal.
5. R is considered to be a 1eft idea1 of itse1f, and an idea1 of itse1f.
However, R is not a maximal (left) idea1 of itself.
6. Interna1 direct surns are denoted by A + B or ~ [3A[3' Externa1
direct surns are denoted by Aê B or œ[3A[3' Direct products are denoted
by rr [3A[3.
7. Except where otherwise indicated, a11 R-modu1es are 1eft R-modu1es, 2
() and the theorems are stated for "left" properties. By symmetry, the
corresponding theorems for "right" properties hold.
() 3 o 1. Simple, Primitive, and Prime Rings
In this section we show that a ring is simple, primitive, or prime
if and on1y if the matrix ring has the same property. We define three
radica1s of a ring and show that any of these radica1s is 0 if and on1y
if the corresponding radical in the matrix ring is O. In order to prove
these resu1ts, we make use of a one-to-one correspondence between the
idea1s of a ring and the idea1s of the matrix ring.
Theorem 1.1. l is an idea1 of R(n) if and on1y if l = J(n) for some idea1 J of R.
Proof. If J is an idea1 of R, then J( , is certain1y an idea1 of D)
() Let l be an idea1 of R(n) and let J = [aU: ~aijeij € Ij. To show
that J is an idea1 of R, let a,b € J and r € R. Then there exist
i aijeij € l with aU = a and ~ bijeij € l with bU = b •. reU € R(n)
and e € R(n)' If f.a .. e .. + f.b .. e .. = "Lc .. e .. € l, then a + b = c 11 ~J ~J ~J ~J ~J ~J 11
€ J.
= (ra)e € 1. Therefore ra € J. reU (~ aijeij)eU U
= (ar)e € 1. Therefore ar € eU (~aijeij)reU U J. Rence J is an idea1 in R.
To show that l ~ J (n) let ~ aijeij € 1. ak ,eU = e1k( i. aijeij)e Il
€ 1. Therefore a € J for aU k,l such that 1 ~ k ~ n, 1 ~ 1 ~ n. kl
Rence l ~ J(n)'
To show that J(n) ~ l observe that J(n) = ~Jeij' It is sufficient
to show for any k,l, that Je \ ~ 1. Let a € J. Then there exists o k i.aijeij € l with aU = a. aek\ = ek1 (~ai/ij)ell € 1. Therefore J (n) 4
() CI. This completes the proof.
Definition. A ring R is simple if 0 is a maximal ideal of R.
Theorem 1.2. 1 is a maximal ideal of R if and only if l(n) is a
maximal ideal of R(n).
Proof. Let 1 be a maximal ideal of R. Then l(n) is an ideal of
R(n). Suppose there exists an ideal K (:f: l(n» of R(n) with l(n) C. K. By Theorem 1.1, there exists an ideal J of R suchthat K = J(n). Then
1 CJ and 1 :f: J. Since 1 is a maximal ideal of R, J = R. Therefore
K = J(n) = R(n)' and l(n) is a maximal ideal of R(n).
Let l(n) be a maximal ideal of R(n). Then 1 is an ideal of R. If () there exists an ideal J of R such that 1 C J and 1 :f: J, then l(n) C J(n) and l(n) :f: J(n). Therefore J(n) = R(n) and J = R. Therefore 1 is a maximal ideal of R.
Corollary 1.2. R is a simple ring if and only if R(n) is simple.
Definitions. An ideal 1 of R is prime if for aIl ideals J,K of
R, JK c;. 1 ~J ç. 1 or K <.;;,.1. A ring R is prime if 0 is a prime ideal
of R.
Lemma 1.3. Let J and K be ideals of R. Then J(n)K(n) = (JK)(n).
Proof~ Clearly J(n)K(n) C (JK)(n). It is also clear that for any
" " "th 1 L.. "..t:.. 1 ~ "~ JK - J K c. J K S" ~,J w~ - ~ - n, - J - n, eij - en e lj - (n) (n). ~nce o (JK) (n) = I~ JKe ij, (JK) (n) ~ J (n)K(n)· 5
Theorem 1.4. l is a prime idea1 of R if and on1y if l(n) is a
prime idea1 of R(n). Proof. Suppose l is a prime idea1 of R. Let J and K be idea1s of
R sueh that J(n)K(n) ç l(n). (By Theorem 1.1, a11 idea1s of R(n) are
of this form.) By Lemma 1.3, (JK)(n) G l(n). Renee JK c... l. Sinee l
is a prime idea1 of R, either J ~ l or K ~ l. Therefore J(n) G l(n)
or K(n) ~ l(n)' and l(n) is a prime idea1 of R(n). Suppose l(n) is a prime idea1 of R(n). Let J,K be idea1s of R
sueh that JK ~l. Then J(n)K(n) = (JK)(n) ~l(n). Sinee l(n) is a
prime idea1 of R(n)' either J(n) ~ l(n) or K(n) c... l(n). Therefore
J ~ l or K ~ l, and l is a prime idea1 of R.
Coro11ary 1.4. R is a prime ring if and on1y if R(n) is a prime
() ring.
Definitions. An idea1 l of a ring R is primitive if l is the annihilator of R/M (1 = Ann(R/M» for some maximal 1èftrHdea1 M of R. A ring R is primitive if 0 is a primitive idea1 of R.
Lemma 1.5. Let M be a maximal 1eft idea1 of R. l = Ann(R/M) if and on1y if l is an idea1 of R, l , M, and if J is an idea1 of R sueh
tha t J ~ M, J S 1.
Proof. Let l = Ann(R/M). Let a,b € land r,x € R.
(a +. b)(x + M) = a(x + M) + b(x +. 'M) = 0 + M. Therefore a + b € 1.
ra(x + M) = r(O + M) = 0 + M. Therefore ra € 1. o ar(x + M) = a(rx + M) = 0 + M. Therefore ar € 1. Therefore l is an idea1 of R. C1early l ~ M. Let J ~M be an idea1 of 6 o R and let y € J. Then yx € J ~M. Renee y(x + M) = 0 ~ M for aIl x € R and y € I. Therefore J ~ I.
Let l be an ideal of R satisfying the two conditions above. Since
IR = l ~ M, l ~ Ann(R/M). Sinee Ann(R/M)~ M is an ideal of R; Ann(R/M)
~I.
Lemma 1.6. R is a primitive ring if and only if there exists a
faithful simple left R-module.
Proof. Let R be a primitive ring. Then there exists a maximal
left ideal M of R sueh that Ann(R/M) = O. This means that R/M is a faithful left R-module. R/M is simple sinee M is a maximal left ideal
of R.
Let A be a faithful simple left R-module and let a €1\A, a ~ O. Sinee A is simple, Ra = A. Renee the mapping n::R --:r A where n:(r) = ra
is an epimorphism. Therefore A ~R/Ker(n:). To show that Ker(n:) is a
maximal left ideal of R, suppose that M is a left ideal of R, M ~ R,
and Ker(n:) ~ M. Then M/Ker(n:) is a submodule of R/Ker(n:). Sinee A is simple, so is R/Ker(n:), and M/Ker(n:) = O. Therefore M = Ker(n:) and
Ker(n:) is a maximal left ideal of R. Now let x € Ann(R/Ker(n:)). Then
for aIl r € R, xr € Ker(~) and xn:(r) = n:(xr) = O. Since n: is an
epimorphism, x € Ann(A) = O. Therefore Ann(R/Ker(n:)) = 0, and R is a primitive ring.
Lemma 1.7. If R is a primitive ring and e is an idempotent in R,
then eRe is a primitive ring.
Proof. Let R be a primitive ring and let e be an idempotent in R.
By Lemma 1.6, there is a faithful simple left R-module A. We sho~ that 7
() eA is a faithful simple left eRe-module.
~A is clearly a left eRe-module. To show that eA is faithful, let
exe € Ann(eA). For aIl a € A, (exe)a = (exe)ea = O. Therefore exe €
Ann(A) = O. To show that eA is simple, let a,al € A, ea f O. Since A
is simple, ea € A generates A. Therefore there exists r € R such that rea = al. ea l = e(rea) = (ere)(ea). Rence ea generates eA, and eA is
simple.
Theorem 1.8. R is a primitive ring if and only if R(n) is a
primitive ring.
Proof. Suppose that R is a primitive ring. Let M be a maximal
left ideal of R such that Ann(R/M) = O. M(n) is clearly a left ideal
of R(n)' which by Zornls Lemma is contained in a maximal left ideal MI.
By Lemma 1.5, Ann(R(n)/M' ) is an ideal of R(n)' and by Theorem 1.1, there
exists an ideal J of R such that Ann(R(n)/M' ) = J(n). By Lemma 1.5, if J f 0, then J $ M. Since M is maximal, J + M = R. Therefore R(n) =
(J"+ M) (n) = J (n) + M(n) ~ MI. This is a contradiction. Hence J = 0
and J(n) = O. Therefore R(n) is a primitive ring.
Suppose that R(n) is a primitive ring. e±i is an idempotent in
R(n). By Lemma 1.J, ellR(n)ell is a primitive ring. Since R ~ ellR(n)ell
as rings, R is a primitive ring.
Corollary 1.8. l is a primitive ideal of R if and only if l(n) is
a primitive ideal of R(n).
Proof. First we show that l is a primitive ideal of R if and only o if R/l is a primitive ring. Let M be a maximal left ideal of R such
that l = Ann(R/M). Since aIl left ideals of R/l are of the forro KIl 8 () where K is a left ideal of R with l ç K, MIl is a maximal left ideal of R/l. Suppose there is an ideal J/l of R/l such that J/l ~M/l. Then
J isan ideal of Rand l G J ~ M. By Lemma 1.5, J ~ 1. Rence J/I = 0
and by Lemma 1.5, R/l is a primitive ring. Now suppose that R/l is a
primitive ring, .where l is an ideal of R. There exists a maximal left
ideal M of R such that Ann(~RtI)/(M/l» = O. Since l ~ M, lÇ Ann(R/M)
= J. J/l ~M/l is an ideal of R/l. By Lemma 1.5, J/l = O. Therefore
Ann(R/M) = l, and l is a primitive ideal of R.
To show that (R/l)(n) ~ R(n)/l(n) as rings, consider the ring
homomorphism rr:R( ) ~(R/l)( ) where rr(z.r .. e .. ) = z..(r .. + I)e ... n n 1.J 1.J 1.J 1.J Clearly rr is an ep imorphi sm, and Ker(rr) = l(n)' Therefore (R/l)(n)
~R(n)/l(n) as rings. lt follows that l is a primitive ideal of R if
and only if l(n) is a primitive ideal of R(n)' ( )
Definitions. The Jacobson radical of a ring R is the intersection
of aIl the primitive ideals of R. R is semi-primitive if the Jacobson
radical of R is O.
The prime radical of a ring R is the intersection of aIl the prime
ideals of R. R is semi-prime if the prime radical of R is O.
The Brown-McCoy radical of a ring R is the intersection of aIl the
maximal ideals of R. R is semi-simple if the Brown-McCoy radical of R
is O.
Theorem 1.9. A ring R is semi-simple, semi-primitive, or semi-prime
if and only if R(n) has the same property.
Proof. The theorem follows immediately from Theorem 1.2, Theorem
o 1.4, Corollary 1.8, and the definitions. 9 "c) 2. Chain Conditions
In this brief section we show that a ring is artinian or noetherian
if and only if the matrix ring has the same property.
Definitions. A module A over a ring R is artinian if every strictly
decreasing sequence of submodules of A is finite. R is a left artinip.~
ring if the left R-module R is artinian.
A module A over a ring R is noetherian if every strictly increasing
sequence of submodules of A is finite. R is a left noetherian ring if
the left R-module R is noetherian.
Lemma 2.1. If M and N are artinian or noetherian modules over R, so ( ) is M ~ N.
Proof. Let A = M El) N and B = M e O. Then B <;;.A, M ~ B, and N ;;- A/B.
It is sufficient to show that if B ~ A and Band A/B are artinian
(noetherian), then so is A.
Suppose Band A/B are artinian. Let A = AO ~ Al"2 A2 '2 .•. be a
decreasing sequence of submodules of A. Then B ;2 Al B ~ A ("\ B 2- n 2 is a decreasing sequence of submodule"s of B, and A/B '2. (Al + B) /B 2
(A + B)/B ~ ... is a decreasing sequence of submodules of A/B. Since 2 Band A/B are artinian, there exists k such that for i, j >. k, A. n B 1.
= A. () B and (A. + B)/B = (A. + B)/B. Suppose i < j. Then A. ;2 A .. To J 1. J 1. J show that A. E. A., let a. € A.. Since (A. + B)/B = (A. + B)/B, A. + B 1. J 1. 1. 1. J 1.
= A. + B. Therefore there exists a. € A. such that a. + B = a. + B. J J J 1. J
a. - a. € A. () B = A. () B. Rence a. = (a. - a.) + a. € A.. Therefore 1. J 1. J 1. 1. J J J c> Aj = Ai' and A is artinian. 10
Suppose Band A/B are noetherian. Let 0 f: Al ~ A c,; ••• be an 2
increasing sequence of submodules of A. Then 0 ~ Al" B S A2 '" B ~
is an increasing sequence of submodules of Band 0 ~ (Al + B)/B ~
(A + B)/B ~ ... is an increasing sequence of submodules of A/B. The 2 remainder of the proof is similar to the artinian case.
Corollary 2.1. If R is artinian or noetherian, then R(n) is an
artinian or noetherian left R-module. 2 Proof. R(n) is a direct SUffi of n copies of R. The proof is by
induction.
Theorem 2.2. R is a left artinian or left noetherian ring if and
only if R(n) has the same property.
Proof. Any sequence of left ideals of R(n) is a sequence of
submodules of the left R-module R(n)' Renee, if R is left artinian or
left noetherian, then so is R(n)'
If l is a left ideal of R, then I(n) is a left ideal of R(n)'
Renee any chain of left ideals of R gives rise to a chain of left ideals
of R(n)' Therefore if R(n) is a left artinian or left noetherian ring,
then so is R.
o Il o 3. A One-to-one Correspondence between Left R-modules and Left R(n)-modules
In this section we define two exact covariant functors, .one from
the category of left R-modules to the category of left R(n)-modules,
and one in the opposite direction. These functors are mutual inverses,
which take projective modules into projective modules, injective
modules into injective modules, and finitely generated modules into
finitely generated modules. As a result, R is left he~èditary, left
semi-hereditary, left self-~njective, left perfect, or left semi-perfect
if andonly if R(n) has the same property. We define a natural isomorph
ism between the tensor product of a right and a left R-module, and the
tensor product of the corresponding R(n)-modules, and deduce that the
functors take fIat modules into fIat modules. It follows that R is
(Von Neumann) regular if and only if R(n) is. We conclude the main
part of this section with a discussion of homological dimension. It is
easily seen that the functors preserve projective, injective, and weak
homological dimension. It follows immediately that the global and
finitistic projective, injective and weak dimensions of R(n) are equal
to those of R.
In the appendix to this section we provide an alternative proof to
Theorem 3.10 which is an exercise in (4J.
Definitions. A category e consists of the following:
(a) A class of objects A,B,C,D ••• We may write A € e. (b) A disjoint family of sets Hom(A,B), one for each pair of
objects. The elements of Hom(A,B) are called morphisms from A to B. 12 () If 0( € Hom(A,B), we write O<:A~B or A~B. (c) A ru le which assigns to each ~ € Hom(A,B) and ~ € Hom(B,C)
an e1ement ~o( € Hom(A, C) •
(d) A ru le which assigns to each object A an e1ement lA € Hom(A,A).
The following two axioms are satisfied: (i) If O(€ Hom(A,B),
~ € Hom(B, C), and y € Hom(C,D), then (~) 0( = y(~O(). (ii) If 0{ €
Hom(A,B) then OC1A = 0{ = 1B~'
A covariant functor F from a categoryC to a category ~ is a pair of functions F, one which assigns to each object A of e an object F(A)
of ~~ and one which assigns to each ~€ Hom(A,B) a morphism F(~) €
Hom(F(A),F(B». Two axioms are satisfied: (i) F(l ) = 1 (A)' A F (H) F(~O{) = F(~)F(O(). () Let F and G be two covariant functors from a category e to a category e. A natura1 transformation y from F to G consists of morphisms
YA € Hom(F(A),G(A», one for each object A of e, such that if B € e and
~ € Hom(A,B) then the fo11owing diagram is commutative:
F(~) ) F(B) h G(cf) ., G(B)
y is a natura1 equiva1ence if each YA is an isomorphisme
In a certain restricted c1ass of categories, inc1uding the category () of 1eft R-modu1es for any ring R, and the category of abe1ian groups, 13
the concept of exactness may be defined.
The sequence A~B~C is exact at B if Im(<<) = Ker (13). An
exact sequence is a sequence of objects andmorphisms which is exact
at every object where the term is defined.
A covariant functor E fro~ e to ~ is exact if for every exact sequence A~B~C in e, the sequence F(A) F(C(») F(B) F(ê) )F(C)
is an exact sequence inb.
An exact sequence O~A~B~C~O splits if there exists
y € Hom(C,B) such that ~y = lc.
Theorem 3.1. Let R be a ring. Let?1l, be the category whose objects
are the left R-modules, and whose morphisms are the left R-module
homomorphisms. Let ~n be the category of left R(n)-modules and left
R(n)-module homomorphisms. There are exact covariant functors S from to 7l'/ and T from to These functors are inverses in the sense 'm n m 'YI!.n that there are natural equivalences ~ from the identity functor on ~
to the functor ST, and V from the identity functor on ~ to the functor n TS.
Sand T preserve actual inclusions (not simply inclusions up to
isomorphism), sums, and intersections. Also, if M,N €~ and N ç M,
then ~N(x) ~M(x) for all x € N. If M',N € ~ and NI ~ MI, then = ' n VN,(x) = YM,(x) for all x e NI. Sand T also take finitely generated modules into finitely generated modules.
I o Proof. Definition of S. Let MI €~. S(M ) = ellM~ For r e R and ellm e ellM'define r(ellm) = (rell)(ellm). This is defined since 14
() rell € R(n)· Also (rell)(ellm) = e1l(rell) (ellm) € ellM~ Thus ellM'is an object of 'm. (Note that (freii)(ellm) = (rell)(ellm) = r(ellm).)
Let N' € ~n and let ~' € Hom(M',N'). Let m € M'. Then ellmE ellM'.
~'(ellm) = ell~'(m) E ellN'. Define S(~') € Hom(S(M'),S(N'» by
S(cIl')(ellm) = 4"(ellm). Cle.arly S(t'41') = S(t')S(CP') for any P' E?rln and t' € Hom(N',P'), and S(lM') = lS(M'). Therefore S is a functor from'»1 to n m. cP' v~ ') N' --.;...;;...~) P' be an exact sequence in 'Y'Il. Consider Let M' n the sequence S(M') S(cfJ'» S(N') S(i"» S(P') in 'Yil. S(i')S(cfl') =
S(t'~') = S(O) = O. Hence Im(S(~'» ~ Ker(S(t'». To show that
Ker(S(y'» ~ Im(S(f'» let x € N' be such that ellx € Ker(S(y'». Then
~'(ellx) = S(y')(ellx) = O. Therefore there exists m € M' such that
~'(m) = ellx. S(f')(ellm) = ~'(ellm) = ell~'(m) = el1x. Rence S is () exact. n Definition of T. Let M €~. T(M) = M , a direct sum of n copies
of M. By treating the elements of T(M) as column vectors, T(M) €~ • n
Explicitly, (~r .. e .. )(ml, ••• ,m) = (~rl.m., ••• ,~r .m.). Let N € 'hl i,j ~J ~J n J J J J nJ J and let
T(f) € Hom(T(M),T(N» it is sufficient 'to show for any e .. and any ~J (ml, ••• ,m ) € T(M), that T(~)(eij(ml, ••• ,mn» = e .. T(.p)(m , ••• ,m ). n ~J l n e .. T(~)(ml, ••• ,m) = e .. ({>(ml), ••• ,cf)(m » ~J n ~J n th = (O, ••• ,~(m.), ••• ,O) where f(m.) is the i entry J J th = T(~)(0, ••• ,mj, ••• ,9) where mj is the i entry = T(cp)(e .. (m1, ••• ,m ». ~J n () Therefore T(f) € Hom(T(M),T(N). Now let P € 'rll and let t€ Hom(N,P). 15 " = (i'tf(m ),···, tf(m » O·. ~ 1 n = T(tep) (~, • • • ,mn)· Hence T(tf) = T(y)T(~). Clearly T(lM) = lT(M). It follows that T is a functor from. '»l to 'ni. • n Let M~N~ P be an exact sequence in 'm. Consider the sequence
T(M) T(cp) )T(N) T('ljI), T(P) in o/t[. T(t)T(CP) = T(tCP) = T(O) = O. Hence n
Im(T(f» G Ker(T(y». To show that Ker(T(t» ~ Im(T(~» let T(y)(nl , ••• ,nn) = o. Then 'Ijr(~) = 0 for aU k, 1 ~ k ~ n. Since M~ N ~P is
exact, there exist ~ € M such that ~(~) =~, 1 ~ k - n.
T(f)(ml, ••• ,mn) = (~(ml), ••• ,.p(mn» = (nl, ••• ,nn). Therefore T is an exact functor.
Definition of t and ~. Let M €m and let m € M. ST(M) = ell~ is
the direct swn of M with n-l copies of O. Let fM(m) = (m,O, ••• ,O). }J-M t) is clearly an isomorphisme Let N € mand let 4' € Hom(M,N), ST (q»h1(m)
= T(~)(m,O, ••• ,O) = (~(m)~O, ••• ,O) = fN~(m). Therefore ST(~)~M = fNCP.
That is, ~ is a natural equivalence from the identity functor on mto the functor ST.
Let M' € and let m € M'. TS(M') (eUM,)n. Let »M' (m) mn = = (eUm~, •• ,elnm). 'YM'(m) € TS(M') since for every k, 1 ~ k ~n, elkm =
e1lelkm € e1lM'. v ' is clearly an R-module homomorphisme To show that M v ' € Hom.(M',TS(M'» it is sufficient to show for any e .. that e ..VM,(m) M J.] J.] = V , (e ..m). M J.] eijvM,(m) = eij(el1m, ••• ,elnm) . h . th = (0 , ••• ,e1jm, ••• , 0) wh ere eljm J.S t e J. entry
= (e 'l-·e ..m, ••• ,el e ..m) 1 J.] n J.] = v'M' (e ..m). o J.] Therefore ' € Hom(M',TS(M'». If vM,(m) 0 then elkm 0 for aIl k, vM = = 16 ., C) 1 ~ k ~ n. Therefore m = f.,eklelkm = O. Let ~ € M', 1 ~ k ~ n. VM'(tekl~) = (eUml,···,eUmn)· Therefore "M' is an isomorphisme Let
NI € ~ and let ~I € Hom(M',N'). n
TS(~')vM,(m) = TS(~')(ellm, ••• ,elnm)
= (~'(ellm), ••• ,~'(elnm»
(e 4>'(m), ••• , e
Therefore TS(~')~M' =~N'~" that is, v is a natural equivalenee from
. ' ~ ,- the identity funct6r on ~ to the functor TS. Renee Sand T are inverses n in the required sense.
Further Properties of Sand T. It is clear that T preserves
. actual inclusions, surns, and intersections. It is also clear that.S
preserves actual inclusions and suros. Let M',N' €~. Clearly ( ) S(M' n N') G S(M') Il S(N') since S preserves inclusions. Since ellM'
C.M'and ellN' c. N', ell(eU M' 11 eUN') GeU(M' () N'). But ell(eUM' Il ellN') = ellM' n eUN'. Therefore S(M') fl S(N ' ) ~ S(M' n N'). It is clear from the definitions of r and V that if M,N €~, M',N' €~n'
N ~ M and N' ~ M', then J'lN(X) = fM(x) for aU x € N and vN, (y) = \lM' (y) for aU y € NI.
Let l be an index set and let tX13:13 € l } be a set of generators
for M €~. Then f(x ,O, ••• ,O):13 € Il is a set of generators for T(M). 13 Let [YI3:13 € Il be a set of generators for MI €~n. Then ielkYI3:13 € 1,
1 ~ k ~ nJ is a set of generators for S(M'). Therefore Sand T take
finitely generated modules into finitely generated modules.
This completes the proof of Theorem 3.1. () Definitions. Let ~ be a category. An object P of ~ is a projective 17 o object if every diagram in e of the following type, with the row exact: p l A -----"'1 B --~~, 0
may be embedded in a commutative diagram ine:
A ---~>/1 B ---~)o
A projective left R-module is a projective object of m. An object Q of e is an injective object if every diagram in e of o the following type, with the row exact: Q 1 o ---~) A -----:) 8
may be embedded in a commutative 4iagram in e:
o----?) f~A '8
An injective left R-module is an injective object of '»l.
Lemma 3.2. Let F be an exact covariant functor from a category e o to a category~, which has an inverse G in the sense of Theorem 3.1. Let M € e. If F(M) is projective, then so is M. If F(M) is injective, then 18 () so is M. Proof. Let y be a natural equivalence fDom the identity functor on e to the functor GF. Let M e: e and let F (M) be projective. Let
A~B ~O be an exact sequence in ~ and let ~ e: Hom(M,B). Since F
is exact and F (M) is projective, there exists 'Ijr e: Hom(F(M) ,F(A» making
the following a commutative exact diagram in ~.
F(A) -----') F(B) --~) 0 t=' Il)
The proof for the injective case is sünilar, but with the arrows
reversed.
Theorem -3.3. Let M e: ,»/, MI e: '»l. T(M) is projective ilr injective n if and only if M has the same property. S(MI) is projective or injective
if and only if MI is.
Proof. The theorem follows from Lemma 3.2 and the fact that M ~
ST(M) and MI ~ TS (MI).
Definition. A ring R is left self-injective if the left R-module R
is injective.
Theorem 3.4 was apparently first proved by Y. Utumi [13j; though by
an entirely different method. 19 () Theorem 3.4. R is 1eft self-injective if and on1y if R(n) is 1eft self-injective.
Proof. Since Rn is a direct pnoduct of n copies of R, R is 1eft
self-injective if and on1y if Rn is an injective 1eft R-modu1e. Rn ~
S(R(n» under the isomorphism (r , ••• ,r ) ~frke1k. The theorem 1 tl fo110ws from Theorem 3.3.
Definitions. A ring R is herèditary if every 1eft idea1 of R is
projective. R is semi-hereditary if every finite1y generated 1eft idea1
of R is projective.
Theorem 3.5. If R is (semi-) hereditary, then every (finite1y
generated) submodu1e of Rn is the direct sum of n modules each isomorphic
to a (finite1y generated) 1eft idea1 of R.
Proof. The proof is by induction on n. The theorem ho1ds trivia11y
for n = 1. Suppose it is true for n-1. Let {x , ••• ,x 1 be a basis for 1 . n 1\-, ,.., n-1 Then B t..Rx R • Let A be a (finite1y generated) ssubmodu1e of = I and r € R. The mapping a ~ r is a mapping from A onto a (finite1y generated) 1eft idea1 l of R, whose kerne1 is A n B. Since R is (semi-) hereditary, l is projective, and the exact sequence o --"A 1) B~ A~I~O splits. Therefore A = (A n B) + C where C ~ I. (If A is finite1y generated, so is A n B.) By assumption, A 11 B satisfies the conclusion of the theorem. Therefore, so does A. Corollary 3.5. If R is (semi-)heredit~ry, then every (finitely o generated) submodu1e of Rn is projective. 20 () Theorem 3.6. R is (semi-) hereditary if and only if R(n) is (semi-) hereditary. Proof. Let R be (semi-) hereditary and let l be a (finitely generated) left ideal ·of R(n). Then 8(1) is a (finitely generated) submodule of 8(R(n)). In the proof of Theorem 3.4 it was shown that Rn,~ 8(R(n)). By Corollary 3.5,8(1) is projective. By Theorem 3.3, l is projective. Therefore R(n) is (semi-) hereditary. Let R(n) be (semi-) hereditary and let l be a (finitely generated) left ideal of R. Then T(l) is a (finitely generated) left R(n)-module. T(I) ~ I(n)ell, a left ideal of R(n)' under the R(n)-isomorphism (xl'·· .,Xn) ~fxkekl· Byassumption, I(n)eU is projective. Therefore T(I) is projective. By Theorem 3.3, l is projective. Rence R is (semi-) hereditary. o Definitions. Let M be a left R-module and let N be a submodule of M. N is superfluous in M if for every submodule S of M, N + 8 = M ==r 8 = M. A homomorphism cf':M ~ N is a minimal homomorphism if Ker ( CP) is superfluous in M. A projective cover of a module M is a minimal epimorphism from_~a projective module to M. A ring R is left perfect if every left R-module has a projective cover. R is left semi-perfect if every finitely generated left R-module has a projective cover. () In order to gain more insight into the nature of perfect and semi- perfect rings, we state without proof two theorems of R. Bass [3J. 21 () Theorem B1. Let R be a ring and Nits Jacobson ~adica1. Then the fo11owing are equiva1ent: (1) N is 1eft transfinite1y nilpotent and R/N is artinian. (2) R is 1eft perfecto (3) Every 1eft R-modu1e has the same weak as projective dimension. (4) A direct 1imit of 1eft R-modu1es of projective dimension ~ n has projective dimension ~n. (5) A direct 1imit of projective 1eft R-modu1es is projective. (6) R satisfies the descending chain condition on principal right idea1s. (7) R has no infinite sets of orthogonal idempotents, and every non-zero right R-modu1e has non-zero socle. o Theorem B2. Let R be a ring and Nits Jacobson radical. Then the fo11owing are equiva1ent: (a) R is 1eft semi-perfect. (al) R is right semi-perfect. (b) R/N is artinian and idempotents can be 1ifted modulo N. (c) Every cyc1ic 1eft R-modu1e. has a projective cover. (Cl) Every cyc1ic right R-modu1e has a projective cover. Theorem 3.7. R is 1eft (semi-) perfect if and on1y if R(n) is 1eft (semi-) perfecto Proof. Let R be 1eft (semi-) perfect and let MI be a (finite1y generated)left R(n)-modu1e. Then S(M' ) is a (finite1y generated) 1eft cS; R-modu1e. By assumption, there exists an exact sequence o----7K----?P~S(M,)----?O in 'mwith P projective and K superfluous 22 ~:-IT(:n:) o in P. By Theorem 3.1, O~T(K)~T(P) tI ) M' ~ 0 is an exact sequence in ~ , and T(K) E T(P). By Theorem 3.3, T(P) is projective. n Suppose there exists a submodule N'of T(P) such that T(K) + N' = T(P). -Ir, " . _1[ , -Ir, ]. Then Pp LST(K~ + f p S(N'~ = f p LST(P); 1n other words, K + fp -1[S(N')] = P. Since K is superfluous in P, fp-l[S(N')] = P. Since ~ is an epimorphism, SeN') = ST(P). Hence TS(N') = TST(P.). Since N' ~ T(P), N' = 'YT(p)-I[TS(N')] = 'YT(p)-l[TST(P)] = T(P). Therefore T(K) is superfluous in T(P), and R(n) is left (semi-) perfect. The proof of the converse is similar, but with Sand T interchanged. Definition. A ring R is regular if for every a € R there exists X € R such that axa = a. o Regular rings are characterized by the property that every left module over the ring is flat.[l]. It is this characterization which motivates the present discussion. Theorem 3.1 holds for right modules as well as left modules. . l.et nto represent the category of right R-modules '}\10 000. 0 and,,~ the category of right R(n)-modules. Let S ,T 'f ' and v correspond to S,T,~, andv. Definitions. Let M €~o and N € m. Let MX N be the free abelian group generated by aIl ordered pairs (m,n) with m € M and n € N. Let G be an abelian additive group. A mapping .p:M X N ~ G is R-bilinear if (a) ~(m+ m'~n) = ~(m,n) + f(m',n) (b) ~(m,n + n') = ~(m,n) + ~(m~n') o (c) ~(m,rn) = f(mr,n) for aIl m,m' € M, n,n' € N, and r € R. 23 o The tensor product of M and N is an R-bilinear map ~ from M X N onto an abelian group M®RN such that if 9 is any R-bilinear map from M X N to an ab.elian group G, then there exists a unique group homomorphism 1jr:M®RN ~G such that te{>= B. cf>(m,n) is denoted by m®Rn. Let M,M' e ~o and N,N' em. Let ~ e Hom(M,M') and t e Hom(N,N'). The tensor product of 4> and'" is a group homomorphism ~®R1jr:M®RN ~ M'®RN' where (CP®Rt)(m®Rn ) = 4>(m)®R'\jr(n). Let M e m. M is a fIat module if for every exact sequence 1 O---""N ~P in 'm0, the sequence O~N®RM 4>®1I. M ~PQl)RM is an exact sequence of abelian groups. Theorem 3.8. Let M e 0 and let N e There is a group ~n ~.n o isomorphism ~ N:M®R N -"?' SO(M) ®RS(N). If M' e 'i'l/n and N' e ~n' , (n) o and if cp e Hom(M,M') and Ve Hom(N ,N' ), then X. connnutes with cp®y; in o S symbols, XM' ,N' (~®R(n:) = (S (4)) ®R (1jr»XM,N· Proof. SO(M) = Men. S(N) = enN. Define 9:M X N ~Men ®RenN .. by B( ~(m.,n.» = ~ ~(m.ekl®Relkn.). e is clearly an R-bilinear map. i ~ ~ l "si ~ ~ To show that e is R(n)-bilinear, it is sufficient ta show for any m. e M, n e N and any e .. , that e (me .. ,n) = B(m;e .. n). ~J ~J ~J n e(meij,n) = t-(mei / kl ®Relk ) = me .. e .1®Rel.n ~J J J me·I®Rel·e .. n = ~ ~ ~J = ~(mekl ®R~lkeijn) = 8(m,e .. n). ~J Therefore & is R(n)-bilinear. From the definition of M®R N, there (n) exists a unique group homomorphism"M N:M®R N --'>-Me ®RenN such that C) , (n) n ~ N(m ®R n) = 9 (m, n) for an m e M and n € N. , (n) To show that XM,N is an epimorphism, it is sufficient to show that e 24 o is surjective. A genera1 e1ement of Me11®Re11N is of the form f (mieU f(mieU ®ReUni) = El ( r(mieU ,eUni». Therefore 9 is surjective. To show that 1L-,N is a monomorphism, let ')C.M N( ~ m. ®R n.) = O. ·~M , 1 ~ (n) ~ Then r t (mi ek1 ~ Relkni) = O. (*) tm·®R ni = ~i(m.ek1elk®R n.) 1 ~ (n) 1 k ~. . (n) ~ = ~2(m.ek1®R e n.) 1 k ~ (n) 1k ~ 0 from since R G R(n) , e N c N, and Me ~M. = (*) l1 1l Therefore X , N is an isomorphisme M To show that XM,N counnutes with cf>®,y, it is sufficient to consider e1ements of the form m~R n with m € M and n € N, since aU maps (n) concerned preserve suros. XM' NI (f) ®R y)(m®R n) = XM' NI (4)(m) ® R t(n» , (n) (n) , (n) o = ~ (~(m)ekl ® Relk"'\r(n» = ~ (cf>(mek1)® Ry(elkn» e n = t(so(cp)(mek1) ® RS("l\r)(elkn» = f (So ( o CoroUary 3.8. Let M € nt and N €?il. Then there is a group isomorphism "lM N:M~RN ~To(M)®R T(N). If MI € 'ff[0 and NI € 1rt, and .' (~ if 4> € Hom(M,MI ) and l' € Hom(N,N I), then"Yl counnutes with I{)®t. Proof. )A °M®RfN:M®RN ~ SOTo(M)®RST(N) is an isomorphism whose . . (0 )-1 -1. o( ) ( ) ~nverse ~s fA" M . ~R~ • Def~ne '7 N:M~RN --=>T M ®R ..,T N by . M, (n) -1 0 ' '7M,N = (XTo(M),T(N» (r M®R~)· It is easily checked that l'[ has the desired properties. o Theorem 3.9. Le t M € and MI € • If T (M) is fla t then so is M. 'n1 '»rn If S(M I) is f1at then so is MI. 25 'V .. 'i''1) Proof. Stippo.se that T(M) is f1at. Let 0 ~N ~P be an exact se quence 1.n• 'HI"\ o • By Theorems 3.1 and 3.8, the following is a connnutative exact diagram an abe1ian groups: monomorphisme The proof that S(M' ) is f1at ~ MI is f1at is simi1ar. Theorem 3.10. R is regu1ar if and on1y if R(n) is regu1ar. (J Proof. By Theorem 3.9, every 1eft R-modu1e is f1at if and on1y if every 1eft R(n)-modu1e is f1at. We turn now to a discussion of homo1ogica1 dimension. Theorem B1, stated on page 21, is one examp1e of the importance of dimension theory in the theory of rings and modules. Definitions. A reso1ution of a module M is an exact sequence 6,., P 60 M 0 f h ••• --'t Pm ')' m-1 ~ ••• ~ Po ., ~. l eac Pm is projective, the reso1ution is a projective reso1ution. A reso1ution has 1ength m if P ~ 0 and P. 0 for a11 i ~ m. m 1. = An injective coreso1ution of 1ength m is definedsimi1ar1y to a o projective reso1ution of 1engthnn, but with the arrows reversed and each P. injective. 1. 26 The projective dimension of a module M, denoted by PD(M) , is the o length of the shortest projective resolution of M. If there is no projective resolution of M of finite length, PD(M) =00. PD(O) = -1. The injective dimension of a module M, denoted by ID(M), is the length of the shortest injective coresolution of M.lf there is no injective coresolution of M of finite length, ID(M) =~. ID(O) = -1. IGFD(R) = sup{PD(M):M is a left R-module}. This is called the left global projective dimension of R. lFPD(R) = sup{PD(M):M is a left R-module and PD(M) <~}. This is called the left finitistic projective dimension of R. lfPD(R) = sup{PD(M):M is a finitely generated left R-module and PD(M) < cO}. The injective dimensions lGID(R), lFID(R), and lfID(R) are defined o similarly. Under certain conditions, some of these dimensions may be equal. We will not consider this problem. However, it is weIl known that lGPD(R) = lGID(R) always (7). M. Auslander (2] has shown that lGPD(R) = sup{PD(M): M is a cyclic left R-module l = sup lID(M):M is a cyclic left R-moduleJ. Lemma 3.11. Let M € ~ and M' € ~. Then PD(T(M» PD(M), n = PD(S(M'» = PD(M'), ID(T(M» = ID(M), and ID(S(M'» = ID(M'). Proof. We prove only the case 0 ~ PD (M) .:::; m.< Cie), .:: Le t ô ... o O ~P --~~ ••• ----, Po Ô ) M ~ 0 b e a pro Ject~ve" reso l'ut~on 0 f M. m T(ô..,) T(ô ) Then 0 ~T(Pm)-~~~') ... ~T(PO) 0 )T(M) ~o is a projective resolution of T(M). Therefore PD(T(M») ~ PD(M). Similarly PD(S (M'» ~ o PD(M'). Therefore PD(M) = PD(ST(M» ~ PD(T(M» ~PD(M). The prodfs for the other cases are similar. 27 0 Theorem 3.12. 1GPD(R(n» = 1GPD(R). 1FPD(R(n» = lFPD(R) • 1fPD(R(n» = lfPD(R) • 1FID(R(n» = 1FID(R) • lfID(R(n» = 1fID(R). Proof. The theorem foUows immediate1y from Lemma 3.11. We now define the weak dimension. Let D Ô ) N ~o be a projective reso1ution of the o 1eft R-modu1e N. Let M € 'ln • Then the sequence l®ôm ••• ~M~RPm--~~ ••• 1~ Ôl ) M~RP O~o is not in genera1 an exact sequence, but for aU m, (1 ®ôm)(l ®ômf-1) = 1 ®ômômf-1 = 1 ® 0 = O. In other words, Im(l®ômf-l) ~ Ker(l@ôm). () Definition. Torm(M,N) = Ker(l ® Ôm) /Im(l ®Ômf-l). It is well known that Tor (M,N) is independent of the particular projective resolution m chosen for N, and that it may be obtained either as above or by taking a projective resolution of M and tensoring with N [7J. Definition. The weak dimension of N is defined to be suptm:there o exists M such that Tor (M,N) ~ It is denoted by WD(N). WD(O) €m m 01. = -1. The weak dimensions lGWD(R), lFWD(R), and 1fWD(R) are defined similarly to the projective dimensions. M. Auslander [2J has shown that if R is noetherian, then lGWD(R) = IGPD(R). It is clear by symmetry that lGWD(R) = rGWD(R), the right weak global dimension of R. J. Ril1el [12] has shown that lGWD(R) = suptWD(M): o M is a cyclic left R-modulej. ------~------~------ 28 o 0 Lemma 3.13. Let M €"l , N €?YI, M' €?11 n ,and N' € 1Yn."(n Then o o Tor (T (M),T(N» ~ Tor (M,N) and Tor (So(M'),S(N'» ~ Tor (M',N') for m m m m aH m ). O. Proof. Throughout the proof of this le~, subscripts are omitted where no confusion is likely to arise. Let ... ~Pm 5", ) ... ~PO 50 lN ~O be a projective resolution of N. T(5",) T(5 ) Then ... ~T(Pm) , ... ~T(PO) \ 0) T(N)~O is a projective resolution of T(N). Let P-1 = 0 and replace 1 ® 50 by O. By Theorem 3.8, for any m ~ 0, the following diagram is commutative: ___1_®_5...:;"':;.;:;+;:..' __ ~, M®P ___1_~_5_.., __~> M®P 1 f . F- 1®T(5",+,» TO(M) ®T(P) 1 ®T(5..,) ) TO(M) ~T(P ) o m ~l Since ~ is an isomorphism, it is sufficient to show that Consider the following commutative diagram: D -----'-----')) E -----:...---~) F Let ~,r ,y be isomorphisms. Then it is sufficient to show that o ~[Ker(>')] = Ker('1\') and ~[Im(e)1 = Im(cf\). b € Ker(~) ~ '" (b) =: 0 ~ -r~(b) = )'>'(b) = 0 =9~(b) € Ker(v). e € Ker(y) ~ e = ~(b')' for some 29 O· b' e Band 'tP(b') = 1jr(e) = 0 ~ y'>l(b') = i'P(b') = 0 ~X(b') = 0 ~ b' e Ker(~).~ e e p[Ker(~)]. Therefore ~IKer("')] = Ker(t). ~[Im(8)1 = Im(p9) = Im(41") = Im(cf» sinee 0<. is an isomorphisme Theproof that Torm(So(M'),S(N'» ~ To~(M',N') is simi1ar. Coro11ary 3.13. WD(T(N» = WD(N). WD(S(N'» = WD(N'). Theorem 3.14. 1GWD(R(n» = 1GWD(R). lFWD(R(n» = 1FWD(R). 1fWD(R(n» = 1fWD(R). Proof. The theorem fo11ows immediate1y from Coro11ary 3.13. o o 30 C) Appendix to 3. [4, page 76, exercise 16] gives an alternative proof for Theorem 3.10. Lemma. Let R be a regular ring. Then every finitely generated submodule M of Rn is a direct summand of Rn. Proof. Let M be a finitely generated submodule of R.n Define rr:M ~R by rr(rl, ••• ,rn ) = f. n N = Im(rr) is a finitely generated left ideal of R. Since R is regular, N is principal. Therefore there exists a left ideal N' of R such that N + N' = R. It follows that N is projective. S~nce. Ker ( rr ) = M n Rn-l , t h e f 0 11 owing is a sp 1 it exact sequence of left R-modules: o Therefore M n Rn-l is finitely generated. By induction, there exists a submodule M' of Rn-l such that (M n Rn-l) + M' = Rn-l. We show that (M'-:œ N') + M = Rn. Let (rl, ••• ,r ) e (M' E9 N') Iî M. Then r € N {'\ N' = O. Hence n n (rl, ••• ,r ) e M' (M n Rn-l) O. n n = Since N + N' R, there exist a € N and = n b € N' such that r = a + b. Since rr is an epimorphism, there n n n n existsal, ••• ,a _ € R such that (al, ••• ,a ) € M. Let (bl, ••• ,b ) n l n n = (rl, ••• ,r ) - (al, ••• ,a ). Since (M ('\ Rn-l) + M' = Rn-l, there exist n n (cl, ••• ,c _ ) € M {'\ Rn-l and (dl, ••• ,d _ ) € M' such that (cl' ••• n l n l '~n_l) + (dl,···,d _ ) = (b ,···,b 2l). (al, ••• ,a ) + (cl, ••• ,cn_1,0) € M and o n l l n n (d1,··.,dn_l,bn) € M' œ N'. Also (al, ••• ,an) + (cl, ••• ,cn_l,O) + 31 (d ,···,d _ ,b ) = (a ,···,a ) + (b ,· •• ,b ) = (r , ••• ,r ). Therefore o l n 1 n 1 n 1 n 1 n Rn = M+ (MI ~ NI). n n Proof of Theorem 3.10. Let ~ € R(n) = Hom(R ,R). Im(f) is a finitely generated submodule of Rn. By the lemma, there exists a n • n submodule M of R such that M+ Im(~) = R. Therefore Im(~) is projective, and there exists t:lm(.p) such that l (cfI). ~Rn cf>V= Im y may be extended to an element of R(n) by letting tÜ1J = o. It is clear that cfhjroP= CP. Therefore R(n) is regular. The converse is trivial. R ~ e1lR(n)el1 as rings. Let R(n) be regular and let ellre1l € ellR(n)ell• Then there exists x € R(n) such that (ellrell)x(ellrell) = ellrell• But then (ellrell)(ellxell)(ellrell) = e re and R is regular. () l1 ll o 32 o 4. Coherent Rings Definitions. A left R-module M is finitely presented if there exists an exact sequence o-----l)K~F--.-,M~O of left R-modules with F free and finitely generated and K finitely generated. (S.'U. Chase uses the term "finitely rela.ted".) A left R-module M is pseudo-coherent if every finitely generated submodule of M is finitely presented. M is coherent if M is both pseudo-coherent and finitely generated. A ring R:is left coherent if the left R-module R is coherent. S.u. Chase [8] has shown that left coherent rings are characterized by theproperty that the direct product of any family of copies of the o ring is a flat right module over the ring. Theorem 4.3 is a modification of this theoremo In [7, page 122, exercise 4J it is shawn that every left noetherian ring is left coherent. However, M.E. Harris [11] exhibits a right coherent ring which is not a direct limit of right noetherian rings. In this section we show that R is left coherent if and only if R(n) is. Proposition 4.1. Let R be a ring and let O~K~F~A~O be an exact sequence of right R-modules where F is free, K = Ker(~), and ~ is the natural injection. Then the following statements are equivalent: " (a) If ~,akrk = 0 where ak € A and r k € R, 1::- k ~ 1', then there exist bl, ••• ,bm € A and ~sik:l ~ i ~ m, 1 ~ k ~ rj ç R such that '" ~ a = ~bosok and l.sokrk = 0 for a11 i, l ~ i~ m. k 1:1 l. l. Ie./ l. o (b) Given any u € K, there exists a homomorphism 8:F ~K such that 9(u) = u. 33 ( e) Given any u ' ••• , Us € K, there exis ts a homomorphism 9:F ~ K o 1 such tha t 8 (u 0) = u 0' 1 ~ i ~ s. 1.,. 1. (d) If f.,akr = 0, a € A, r € R, 1 6 j ~ s, then there exist kj k kj ... 1 ~ k -5 rJ ~ R such that a = ,ib 0 s ok k 1" 1. 1. l' and Z:sokrko = 0 for aH i,j with 1 60 i ~ m, 1 ~ j ~ s. k., 1. J Proof. (a) ~ (b). Let F have a basis fX~:13 € Il and let u = \" ,. ~x r € K. Let a 'o/(x(3k)' 1 ~ k ~ r. Then ~,akrk v(u) O. By 10;=, (3k k k = = = assumption, there exist b1, ••• ~bm € A and fsik:1 6 i ~ m, 1 ~ k ~ rJ ~ R M ,. such that a = ~bosok for aH k, 1 ~ k ~ r, and 2.sokrk = 0 for aH i, k ,:, 1. 1. ,,"c, 1. 1 ~ i ~ m. Sinee 1Jr is an epimorphism, there exist zl' ••• ,zm € F such that 'l/r(z 0) = b 0' 1 ~ i :G m. Define B:F ~ Fto be a homomorphism such 1. 1. ... that 8(x(3 ) x(3 - ~zosok for 1 :!:: k ~ rand e(x(3) 0 if (3 ;1: (3kofor any = 1:, 1. 1. = k k .., k. Then '1jr9(X(3 ) = a - ~boSok = 0 for 1 6. k 6 rand t 9(x(3) = 0 if k k ~, 1. 1. o (3 ~ (3k. Renee Im(e) ~ Ker(t) = K and we may eonsider 8:F ~ K. ,. 1" r ...... 1'" e(u) = 9(~x(3 r ) = ~X r - ~~zosokrk = u -:i. zo("Z:sokrk) = u. 1:', k k 1:., (3k k "., '1'=1 1. 1. I~' 1. k., 1. (b) ~ (e). Let u , ••• ,u € K. If s 1, then (b) ~ (e) 1 s = trivial1y. Assume s > 1 and (c) ho1ds for s - 1. Let :F ~K be such es that (u ) us. Let v for 1 k 6 s - 1. By assumption 8s s = k = ~ - as(~) ~ there exists e':F ~ K sueh that 9' (v ) v ' 1 ~ k ~ s - 1. Define k = k &:F---'1'K by 6(x) x - (1 - 6')(1 - )(x). If k s, then = e s < e(~) = ~ - (1 - e')(~ - e s(~» = ~ - (1 - e')(vk ) = ~ - (vk - e'(vk» = ~. 9(u ) s = Us - (1 - e')(us - 9 s (u s » u - (1 - 9')(0) = s = u s • 34 .. a € A, r € R, 1 j ~ s. Choose o (c) ~ (d). Let z.akrk . = 0, k kj = "" J .. x , ••• ,x € F such that~(xk) = a , 1 ~ k~ r. Let u. = i.xkr ·, 1 ~ j ~ s. 1 r k J ~., kJ r "Ijr(u.) = ~akrk' = O. Therefore u. € K. By assumption, there exists a J k., J J homomorphism 9:F --"7K such that B(u ) u ' 1 ~ j ~ s. x - 8(x ) € F. j = j k k Therefore there exist z., 1 ~ i ~ m, forming part of a basis for F, and 1. ... .-!.. • ..::. - 1. - m, 1:6 k ~ rJ ~ R such that x -9(x ) = 'f.z.s. , 16 k 6- r • k k j-, 1. 1. k ... = i:b.Sek' 1~ k ~ r. i~1 1. 1. r .. lOt o = u. - S(u.) = tex. - 8(x »r . = if.z.s.kr ., 1 !, j 6 s. Since J k·, le k k k:, i~1 1. 1. k J J ... J the z. form part of a basis for F, of s 'krk' = 0, 1 l. i ~ m, 1 6 j 6: s. 1. kSI 1. J (d) ~ (a) trivially. c) Lemma 4.2. Let A be a finite1y presented 1eft R-module and let o ~N~M~A~O be an exact sequence of 1eft R-modu1es. If M is finite1y generated, then so is N. Proof. Let A be finite1y presented. Then there exists an exact sequence 0 ~K~F ~A ---+0 with F free and finite1y generated and K = Ker(9) finite1y generated. Thefo110wing (without the dotted arrows) is an exact diagram of 1eft R-modu1es. tt 8 0 ~ K )F >A ~O 1 1 1 1 :y :t3 1 1 liA "V Since F is free, and hence projective, there exists a homomorphism o t3:F --"M such that 'l't3 = lAS = e. If k € K, e(k) = 0 and hence 'o/t3(k) = O. Therefore t3(k) € Ker(t) ~ N. Therefore there exists a homomorphism 35 o y:K ~N such that o = Ker(lA)~N/lm(y) ~M/lm(13) ~A/lm(lA) = 0 is an exact sequence of left R-modules. Therefore N/lm(y) ~ M/lm(13) which is finitely generated since Mis. Im(y) is finitely generated since K is. Since 0~Im(y)----"'N~N/lm(Y)----70 is an exact sequence, N is finitely generated. Theorem 4.3. The following conditions on a ring Rare equivalent: (a) R is left coherent. (b) Any direct product of copies of R, when viewed as a right R-module, satisfies the equivalent conditions on A in Proposition 4.1. (c) Any free left R-module is pseudo-coherent. o Proof. (a) ~(b). Let J be any set and for each 13 € J let R13 be a copy of R. TTRA is the right R-module consisting of all functions from 'i:r 1-' J to R, with multiplication by elements of R defined componentwise. Call it A. l' Suppose t:,akrk = 0, where ak € A and r k € R. Let l be the left ideal of R generated by the r • Let F be a free left R-module with basis k tXl' ••• 'xr.~. Define an epimorphism rc:F~I by rc(xk) = r k, 1'" k 6 r. By Lemma 4.2, K = Ker(rc) is finitely generated, by [zl, ••• ,zm1• Since t' K S F, each z 0 may be expressed uniquely a,'J Z 0 = z. s 0kxk with s Ok € R ~ ~ k .. , ~ ~ \" for all i,k. Define a function u from J to F by u(13) = ~,ak(13)xk. For .. .. all 13 € J, rc(u(13» = f,ak(13)rk = (~/akrk)(13) = 0, and hence u(13) € K. '" Therefore there exist b 0(13) € R, 1 ~ i ~ m, such that u(13) = ~ b 0(13)z. ~ ,., ~ ~ ... r = .~ibo(13)sokxk. b o € A for all i, 1 ~ i ~ m. Since txl' .... ,xrl is a ,=\ Ic., ~ ~ ~ .. basis for F and u(13) = i: a (13)x··, it follows that ~'" b 0(13) s"'k = a (13) for k·, k k ,,,, ~ ~ k ,.. r all 13 € J and for 1 !: k ~ r.· Therefore a = bis • i s 0krk = k *' ik \c., ~ 36 .. 1!( t. s 'kxk) = 1!(z,) = 0 for 1 6 i ~ m. Hence A satisfies condition (a) of o \<., 1 1 Proposition 4.1. (b) =9 (c). Let G be a free left R-module and let L be a finitely s generated submodule of G. We may asswne that G= R for some positive integer s. Let t~ = (rkl, ••• ,rks):l ~ k ~rJ be a set of generators for L and let F be a free left R-module with basis txl, ••• ,xr1. Dëfine an epimorphism 1!:F ~L by 1!(xk) = ~, 1 ~ k ~ r. For each t3 € K = Ker(1!) let Rt3 be a copy of R. A = TTRR is the right R-module consisting of aIl ~E.\C. 1-' functions from K to R. For aIl t3 € K ~ F there exist a (t3) € R, 1 ~ k ~r, k r such that t3 = z'a (t3)x • Each a is a function from K to R, that is, an \<'" k k k \'" .. element of A. For aIl t3 € K, 0 = 1!(t3) = 1Cak(t3)~ • Hence '2.. a (t3) r , = 0 k", .. K. k., k kJ for aU t3 € K and for 1 of,. j ~ s. Therefore ~akrk' = 0 for 1 ~ j f. s. k·, J Since A satisfies condition (d) of Proposition 4.1, there exist b, € A, 1 '" l) 1 .f i ~ m, and i s1'k:l = i ~ m, 1 ~ k ~ r 1 c. R such that a = ~ b ,s Ok' k I~' 1 1 r ~ 1 ~k ~r, and "Z.:'s'krk' = 0, 1 ~ i:f: m, 1 ~j ~ s. Let z, = t:s'kxk € F, ~,1 J 1 ~,1 r 16 i:!f: m. 1!(z,) = ~s'k~ = 0 and hence z, € K for aU i. If t3 € K, 1 k., 1 K. 1 rI'''' '" then t3 = i.a (t3)x = ~~b,(t3)s'kXk = ~b,(t3)z,. Therefore Zl' ••• 'Zm k., k k I:.I,~, 1 1 '-1 1 1 generate K, and L is finitely presented. It follows that G is pseudo- coherent. (c) ==9 (a) triviaUy. Theorem 4.4. R is a left coherent ring if and only if R(n) is [eft coherent. Proof. Suppose that R is left coherent. Let M be a finitely generated left ideal of R(n). Let tml, ••• ,mm1 be a set of generators for M and let F be a free left R(n)-module with basis (xl, ••• ,xm\. Define an R(n) -epimorphism 1!:F -..,.M by 1!(xk) =~, 1 ~ k. 6 m. K = Ker(1!) is a left 37 o R(n)-modu1e. [eij~:l 6 i,j 6 n; 1 6 k 6 ml is a set of generators for the 1eft R-modu1e M, which is a submodu1e of the free le ft R-modu1e R(n). fe .. Je. :1 ~ i, j 6 n; 1.:. k ~ ml is a basis for the free 1eft R-modu1e F. 1J K. ~ is an R-epimorphism. By Theorem 4.3, M is finite1y presented, as a 1eft R-modu1e. By Lemma 4.2, K is a finite1y generated 1eft R-modu1e. Rence K is a finite1y generated 1eft R(n)-modu1e. Therefore M is a finite1y presented 1eft R(n)-modu1e, and R(n) is 1eft coherent. Suppose that R(n) is 1eft coherent. Let l be a 1eft idea1 of R generated by [Y1' ••• 'Ym1 and let F be a free 1eft R-modu1e with basis [x1, ••• ,xml. Define an epimorphism ~:F ~I by ~(xk) = Yk' 1 ~ k~ m. R(n) is a free right R-modu1e, and therefore is flat. Rence R(n) ®RI is an R(n)-submodu1e of R(n)~RR ~R(n)' and [1®RYk: 1 " k" mJ is a set of generators for R(n) ® RI. By assumption, R(n) ®RI is finite1y presented. () R(n) ® RF is a free 1eft R(n)-modu1e with basis [1 ®Rxk:1 6 k = mY. l®R~:R(n)®RF ~R(n)®RI is an R(n)-epimorphism. By Lennna 4.2, K' = Ker(l ®R~) is finite1y generated. Let r = ~ r .. e .. € R(n)' k I,) k 1J 1J 1 6 k ~ m. Then ~rk®Rxk € R(n) ®RF is a genera1 e1ement of R(n) ~RF. (l®R~)(~rk®Rxk) = i. ~(rk· .e. ·®R~(xk» = ~(e. ·®R~(~rk· .x »· k k I/j 1J 1J '/J 1J k 1J k Rence frk ~Rxk € K' if and on1y if t-rkijxk € K for aU i, j with 1 ~ i,j ~ n. It fo11ows that K is finite1y generated. Therefore l is finite1y presented, and R is 1eft coherent. 38 () Summary Section 1 of this thesis deals with rings having properties which can be described in terms of ideals, namely simple and semi-simple rings, prime and semi-prime rings, and primitive and semi-primitive rings. The proof that R has any of these properties if and only if R(n) has, depends on a one-to-one correspondence between the ideals of R and the ideals of Section 2 is a brief section in which it is shown that R is artinian or noetherian if and only if R(n) is. Most of the results of Sections 1 and 2 may be found in [41. In Section 3 a tool is developed which enables us to prove quite easily some results concerning projective, injective, and fIat modules. The tool consists of an exact covariant functor from the category of left R-modules to the category of left R(n)-modules, and an inverse functor which is also exact. These funetors preserve projectivity, injectivity, and the property of being finitely generated. Also, sinee there is a natural isomorphism between the tensor product of a right and a lefti.~.",l;, R-module, and the tensor product of the corresponding R(n)-modules, the functors preserve flatness. As a result, R is left self-injective, left hereditary,'left semi-hereditary, left perfect, left semi-perfect, or (Von Neumann) regular if and only if R(n) has the same property. The functors also preserve homological dimension. It follows that the homological dimensions of R(n) are equal to the corresponding homological dimensions of R. The theorems in this section appear to be new, with the following exceptions: the statement and proof of Theorem 3.5 are based on [7, Theorem 5.3 and Proposition 6.1, pages 13-141; Theorem 3.4 is due 39 ' ·'. to Y. Utumi [13] and Theorem 3.10 is due to J. Von Neumann [14] but the O" proofs presented here are new; a computationa1 proof of Theorem 3.10 may be found in [6]'and it is an exercise in [41; Theorems 3.12 and 3.14 are due to Auslander, Eilenberg, et al [1], [2], [9], and [10J but the proofs presented here are new. In Section 4 it is shown that R is 1eft coherent if and only if R(n) is. The theorems of this section come from [5], (8], and [11). o 40 c=> Bibliography 1. Auslander, M., On Regular Group Rings, Proc. Am. Math. Soc., Vol. 8 (1957) pp.658-664. 2. Auslander, M., On the Dimension of Modules and Algebras III, Nagoya Math. Jour., Vol. 9 (Oct.,1955) pp.67-77. 3. Bass, H., Finitistic Dimension and a Homological Generalization of Semi-primary Rings, Trans. Am. Math. Soc., Vol. 95 (June,1960) pp.466-488. 4. Bourbaki, N., Algèbre, Ch. 8 (Hermann, Paris, 1958). C) 5. Bourbaki, N., Algèbre Commutative, Ch. 1-2 (Hermann, Paris, 1961). 6. Brown, B. and McCoy, N.H., The Maximal Regular Ideal of a Ring, Proc. Am. Math. Soc., Vol. 1 (1950) pp.165-l7l. 7. cartan, H. and Eilenberg, S., Homological Algebra (Princeton, 1956). 8. Chase, S.U., Direct products of Modules, Trans Am. Math. Soc., Vol. 97, No. 3 (Dec.,1960) pp.457-473. 9. Eilenberg, S., Ikeda, M. and Nakayama, T., On the Dimension of Modules and Algebras l, Nagoya Math. Jour., Vol. 8 (Feb.,l955) pp.49-57. OJ· 41 () 10. Eilenberg, S., Rosenberg, A. and Ze1insky, D., On the Dimension of Modules and A1gebras,VIII, Nagoya Math. Jour., Vol. 12 (Feb.,1955) pp.71- 93. 11. Harris, M.E., Some Resu1ts on Coherent Rings, Proc. Am. Math. Soc., Vol. 17, No. 2 (Apr.,1966) pp.474-479. 12. Hi11e1, J., Flat Modules (Master's Thesis, McGi11 University, 1965) p.49. 13. Utumi, Y., On Continuous Rings and Self-injective Rings, Trans. Am. Math. Soc., Vol. 118 (June,1965) pp.158-173. 14. Von Neumann, J., Continuous Geometry (Princeton, 1960) pp.69-81.