Chapter VII Rings with Identity

Rings with Identity CHAPTER VII

The theory of semisimple artinian rings was brought to a satisfactory conclusion in the preceding chapters. In this and the next two chapters we shall present some aspects of the theory of rings with nonzero radical. The theory developed will not be as complete as in the semisimple case and to avoid complications we shall assume that rings have an identity. If R has the descending chain condition on left ideals, R R is a direct sum of finitely many indecomposable left ideals, i.e. left ideals which cannot be expressed as a nontrivial direct sum of two left ideals. In the first section we shall investigate this decomposition fully. We shall see that, here too, idempotents will play the most important part. The main results obtained will be for the class of semiperfect rings with identity, i.e. the class of rings R such that R/J(R) is artinian and each idempotent of R / J ( R ) can be ‘‘lifted.’’ Every artinian ring is semiperfect but not conversely. For example, the ring of power series in one indeterminate over a division ring is semiperfect but not artinian. If in addition R / J ( R )is a simple ring, R can be represented as a matrix ring over a local ring T , i.e. a ring T such that T / J ( T )is a division ring. Theorem VII. 1.10 deals with the structure of a general semiperfect ring and later i n Theorem VI1. I . 19 we give a proof of the essential uniqueness of the decomposition obtained in Theorem 10. The indecomposable left ideals are the prototypes of the so-called projective modules. These modules will be discussed in Section 2. As an application we shall develop the theory of Asano orders in a quotient ring. This is a generalization to noncommutative rings of the classical theory of orders of the algebraic integers in an algebraic number field. 148

1.

SEMIPERFECT RINGS

149

The lattice of left ideals of a semisimple artinian ring is complemented. This gives rise to the concept of an injective module, i.e. a module which is a direct summand of any containing module. These modules will be dealt with in Section 3 and it will be shown that each module is embeddable in a smallest injective. The theory developed will be applied to self-injective rings R, i.e. rings R such that RR is an injective module. If in addition R has the descending chain condition on both left and right ideals, R is a quasiFrobenius ring. 1. Semiperfect rings

Although a nonsemisimple artinian ring R with 1 is not completely reducible (Theorem IV.1.4), RR is still the direct sum of a finite number of indecomposable left ideals. Definition 1. 1. The direct summands of the R-module M are called components of M . 2. An R-module M If (0) is indecomposable if the only components of M are (0) and M . 3. A decomposition of M into components Mi is a representation of M as the direct sum of the M i . The existence of a decomposition RR = L1 @ ... @ L, is a consequence of the minimal condition on left ideals in R: Let 5 be the family of left ideals of R which are not expressible as a direct sum of a finite number of indecomposable left ideals. Assume by way of contradiction that 9 # 0. Then by the minimal condition on left ideals there exists Lo minimal in 9. Lo is decomposable, say Lo = Lo' L". Since Lo' and L" are properly contained in L o , L,' and L" are not members of 5. Hence each is a finite direct sum of indecomposable modules showing that Lo E 5, a contradiction. Therefore 9 = @ . If the ring has 1 we can establish a connection between the components of RRand the representations of 1 as the sum of a finite number of orthogonal idempotents. Let the ring R, considered as a left R-module, be expressed as a direct sum of left ideals Li, i = 1, ..., r ,

+

The identity 1 of R is then expressible in the form, 1 = el

+ ... + e,

where ei E Li

(2)

150

VII.

RINGS WITH IDENTITY

The e, are idempotent and mutually orthogonal, i.e. e.e. 2 3 = Sije, ,

where

S i j is the Kronecker delta

(3)

The e, generate the L, , L,

ProoJ

=

Re,,

i

=

1, ..., r

(4)

Since e, E Li , Re, C L, . Conversely, if 1, E L,

By the directness of the sum in ( l ) ,

Hence by ( 5 ) , li E Re, and (4) is proved. Substituting e, for li in (6) we get the orthogonality, eiefc= 0 if i # k The idempotence of the ei is proved by observing that e,

+ ... + e . = 1

=

e12 + ...

+ er2

and again using the fact that the sum in ( 1 ) is direct. Conversely, if 1 = e, ... e, , where the e, are mutually orthogonal idempotents, then R R = Re, ... Re,

+ +

+ +

Moreover, this sum is direct since, if a = xlel

+ ... + x,e,

(7)

then aeA = x k e k . Hence the Re, are components of RR. Finally, if L is any component of RR, i.e. R R = L 0L’ for some left ideal L’, then by what we have shown above,

L =Re

and

L’

=

R(l - e)

for some idempotent e. We collect these results in the following:

Theorem 1. Let R be a ring with identity. 1. Each decomposition of R R into a direct sum of finitely many left

1.

151

SEMIPERFECT RINGS

ideals gives rise to a representation of the identity 1 of R as a sum of orthogonal idempotents and conversely: 1

=

e,

+ ... + e , o RR = Re, @ ... @ Re,

2. The left ideal Re is an indecomposable component of RR if and only if e cannot be written as the sum of two nonzero orthogonal idempotents. Such idempotents are called primitive. 3. If R is artinian, R R may be decomposed as a direct sum of indecomposable left ideals. I The primitivity of an idempotent of a ring R may be equivalently expressed as follows:

Definition 2. An idempotent e of a semigroup is primitive if for any idempotent f, the equalities f = ef = f e imply e =f: Theorem 2. An idempotent E # 0 in a ring R is primitive in the ringtheoretic sense if and only if it is primitive in the multiplicative semigroup S of R . ProoJ Let e = g + h, with g # 0, h # 0 orthogonal idempotents. Then ge = g 2 gh = g and eg = g 2 hg = g . Therefore ge = eg = g but e # g showing that e is not primitive in the multiplicative semigroup of R . Conversely, if there exists g E R such that 0 g2 = g # e and g = ge = eg, then g and e - g are nonzero idempotents whose sum is e . Therefore e is not primitive in the ring theoretic sense. I

+

+

+

Let J = J(R) be the radical of the ring R with 1 . Suppose that RIJ is artinian and hence completely reducible. The indecomposable left ideals ill RIJ are the minimal left ideals. Denote the coset u J by ii. Then a = R / J = Re, ... &$ where the 2% are primitive orthogonal idempotents whose sum is 1.If we can ‘‘lift’’ this set of idempotents { & , ..., 2), to a set { e l , ..., e,} of orthogonal idempotents in R with Zi = e, + J and such that e, ... e, = 1, then we can deduce that the ei are also primitive; for if e, = g h, g and h nonzero orthogonal idempotents, then 5 = S A. But neither g nor h can be zero since the radical can contain no idempotent. In fact, if 0 # g = g 2 E J and g y - y g = 0, then g2 y g - y g 2 = 0 and hence g = 0, a contradiction. In the following we shall show that if for each idempotent ii E RIJ we can find an idempotent e E R such that 2 = ii, then we can lift any countable family of mutually orthogonal idempotents

+

+ +

+ +

+

+

+

+

152

VII.

RINGS WITH IDENTITY

of R to a family of orthogonal idempotents of R . This discussion motivates the following definition: Definition 3. A ring R with identity 1 and radical J is semiperfect if 1 . RIJ is artinian and 2. each idempotent U E can be lifted to an idempotent e E R. Theorem 3. Let M C J = J ( R ) be an ideal of R and suppose idempotents can be lifted from RIM. Let g = g2E R and suppose u2 = u(mod M ) . If 0 = ug = gu(mod M ) , there exists an idempotent e E R such that e = u(mod M ) and eg = ge = 0. Proof. By hypothesis there exists f E R such that f =f = u(mod M ) . Therefore fg = 0 mod A4. Since 1 E R and fg E J, 1 -,fg is a unit. Let f ’ = (1 -fg>-’f(l -fg). f ’ is also idempotent and since

f ( l -fg)g

=fg -f2g2

=

0,

f‘g

= 0.

Furthermore, f ’ = (1 - f g ) f ’ = f ( l -fg) = f (mod M ) . Let e = (1 - g)f ’. Then ge = eg = 0 and e2 = e(l - g ) f ’ = ef ’ = (1 - g)f’ = e. I The proof that every countable family {U1 , U 2 ,...} of mutually orthogonal nonzero idempotents can be lifted follows easily: Theorem 4. Let M C J, R a ring with 1, A4 an ideal. Assume idempotents modulo A4 can be lifted. Then, given any countable family (ill ,U, , ...} of orthogonal nonzero idempotents in RIM, there exists an orthogonal family (el ,e 2 ,...} of idempotents in R such that ei = uimod M , i = 1,2, ... . Proof. By induction. Assume that the idempotents U1 , ..., U,-l have been lifted to the mutually orthogonal idempotents e l , ..., ePp1.Let g = el

Then u,g

T-1

= i=l

urei = f

+ ... + erp1 r-1

u,ui i=l

= 0 (mod M )

and similarly gu, = O(mod M ) . By Theorem 3, there exists e, er2= e, = u,(mod M ) and erg = ge, = 0. But e,ei = ergei

=0 =

eieT

E

R such that

I

In particular Theorem 4 implies that in a semiperfect ring R any finite

1.

SEMIPERFECT RINGS

153

family of orthogonal primitive idempotents {El , ..., U,} in R/J such that Ul + ... + U, = T can be lifted to orthogonal idempotents el , ..., e, in R . We can show under these circumstances that e, e, = 1. To this end set e

= el

+ +

+ ... + e, . Then

Hence 1 - e is an idempotent in J showing that 1 fore proved the following:

-

e

= 0.

We have there-

Theorem 5. A semiperfect ring R contains a family of orthogonal primitive idempotents whose sum is 1. R R is then expressible as the direct sum of finitely many indecomposable left ideals generated by these idempotents. I Theorem 6. Every artinian ring with identity is semiperfect. This is a corollary to the following: Theorem 7. Let M be a nil ideal in the ring R with 1. Let X idempotent:

x2 = x mod M

=x

+ A4 be (8)

Then there exists an idempotent e E R which is expressible as a linear combination of the elements x, x2, x3,..., with coefficients in Z and such that e=xmodM. Proof (due to Herstein). Since x2 - x E M and M is nil, there exists a natural number n such that (x2

- x)" = 0

(9)

The various powers of x commute with each other and we may therefore apply the binomial theorem to get

where

Let e

=

xngn

154

VII.

RINGS WITH IDENTITY

From (10) we obtain the following equalities: e2

= X2ng2n = Xn-1Xn+lgg2n-l - X2n-1

g

= Xn-l

=

2n-1

xng2n-1 .'. = xngn = e

The idempotent e is congruent to x mod M as may be seen from the following:

> 0. But

since xs = x mod M for all s

0

=

(1 - 1)" = 3=0

=

1

(-1)j

-

j=l

(-1)j

n (.) .I

n (J

Therefore

5

(?) = 1 .I

(-1)j-1

j=1

Hence xg

3

x mod M and xrLgn= xn mod M from which it follows that e =xmodM

since x2 = x m o d M

I

Theorem 8. Let R be a ring with identity. Let e be an idempotent in R . Then J(eRe) = eRe n J(R) = eJ(R)e

Proof. Let a E J(eRe). Then a = eae and exa E J(eRe) for all x E R. Let y be a quasi-inverse of exu in eRe so that (e - y)(e - exa) = e. It follows that yxa - y = exa. Therefore (1 - y)(l - xa)

=

1 -y

-

xa

+ yxa = 1 - (1 - e) xa

+

Since (1 [l - el xa)(l - [l - el xu) = 1 and a(l - e) = 0, 1 - xa has [I - el xa)(l - y ) as a left inverse. Hence for all x E R , 1 xa has a left inverse proving that a E J(R). Therefore u = eae E eJ(R)e. Conversely, let a E eJ(R)e and let x E eRe. Since eJ(R)e C J(R), 1 - xu has a left inverse 1 - y E R . Hence

(1

+

~

(e

-

eye)(e - xa)

=

e(l

-

y) e(e - xu)

=

e.I

=e

1.

155

SEMIPERFECT RINGS

Theorem 9. Let e # 0 be an idempotent in the semiperfect ring R. Then eRe is also semiperfect. Proof. By the isomorphism theorem and the previous result eRe eJe

eRe f J J

- B

+

Let LleJe C L’leJe be left ideals in eReleJe. Consider the left ideals (RL J ) / J and (RL’ J ) / J in RIJ. Suppose if possible that RL J = RL’ J. Then e(RL J)e = e(RL‘ J)e. Hence L eJe = L‘ eJe since eRLe = L and eRL’e = L’. Therefore L = L’, a contradiction proving that ( R L J ) / J C (RL’ J)/J. Hence eRe/eJe is artinian. Now let u E Re and u2 - u E eJe C J . Since R is semiperfect and u(l - e) = (1 - e)u = 0, there exists by Theorem 3 an idempotent f E R with f - u E eJe and f (1 - e) = (1 - e)f = 0. Thus f =f e = ef =f and it follows that f E eRe. I

+ +

+

+ +

+

+ +

+

Since the factor ring RIJ of a semiperfect ring R is artinian semisimple

R, 0 ... @ R, where the R, are simple rings Let 1 = gl + ... + g, , gzE R, . The set (8, ,gz , ...,g,} of orthogonal idempotents of RIJ may be lifted to a set of orthogonal idempotents { g , , ..., g,} in R . Furthermore, g = g, + ... + g, is the identity of R since 1 - g is an RIJ

=

idempotent contained in J. By Theorem 9, the rings g,Rg, are semiperfect w 8% where R, is simple. We therefore obtain a decomposiand g7Rg2/gZJg2 tion of R as follows:

R

=

1R1

=

1

lGZC4

g,&

+

zf 3

g,Rg,

The first sum on the right-hand side is a direct sum of rings g,Rgz since g,g, = &,g, . The second sum is a subgroup M of the additive group of the radical since g,Rg, = g,g3R zz 0 mod J. In general M i s not an ideal because, for example, g,Rg, . g,Rg, C g,Rgl . We have proved the following important theorem: Theorem 10. Let R be a semiperfect ring. Then considered as an additive group, R = S @ M , where M is a subgroup of the additive group of the radical of R and S is a subring of R with the following properties: S , considered as a ring, is a finite direct sum of semiperfect rings R1 , ..., R, whose factor rings R,/J,, i = 1, ..., q are simple. Furthermore, Ri = g,Rg, , J , = giJ(R) g, , M = Cizi giRgj and { g , , ..., g,} is a family of orthogonal idempotents with sum 1, lifted from the set of identities of the rings R, in the decomposition of R / J as a direct sum of the simple rings Ri . I

156

VII.

RINGS WITH IDENTITY

The structure of the rings Ri above will be further investigated in Theorem 21. We shall also show that the subring S and the subgroup of {J, +} are uniquely determined up to an inner automorphism of R. To this end we first prove that if e is an idempotent in R, the ring eRe is antiisomorphic to the endomorphism ring of the R-module Re. More generally we have:

Theorem 11. Let e be an idempotent of R and let a E eRS, the mapping +a:

xe + xea

R. Then for each

for all x E R

is an element of Horn,(&, Rf) and for all a ~ e R f

q: a+

is an isomorphism of the additive group of eRf onto the additive group of Hom,(Re, Rf). If e =f,q is a ring antiisomorphism. Proof. Let 4 E Horn,(&, Rf).4 is uniquely determined by the image of e under 4. Since $(e) E Rf, 4(ez) = e$(e) = euf for some a E R. q clearly preserves addition and therefore eRf and Horn,(&, Rfj are isomorphic as groups. The second assertion follows from [q(ab)]xe

=

y5ab(xe) = xeub

= +bSba(Xe) =

= (xea)b

M b ) d 4 l xe

I

In a semiperfect ring the primitivity of an idempotent e is equivalent to eRe being a local ring: Definition 4. A ring with identity is local if RIJ is a division ring. Theorem 12. Let e be a nonzero idempotent in the ring R with identity. Then the following two assertions are equivalent: 1. e is primitive 2. Re is an indecomposable left ideal Furthermore, Condition 3 implies Assertion 1 : 3. eRe is a local ring If R is semiperfect, Assertions 1 , 2, and 3 are equivalent. Proof: The equivalence of 1 and 2 was proved in Theorem 1. Now assume Re = L1 0L, and let ri be the projection of L, @ L, onto Li. Then r1 and r2 are nonzero orthogonal idempotents in Hom,(Re, Re)

1.

157

SEMIPERFECT RINGS

whose sum is the identity of the ring. By Theorem 11, eRe contains two nonzero idempotents el and e2 such that e, e2 = e. Since the radical eJe of eRe contains no idempotent, e, eJe and e2 eJe are nonzero orthogonal idempotents in eReleJe. Hence this ring is not a division ring. Conversely, if R is semiperfect and if eReleJe is not a division ring, it contains two orthogonal idempotents with sum C since eRe is semiperfect by Theorem 9. These idempotents may be lifted to orthogonal idempotents el and e2 with sum e. I

+

+

+

Theorem 13. Let e be an idempotent in the semiperfect ring R with radical J. Let v: x X = x J be the natural homomorphism of R to RIJ. Then Re is an indecomposable left ideal in R if and only if Rt? is a minimal left ideal in = RIJ. -+

+

Proof. C&? = ZZ = (eRe + J ) / J m eRe/eJe. By Theorem 12, Re (resp. Re) is indecomposable if and only if eRe (resp. ZRZ) is local. Since

ERC FX eRe/eJe and the latter ring is semisimple artinian, so is I%?. Hence eRe is local if and only if 2 8 2 is. The indecomposable left ideals in are minimal left ideals since the lattice Vj&R) is complemented, by Theorem IV. 4.1. Therefore Re is indecomposable if and only if RC is minimal. 1

w

Theorem 14. Let e and f be idempotents in the ring R. The R-modules Re and R f are isomorphic if and only if there exist u, v in R such that uv = e and vu = f .

Proof. Suppose Re FX R f and let xe -+ xeulf, x E R, be the isomorphism (cf. Theorem 11). Let y f -+ yfv’e, y E R be its inverse. Then e . eu’ffv‘e = e and f fv‘eeu’j’ = f . Let u = eu‘f and v =fv’e. Conversely, suppose uv = e and vu = J Then the homomorphism xe + xeu = xuv . u = x u . vu = xuf E R f for xe E Re has the mapping y f yfv for y f E R f as inverse. I -+

If Re w R f for idempotents e and f in R it follows that RZ m Ef in RIJ. The converse is a consequence of the following: Theorem 15. Let e and f be idempotents in the ring R with 1. Let J be the radical of R. Suppose u’u’ = e mod J and v’u’=f mod J. Then there exist elements u and u in R such that uv = e and vu = f .

Proof. fv’ G u‘u’v’ = v’e and eu‘ = u‘v’u’ = u’f mod J imply eu’jiv’ E u‘jiv‘ = u’v’e = e2 = c mod J. Therefore x = e - eu‘fv’ E J and since x = ex, eu’jiv’ = e(l - x). Now x E J and hence there exists y E J such that (1 - x)(l - y ) = 1. Let v =fv’(1 - y ) and u = e d f . Then

158

VII.

RINGS WITH IDENTITY

uu = eu’fv’(1 - y ) = e(l - x )( 1 - y ) = e and hence ( U U ) ~= vuuu = veu = vu. This implies ( f - V U ) ~= f - fuu - uuf uu = f - uu. But since uu =fv’(1 - y ) eu’f =fu‘eu‘f =fu’u’f =f 3 =f (mod J), f - uu E J. Hence f = uu. 1

+

Theorem 16. Let e a n d f b e idempotents in the ring R with identity and let J be the radical. The left ideals Re and Rf are isomorphic R-modules if and only if RZ and Rf are isomorphic R/J-modules. Proof. Theorems 14 and 15.

1

By Theorem 13 the indecomposable left ideals in a semiperfect ring R are precisely those whose homomorphic images in RIJ under the natural homomorphism are minimal left ideals of the artinian ring R/J. By Theorem IV.3.3, these are isomorphic to the irreducible R-modules and therefore as a simple consequence of Theorem 16 we have:

Theorem 17. Let R be semiperfect. Then there is a one-one correspondence between the isomorphism classes of irreducible R-modules and the isomorphism classes of indecomposable left ideals. I In Theorems 11, 14, and 15 we did not assume that R is semiperfect and in Theorem 12 we only needed the assumption to prove the implication 1 3. We may therefore apply these results in the proof of the following theorem:

Theorem 18. Let {el , ...,em} and {fi, ...,f n } be two orthogonal families of local idempotents in the ring R. Let Cz, ei = Zy=lh= 1. Then m = n and there is an inner automorphism of R mapping the set {el, ..., em}onto ...,fin}. the set {fif

+

Proof. 1. Let x X = x J be the natural homomorphism of R onto R/J. We first show that RZi is a minimal left ideal in R = R / J (cf. Theorem 13). Set e = ei and assume m > 1. Since R = K(T - a) Re, it follows that R(T - 2) # R . Applying Zorn’s lemma we obtain a left ideal Z C R maximal subject to containing R(T - a). Clearly R = L Be. This sum is also direct: For suppose (Zn R2) # (0). By semisimplicity of R, (Ln # (0). Let x in R be such that Ti2 E L and & . X2 # 8. The element a22 # 0 has an inverse j i n the division ring ZRZ. Hence j .Z%Z = 2 E 1. Therefore R = L, a contradiction. As the complement of a maximal left ideal in R, Rt; is minimal. ---f

+

+

1.

SEMIPERFECT RINGS

159

2. By hypothesis and what'we have just shown in Part 1,

Re,

+ ... + Re,

=

Rfl

+ ... + Rfn

are two decompositions of as direct sums of minimal (and hence irreducible) R-modules. By the Jordan-Holder Theorem (Theorem V.4.2), m = n and there is a permutation 7~ of the numbers 1, 2, ..., n such that R Z i m Rfni, i = 1, 2, ..., n. Theorem 14 guarantees the existence of elements Ui, iii in R with UiVi = 2, and ViBi = f n i . By Theorem 15 we may assume that uivi = ei and u p i =fni , i = 1, ..., n. It now follows that vieiui= viuiviei =fni and uiLivi = ei . Set

u

=

C eiuifmi

and

v

fniviei

=

i

i

xi

Then uv = ei = 1 =fmi = vu and vej =fnjvjej=.fnjv. The element v is therefore a unit in the ring R and it induces the inner automorphism x + vxv-I which maps ej onto f n j ,j = 1, 2, ..., r. 1 As a first application of this theorem we prove:

Theorem 19. Let R be a semiperfect ring and let R = S @ M where S and M are as in Theorem 10. Then this representation of R is unique up to an inner automorphism of R. Proof. Each idempotent gi in Theorem 10 is the sum of an orthogonal family of primitive (and hence local) idempotents gij . The g i j are mutually orthogonal for all i, j and gij = C i g i = 1. Let { g , , ..., g,} and {h, , ..., h,} be two sets of idempotents which give rise to a decomposition of R as in Theorem 10. By Theorem 18 there exists an inner automorphism x -+ vxu-l mapping { g i j }onto {hij). This mapping takes the summands g,, and gi, of gi onto the summands h,, and hrt of h, since g,gij = S,,gij . This shows the existence of a permutation 7~ of the numbers 1, 2, ..., n such that vg,v-l = h m i . I

xi,$

A further application of Theorem 8 is given in the following characterization of semiperfect rings. Theorem 20 (B*Muller). A ring with 1 is semiperfect if and only if 1 is the sum of a finite number of orthogonal, local idempotents. Proof. I . We have already shown the necessity in Theorem 5. 2. Conversely, let 1 = e, + ... + en , ei local orthogonal idempotents.

160

VII.

RINGS WITH IDENTITY

As was shown in the first part of the proof of Theorem 18, RE$, i = 1, ..., n is a minimal left ideal in R . R therefore has a finite maximal chain of left ideals RZ, @ ... @ RZi, i = 1, ..., II. By Theorem V.4.2 (Jordan-Holder) all such chains have the same length, or, in other words, R is artinian. 3. L e t f = f 2 m o d J a n d f = f l + . . . + f , , T - f = f , + l + * . . + f n bea decomposition of T in R as a sum of orthogonal primitive idempotents in the artinian ring R = R/J. By Theorem 18 there is an inner automorphism X iiXV with UV = Vii = T which maps the set of Zi onto the set of . By , i = I , 2 , ..., n and therefore renumbering we may assume that iieiV ii(Zl ... Zr)V =f. Set e = el ... e, . If u and u are representatives of the cosets U and V respectively, uv = 1 - o for some o E J. Therefore (1 - w’) uv = 1 where (1 - w’)(l - w) = I . Let u1 = (I - w’)u and v1 = v. Then u1 = u mod J and ulul = I . It follows that ululvlul= vlul _= vu = I mod J and hence I - ulul = (1 - ul@ E J. This implies vlul = 1. The mapping x -+ulxvl is therefore an inner automorphism of R and in particular ulevl = ul(el ... e,) u1 E f mod J. Hence idempotents may be lifted modulo J. I ---f

=Ti + +

+ +

+ +

The structure of the direct summands Ri of S in Theorem 10 may be described in terms of the following concept: Definition 5. An R-module M is free if it is a direct sum of modules each isomorphic to RR. Theorem 21. Let R be a semiperfect ring, J its radical and R/J a simple ring. Then R is isomorphic to the full endomorphism ring Hom,(M, M ) of a finitely generated free right T-module M over the local ring T = eRe, where e is a suitable primitive idempotent of R. Proof. Let I be the image of the element x of R under the natural homomorphism of R onto R = RIJ. Theorem 5 applied to R, guarantees the existence of an orthogonal family of primitive idempotents {el, ..., eTZ}such that el ... en = 1. Let e = el . By Theorem 13, ZR is a minimal right ideal of R. By hypothesis, is a simple artinian ring and therefore the ZiR, i = 1, ..., n are isomorphic to each other (Theorem IV.3.3). By Theorems 14 and 15 there exist elements ui , vi in R such that uivi = ei , u p i = e. Since 1 E R, R is isomorphic (as a ring) to HomR(RR, RR), the ring of left multiplications of R by elements of R (cf. Theorem 1.2.3). We first show

+ +

H o m R ( R R RR) , m HomPRe(Re, Re)

(12)

where eRe is the centralizer of the R-module Re and Re is considered as a right eRe-module. (If J = 0, (12) has already been proved in Theorem IV.3.2.)

1.

SEMIPERFECT RINGS

161

Proof of (12). Let 4 E HomR(RR, RR) and set d'(xe) = +(xe) = $(xe2) = $(xe)e E Re for x E R. 4' is clearly an element of Hom(Re, Re). Furthermore, $'(xe . eye) = +(xe) eye = +'(xe) eye for all y E R and hence 4' E HomeRe(Re,Re). Let 7 be the mapping defined by 7: --sr 4' for 4 E HOmR(RR ,RR). We show 7 iS an isomorphism onto HOm,R,(Re, Re). Clearly 7 is additive and preserves multiplication. Given 4' as the image of 4, we may recover q5 as follows: uiui = e implies uieui = uiui ' uiui = ei2 = ei for *i = I , 2, ..., n. Let a E R. Then a = e,a ... e,a = uievia. It follows that C$

+ +

+a =

4'

C y5(uievia) = C (4(uie))via = C +'(uie) via

= 7 4 = 0 we therefore have

4 =0

and 7 is one-one. Now let C $(uie) uieia for a E R is e,R by the orthogonality of the e, . Since ax = C eiax and eiax E eiR, #o is clearly an endomorphism of RR . To show that ~ q =5 $,~ let a = ae = C eiae. Since $ is an endomorphism of the right eRe-module Re and $(uie) = $(uie2) = #(u,e)e, the product $(uie) uieiae = $(uievieiae) = $(eiae). Therefore (740)(ae)= q50(ae) = $(ae) and 7 is an epimorphism. Hence 7 is an isomorphism. We therefore have If

$ E HomeR,(Re, Re). The mapping well-defined since RR = e,R + ...

R

%

HOmR(RR, RR)

%

c$~:a

+

= eia -+

HOm,R,

(01eiRe, @ 1eiRe) i

i

(13)

We must now show that the eiRe are isomorphic as right eRe-modules to T, , where T = eRe. The mappings

eixe + uieixe,

x E R,

i = 1, 2,

..., n

(14)

are the required isomorphisms since uieixe

=

uiuiuixe = evixe E eRe

In other words, we have proved that R is isomorphic to the full ring of eRe-linear mappings of a direct sum of n copies of the ring eRe, considered as a right eRe-module. A free right T-module M with generators m, , ..., m, is the direct sum of the right T-modules miT. A T-linear mapping A of M into itself is (as in the case of vector spaces) uniquely determined by the images Am,, k = 1 , ..., n. Let Am, = mitik . Then Hom,(M, M ) is isomorphic to the ring of n x n matrices, T n x n ,over T. I

xi

We now prove:

Theorem 22. Let P be a ring with 1 and let J be its radical. Let P,,, be the

162

VII.

RINGS WITH IDENTITY

ring of n x n matrices over P and let Cik denote the matrix with I in the ith row and kth column and zeros elsewhere. Then the radical J(Pnxn)

=

C J(P) C i k

i.k

+ +

ProoJ Let E = C,, ... C,, be the identity of R = P,,, . Let Z k = zikCik where zilCE J , i = 1 , ..., n. In particular there exists (1 - Z k k ) - l E P. Set z k ' = - z k ( 1 - Z k k ) - l . Then

xi

( E - Zk')(E - 2,) = E =

2i

+C

~ i k ( l- ? k k ) - l C i k

zik(l

- Zkk)-lZkkCik

-

i

C

~ikCik

i

since c i k Cjk= S k j C i k .Therefore ( E - Z,')(E - 2,) = E and the left ideals JCik are quasiregular for each k = 1, ..., n. Hence x i , k JCik is contained in the radical J(R). Conversely, let 2 = Z&k E J(R). For each a E P, J(R) also contains aE . Z = azikCik.Then

xi

9

2 h

There exists Z' E

=zikc(jc E =

'hi

c

i,k

xi,k

xi,k =

azikcikcLh

azLhE

R such that

( E - Z')(E - az&)

=

( E - Z')(I - azjz).

Therefore 1 = ( 1 - zlJ1 - a z j z ) for all a E R and azj is quasiregular. JCil, . I Hence J(R) C

xi,k

Now let Tnxndenote the ring of n x n matrices over the local ring T and let J ( T ) be the radical of T. By Theorem 22, the radical of Tnxn consists of those matrices all of whose entries lie in J ( T ) . It follows that the factor ring Tnx,/J(Tnxn) is isomorphic to the ring of n x n matrices over the division ring T/J(T)and is therefore a simple ring. Moreover, the identity C1, ... C,, of T,,, is the sum of the orthogonal family of local idempotents Cii , i = 1, ..., n. By Theorem 20, Tnxnis therefore a semiperfect ring with simple factor ring Tnxn/J(Tnx,). In conjunction with Theorem 21 we have the following characterization of these rings:

+ +

Theorem 23. R is a semiperfect ring with R/J simple if and only if it is isomorphic to the ring of n x n matrices over a local ring. I As an illustration of this theorem we prove:

Theorem 24. Let @ be a field and @ *

= @[[w]]

the ring of formal power

1. SEMIPERFECT RINGS

163

series in the indeterminate w . Then the ring @$xn of n x n matrices is semiperfect and factored by its radical w@zxn is the simple ring G n X n . However, , @ ,; is not artinian.

@zxn

Proof. The radical of @* is w @ * and @ * / w @ * = @. However, @* is not artinian since urn@*, m = 0, 1, ..., is an infinite, strictly descending chain of ideals and this gives rise to the descending chain in . I

@zxn

In Theorem 18 we proved that if we are given 2 families of orthogonal, local idempotents whose sum is 1, we can find an inner automorphism which maps one family onto the other. We now use this fact in the proof of the following theorem due to Krull and Schmidt which we shall need in the next section: An R-module M is said to be indecomposable if M has no nontrivial direct summands. Now let M = Ml @ ... @ M , and M = Nl @ ..* @ N , be two representations of M as direct sums of indecomposable submodules. It will be shown that m = n and that for a suitable permutation ?T of the numbers 1, 2, ..., n, M im Ngi if M is both artinian and noetherian. The proof depends on the fact that such a module is local if and only if M is indecomposable. Theorem 25. Let 4 E Hom,(M, M ) . 1. If M is artinian, 4 is an epimorphism if is a monomorphism. 2. If M is noetherian, 4 is a monomorphism if 4 is an epimorphism.

+

Proofi 1. There is a natural number n such that M 3 Mq5 3 ... 3 M$n= M+"+l. Let x E M . Then xqP = y+"fl for some y E M . But 4%is a monomorphism since 4 is and therefore x = y+. 2. Similarly O C 04-1 C ... C 04-. = O+-n-l for some n. Let x+ = 0 for some EM. is an epimorphism since 4 is. Hence there exists y e M such that y4" = x . Then y+"+l = X#J = 0 and y E O+-"--l = O$-.. This implies x = y+n = 0. I

Every linear transformation T of a finite-dimensional vector space gives rise to a decomposition of the space as a direct sum of the kernel of the transformation and a subspace isomorphic to the image of T. A similar situation obtains when M is artinian and noetherian Theorem 26 (Fitting). Let A4 be an artinian and noetherian R-module and let 4 E Hom,(M, M ) . Then for some natural number n,

164

VII.

RINGS WITH IDENTITY

Proof. Since M is artinian, there exists n with M 3 A443 ... 3 M4" = M P + l = ... - M # P . Thus c#Pinduces an epimorphism of the noetherian module M4". By Theorem 25 this restriction of 4" to A44n is a monomorphism. Hence the sum MqP @ OC#-. is direct. Moreover, if X E M , x$" = yq52n for some Y E M . Therefore x - yqP E O C # - ~ and x = y+" (x - y4") proving that M~#P0O+-" = M . I

+

Theorem 27. Let M be an artinian and noetherian R-module. Its endomorphism ring Hom,(M, M ) is local if and only if M is indecomposable. In this case each endomorphism is either an automorphism or nilpotent.

Proof. Let 4 € H o m R ( M , M ) and let M be indecomposable. By Theorem 26 either M 4 n = 0 or M+" = M . In the latter case 4" is onto and therefore 4 is a monomorphism by Theorem 25. In the former case, c $ ~ is nilpotent. If C$ is not an automorphism, neither is a 4 for any a E Hom,(M,M). Therefore every element in the left ideal of Hom,(M, M ) generated by 4 is nilpotent. This implies 4 is in the radical of the endomorphism ring. On the other hand no automorphism of M is in the radical of Hom,(M, M ) since the automorphisms are units in Hom,(M, M ) . Therefore the radical consists precisely of the nilpotent elements of the endomorphism ring. It follows that each nonzero element of Hom,(M, M)/J(Hom,(M, M ) ) is a unit and therefore Hom,(M, M ) is local. Conversely, if M is decomposable, say M = M I 0M , , then the projections x1 and x 2 of M onto M , and M, , respectively, are two orthogonal idempotents in Hom,(M, M ) , showing that this ring is not local. I Theorem 28 (Krull-Schmidt). Let M be an artinian and noetherian R-module. Let M = M , @ ... @ M , and A4 = Nl 0 ... @ N,s be two decompositions of M as the direct sum of indecomposable submodules. Then r = s and for some permutation x of the numbers I , ..., r, M i m M T i , i = 1 , 2,..., r .

Proof. Set E = Hom,(M, M ) . Let x , , i = 1, ..., r be the projections of M onto M i , qj , , j = 1, ..., s the projections of A4 onto N j . Then { x l , ..., 7 ~ and {vl, ..., v,) are two orthogonal families of idempotents of E, each with sum 1, where 1 denotes the identity automorphism of M . By Theorem 27 the endomorphism rings of the M iand N , are local. To apply Theorem 18 we must first show that these rings are isomorphic to the x i E x i and q j E q j , respectively. Since x i M i = M i , for each 4 E E, r i + x i induces an endomorphism of M i .On the other hand let 4, E Hom,(M, , M i ) . Set ('& m k )$i= tn& . Then $iE Hom,(M, M ) and x i $ i x i = q& . Therefore the mapping +%+ $i is an isomorphism of Hom,(M, M ) onto the subring

~ )

1.

SEMIPERFECT RINGS

165

riErTTi of E and hence niEri is local. Similarly yjEyj is local. By Theorem 18, r = s a n d there is a unit 5 E E (i.e. an automorphism of M ) and a permutation r of 1, 2, ..., r such that c-'vic = qni for i = 1,2, ..., r. Since Nni = Mqni = M[-%ri< = Mri5 = Mi[, M iis mapped onto Nniunder this automorphism f o r i = 1 , 2 ,..., r.

a

Theorem 28 was proved under the assumption that the R-module is artinian and noetherian. In applications the following is often useful:

Theorem 29. Let R be artinian with identity and M a unital R-module. Then M is artinian if and only if it is noetherian. Proof. Consider the following chain of submodules:

M 3 J M 3 ...3 JNM

=

0,

where J is the radical of R. The factor modules JiM/Ji+lMare RIJ-modules and hence completely reducible. If M is artinian (noetherian) so is JiM/Ji+lM. A completely reducible module is artinian (noetherian) if and only if it is a direct sum of finitely many irreducible modules. Therefore if M is artinian (noetherian), M has a finite composition series and by Theorem V.4.2 (Jordan-Holder), M is noetherian (artinian).

a

Theorem 30. Let R be a ring with 1 and U a submodule of M . Then 1. M is artinian (noetherian) if and only if both U and M / U are. 2. Let M be a finitely generated R-module. Then M is artinian (noetherian) if and only if R is.

Proof. l(a) Clearly if M is artinian (noetherian), so are U and MlU. l(b) Let U and MIU be artinian and

M

=

Mo1M,2M,2...

be a descending chain of submodules of M . Then U = M , n U 2 M I n U 2 M2 n U 2 . * . and MIU = ( M , U ) / U2 (M1 U ) / U 2 *.. are descending chains of submodules of U and MIU, respectively. By hypothesis there exists N such that MN n U = M , n U and ( M N U ) / U = ( M , U ) / U for all n 2 N . Since M , 2 M , and since the lattice of R-submodules is modular we have M , = M , (U n M,) = M , ( U n MN)= ( M , U ) n MN = (M, U ) n M , = M , for all n 3 N . Similarly we may show that if U and MIU are noetherian so is M . 2. Let M be finitely generated, say M = Rm, Rm, ... Rm, and let R be artinian (noetherian). If n = 1, M is cyclic and therefore isomorphic

+

+

+

+

+

+ +

+

+

+ +

166

VII.

RINGS WITH IDENTITY

to aR/L, where L = {x E R 1 xm, = O}. Since RR is artinian (noetherian) so is every factor module. Assume inductively that the theorem holds for modules which can be generated by n - 1 or fewer elements. Then Rm, is artinian (noetherian) as we have just shown and MIRm, w (Rm,

+ ..' + RmJRm,

n (Rm, i...

+ Rm,)

is artinian (noetherian) by induction hypothesis. Hence M is artinian (noetherian) by the first assertion of the theorem. I

2. Projective modules and Asano orders

In the last section the indecomposable direct summands of RR and in Theorem 1.21 the free R-modules played the most important roles. All these modules have the property expressed in Definition l(1) below. Definition 1. 1. Let ... -+ A BG C %D ... be a sequence of modules and homomorphisms. The sequence is said to be exact at B if image f = kernelg. The sequence is said to be exact if it is exact at each module. 2. Let R be a ring with 1 and M an R-module. M is called a projective R-module if the left diagram can be embedded in the right diagram -+

M I4

A

2 B +o

n

exact

A-B-0

in such a way that T $ = +. Roughly speaking M is projective if homomorphisms of M can be lifted from factor modules B to modules A. Theorem 1. The R-module RR is projective. Proof. Let 16 = b E B and b = UT for some a E A. Then x+ = x b and ( x a ) = ~ xb for X E R. Set xy5 = xu. Then $ E HomR(R, A ) and X$T = ( x u )= ~ x b = ~ ( 1 4=) ~ 4 I . Theorem 2. Let M be the direct sum of the R-modules M i ,i E I . A4 is projective if and only if the Mi are. Proof 1. Assume each M iis projective and consider the diagram M

I+

A L M - 0

2.

PROJECTIVE MODULES A N D ASANO ORDERS

where the row is exact. Let 4 I M Mi = 4ibe the restriction of projectivity of M i there exists $i E HomR(Mi, A ) with # E HomR(M, A ) . Then #T = 4 and M is projective. $= 2. Conversely, if M is projective and we have the diagram

xi

with exact row, extend to 4 E Hom,(M, B ) by defining $T = 4 for some $ E HomR(M,A ) . Then the restriction #A

=

$ IMA

makes the diagram commutative, i.e. T$,,

9

=

4 to i

= ~

167

M i . By

di. Let

4=

Let

A)

4A. I

The following is an immediate consequence of Theorems 1 and 2:

Theorem 3. Every free R-module and every indecomposable left ideal in a ring with 1 is projective. I Theorem 4. Every R-module is the homomorphic image of a free (and hence projective) R-module P . Proof. Let {bi I i E I}be a set of generators of the R-module B. Let Ri , i E I be a family of R-modules each equal to RR and let P = @ C Ri . The mapping xi -+ C xibi is an epimorphism of the free R-module P onto B. I

xi

Theorem 5. An R-module M is projective if and only if for every exact sequence A + M 0 there exists a submodule U of A such that A = U 0M ’ , where M’ w M . ---f

Proof. 1. Sufficiency: Let P be the module in Theorem 4. Then M is isomorphic to a direct summand of the projective module P and therefore by Theorem 2, M is projective. 2. Necessity: Take = 1 , the identity of M in Definition 1. Then there exists a monomorphism $ from M to A such that #T = l,,,, . Moreover, T$T$ = T $ and the mapping T $ from A onto M$ is an idempotent endomorphism of A . Hence A = M T $ 0A(1, - T $ ) since a = UT$ a(1, - T # ) for all a E A I

+

+

168

VII.

RINGS WITH IDENTITY

Theorems 2, 4,and 5 imply the first assertion of the following: Theorem 6. 1 . An R-module is projective if and only if it is isomorphic to a direct summand of a free module. 2. Let R be a semiperfect ring and let M be a finitely generated R-module. M is projective if and only if M is isomorphic to a direct sum of finitely many R-modules each isomorphic to some indecomposable left ideal of R . Proof. All has been proved except for the necessity of the second assertion.

R = M / J M is a finitely generated R = R/J-module and is therefore a direct sum li? = C @ U, of irreducible R-modules U, since R is semisimple artinian. These are images of indecomposable left ideals Re,in Runder epimorphisms 4%.

+

The epimorphism whose restriction to Rei is +t therefore maps the direct sum P = 0Re, onto M . Since P is projective by the first assertion of the theorem, there exists $ E HomR(P, M ) with i,!m = 4, where T :M -+ Since both 4 and $ 7 ~ are epimorphisms, we have M = P# JM. By the following lemma, J M is small in M , i.e., M = JM N implies M = N. This proves P$ = M . Since M is projective, ker t,h is a direct summand of P by Theorem 5. Moreover, #I = $T and hence ker $ C ker #I. By the definition of 4, ker = C @ J e t = JP and is therefore small in P. Hence ker t,h is small in P and, as a direct summand of P, is the zero module. In other words, $ is an isomorphism of P onto M.

+

+

a.

+

Lemma. (Nakayama). Let R be a ring with identity and J its radical. Let M be a finitely generated R-module. Then J M is small in M , i.e., if A4 = J M N for some submodule N of M , then M = N .

+

+ + + + + +

+

... Rm, and M = JM N i t follows that Proof. From M = Rm, m, = wlml + ... w,m, v , where w i E J and v , E N . Therefore (1 - w,)m, = wlml ... ~ , - ~ m , - , us . Multiplying this equation on the left by the inverse (1 - w,’)of (1 - w,),we get

+

+

m,

=

wl’ml

+ ... + ~ ~ - ~ m +, -us’ , + N . Applying the same process to m,-l

Thus A4 = J M , ... 4-Jnz,-, as we applied to m, and continuing in this fashion, we eventually arrive at M=N. I A vector space M over a division ring R is a free R-module and, given a basis {b, I i E I } each m E M is expressible in the form M = C(rn$,)6, , where $, E Hom,(M, ,R) maps the vector m onto its ith coordinate relative to the basis {b, 1 i E 11. The following is a generalization of this situation: Theorem 7. Let R be a ring with 1 and M and R-module generated by

2.

PROJECTIVE MODULES AND ASANO ORDERS

169

{bi I i E Z } . Then M is projective if and only if there exists a set 1 $iE HomR(M,RR), i E I } such that

{$t

m where m$i

=0

=

1 (m$J bi

for all m E M

i

for all but a finite number of

(1)

&.

xi

Proof. Let P = 0 Rei be the free R-module with basis lei 1 i E I } and rr an epimorphism of P onto M taking ei to 6, (Theorem 4). M is projective if and only if it is isomorphic to a direct summand of P. As in Theorem 5, this is the case if and only if there exists a homomorphism $ of M to P such that $rr is the identity automorphism 1 on M . If $ exists, each image

m$ is uniquely expressible in the form

m*

=

1 (m*,>ei

(2)

where m$i is an element of R depending on m and i. The mapping Qi:m -+my$ is an R-module homomorphism of M to RR. Applying rr to both sides of (2) we get (1) since $7r = l Mand e p = bi . Conversely, if the condition is satisfied, then we may define an R-module homomorphism $ of M to P by m$ = (m$) ei for all m E M . Applying rr to both sides of (1) we get m$n = (m$J bi = m for all m E M since ein = bi . I

xi

xi

We may now base the theory of noncommutative Asano orders (due to Robson) on this theorem. We start with the following: Definition 2. An element a of a ring R is regular if it is neither a right nor a left zero divisor. A ring Q with 1 is a quotient ring if each regular element has an inverse. The subring R of Q is an order (more precisely, a left order) in the quotient ring Q if each element of Q is expressible in the form a-lb, where a, b E R and a is regular. An additive subgroup I of Q is a left R-ideal provided (i) RZC I, (ii) I contains a regular element, and (iii) there exists a regular element b E R such that Ib C R. We define right R-ideals and (twosided) R-ideals in Q analogously. An R-ideal is integral if it is contained in R. Definition 3. The order R with 1 in the quotient ring Q is an Asano order if the R-ideals in Q form a group under multiplication. In this case the group is abelian, the prime ideals in R are the maximal ideals of R and each ideal in R is uniquely expressible as a product of powers of prime ideals (cf. Theorem 11). The integers are an example of an order in the quotient ring Q of rationals. The Z-ideals in Q are of the form Za/b where a # b. The aim of the following discussion is to obtain a characterization of the Asano orders R in Q. It will be shown that they are precisely

170

VII.

RINGS WITH IDENTITY

the “maximal orders” in Q with the property that each integral ideal in R is projective as a left R-module. The expression “maximal order” is precisely defined as follows: Definition 4. An order S in the quotient ring Q is equivalent to the order R in Q if there exist regular elements c, d , e , f in Q such that cRd C S and eSf C R. An order R in Q is maximal if it is maximal (with respect to set theoretic inclusion) among all orders equivalent to R. Theorem 8. 1. Let R be an order in the quotient ring Q. Then for each pair (a, b) of elements in R with a regular, there exists a pair ( a , , b,) in R with a, regular, such that alb = b,a is a common left multiple of a and b. Note: In Chapter XI, we shall show that this condition (due to Ore) is a sufficient condition for the existence of a quotient ring Q containing R as an order. Proof. The product ba-l is in Q . Hence there exist a, ,6, E R such that ba-l = ay’b, . 1

Addition and multiplication in Q can be reduced to the corresponding operations in R once we have established the existence of common denominators. Theorem 9. Let a , , a 2 ,..., a, be regular elements in the order R of the quotient ring Q. Then there exist regular elements a, 6, , ..., b, in R such that a;’ = a-lb, ,i = 1, ..., n. Proof. By induction on n: When n = 1, take a = a12,b, = a, . Assume inductively that regular elements a,, b,’, ..., bhPl can be found so that a;’ = aklb,’, i = 1, ..., n - 1. By Ore’s condition (Theorem 8) there exist elements b and b, in R with b, regular such that bnan = ba, . Solving for b we get b = bnana;‘ and as a product of three units in Q, b is also regular. It follows that a = bnan = ba, is regular and a;’ = a-lb, a;’ = a-lb, Finally, set bi = bbi’, i = I , 2, ..., n - 1. Then a;’ = aG1bi’ = a-lbb,‘ = a-lbi . 1

.

Let I be an R-ideal of Q. It may happen that Z is also a left ideal with respect to an order R’ properly containing R and contained in Q. To understand this we need the following: Definition 5. Let R be an order in the quotient ring Q and Z an R-ideal in Q. Then the set Ol(Z) = {q E Q I qZ C Z} is the left order ofZ and the set Or(Z) = {q E Q I Zq C I } is the right order of I . Finally, we define I* = { q E Q 1 ZqZ L I } as the inverse of I.

2.

PROJECTIVE MODULES AND ASANO ORDERS

171

Since R is contained in both Ol(Z) and Or(Z), each is an order in Q. Moreover, they are equivalent to R: In fact, let x be a regular element of I and b be a regular element in R with Zb S R. (x and b exist by definition). Then for each q E Ol(Z), I . q . xb C Zb C R and 1 . R . 1 C Ol(Z). Similarly we may show that Or(Z) is equivalent to R. Since

I*

= {q E

Q 114 C Ol(Z))

= (4 E

Q I qZC Or(Z))

(3)

and since &(I) and Or(Z) are subrings of Q, I* is a subgroup of the additive group of Q for which Or(Z) . I*

C I*

and

I* . Ol(l)

L I*

Let b be a regular element with Zb C R C Ol(Z). Then b is in I*. Let x be a regular element in I. Then Z*x C Or(Z) and xZ* C Ol(Z). Hence I* is a left Or(Z)-ideal and a right OI(Z)-ideal. It follows from (3) that Z*I C Or(Z) for arbitrary I. The following theorem characterizes those R-ideals Z for which Z*Z = Or(Z).

Theorem 10. Let Z be an R-ideal in the order R of the quotient ring Q. Then Z*Z = Or(Z) if and only if Z is a projective left O1(I)-module. In this case Z is a finitely generated left Ol(Z)-ideal. If, in addition, R is a maximal order in the quotient ring Q then Or(Z) = Ol(Z) = R. Proof. 1. Since 1 E Or(Z) and Z*Z = Or(Z), there exist finite sets {q, E I* 1 i E J } and {b, E I I i E J } such that C qibi = I . Therefore, for each a E I, C aq,bi = a. Set a+, = aqi for a E Z and i E J. By (3), the form a set of homomorphisms of the left O1(Z)-module Z to the left Ol(I)-module Ol(Z) such that C (a#,) bi = a for all a E I. It follows from the criterion established in Theorem 7 that I is projective as a left OI(Z)-module and is generated by finitely many 6, . 2. Conversely, let Z be a projective Ol(Z)-module. Again, by Theorem 7, let {bi I i E J } be a set of generators of Z considered as a Ol(Z)-module and let {+, j i E J } be a set of Ol(Z)-module homomorphisms of Z to Oi(Z) with the property that Xi(a+,) b, = a for all a E I. Since 1 E Q , the endomorphism ring of ,Q consists of the set of right multiplications by elements q of Q. The mapping x xq for all x E Q induces an O1(Z)-module homomorphism of I to Ol(Z) if and only if Zq C Ol(1); in other words, if and only if q E I*. Since Z contains a regular element, it follows that distinct elements of I* give rise to distinct elements of Homo,c,)(Z, Ol(Z)). Moreover, this monomorphism of the additive group of I* to the additive group of Homo,c,)(Z, Oi(Z)) is an epimorphism as can be seen from the following: Since Z contains a

+,

---f

172

VII.

RINGS WITH IDENTITY

regular element of Q (and hence a unit of Q), QZ = Q. Furthermore, Ol(Z) is an order in Q and therefore each element of Q is of the form c-lx, where c, x E Ol(Z) and c is regular. Now Q = QZ = {CZ1c-lxiai I ai E I , ci , xi E Ol(Z)}. The existence of common denominators proved in Theorem 9 gives Q = {c-la 1 c E Ol(Z), c regular, a E I}. Now let 8 be an Ol(I)-module homomorphism of Z to Ol(l). Then

8*: c-la

-,c-l(a8)

for c E Ol(Z), a E I

is a well-defined mapping of Q into itself. Proof. Suppose c-la = cylal. Then a = ccyla, . Since Ol(Z) is a (left) order in Q, there exist elements d and dl in Ol(Z) such that cc;" = d-ld 1 . Then a = d-ld1a, and hence du = dlal . It follows that d(a8) = (da)8 = (dlal)O = dl(alO) and therefore a0 = d-ld1(al8) = ccT1(a,8) proving that c-l(a8) = c;l(a18). Hence 8* is well-defined. That 8* is additive again follows from the existence of common denominators. Moreover, 8* is a Q-module homomorphism. Proof: cy1al . c-la = c;' * d-lb . a, where b E Z,d E Ol(Z) with a1c-l = d-lb. Then (cyla, . c-la) 8* = (dc,)-l ((ba) 8) = cy1d-lb(a8) = cy1a1c-l(a8) = ~;'a,((c-~a)8*). The mapping 8 E Homol(,,(Z, Ol(Z)) is the restriction of 8* E Hom,(,Q, ,Q) to Z since (1-la) 8* = a8 for all a E I. It follows that to each 8 E Homol(,)(Z, Ol(Z)), there exists q E I* such that a8 = aq for all a E Z. Let q i E I*, i E J correspond to the +i, i E J occurring in the equation a = (a&) b, for all a E I

c i

For a given a E I , aq, = = 0 for all but a finite number of i. Since Z contains a regular element this implies that q i = 0 for almost all i and hence almost all are the zero map. It follows that since

+<

a

=

1(a+i) bi = 1aqibi = a c Bibi 2

for all a E I

z

I is generated as an O1(I)-module by the finitely many b, for which q i # 0. Moreover, if we choose a regular in Z, we get 1 = C Bib, E I*I and hence Z*Z C Or(Z) C Z*ZOr(Z) = Z*Z. I

Theorem 11. Let R be a maximal order in the quotient ring Q. Let R contain the identity of Q and let each integral R-ideal in Q be projective as an R-module. Then the lattice of integral R-ideals in Q is noetherian and we have the following: 1. To each integral R-ideal Z in Q there corresponds an R-ideal I-1 in Q such that Z-lZ = ZZ-1 = R.

2.

PROJECTIVE MODULES AND ASANO ORDERS

173

2. Each integral R-ideal in Q is a product of maximal ideals in R. In particular, each prime ideal is maximal. 3. The R-ideals in Q form an abelian group under multiplication and therefore R is an Asano order in Q.

Furthermore, each R-ideal I in Q is expressible as a product of powers of maximal ideals in R: I

=

JLFJL:z

A,:

k, E Z

This product is unique up to order.

Proof. 1. By Theorem 10, Ol(I) = O,(I) = R for each R-ideal I in Q since R is a maximal order in Q. The lattice of R-ideals in Q is noetherian since each R-ideal is finitely generated, by Theorem 10. Hence, a fortiori, the lattice of integral R-ideals is noetherian. In Theorem 10 we have already shown 1*1 = R. We now prove that if I is a maximal ideal in R, then 11* = R. Assume R = I*. Then R = 1*1 = R I = I, contradicting the maximality of I. Hence R # I*. Since 1 E I*, R C I* and hence I C 11* C R. Assume I = II*. Then R = 1*1 = RI* = I*, a contradiction. Now let I be an arbitrary integral R-ideal. There exists a maximal R-ideal A, containing I . If I = I&,,* C &&* = R then &* = R&* = I*I&* = 1*1 = R, a contradiction. Therefore I C I&*. If I&,* # R, I&,* is contained in a maximal R-ideal A,. Again, as before, we have I&,*&,* _C R and I C I&,* C IJL,*jlt,* C R. After a finite number of steps we get IGl*JL2* ..*A,*

=R

Set I-, = A,* ... A,*. Then 11-1 = I-lI = R and the first assertion is proved. Multiplying (4)on the right by A, ... A, we get I = A,, ... A, . This proves the second assertion since if P is prime, P = JL, ..* JL, implies that P must be one of the A+. 2. If A, and A, are maximal ideals, JL1JL2= A, n A,. Proof JCl n A, C A, if JL, A, . Therefore, as we have already shown in 1 , A, n A, = A1S for some integral R-ideal S . This implies JLilc,GC A, and therefore 'S C A, since A, JL, and A, is prime because it is maximal. Hence A, n A, = A19C &,A2C A, n A , . In other words: multiplication of two maximal ideals is commutative and therefore by Part 1 , multiplication of arbitrary integral ideals is commutative. 3. Each integral ideal 'S possesses an inverse R-ideal 'S-l in Q. Let I be an R-ideal in Q. Let J = { x E R 1 Ix C R}. Then J is a subgroup of the additive group of R for which R J = J = JR. By the definition of an R-ideal, R contains a regular element a such that l a C R. Clearly a lies in J and therefore J is an integral R-ideal as well as the product IJ. By what we proved

174

VII.

RINGS WITH IDENTITY

in Part I, Z = IR = IJJ = Z J . J-1and hence I ( J . (ZJ)-l) = (ZJ)(ZJ)-l = R. Therefore each R-ideal in Q has an inverse in the multiplicative semigroup of R-ideals in Q with R as identity. We have therefore proved the first part of the third assertion of the theorem. Finally, if Z J = A1JLG, ... A, and J= ... M r , then

I

=

IJ * J-'

=

A1 .M,,JL;tl

a * -

A;'

It follows that the representation of Z as a product of maximal ideals is uniquely determined up to order since the JLi are prime. I We have just proved that if R is a maximal order, fulfilling the other hypotheses in Theorem 11, in a quotient ring Q it is an Asano order. The converse is also true: Theorem 12. Let R be an Asano order in the quotient ring Q. Then R is a maximal order and each integral R-ideal in Q is a projective R-module.

Proof. 1. We first show that R is a maximal order in Q . Suppose not and let T be equivalent to R with R C T. By Definition 1, there exist regular elements a-lb and c-ld in Q with a-lbTc-ld C R where a, b, c, and d are regular elements of R. It follows that

+

bTd

=

bTcc-ld C bTc-ld C aR C R

TdR is a subring of T containing R and therefore U is Hence U = R an order in Q . Moreover, bU = bR bTdR C R. If R # U , set S = U . If R = U , then Td C TdR C U. Set S = T in this case. In either case S is an order in Q which is equivalent to R. Furthermore, R C S and bS C R (or Sd C R) for a suitable element b (resp. d) in R. Suppose bS C R. Then RbS is an integral R-ideal. By hypothesis, there exists an R-ideal C in Q which is the inverse of RbS. Therefore C * RbS = R. It follows that S = R S = C . RbS . S = CRbS = R, a contradiction. 2. Now let I be an R-ideal in Q. As we have just shown in Part I, R is a maximal order in Q and therefore by (3), I*ZC Or(Z) = R . By hypothesis, Z has an inverse R-ideal I-1 in Q. Now 11-1 = R implies ZZ-lZ = I from which it follows that Z-l C I*. Therefore, R = I-ll C I*I C R and hence Z*Z = R = Or(Z). By Theorem 10 this proves that Z is projective as an R-module. I

+

3. Injective modules and self-injective rings The following definition is the "dual" of the definition of a projective module given in Section 2:

3.

INJECTIVE MODULES AND SELF-INJECTIVE RINGS

175

Definition 1. Let R be a ring with identity and let B be an arbitrary R-module. Then an R-module M is injective if homomorphisms of submodules of B into M can be lifted to homomorphisms of B into M. In terms of diagrams, M is injective if the diagram

where i is the embedding of A into B, can be embedded in the commutative diagram

O + A ~ - B

M

If M is injective, take L

=A,

L a left ideal of R,and take R

= B.

Then

xt,h = (x . l)$ = x( 13) for all x E R. Hence there exists m, E M (namely, 13)

such that 14 = Im, for all I E L. The converse of this remark is also true and hence we have the following characterization of injectivity: Theorem 1. The R-module M is injective if and only if for each left ideal L in R and each R-module homomorphism 4 of L into M , there exists m, E M such that I+ = Im, for all I E L.

Proof. Suppose the condition is satisfied. Let B be an R-module and let A C B be a submodule of B. Let 4 E Hom,(A, M ) . Consider the family 3; of pairs (A’, +’), where A’ is a submodule of B containing A and 4‘ E Hom,(A’, M ) is an extension of 4 to A ’ . Define a relation C on 3; by (A’, 4’) C ( A ” , 4”) if A’ C A” and 4’’ is an extension of 4’. Clearly C partially orders 3;. Moreover, if { ( A , , 4A)}AEA is a chain in 3; and 4 is the homomorphism defined on UAEA A,, by

$:a+a+,

if a E A , ,

PEA

the pair ((JAG,, AA ,4) E 3; is an upper bound for the chain. By Zorn’s lemma, there exists a maximal element (B, , 3,) in 3;. We must now show that B, = B. Suppose not and let b, E B\B, . Then the set { x E R I xb, E B,} is a left ideal L of R and the mapping x + (xb,) 3, for x E L is an R-module homomorphism of L to M. By hypothesis there exists m, E A4 such that (xb,) 3, = xm, . Set 3,: b, ybl +b0$, ym, for b , B,~ and Y E R. We show is well

+

+

176

VII.

RINGS WITH IDENTITY

+

+ +

defined: Suppose b, yb, = b,’ y’b, . Then b, - b,’ = (y‘ - y ) b, and therefore y’ - y E L. It follows that (b, - b,’) #, = ((y’ - y ) b,) I), = (y’ - y ) m, and hence b,+, ym, = b0’#, y‘m, , showing that #, is well defined. Rb, , M ) extends #, and B, Rb, 3 B, , contraClearly I), E Hom,(B, dicting maximality of (B, , #,). Therefore B, = B. I

+

+

+

The following is analogous to Theorem VII.2.2: Theorem 2. The complete direct sum M of modules M i, i E I is injective if and only if each M iis injective.

Proof. Suppose each M , is injective and let 4E Hom,(A, Cc M,) where A is a submodule of a given module B. Let rz be the projection of M onto k f z . Now $rzE Hom,(A, M,) and by injectivity of M , there exists E Horn,(& M,) such that #%I A = 4r,. Then # = C @ I),E Horn,(& M ) is an extension of 4 to B and hence M is injective. Conversely, suppose M is injective and let $ z Hom,(A, ~ M,). Denote the embedding of M , in M by p, . Then = +%pZ E Hom,(A, M ) . By hypothesis there exists #€Hom,(B, M ) with # ( A = 4 = dZpt and I)wZ is an extension of = (bZPL,T, to B. I

4

4,

Injective P-modules (i.e. abelian groups considered as modules over the integers) are characterized by the following: Theorem 3. An Abelian group M is injective as a Z-module if and only if it is divisible, i.e. for each m E M and each natural number N , there exists m, E M such that Nnz, = m.

Proof. Let L = PI, # 0 and let 4 be an additive map of L to M . Let lo+ = m E M . If M is divisible there exists m, E M such that m = lorno. Clearly /I#I = lm, for all 1 E L and by Theorem 1, M is injective. Conversely, suppose M is Z-injective. The mapping (b from Zl, to M defined by 4:z/, + zm is additive for arbitrary m E M , I, E Z.If M is injective there exists I):Z M which is an extension of 4.Let 14 = m, . Then lm = 1,(1#) = (I,#) = /,$ = m. Hence M is divisible. I ---f

The rational numbers under addition form an abelian group Q containing

as a subgroup the additive group of integers H. The group Q/Z is divisible

and hence Z-injective. Using this fact we may construct more injective modules. We begin with the following:

Definition 2. Let M be a right R-module. Then the additive group

3.

INJECTIVE MODULES AND SELF-INJECTIVE RINGS

177

Hom,(M, O/Z) can be made into an R-module M * by defining for a E R and x E M*

m(ax) = (ma)x

for all m E M

(1)

The R-module M* is called the character module of M . The R-module property of M * under (1) follows from

m((ab)x )

=

(mab) x

=

((ma)b) x

= (ma)(bx) =

rn(a(bx)) for all m E M (2)

Theorem 4. If F is a free, right R-module, its character module F* is an injective left R-module. Proof. Let L be a left ideal in R and let 4 E Hom,(L, F*). By Theorem 1 , it is sufficient to show that 4 can be extended to ,R. Let { f i I i E Z} be an R-basis of the free right R-module F. Then F L = {hill .’. +&,I, I li E L} is a subgroup of the additive group of F. In the representation of an element of FL as a sum of fili , the li are uniquely determined since {fi I i E Z} is an R-basis of F. Define a mapping # from F L to Q/Z by

+

Clearly $ E Hom,(FL, OjZ). Since by Theorem 3, O/Z is injective as a Z-module, there exists x E Hom,(F, Q/Z) = F* which extends $. By the definition of F* the mapping

q:a+a#

for U E R

is an R-module homomorphism of R R to F* and hence an element of Hom,(R, F*). By (3) the mapping q is such that

f(b> = f ( l x >= (f0x = (fl># Therefore q extends 4 to the whole of RR. I

=

fU4>

In Theorem 7 we shall show that an R-module is injective if and only if it is a direct summand of the character module of some free right R-module. To this end we first prove:

Theorem 5 (Baer). Every right R-module M is isomorphic to a submodule of the character module of some free module. Hence every right R-module is a submodule of an injective module.

178

VII.

RINGS WITH IDENTITY

Proof. 1. We first prove if 0 # m, E M , there exists x E M* such that moX # 0. Suppose first that zmo # 0 for all 0 Z z E Z.Set (zmo) = (z/2)rr where 77 is the natural homomorphism of (a,+) onto Q/Z. Since c$ E Hom,(Zm, , Q/Z), 4 can be extended to x E Hom(M, Q/Z) = M * and mox = mo$ = &r # 0. Next suppose zm, = 0 for some 0 f z E H. Let zo be the natural number for which z,m, = 0. Define by

4: (zmo) (zlz,) +

77

Clearly 4 E Hom,(Zm,, Q/Z) and again it can be extended to x E M*. Moreover, m,X = (I/z,)rr f 0. 2. M is isomorphically embedded as a submodule of M**. Proof For each m E M define the mapping & by

&:X+mX

for X E M *

Clearly & E Hom(M*, Q/Z) and the mapping m + E% is additive. E% implies m x = 0 for all x E M * and hence by Part 1, m = 0. Moreover, A

max

=

=

6

(ma)x = rn(ax) = &(ax) = (&a)x

for all x E M* A Therefore ma = h a and m & is an R-module monomorphism of M into M**. 3. Let r#~ be an R-module homomorphism of the right module U to the right module Wand let +* be the mapping of W* to U* defined by ---f

X+*:U-(+U)X

for

U E

U

+* is an R-module homomorphism since for U E U and a E R we have u((ax) $*) = (&)(ax) = (+(ua)) x = u(a(x$*)). If 4 is an epimorphism, c$* is a monomorphism. Proof Let x E W* and suppose XC$* = 0. Then ( 4 U ) x = 0 and hence Wx = 0 proving that x = 0. 4. As we showed in Part 2, M is isomorphic to a submodule of M** = (M*)*. But M* is the homomorphic image of a free module RF. By Part 3 it follows that ( M * ) * is isomorphic to a submodule of F*. I Theorem 6. The R-module M is injective if and only if A4 is a direct summand of any containing module.

Proof. (Cf. Proof of Theorem 2.5.) 1 . Suppose A4 is injective and let M C A . By injectivity of M the identity map 1 on M extends to a homo- #) = A#(# - 1,) = 0 and hence # is morphism # of A to M . Now idempotent. Therefore A = A # @ A(1, - #)

,

3.

INJECTIVE MODULES AND SELF-INJECTIVE RINGS

is a representation of A as a direct sum of the submodules A# 41.4 -

$1.

=M

179 and

2. If M satisfies the condition of the theorem then in particular M is a direct summand of the module F* of the previous theorem. By Theorem 2, M itself is injective. I

As an immediate corollary to Theorem 6 we have: Theorem 7. The R-module M is injective if and only if it is isomorphic to a direct summand of the character module of a free right R-module. I This theorem is analogous to the first assertion of Theorem 2.6 which states that an R-module is projective if and only if it is isomorphic to a direct summand of a free module. If R is an algebra over the field @ and if M is a right algebra R-module we can substitute @ for Q/Z in Definition 2. The elements of Hom,(M, 0) are called linear functionals on M . They form an R-module, again denoted by M * , if we define for a E R and x E Hom,(M, @)

m(ax) = (ma) x

for all m E M

Except for minor alterations in the terminology, Theorem 4 remains unchanged and the same proof carries through since the field @ is an injective @-module. The same is true of Theorems 5 and 6. In the context of @-algebras, Theorem 7 becomes: Theorem 7’. Let R be an algebra with identity over the field @. The R-(algebra)-module M is injective if and only if it is isomorphic to a direct summand of the dual module F* of a free R-(algebra)-module F. I By Theorem 5 every R-module can be embedded in an injective R-module. Our next objective is to prove that an R-module can be embedded in a minimal injective and that any two minimal injectives are isomorphic. For this purpose we introduce the following concept: Definition 3. The R-module U is an essential extension of the submodule M if every nonzero submodule of U intersects M nontrivially. The following theorem shows that every injective extension I of M isomorphically contains every essential extension U of M . Theorem 8. Let Z be an injective extension of the R-module M and let U be an essential extension of M . Then there exists a monomorphism of U into Z whose restriction to A4 is the identity 1 .

180

VII.

RINGS WITH IDENTITY

Proof. From the exactness of the sequence 0 + M U and the injectivity of Z follows the existence of 4 E Hom,(U, I ) with 4 , 1 = 1,. This implies that ker 4 n M = (0). Since U is an essential extension of M it follows that ker 4 = (0). Hence 4 is a monomorphism. I ---f

Theorem 9. The R-module M is injective if and only if it has no proper essential extension, i.e. no essential extension properly containing M . Proof. 1. If M is injective and U is an extension of M , then M is a direct summand of U. Hence M = U if we assume U is essential over M . 2. Suppose M has no proper essential extensions and let I be an extension of M . Let M’ be a submodule of I , maximal subject to M n M‘ = (0.) (M’ exists by Zorn’s lemma.) Now I / M ’ C ( M M’)/M’ = M . Moreover, if M’ C U C Z and ( U / M ’ )n ( M M’)/M’ = (0), then U n ( M M’) C M’ and hence U n M C M‘ n M = (0). By the maximality of M ’ this implies U = M‘ proving that I/M’ is an essential extension of ( M M’)/M’. By hypothesis this implies that I = M @ M‘. If we choose I to be an injective extension of M , M will be a direct summand of I and hence, by Theorem 2, M is injective. I

+

+

+

+

Now let M be a submodule of the module I . The set theoretic union of an ascending chain of essential extensions of M in I is itself an essential extension of M . By Zorn’s lemma there exists a submodule U of Z maximal subject to containing M essentially. If U‘ is an essential extension of U contained in I then any nonzero submodules S of U’ intersects U nontrivially. Since U is an essential extension of M and since S n U # (0),

S n U n M = S n M # (0)

Hence any essential extension of U in I is also an essential extension of M . Therefore U has no proper essential extensions in I. Moreover, if I is injective, we can show that U has no proper essential extensions in any containing module: Let U‘ be any essential extension of U. By Theorem 8 applied to the essential extension U‘ of U , there exists a monomorphism 4 of U ‘ into I such that U C Or‘+. Clearly U‘4 is an essential extension of U in I. But U is maximal essential in Z and therefore U = U‘4. Since U has no proper essential extension it is an injective R-module by Theorem 9. We have therefore shown that if I is an injective extension of M , each maximal essential extension U of M in Z is an injective R-module. On the other hand, by Theorem 9, no proper submodule 0of U containing M is injective since U is a proper essential extension of 0.Therefore U is a minimal injective extension of M and is uniquely determined (up to isomorphism) by M : In fact, if U‘ is any essential extension of M , then by Theorem 8 there exists

3.

INJECTIVE MODULES AND SELF-INJECTIVE RINGS

18 1

a monomorphism C$ of U‘ into U . If U ‘ is also a maximal essential extension of M , so is U’C$.Hence U‘C$ = U . Before collecting the results proved in the above discussion we make the following: Definition 4. The R-module Z is an injective hull of its submodule M if 1. Z is injective 2. M C L C Z,L injective implies L = I.

Theorem 10. 1. Every R-module M has an injective hull. 2. Every injective extension of M contains an injective hull of M . 3. An R-module U containing M is an injective hull of M if and only if it is a maximal essential extension of M . 4. If Z and I’ are injective hulls of M then there exists an isomorphism of Z onto I’ restricting to the identity on M . I Since the injective hulls of a given module M are isomorphic we speak of “the” injective hull H ( M ) of M . This last theorem can be used as the basis of an investigation of those rings R with are injective considered as R-modules over themselves. Let us assume that 0is an indecomposable component in such a ring and suppose 0 contains at least one minimal left ideal M of R. By hypothesis RR is injective and since 0 is a direct summand, 0 itself is also injective. By Theorem 10, 0contains an injective hull H ( M ) of M . By Theorem 6 , H ( M ) is a direct summand of 0and hence H ( M ) = 0since 0was assumed to be indecomposable. Again by Theorem 10, 0 is an essential extension of M . If 0 contained another minimal left ideal M‘, then M‘ n M = (0) contradicting the fact that 0 contains M essentially. Therefore 0contains exactly are two isomorphic indecomposable one minimal left ideal. Now if 0and 0’ components of RR each containing minimal left ideals M and M‘, respectively, then M and M‘ are uniquely determined by 0and Or,respectively, and since 0 rn 0, M m M‘. Conversely if M w M‘ it follows that 0 w 0’since the injective hulls = H ( M ) and 0’ = H ( M ’ ) are uniquely determined up to isomorphism by Theorem 10. Finally, let M be a minimal left ideal in R and let H ( M ) be an injective hull of M contained in RR. Then H ( M ) = S @ T, S # (0) implies S n M # (0) since H ( M ) contains M essentially and hence M C S by minimality of M . Since S is a direct summand of an injective it is injective (Theorem 2). Therefore T = (0) since H ( M ) is minimal injective and hence H ( M ) is indecomposable. If the ring R is artinian with identity and RR is injective, then there is a one-one correspondence between the isomorphism classes of indecomposable left ideals and the isomorphism classes of minimal left ideals. On the other

182

VII.

RINGS WITH IDENTITY

hand, we showed in Theorem 1.17 that there is a one-one correspondence between the isomorphism classes of irreducible R-modules and the isomorphism classes of indecomposable submodules of RR. We have therefore proved the following:

Theorem 11. Let R be artinian with identity and let RR be injective. Then there is a one-one correspondence between the classes of irreducible R-modules and the classes of indecomposable left ideals in R. Moreover, each indecomposable left ideal contains exactly one minimal left ideal. I Theorem 12. Let R be artinian with identity. Then the following are equivalent: 1. RR is an injective R-module 2. A finitely generated R-module M is injective if and only if it is projective. Definition 5. A ring R is self-injective if RR is an injective R-module. Proof of Theorem 12. 1, Assume Part 2. By Theorem 2.1 RR is projective and therefore R Ris injective. 2. Conversely, suppose RR is injective and let M be a finitely generated projective R-module. By Theorem 2.6, M is a direct sum of a finite number of submodules, each isomorphic to some indecomposable left ideal of R. Since each of these is a direct summand of the injective module RR, each is injective and therefore by Theorem 2 M is injective. 3. Let RR be injective and let M be a finitely generated injective module. Since R is artinian and M is finitely generated the lattice VR(M)of submodules of M satisfies the minimal condition by Theorem 1.30. The module M is therefore the direct sum of a finite number of indecomposable submodules. Let U be one of these, As a direct summand of M , U is also injective. Let W be a minimal submodule of U . Then H ( W ) , the injective hull of W, is contained in U . As above, U = H( W ) .Since W is an irreducible submodule of M , by Theorem 11 it is isomorphic to a minimal left ideal L in R whose injective hull H ( L ) is an indecomposable left ideal of R. Since W m L it follows from Theorem 10 that H ( W ) m H(L). Now H(L) is an indecomposable left ideal and hence is a direct summand of RR. Therefore H ( L ) is projective proving that U = H( W ) is also projective. Hence M is projective. I The lattice V(RR) of left ideals L of R may be mapped into the lattice of right ideals V(RR) by means of the right annihilator mapping r defined by L

---f

r(L)

= {x E

R 1 Lx

=

0)

3.

INJECTIVE MODULES AND SELF-INJECTIVE RINGS

183

Similarly the lattice V(R,) of right ideals W is mapped into V(RR) by the left annihilator I defined by

In (5) the inclusions could be strict. It is an open question whether the mappings are lattice isomorphisms between V(RR)and V(RR). Let R be a ring with identity. In the following it will be shown that the inclusions in (5) may be replaced by equalities if R is a self-injective left artinian and right noetherian ring. We first prove that for principal right ideals aR of R we have (6)

r(/(aR))= aR

Proof. x E l(aR)implies xa = 0. Hence a E r(x)and therefore aR C r(/(aR)) always holds. Conversely, let b E r(l(aR)). Then xb = 0 for all x E I(aR). Since I(aR) C I(bR), the mapping for X E R

+:xa+xb

is an element of HomR(Ra,Rb). Since RR is injective by Theorem I there exists c E R such that xb = XU+ = xac for all x E R. Therefore b = ac E aR and ( 6 ) is proved. To sharpen the first inclusion in (5) to an equality let b E r(L, n L,). Define +iE HornR(& , RR), i = 1 , 2 as follows:

11+,

=

I, for I,

E

and

L1

+ + +

=

/,(1

-

b) for I,

E

L,

+

+

The mapping (11 I,) = I,+, /2+2 is well defined: ll l2 = I,’ I 12‘ implies - I,‘ = --I, 1,’ E L, n L, . But b E r(L, n L,) and therefore Z,b = /,‘b showing that (1, I,) = (I,’ 12’) 4. Since RR is injective, by Theorem 1 there exists c E R such that ( I , 12) 4 = ( I , I,) c. This implies

I1 I /2(1

+

+

-

b)

=

+ +

(I1

+

+ + 12)

+

=

(11

I 12) c

and therefore 11(1 - c) /,(1 - b - c) = 0 for all I, E L, , 1, follows that 1 - c E r(L,) and 1 b c E r(L2).Therefore ~

b = ( l -c)-(I

~

-b-c)Er(L,)+-r(L,)

E

L, . It

184

VII.

RINGS WITH IDENTITY

This shows r(L, n L,)

=

r(Ll)

+ r(LJ

for L1 , L, E VLR)

(7)

and the mapping L + r(L) is a lattice homomorphism of V(RR) into V(RR). If in addition R is right noetherian, i.e. if V ( R R )satisfies the maximal condition, the mapping L r(L) is also an epimorphism. Proof: Each right ideal W in R is the sum of a finite number of principal right ideals: W = a,R ... a,R. By (4), (7), and (6) it follows that --f

+ +

+ + + +

r(l(W)) = r(l(a,R ... a,R)) = r(/(a,R) n ... n /(a,$)) = r(/(alR)) ... r(/(a,R)) = a,R ... a,R = W

+ +

Hence W E V(RR) is the image of the left ideal I(W) under the mapping L r(L). Finally this mapping is a monomorphism if R is also artinian. In this case we apply Theorem 11 as follows: Assume L C /(r(L))for some left ideal L. Then there exists a left ideal Lo _C /(r(L))such that L, contains L and Lo/L is an irreducible R-module. By Theorem 11, L,/L is isomorphic to a minimal left ideal in R. Therefore there exists 4 E HomR(Lo, RR) with L 4 = 0 and Lo minimal in V(,R). Since RR is injective there exists c E R such that I,,+ = /,c for all 1, E L o . In particular, Lc = 0 and therefore c E r(L). But Lo _C /(r(L))and hence Loc = 0, a contradiction, since L,c = L, . Therefore I(r(L))= L for all L E V(,R) and the mapping L + r(L)is one-one. We gather these results in the following: --f

Theorem 13. Let R be a self-injective left artinian and right noetherian ring. Then the mapping L r(L), L a left ideal, is a lattice antiisomorphism of the lattice V(,R) of left ideals in R onto the lattice V ( R R )of right ideals in R. The inverse of this mapping is the mapping W I( W ) , W E V(RR).In particular we have /(r(L)) = L for all L E V(RR) and r( /( W) )= W for all --f

---f

WE

V(RR).

I

Definition 6. A left and right artinian ring with identity whose left and right ideals satisfy /(r(L)) = L

and

r ( / ( W ) )= W

(8)

respectively is called a quasi-Frobenius ring. Theorem 13 shows that a self-injective left and right artinian ring with identity is a quasi-Frobenius ring. The converse is also true [cf. Curtis and Reiner (1962), Theorem 58.61. In the next chapter, as a corollary to Theorem V111.2.11, we shall prove the converse in the special case when R is an algebra of finite dimension over a field.