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Extensions of Injectivity and CoherentRingsJianlong Chen a & Yiqiang Zhou ba Department of Mathematics , Southeast University , Nanjing, Chinab Department of Mathematics and Statistics , Memorial University ofNewfoundland , St. John's, CanadaPublished online: 03 Sep 2006.

To cite this article: Jianlong Chen & Yiqiang Zhou (2006) Extensions of Injectivity and Coherent Rings,Communications in Algebra, 34:1, 275-288, DOI: 10.1080/00927870500346263

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Communications in Algebra®, 34: 275–288, 2006Copyright © Taylor & Francis, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870500346263

EXTENSIONS OF INJECTIVITY AND COHERENT RINGS

Jianlong ChenDepartment of Mathematics, Southeast University, Nanjing, China

Yiqiang ZhouDepartment of Mathematics and Statistics, Memorial University ofNewfoundland, St. John’s, Canada

For a ring D and a subring C, let ����D�C� = ��d1� � � � � dn� c� c� � � � � � di ∈ D� c ∈C�n ≥ 1, a subring of

∏∏∏�i=1 D, and let D ∝ D = �

(a b0 a

)� a� b ∈ D, a subring of

�2�D�. In this article, various conditions are obtained for ����D�C� and D ∝ D

to be right �m� n�-injective �in particular, P-injective, f -injective, FP-injective�, rightcoherent, and right FC. Right nonsingular right �m� n�-injective rings are characterizedusing their maximal right quotient rings.

Key Words: Annihilator; Coherent ring; Injectivity; Trivial extension.

Mathematics Subject Classification: Primary 16D50, 16S99.

1. INTRODUCTION

Throughout, R is an associative ring with identity. We write Rm×n for the set ofall m× n matrices over R, and let Rn = R1×n and Rn = Rn×1. Following Chen et al.(2001), a right R-module M is called �m� n�-injective if every right R-homomorphismfrom an n-generated submodule of Rm to M extends to one from Rm to M , andthe ring R is a right �m� n�-injective ring if RR is �m� n�-injective. The ring R iscalled right n-injective (respectively, right P-injective) if R is right �1� n�-injective(respectively, right �1� 1�-injective). If R is right n-injective for every positive integern, then we call R right f -injective. The ring R is called right FP-injective if R is right�m� n�-injective for all positive integers m and n.

Let R be a ring. The trivial extension of R and R is R ∝ R = �(a b

0 a

)� a� b ∈ R�

with usual addition and multiplication of matrices. It is well known that R ∝ R isright self-injective iff so is R (see Faith, 1979). In Section 2, we consider when R ∝ Ris right P-injective, right f -injective, right FP-injective, and more generally, right�m� n�-injective, respectively. Various sufficient conditions are obtained for R ∝ R tohave these injectivities. It is also proven that R ∝ R is right coherent (respectively,right FC) iff so is R.

Received September 13, 2004; Revised and Accepted March 31, 2005. Communicated byI. Swanson.

Address correspondence to Yiqiang Zhou, Department of Mathematics and Statistics,Memorial University of Newfoundland, St. John’s, A1C 5S7, Canada; Fax: (709) 737-3010; E-mail:zhou@math.mun.ca

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276 CHEN AND ZHOU

For a ring D and a subring C, let ��D�C� = ��d1� � dn� c� c� � �di ∈ D� c ∈ C� n ≥ 1�. With addition and multiplication defined componentwise,��D�C� is a ring. Though ��D�C� is never a right self-injective ring, it is a richsource of rings with certain weaker injectivity. In Section 3, necessary and sufficientconditions are obtained for ��D�C� to be right �m� n�-injective, right FP-injective,right f -injective, right P-injective, right coherent, respectively. As an application,right nonsingular right �m� n�-injective rings are characterized using their maximalright quotient rings.

If M is a right module over R, and if X ⊆ M and I ⊆ R, we let rR�X� = �r ∈R � xr = 0 for all x ∈ X� and let lM�I� = �x ∈ M � xr = 0 for all r ∈ I�. If M is a leftmodule over R, one defines lR�X� and rM�I� similarly.

2. INJECTIVITY OF TRIVIAL EXTENSIONS

Let R be a ring and M a bimodule over R. The trivial extension of R andM is R ∝ M = ��a� x� � a ∈ R� x ∈ M� with addition defined componentwise andmultiplication defined by �a� x��b� y� = �ab� ay + xb�. In fact, R ∝ M is isomorphicto the subring �

(a x

0 a

)� a ∈ R� x ∈ M� of the formal 2× 2 upper triangular matrix ring(

R M

0 R

), and R ∝ R � R�x�/�x2�. For convenience, we let I ∝ X = ��a� x� � a ∈ I� x ∈

X�, where I is a subset of R and X is a subset of M .Recall that a ring R is right �m� n�-injective iff, for any C ∈ Rm×n, every right

R-homomorphism from CRn to R extends to one from Rm to R iff lRnrRn�A� = RmA

for all A ∈ Rm×n (Chen et al., 2001).

Theorem 1. Suppose that, for any A�B ∈ Rm×n, every right R-homomorphism fromARn + B · rRn

�A� to R extends to one from Rm to R. Then R ∝ R is a right �m� n�-injective ring.

Proof. LetS = R ∝ R. Supposef � A1S + · · · + AnS → S isa rightS-homomorphism,

where Ai =(

u�i�1

u�i�m

)∈ Sm �i = 1� � n� with u

�i�j = �a

�i�j � b

�i�j � ∈ S �j = 1� � m�. Let

A =[

a�1�1 ··· a

�n�1

a�1�m ··· a

�n�m

], B =

[b�1�1 ··· b

�n�1

b�1�m ··· b

�n�m

], and write f�Ai� = �pi� qi�. Define ARn +

BrRn�A�

g→ R via

A

d1dn

+ B

c1cn

�→ �p1� � pn�

d1dn

+ �q1� � qn�

c1cn

If A

( d1dn

)+ B

( c1cn

)= 0 with A

( c1cn

)= 0, then

∑ni=1 a

�i�j ci = 0 for all 1 ≤ j ≤ m

and∑n

i=1�a�i�j di + b

�i�j ci� = 0. So

∑ni=1�a

�i�j � b

�i�j ��ci� di� = 0 for all 1 ≤ j ≤ m. Let

vi = �ci� di�. It follows that∑n

i=1 u�i�j vi = 0 for all 1 ≤ j ≤ m. Thus, Aivi =

�∑n

i=1 u�i�1 vi� �

∑ni=1 u

�i�m vi�

T = 0. Since f is a right S-homomorphism, f�Ai�vi = 0,i.e., �pi� qi��ci� di� = 0. Hence pidi + qici = 0. So g is a well-defined right

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EXTENSIONS OF INJECTIVITY AND COHERENT RINGS 277

R-homomorphism. By hypothesis, there exist ai ∈ R�i = 1� � m� such that g�w� =�a1� � am�w for all w ∈ ARn + BrRn

�A�. In particular, �a1� � am�A� = g�A�� =�p1� � pn�� for all � ∈ Rn. It follows that �p1� � pn� = �a1� � am�A.

Note that if A = 0, then rRn�A� = Rn. So the hypothesis implies that, for any

C ∈ Rm×n, every right R-homomorphism from CRn to R extends to one from Rm

to R, i.e., R is right �m� n�-injective. Now define

ARn

�→ R�

A� �→ ��q1� � qn�− �a1� � am�B��� �� ∈ Rn�

If A� = 0, then � ∈ rRn�A�. By the definition of g� �a1� � am�B� =

g�B�� = �q1� � qn��, so ��q1� � qn�− �a1� � am�B�� = 0. Hence � is a well-defined R-homomorphism. Thus, there exist bi ∈ R for i = 1� � m such that�b1� � bm�Au = ��Au� = ��q1� � qn�− �a1� � am�B�u for all u ∈ Rn. So�q1� � qn� = �a1� � am�B + �b1� � bm�A. Let wi = �ai� bi�. Then

�w1� � wm�Ai = �w1� � wm�

u

�i�1

u�i�m

= wju

�i�j

= �aj� bj��a�i�j � b

�i�j �

= �aja�i�j � �ajb

�i�j + bja

�i�j ��� so

�w1� � wm��A1� � Am� = ��a1� � am�A� �a1� � am�B + �b1� � bm�A�

= ��p1� � pn�� �q1� � qn��

This shows that �w1� � wm�Ai = �pi� qi� = f�Ai�. So f is the left multiplicationf = �w1� � wm�·. Hence S is right �m� n�-injective. �

Theorem 2. If S = R ∝ R is right �m� n�-injective, then so is R.

Proof. Let f � ARn → R be a right R-homomorphism where A = �aij� ∈ Rm×n. LetA = ��aij� 0�� ∈ Sm×n. Define

ASng→ S� via A

s1sn

�→ �d1� � dn�

s1sn

�

where di = �ci� 0� and ci = f�Aei�. If A

( s1sn

)= 0 with sj = �aj� bj�, then 0 =

∑nj=1�aij� 0��aj� bj� = �

∑nj=1 aijaj�

∑nj=1 aijbj�, showing that A

( a1an

)= 0 and

A

( b1bn

)= 0. Since A

( a1an

)= Ae1a1 + · · · + Aenan� 0 = f�Ae1a1 + · · · + Aenan� =

f�Ae1�a1 + · · · + f�Aen�an = c1a1 + · · · + cnan, where ei is the transpose of

�0� � 0�i

1� 0� � 0� �i = 1� � n�. Similarly, c1b1 + · · · + cnbn = 0. Thus, we

have �d1� � dn�

( s1sn

)= disi = �ci� 0��ai� bi� = �ciai� cibi� = 0. So g is a

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278 CHEN AND ZHOU

well-defined S-homomorphism. Since S is right �m� n�-injective, there exist ti ∈ S for

i = 1� � m such that �d1� � dn�

( s1sn

)= g

(A

( s1sn

))= �t1� � tm�A

( s1sn

)for all( s1

sn

)∈ Sn. Thus, �d1� � dn� = �t1� � tm�A. Write ti = �ki� li�. Since di = �ci� 0�,

it follows that �c1� � cn� = �k1� � km�A. For any � =( x1

xn

)∈ Rn,

f�A�� = f�Aeixi� = f�Aei�xi = �c1� � cn�

x1xn

= �k1� � km�A��

so f = �k1� � km�· is a left multiplication. So R is a right �m� n�-injective ring. �

Corollary 3. Suppose that, for any positive integers n and m and for any A�B ∈ Rm×n,every right R-homomorphism from ARn + B · rRn

�A� to R extends to one from Rm to R.Then R ∝ R is a right FP-injective ring.

Corollary 4. Let n > 0 be a fixed integer. Suppose that, for any ai� bi ∈ R�i = 1� � n�, every right R-homomorphism from

∑ni=1 aiR+ �b1� � bn��a1� � an�

⊥

to R extends to one from R to R, where �a1� � an�⊥ =

{( x1xn

)∈ Rn �

ni=1aixi = 0

}.

Then R ∝ R is a right n-injective ring.

Corollary 5. Let m > 0 be a fixed integer. Suppose that, for any ai� bi ∈ R

�i = 1� � m�, every right R-homomorphism from

( a1am

)R+

( b1bm

)( a1am

)⊥to R

extends to one from Rm to R, where

( a1am

)⊥= �x ∈ R �

( a1am

)x = 0�. Then every

m-generated left ideal of R ∝ R is a left annihilator.

Corollary 6. Suppose that, for any a� b ∈ R, every right R-homomorphism fromaR+ br�a� to R extends to one from R to R. Then R ∝ R is a right P-injective ring.

The converse of Corollary 6 does not hold, as shown by the next example.

Example 7. Let K be a field and let L be a proper subfield of K such that � K → L is an isomorphism. Let K�x � � be the ring of twisted right polynomialsover K where kx = x �k� for all k ∈ K. Let R = K�x� �/�x2�. Then R = �k+ xl �k� l ∈ K� with x2 = 0 and kx = x �k� for all k ∈ K.

(1) S = R ∝ R is a left P-injective ring.(2) If K = ��x1� x2� � xn� � and L = ��x21� x

22� � x

2n� �, then there exists

a� b ∈ R such that some left R-homomorphism from Ra+ l�a�b to R does notextend to R.

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EXTENSIONS OF INJECTIVITY AND COHERENT RINGS 279

Proof. (1) By Rutter (1975, Example 1, p. 208), R is a left P-injective ring. Wefirst prove that S is left P-injective.

Suppose �a� b� ∈ S and we need to show that rl�a� b� = �a� b�S. Clearly,rl�a� = aR implies rl�a� 0� = �a� 0�S and rl�b� = bR implies rl�0� b� = �0� b�S. Sowe can assume that a = 0 and b = 0. Write a = a1 + xa2� b = b1 + xb2 witha1� a2� b1� b2 ∈ K. If a1 = 0, then a is a unit of R, so �a� b� is a unit of S and hence�a� b�S = rl�a� b�. So we can further assume that a1 = 0. Then a2 = 0.

Case 1. b1 = 0. We claim that �a� b�S = r�−xa2b−11 � 1�. It is easy to see that

�a� b�S ⊆ r�−xa2b−11 � 1�. Let �m1 + xm2� n1 + xn2� ∈ r�−xa2b

−11 � 1�. Then

0 = (− xa2b−11 � 1

)�m1 + xm2� n1 + xn2�

= (− xa2b−11 m1�−xa2b

−11 n1 +m1 + xm2

)⇒ m1 = 0�m2 = a2b

−11 n1

So r�−xa2b−11 � 1� = ��xa2b

−11 n1� n1 + xn2� � n1� n2 ∈ K�. Let c = a−1

2 �n2 − b2b−11 n1�.

Then

�a� b�(b−11 n1� c

) = �xa2� b1 + xb2�(b−11 n1� c

)= (

xa2b−11 n1� xa2c + n1 + xb2b

−11 n1

)= (

xa2b−11 n1� n1 + x�n2 − b2b

−11 n1�+ xb2b

−11 n1

)= (

xa2b−11 n1� n1 + xn2

)

So �a� b�S ⊇ r�−xa2b−11 � 1�, and hence �a� b�S = r�−xa2b

−11 � 1�. It follows that

�a� b�S = rl�a� b�.

Case 2. b1 = 0. Then �a� b� = �xa2� xb2� with a2 = 0 and b2 = 0. We nextshow that �a� b�S = rS�x� x�, implying that �a� b�S = rl�a� b�.

Clearly, rS�x� x� = ��xk� xl� � k� l ∈ K�, so rS�x� x� ⊆ �a� b�S since �xk� xl� =�xa2� xb2��a

−12 k� a−1

2 �l− b2a−12 k�� ∈ �a� b�S. It is clear that �a� b�S ⊆ rS�x� x�. So

�a� b�S = rS�x� x�. Therefore, S is left P-injective.

(2) Let e1� e2� be a basis of the L-vector space K. Choose a = x and b = 1.Then Ra+ l�a�b = Rx + l�x� · 1 = l�x� = xK. Define

� � Ra+ l�a�b → R

x�l1e1 + l2e2 + l3e3 + · · · � �→ x�l2e1 − l1e2 + l3e3 + · · · �

It can easily be checked that � is a left R-homomorphism. Suppose that � is given byright multiplication by some c + xd ∈ R. Then −xe2 = ��xe1� = xe1�c + xd� = xe1cand xe1 = ��xe2� = xe2�c + xd� = xe2c. It follows that −e2 = e1c and e1 = e2c, soe1 = −e1c

2. Since e1 = 0� c2 = −1. So, if we take K = ��x1� x2� � xn� �, then,since c must not be in K, the homomorphism � cannot be extended to R. �

We have been unable to answer the question whether R being right P-injectivealways implies R ∝ R being right P-injective.

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280 CHEN AND ZHOU

Following Nicholson and Sánches Campos (2004), a ring R is calledleft morphic if, for every a ∈ R�R/Ra � l�a�. Right morphic rings are definedanalogously. A left and right morphic ring is called a morphic ring. Note that, byNicholson and Sánches Campos (2004, Lemma 1), R/Ra � l�a� iff, there exists b ∈ Rsuch that l�a� = Rb and l�b� = Ra.

Lemma 8. If R is right P-injective, right morphic, then R is 2-injective, morphic.

Proof. We first prove that R is right 2-injective. For a ∈ R� aR = r�c� for somec ∈ R (since R is right morphic). Also Rc = lr�c� and Ra = lr�a� (since R is rightP-injective). So, by Nicholson and Yousif (1995, Lemma 1.1),

l�aR ∩ bR� = l�r�c� ∩ bR�

= Rc + l�b� �since R is right P-injective�

= lr�c�+ l�b� = l�a�+ l�b�� for all b ∈ R

Thus, R is right 2-injective by Chen et al. (2001, Corollary 2.10). Thus, byNicholson and Sánches Campos (2004, Proposition 27), R is left morphic. SinceR is right morphic, R is left P-injective by Nicholson and Sánches Campos (2004,Theorem 24). So R must be left 2-injective as above. �

Corollary 9. Let R be right morphic. Then R is right P-injective if and only if R ∝ Ris right P-injective.

Proof. One implication is by Theorem 2. Suppose that R is right P-injective. SinceR is right morphic, for any a ∈ R� r�a� = cR for some c ∈ R. So aR+ br�a� = aR+bcR. Since R is right 2-injective by Lemma 8, every R-homomorphism from aR+br�a� to R extends to one from R to R. So R ∝ R is right P-injective by Corollary 6.

�

A ring R is called right P-coherent if for any a ∈ R� r�a� is a finitely generatedright ideal.

Corollary 10. If R is right P-coherent and right f -injective, then R ∝ R is rightP-injective.

Proof. For a� b ∈ R, since R is right P-coherent, r�a� = b1R+ · · · + bnR, soaR+ br�a� = aR+ bb1R+ · · · + bbnR. Since R is right f -injective, every R-homomorphism from aR+ br�a� to R extends to one from R to R. So R ∝ R is rightP-injective by Corollary 6. �

Corollary 11. If every matrix ring �n�R� is morphic, then R ∝ R is FP-injective.

Proof. For any n > 0��n�R� is morphic, so �n�R ∝ R� � �n�R� ∝ �n�R� isP-injective by Corollary 9. Thus, R ∝ R is FP-injective. �

Theorem 12. A ring R is left coherent if and only if R ∝ R is left coherent.

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EXTENSIONS OF INJECTIVITY AND COHERENT RINGS 281

Proof. Let S = R ∝ R.

‘⇐’. We first show that R is left P-coherent, i.e., for any a ∈ R� l�a� is finitelygenerated. Note that lS�0� a� = lR�a� ∝ R. Since S is left coherent, lR�a� ∝ R is afinitely generated left ideal of S. Write lR�a� ∝ R = S�a1� b1�+ · · · + S�an� bn� withall �ai� bi� ∈ S. It follows that l�a� = Ra1 + · · · + Ran. So R is left P-coherent.

Now since R ∝ R is left coherent, �n�R� ∝ �n�R� � �n�R ∝ R� is leftcoherent for all n > 0. Thus, as above, �n�R� is left P-coherent, and so R is leftcoherent by Chen and Ding (2001).

‘⇒’. Suppose that R is left coherent. We first show that for any a� b ∈R� �Ra � b� �= �t ∈ R � tb ∈ Ra� is a finitely generated left ideal. Define � � R2 →Ra+ Rb� �r� s� �→ ra+ sb. Then, since R is left coherent, ker��� is finitely generated.Write ker��� = R�r1� s1�+ · · · + R�rn� sn�. Since ria+ sib = 0, si ∈ �Ra � b�. For s ∈�Ra � b�� sb ∈ Ra. Write sb = −ra with r ∈ R. Then �r� s� ∈ ker���, so �r� s� =a1�r1� s1�+ · · · + an�rn� sn� with all ai ∈ R. It follows that s = a1s1 + · · · + ansn. So�Ra � b� = Rs1 + · · · + Rsn is finitely generated.

Next we show that S is left P-coherent, i.e., for any �a� b� ∈ S� lS�a� b� is finitelygenerated. Since R is left coherent, l�a� is finitely generated. Thus, �Ra � b� ∩ l�a� isfinitely generated. Write �Ra � b� ∩ l�a�=Ra1 + · · · +Ran with ai ∈R. Thus, aia= 0and aib ∈ Ra. Write aib = −bia with bi ∈ R (i = 1� � n). Then �ai� bi� ∈ l�a� b�.If �r� s� ∈ lS�a� b�, then r ∈ l�a� ∩ �Ra � b�, so r = r1a1 + · · · + rnan, where ri ∈ R.Thus, �s − �r1b1 + · · · + rnbn��a = sa+ �r1a1 + · · · + rnan�b = sa+ rb = 0. Since l�a�is finitely generated, write l�a� = Rc1 + · · · + Rcm with cj ∈ R. Then s − �r1b1 + · · · +rnbn� = t1c1 + · · · + tmcm, where all ti ∈ R. So

�r� s� = �r1a1 + · · · + rnan� r1b1 + · · · + rnbn + t1c1 + · · · + tmcm�

=n∑

i=1

�ri� 0��ai� bi�+m∑j=1

�tj� 0��0� cj�

Therefore, lS�a� b� =∑n

i=1 S�ai� bi�+∑m

j=1 S�0� cj� is finitely generated. Thus weproved that R being left coherent implies that R ∝ R is left P-coherent. Since R is leftcoherent, �n�R� is left coherent for each n > 0. As above, �n�R ∝ R� � �n�R� ∝�n�R� is left P-coherent. So R ∝ R is left coherent by Chen and Ding (2001). �

A ring R is called left FC if R is left coherent, left FP-injective.

Theorem 13. A ring R is left FC if and only if R ∝ R is left FC.

Proof. ‘⇒’. By Corollary 10, for any n > 0��n�R ∝ R� � �n�R� ∝ �n�R� isleft P-injective for �n�R� is left FC. So R ∝ R is left FP-injective. By Theorem 12,R ∝ R is left coherent. Thus, R ∝ R is a left FC-ring.

‘⇐’. By Chen et al. (2005, Corollary 10) or by Theorem 2, if R ∝ R is leftFP-injective then R is left FP-injective. It is by Theorem 12 that R is left coherent.So R is left FC. �

The commutative case of the next result was proven in Matlis (1985).

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282 CHEN AND ZHOU

Corollary 14. If R is von Neumann regular, then R ∝ R is a FC-ring.

3. THE CONSTRUCTION ����D�C�

Let D be a ring and C a subring of D with 1D ∈ C. We set

��D�C� = ��d1� � dn� c� c� � � di ∈ D� c ∈ C� n ≥ 1� and

��D�C� = ��d1� � dn� cn+1� cn+2� � � di ∈ D� cj ∈ C� n ≥ 1�

With addition and multiplication defined componentwise, ��D�C� and ��D�C� areboth rings. A ring R is called right CS if every complement right ideal of R is a directsummand of R. Every right self-injective ring is right CS.

Proposition 15. S = ��D�C� can never be right CS.

Proof. Let A = ��a1� � an� 0� 0� � ∈ S � n ≥ 1� ak = 0 if k is even� and let B =��a1� � an� 0� 0� � ∈ S � n ≥ 1� ak = 0 if k is odd�. Then A�B are complementsof each other. But A�B are not direct summands of SS . �

Theorem 16. Let m�n be positive integers. The following are equivalent:

(1) ��D�C� is right �m� n�-injective;(2) D is right �m� n�-injective and Cn ∩DmA = CmA for all A ∈ Cm×n;(3) ��D�C� is right �m� n�-injective.

Proof. Let S = ��D�C�.

�2� ⇒ �1�. Given A ∈ Sm×n, we need to show that lSnrSn�A� = SmA. WriteA = ��ij�� �ij = �a

�1�ij � a

�2�ij � �. There exists a k > 0 such that a

�k+1�ij = a

�k+2�ij =

· · · = aij ∈ C for all i� j. Let u = ��1� � �n� ∈ lSnrSn�A�, where �i = �b�1�i � b

�2�i � �

�i = 1� � n�. There exists an l > k such that b�l+1�i = b

�l+2�i = · · · = bi ∈ C for

i = 1� � n. Now let As = �a�s�ij �� bs = �b

�s�1 � � b�s�n �. Next we show that for all

s > 0� rDn�As� ⊆ rDn

�bs�. If Y =( y1

yn

)∈ rDn

�As�, then AsY = 0, i.e.,∑n

j=1 a�s�ij yj = 0.

Let yj = �0� � 0�sth

yj� 0� � ∈ S. Then j�ij yj = �0� � 0�∑n

j=1 a�s�ij yj� 0� � = 0,

so A

( y1yn

)= 0. Thus,

( y1yn

)∈ rSn�A�. Since u ∈ lSnrSn�A�, we have 0 = u

( y1yn

)=

�jyj = �0� � 0�∑n

j=1 b�s�j yj� 0� �, showing that n

j=1b�s�j yj = 0, i.e., bsY = 0.

Thus, Y ∈ rDn�bs�, so

rDn�As� ⊆ rDn

�bs�� for all s > 0 (3.1)

Then bs ∈ lDnrDn�As� = DmAs (since DD is �m� n�-injective). There exists ds =

�d�s�1 � � d�s�

m � ∈ Dm such that bs = dsAs. Note that, by the choices of kand l, when k+ 1 ≤ s ≤ l, then As = A0 where A0 = �aij� ∈ Cm×n; when l < s,then As = A0 and bs = b0 where b0 = �b1� � bn�∈Cn. Thus, rDn

�A0�⊆ rDn�b0�,

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EXTENSIONS OF INJECTIVITY AND COHERENT RINGS 283

so b0 ∈ lDnrDn�A0�=DmA0 (since DD is �m� n�-injective), and hence b0 ∈ Cn ∩DmA0 =

CmA0. There exists c0 = �c1� � cm� ∈ Cm such that b0 = c0A0.Now let �i = �d

�1�i � � d

�l�i � ci� ci� � ∈ S �i = 1� � m�. Note that

for s = 1� � k� bs = dsAs ⇒ b�s�j = d

�s�i a

�s�ij �

for s = k+ 1� � l� bs = dsA0 ⇒ b�s�j = d

�s�i aij�

for s > l� b0 = c0A0 ⇒ bj = ciaij

So

�j =(b�1�j � b

�2�j �

)= (

d�1�i a

�1�ij � � d

�k�i a

�k�ij � d

�k+1�i aij� � d

�l�i aij� ciaij�

)=

(d�1�i � � d

�k�i � d

�k+1�i � � d

�l�i � ci� ��a

�1�ij � � a

�k�ij � aij� � aij� aij�

)= �i�ij

Thus, u = ��1� � �n� = ��1� � �m�A ∈ SmA. So lSnrSn�A� = SmA, and S is right�m� n�-injective.

�1� ⇒ �2�. Since D is a ring direct summand of S� S being right �m� n�-injective clearly implies that D is right �m� n�-injective. We next show thatCn ∩DmA = CmA for all A ∈ Cm×n. It is clear that CmA ⊆ Cn ∩DmA. Let u =�u1� � un� ∈ Cn ∩DmA. Write A = �aij�, and let ui = �ui� ui� � ∈ S and A =�aij� ∈ Sm×n, where aij = �aij� aij� � ∈ S. If Y =

( y1yn

)∈ rSn�A�, then AY = 0.

Write yj = �y�1�j � y

�2�j � �. Then

0 = AY = jaijyj

= (jaijy

�1�j � jaijy

�2�j �

)

Thus, jaijy�s�j = 0 for all s ≥ 1 and all i = 1� � m. This gives that A

(y�s�1y�s�n

)= 0.

Since u ∈ DmA, we have u

(y�s�1y�s�n

)= 0. So uiy

�s�i = 0. Thus,

�u1� � un�Y = uiyi

= (uiy

�1�i � uiy

�2�i �

) = 0

This shows that rSn�A� ⊆ rSn�u�, where u = �u1� � un�. So u ∈ lSnrSn�A� = SmA

(since S is right �m� n�-injective), and hence u = ��1� � �m�A, where �i =�b

�1�i � b

�2�i � � ∈ S. There exists a k > 0 such that b�k+1�

i = b�k+2�i = · · · = bi ∈ C for

i = 1� � m. Thus,

�uj� uj� � = uj = i�iaij = i�b�1�i � � b

�k�i � bi� bi� ��aij� aij� �

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284 CHEN AND ZHOU

It follows that uj = ibiaij . So u = �u1� � un� = �b1� � bm�A ∈ CmA. Hence Cn ∩DmA = CmA.

�2� ⇔ �3�. Similar to the proof of �1� ⇔ �2�. �

Proposition 17. If C is right �m� n�-injective, then Cn ∩DmA = CmA for allA ∈ Cm×n.

Proof. Let u ∈ Cn ∩DmA. Then u = vA with v ∈ Dm. If AX = 0 with X ∈ Cn, thenuX = vAX = 0, so rCn

�A� ⊆ rCn�u�. It follows that u ∈ lCnrCn

�A� = CmA (since C isright �m� n�-injective). So Cn ∩DmA = CmA. �

Corollary 18. If D�C are both right �m� n�-injective, then ��D�C����D�C� areright �m� n�-injective.

A left R-module V is said to be �m� n�-presented, if there is an exact sequenceof left R-modules 0 → K → Rm → V → 0 with K n-generated. For a submoduleP of a right R-module M , following Zhu et al. (2003), P is called an �m� n�-puresubmodule if the canonical map P ⊗R V → M⊗R V is a monomorphism for every�m� n�-presented left R-module V ; and by Zhu et al. (2003, Theorem 1.5), P ≤ M isan �m� n�-pure submodule iff Pm ∩MnA = PnA for all A ∈ Rn×m. Following Hirano(1994), P ≤ M is called an R-pure submodule if Ma ∩ P = Pa for all a ∈ R. Recallthat P ≤ M is called a pure submodule iff the canonical map P ⊗R V → M ⊗R V isa monomorphism for every left R-module V iff P ≤ M is an �m� n�-pure submodulefor every positive integers m�n.

Corollary 19. The following are equivalent:

(1) ��D�C� is right FP-injective;(2) ��D�C� is right FP-injective;(3) D is right FP-injective, and CC ≤ DC is a pure submodule.

Corollary 20. The following are equivalent:

(1) ��D�C� is right f -injective;(2) ��D�C� is right f -injective;(3) D is right f -injective, and CC ≤ DC is an �n� 1�-pure submodule for all n ≥ 1.

Corollary 21. The following are equivalent:

(1) Every finitely generated left ideal of R�D�C� is an annihilator;(2) Every finitely generated left ideal of R�D�C� is an annihilator;(3) Every finitely generated left ideal of D is an annihilator, and CC ≤ DC is a �1�m�-

pure submodule for all m ≥ 1.

The special case where D is semisimple of the next result was proven by Hirano(1994, Proposition 2).

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EXTENSIONS OF INJECTIVITY AND COHERENT RINGS 285

Corollary 22. The following are equivalent:

(1) ��D�C� is right P-injective;(2) ��D�C� is right P-injective;(3) D is right P-injective, and CC ≤ DC is an R-pure submodule.

The previous discussion can be applied to characterizing right nonsingularright �m� n�-injective rings using purity.

Theorem 23. Let R be a right nonsingular ring, and Q the maximal right quotientring of R. Then the following are equivalent:

(1) ��Q�R� is right �m� n�-injective;(2) ��Q�R� is right �m� n�-injective;(3) R is right �m� n�-injective;(4) RR ≤ QR is an �n�m�-pure submodule.

Proof. Since R is right nonsingular, Q is regular and right self-injective. Thus, theequivalences �1� ⇔ �2� ⇔ �4� follow from Theorem 16. Moreover, (3) implies (4) byProposition 18.

�4� ⇒ �3�. Since QR � E�R�R�QR is right �m� n�-injective. By Zhu et al. (2003,Theorem 2.4), every �n�m�-pure submodule of an �m� n�-injective module is also�m� n�-injective. Thus, (3) follows from (4). �

Corollary 24 (Hirano, 1994, Theorem 1). Let R be a right nonsingular ring with themaximal right quotient ring Q. Then R is right P-injective iff RR ≤ QR is an R-puresubmodule.

Corollary 25. Let R be a right nonsingular ring with the maximal right quotient ringQ. Then the following are equivalent:

(1) ��Q�R� is right FP-injective;(2) ��Q�R� is right FP-injective;(3) R is right FP-injective.

Theorem 26. Let S = ��D�C�.

(1) ��D�C� is left P-coherent iff D and C are left P-coherent.(2) ��D�C� is left coherent iff D and C are left coherent.

Proof. �1��⇐�. Let a = �a1� � an� c� � c� � ∈ S.For x = �x1� � xn� � xm� x� � ∈ lS�a�, we have xiai = 0, �i = 1� � n�,

where an+1 = · · · = am = c, and xc = 0. Since lD�ai� �i = 1� � n�, lD�c� andlC�c� are all finitely generated, there exists an integer k > 0 such that lD�ai� =Dai1 + · · · +Daik �i = 1� � m� and lC�c� = Cc1 + · · · + Cck. So xi =

∑kj=1 dijaij

�i = 1� � m� and x =∑kj=1 c

′jcj with all dij ∈ D� c

′j ∈ C. Thus, x =∑k

j=1 dj aj withdj = �d1j� � dnj� � dmj� c

′j� � ∈ S and aj = �a1j� � anj� � amj� cj� � ∈

lS�a�. So lS�a� =∑k

j=1 Saj is finitely generated. Hence S is left P-coherent.

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286 CHEN AND ZHOU

�1��⇒�. Since D is a ring direct summand of S, S being left P-coherent clearlyimplies that D is left P-coherent.

Let c ∈ C and c = �c� c� � ∈ S. Since S is left P-coherent, lS�c� = S�1 +· · · + S�t, where �i = �b

�i�1 � � b�i�m � ci� � ∈ S for i = 1� � t. Thus, ci ∈ lC�c� for

i = 1� � t. For x ∈ lC�c�, let x = �x� x� �. Then x ∈ lS�c� and x =∑ti=1 �i�i with

�i ∈ S. There exists k ≥ m such that �i = �s�i�1 � � s

�i�k � ti� � for i = 1� � t. Thus,

by computing the (k+ 1)th component of x, we have x = tici ∈ Cc1 + · · · + Cct. SolC�c� = t

i=1Cci is finitely generated. Hence C is left P-coherent.

(2) follows from (1) since �n�S� � ���n�D���n�C�� for all n > 0 and D isleft coherent iff �n�D� is left P-coherent by Chen and Ding (2001). �

Corollary 27. If D, C are left FC-rings, then ��D�C� is left FC.

The converse of Corollary 27 does not hold, as the next two examples show.

Example 28. Let S = ��D�C�, where D = � ∝ �� and C = � ∝ 0. Then C � � is

not P-injective. As known, D is a (commutative) FP-injective ring. S is a FP-injectivering: for any aij� bj ∈ C, if jaijdj = bi� dj = �nj� qj� ∈ D, then jaij�nj� 0� = bi with�nj� 0� ∈ C. This shows that Cn ∩ ADm = ACm, so S is FP-injective by Theorem 16.

Recall that a ring R is called right GP-injective if, for any 0 = a ∈ R, thereexists an n > 0 such that an = 0 and lr�an� = Ran. Note that right GP-injective ringsneed not be right P-injective (Chen et al., 2005).

Example 29. For a field K, let D = �3�K� and C ={ (

a 0 0b a 0c 0 a

)� a� b� c ∈ K

}. Then

C is commutative, but C is not left P-injective. In fact, xC = rl�x� for x =(

0 0 01 0 01 0 0

).

Let S = ��D�C�. By Hirano (1994, Example 2), S is left P-injective, but not rightP-injective. We verify next that S is indeed left FC-, but not right GP-injective.

Proof. To see that S is not right GP-injective, take u =(

0 0 01 0 01 0 0

)∈ C, and let

u = �u� u� � ∈ S. Then u2 = 0. It can be verified that

rS�u� =�d1� � dn� c� c� � � n ≥ 1� di ∈

0 0 0K K KK K K

� c ∈

0 0 0K 0 0K 0 0

�

lrS�u� =�d1� � dn� c� c� � � n ≥ 1� di ∈

K 0 0K 0 0K 0 0

� c ∈

0 0 0K 0 0K 0 0

�

Su =�d1� � dn� c� c� � � n ≥ 1� di ∈

K 0 0K 0 0K 0 0

� c =

0 0 0t 0 0t 0 0

� t ∈ K

So Su = lrS�u�, and S is not right GP-injective.Next, since D�C are left Noetherian, they are left coherent. So S is left

coherent by Theorem 26.

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EXTENSIONS OF INJECTIVITY AND COHERENT RINGS 287

Finally, we show that S is left FP-injective. Since D is semisimple, it is FP-injective. So, by Corollary 19, it suffices to show that, for any positive n�m and anyAij� Bj ∈ C, whenever the following equation is solvable in D, it is solvable in C:

A11 · · · A1m

An1 · · · Anm

X1Xm

=

B1Bn

(3.2)

So assume that (3.2) is solvable in D, and we prove it is solvable in C by inductionon n.

(1) Suppose n = 1 and suppose that

A1X1 + · · · + AmXm = B� Ai� B ∈ C (3.3)

is solvable in D. Thus there exists Di ∈ D such that A1D1 + · · · + AmDm = B. If someAi is invertible, then A−1

i B ∈ C and A10+ · · · + Ai�A−1i B�+ · · · + Am0 = B. So (3.3)

is solvable in C. So we assume that every Ai is not invertible. Then Aj =(

0 0 0bj 0 0cj 0 0

).

Write Dj =(

d�j� ∗ ∗∗ ∗ ∗∗ ∗ ∗

). Thus, noting that B ∈ C,

B = AjDj = 0 0 0bjd

�j� 0 0cjd

�j� 0 0

= Aj�d

�j�I�

with d�j�I ∈ C. So (3.3) is solvable in C. Hence the claim is true for n = 1.

(2) If n > 1 and some Aij in (3.2) is invertible, then, without loss ofgenerality, we may assume that A11 is invertible. Thus, (3.2) can be reduced to thefollowing equivalent equation:

A11 A12 · · · A1m

0 A′22 · · · A

′2m

0 A′n2 · · · A

′nm

X1

X2Xm

=

B1

B′2B

′n

� (3.4)

where A′ij� B

′j ∈ C. By induction hypothesis, there exist Cj ∈ C (j = 2� � n)

satisfying the last �n− 1� equations of (3.4). Moreover, the equation

A11X1 = B1 − A12C2 − · · · − A1mCm (3.5)

has a solution in C (since A11 is invertible), say X1 = C1. Thus, Xi = Ci (i = 1� � m)give a solution in C to (3.2). So we can assume that every Aij in (3.2) is not invertible.Arguing as in the proof of (1), we can show that (3.2) is solvable in C. Therefore, Sis left FP-injective. �

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288 CHEN AND ZHOU

ACKNOWLEDGMENTS

The first author was supported by the National Natural Science Foundationof China (No. 10171011) and the Teaching and Research Award Program forOutstanding Young Teachers in Higher Education Institutes of MOE, P.R.C. Thesecond author was supported by NSERC (Grant OGP0194196), Canada.

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Chen, J., Ding, N., Li, Y., Zhou, Y. (2001). On �m� n�-injectivity of modules. Comm. Algebra29(12):5589–5603.

Chen, J., Zhou, Y., Zhu, Z. (2005). GP-injective rings need not be P-injective. Comm. Algebra33(7):2395–2402.

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theorem. J. Algebra 271:391–406.Rutter, E. A. (1975). Rings with the principal extension property. Comm. Algebra 3:203–212.Zhu, Z., Chen, J., Zhang, X. (2003). On �m� n�-purity of modules. East-West J. Math. 5:35–44.

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