ON RINGS WHERE LEFT PRINCIPAL IDEALS ARE LEFT PRINCIPAL ANNIHILATORS

The rings in the title are studied and related to right principally injective rings. Many properties of these rings (called left pseudo-morphic by Yang) are derived, and conditions are given that an endomorphism ring is left pseudo-morphic. Some particular results: (1) Commutative pseudo-morphic rings are morphic; (2) Semiprime left pseudo-morphic rings are semisimple; and (3) A left and right pseudo-morphic ring satisfying (equivalent) mild finiteness conditions is a morphic, quasi-Frobenius ring in which every onesided ideal is principal. Call a left ideal L a left principal annihilator if L = l(a) = {r ∈ R | ra = 0} for some a ∈ R. It is shown that if R is left pseudo-morphic, left mininjective ring with the ACC on left principal annihilators then R is a quasi-Frobenius ring in which every right ideal is principal and every left ideal is a left principal annihilator. Mathematics Subject Classification (2010): 16L60, 16D50, 16P60, 16U99


Introduction
Unless otherwise noted, every ring R is associative with unity and all modules are unitary.We write the Jacobson radical as J = J(R) and abbreviate to J when no confusion can result, with a similar convention for the unit group U, the left and right socles S l and S r , and the left and right singular ideals Z l and Z r of R. The ring of n × n matrices over R will be denoted by M n (R), and we write the left and right annihilators of a set X as l(X) and r(X) respectively.We denote the ring of integers by Z and write Z n for the ring of integers modulo n.The term "regular ring" means von Neumann regular ring.We write A R to indicate that A is a two-sided ideal of R, and the notations N ⊆ ess M, N ⊆ ⊕ M and N ⊆ max M signify that N is an essential submodule (respectively a direct summand, a maximal submodule) of a module M. We write module morphisms opposite the scalars, and we write M * = hom(M, R) for the dual of the module M. Maps given by right or left multiplication by w will be written •w and w•, respectively.Given a ringtheoretic condition c, a ring will be called a c-ring if it is both a left c-ring and a right c-ring, with a similar convention for elements.
A left ideal L of a ring R will be called a left principal annihilator if L = l(b) for some b ∈ R. In 2005, a ring R is called left morphic [9] if for all a ∈ R

Pseudo-morphic rings
We begin with a characterization of left pseudo-morphic elements.
Lemma 2.1.The following are equivalent for an element a in a ring R : (1) Ra = l(b) for some b ∈ R.
Proof.Call an element a ∈ R a left pseudo-morphic element if it satisfies these conditions.
Hence a ring R is left pseudo-morphic if every element has this property.
Every regular element is left (and right) pseudo-morphic, so regular rings are pseudo-morphic.However, as we shall see, Z n is pseudo-morphic for every n ≥ 2.
A ring R is left semi-hereditary (a left PP-ring) if every finitely generated (principal) left ideal is projective.Lemma 2.1 gives: Proposition 2.2.The following are equivalent for a ring R : (1) R is regular.
(3) R is left pseudo-morphic and left PP. Proof.
As mentioned above, a ring R is called left quasi-morphic [2] if, for every a ∈ R, (1) R R is uniserial of finite length.
(2) R is local and J = Rc where c ∈ R is nilpotent.
(3) R is left morphic, local and J is nilpotent.
These rings are all left pseudo-morphic by (3).However, if we drop "local" or "J = Rc" in (2) then R need not be left pseudo-morphic, even if it is artinian (see

Examples 2.3 and 2.4 below).
Question 1.Let R be a local, left pseudo-morphic ring with J nilpotent.Is R left special?Equivalently is J = Rc for some c ∈ R?
where D is a division ring.Then R is artinian with J 2 = 0, and J = Rγ for some γ ∈ R, but R is neither left nor right pseudomorphic by Proposition 2.6 below.However, R is not local.
where D is a division ring.Then R is a local, artinian ring that is neither left nor right pseudo-morphic.
Proof.The ring R is clearly local and artinian with J 3 = 0.And J = Rγ for any γ ∈ R because, as γ is not a unit, it is impossible that the matrix units e 12 and e 23 are both in Rγ.
To see that R is not left pseudo-morphic, let α = (1) If V contains an element w such that l R (w) = 0, then R is not left pseudomorphic.
(2) If V contains an element w such that r S (w) = 0, then R is not right pseudo-morphic.
Proof.We prove (1); the proof of ( 2 Yang [15,Example 8] proves (1) of the next theorem; we include a proof for completeness.
(2) S is left pseudo-morphic but M 2 (S) is not left pseudo-morphic.
Proof.We use the notation of Example 2.7.
(1) The ring S is left special (and so left pseudo-morphic) because 0, J and R are the only left ideals.To see that S is not right pseudo-morphic we show that tS Suppose on the contrary that tS = r(x) for some 0 = x ∈ S, say x = a+bt, a, b ∈ F.
a contradiction because F = F .
( Example 2.9.If R is left pseudo-morphic and U ⊆ R is a left denominator set, the ring of quotients We are going show that all commutative pseudo-morphic rings are morphic.In fact all "reversible" left pseudo-morphic rings are left morphic, where a ring R is called reversible if ab = 0 implies ba = 0.The Björk example (Example 2.7) is reversible.Indeed, any local ring R with J 2 = 0 is reversible (if xy = 0 in S then either x or y is a unit or x, y ∈ J).In fact, every left special ring is reversible.
Theorem 2.10.If R is a reversible ring, the following are equivalent: (1) R is a left pseudo-morphic ring.
In particular, a commutative pseudo-morphic ring is morphic.
Proof.The last statement and ( 2 The following example shows that the "reversible" hypothesis is essential in Theorem 2.10. Example 2.11.Write M ω (D) = end( D V ) where V is a left vector space on a basis {v 0 , v 1 , . ..} over a division ring D. Then M ω (D) is pseudo-morphic (in fact regular) but it is not left morphic.
Proof.The ring R = M ω (D) is clearly regular.Assume that R is left morphic.

Principally injective rings
The next result, part of [11, Lemma 1.1], identifies a class of rings that are important for this paper.
Lemma 3.1.The following conditions are equivalent for an element a in a ring R : (1) R-linear maps aR → R R all extend to R R → R R .
(2) lr(a) = Ra. ( A ring R is called right principally injective (right P-injective for short) if every element a ∈ R satisfies these conditions.Clearly right self-injective rings are right P-injective, as are all regular rings.The following results of Yang (Theorems 6 (2) and 11 (2) in [15]) will be needed, and we include short proofs for completeness.
(2) R is pseudo-morphic if and only if R is quasi-morphic., so R is quasi-Frobenius (being artinian).In particular The following implications hold for any ring: (1) M α = ker(β) for some β ∈ E.
(2) M/M α ∼ = M β for some β ∈ E.  Thus R is a left pseudo-morphic ring if and only if R R is a pseudo-morphic module.
where π is the projection onto M α with kernel K.In particular Clearly every quasi-projective module is image projective.Proof.Assume that Z l = 0.If L is any left ideal of R we show that L ⊆ ⊕ R R. By Zorn's Lemma choose a left ideal M such that L ⊕ M ⊆ ess R R; we show that r(L ⊕ M ) ⊆ Z l .If a ∈ r(L ⊕ M ) then L ⊕ M ⊆ l(a).It follows that l(a) ⊆ ess R R; that is a ∈ Z l , as required.
there exists b ∈ R such that Ra = l(b) and l(a) = Rb.In 2007 R is called left quasi-morphic [2] if the sets of left principal ideals and left principal annihilators coincide: {Ra | a ∈ R} = {l(b) | b ∈ R}.That same year, the rings R for which {Ra | a ∈ R} ⊇ {l(b) | b ∈ R} were called left generalized morphic rings by Zhu and Ding [17].Our interest here is in the rings R satisfying the other inclusion: For all a ∈ R there exists b ∈ R such that Ra = l(b).These rings were called left pseudo-morphic by Yang [15] who investigated them in 2010.An outline of the paper is as follows: The general properties of left pseudomorphic rings are investigated in Section 2 (the commutative ones are morphic); Their relation with right principally injective rings is outlined in Section 3; Pseudomorphic modules are considered in Section 4 (often with pseudo-morphic endomorphism rings); The semiprime pseudo-morphic rings are characterized in Section 5 (they are semisimple); and finally, in Section 6, it is proved that, in the presence if one of several (equivalent) mild finiteness conditions, the following are equivalent for a ring R : (1) R is a (left and right) pseudo-morphic ring, R is morphic and quasi-Frobenius, and (3) R is an artinian principal ideal ring, (extending an earlier characterization of these rings in [2, Theorem 19]).In fact we obtain a one-sided result: A left pseudo-morphic, left mininjective ring with the ACC on {l(a) | a ∈ R} is a quasi-Frobenius ring in which every right ideal is principal and every left ideal is a left principal annihilator.
we have Ra = l(b) and l(a) = Rc for some b and c in R. If b = c for each a, R is called left morphic [9].These rings are clearly left pseudo-morphic.A ring R is called left special if it satisfies the following equivalent conditions [9, Theorem 9]: ) is similar.As to (1), let α = 0 w 0 0 , and suppose αΛ = r Λ (β) for some β ∈ R. Then βα = 0 so, by hypothesis, β has the form β = 0 v 0 s .But then we have S V 0 0 ⊆ r Λ (β) = αΛ = 0 wS 0 0 .This contradiction proves (1).Proposition 2.6.No upper-triangular matrix ring is left or right pseudo morphic.Example 2.7.(Björk Example) [1, Page 70] Let F be a field with an isomorphism a → ā from F to a proper subfield F ⊂ F, and let S = {a + bt | a, b ∈ F } be the F -algebra on basis {1, t} where t 2 = 0 and ta = āt for each a ∈ F. It is easy tosee that S has a unique proper left ideal F t = St = J(S), so S is left special (and so local and left artinian).Moreover, S may be taken to be right artinian by making the following choices: If p ∈ Z is a prime, take F = Z p (x) to be the field of rational functions, and define ā = a p for all a ∈ F.

Question 2 .
λ) then tx + dtz = 0. Writing x = a + bt and z = a 1 + b 1 t in S, we obtain ta + dta 1 = 0.This implies that (ā + d a 1 )t = 0, so ā + d a 1 = 0.Because d / ∈ F this gives a 1 = 0. Hence z ∈ F t and ta = 0. Thus a = 0 so x ∈ F t too.Similarly y, w ∈ F t, proving that F t F t F t F t .As the other inclusion is clear, the Claim follows.We complete the proof by showing that Rλ = l(µ) with µ ∈ R is impossible.Indeed, it implies that µ ∈ r R (λ).If we write ρ = t 0 0 0 then (by the Claim) ρ ∈ l(µ) = Rλ, say ρ = βλ where β ∈ R. If β = u v w z , we obtain t = ut and 0 = udt.Writing u = m + nt, m, n ∈ F, these become t = mt and 0 = mdt.But then m = 1, so dt = 0, a contradiction since d = 0. Theorem 2.8 leads to the question whether the other "half" of Morita invariance is true for the left pseudo-morphic rings.If R is left pseudo-morphic and e 2 = e ∈ R, is eRe left pseudomorphic?What if ReR = R?The answer is "yes" if R is left morphic [9, Theorem 15], and we also know that every left ideal of eRe is an annihilator[14, Lemma 8.10].But the question remains open, even if R is left quasi-morphic.

) ⇒ ( 3 )
⇒ (1) are clear.Assume (1).Given a ∈ R and Ra = l(b) for some b ∈ R, we prove (2) by showing that l(a) = Rb.Since R is reversible we have l(a) = r(a) = r(Ra) = rl(b) = lr(b) where the reversible hypothesis is used at the first and last steps.But lr(b) = Rb by Lemma 3.1 and Theorem 3.2 below, so l(a) = Rb as required.This proves that (1) ⇒ (2).

Proof. ( 1 )( 2 )Question 3 .Example 3 . 3 .
If a ∈ R and Ra = l(b) then lr(a) = lr(Ra) = lrl(b) = l(b) = Ra.Use Lemma 3.1.Every quasi-morphic ring is pseudo-morphic.Conversely, if R is pseudo-morphic and a ∈ R, let aR = r(c) where c ∈ R. Then l(a) = lr(c) = Rc by (1), so R is left quasi-morphic.Similarly R is right quasi-morphic because it is also left Pinjective-by the analogue of (1).Note.The converse to (1) of Theorem 3.2 is false: Example 3.3 below is a finite commutative P-injective ring (in fact quasi-Frobenius) that is not pseudo-morphic.If both R and M 2 (R) are left pseudo-morphic then Theorem 3.2 and [14, Proposition 5.36] show that R is right 2-injective (maps aR + bR → R R extend to R).If R is left pseudo-morphic and right 2-injective, is M 2 (R) right pseudo-morphic?A ring is called quasi-Frobenius if it is right or left self-injective and right or left artinian (all four combinations are equivalent).These rings grew out of the theory of representations of a finite group as a group of matrices over a field.The group ring R = Z 4 C 2 is a finite, commutative, local, quasi-Frobenius ring that is not pseudo-morphic.Proof.The ring R is local by [3, Example 20].And R is self-injective by a theorem of Connell [4, Theorem 4.1] Left quasi-morphic ⇒ Left pseudo-morphic ⇒ Right P-injective Example 3.3 shows that the converse to the second implication is not true, but (surprisingly) the converse to the first implication is still open: Question 4. [15, Question 10] Does there exist a left pseudo-morphic ring that is not left quasi-morphic? 4. Pseudo-morphic modules It is always instructive to view a ring-theoretic property in an endomorphism ring.Lemma 4.1.Let R M be a module and write E = end(M ).The following are equivalent for α ∈ E :
Part (b) ⇒ (a) in the following theorem gives a condition that an endomorphism ring is left pseudo-morphic, and extends parts of[10, Lemma 31].The proof of (a) ⇒ (b) involves the following notion: We say that a module R M generates its submodule K if K = Σ{M λ | λ ∈ E, M λ ⊆ K}, and we say that M generates its kernels if it generates ker(β) for all β ∈ end(M ).

Remark 4 . 4 .Theorem 5 . 3 .( 1 )( 2 )Example 5 . 5 .Lemma 5 . 6 .
The proof shows that if end( R M ) is a left pseudo-morphic ring then M is always an image projective module.Let R be any left pseudo-morphic ring.If L ⊆ R is any finitely generated left ideal then L = l(b) for some b ∈ R. In this case lr(L) = L.Proof.For (1) the proof of Theorem 4.6 goes through if we take M = R R. As to (2), if L = l(b) then lr(L) = lrl(b) = l(b) = L.Call a ring R left finitely Kasch if it satisfies the following condition: If L is a finitely generated left ideal of R and r(L) = 0, then L = R.The name comes from the fact that R is left Kasch if and only if every left ideal L satisfies this condition (see Lemma 5.1).Thus left Kasch rings are left finitely Kasch, and the converse holds if maximal left ideals are finitely generated.Theorem 5.4.Every left pseudo-morphic ring R is left finitely Kasch.The converse is false.Proof.Let L ⊆ R be a finitely generated left ideal with r(L) = 0.By Theorem 5.3 we have L = l(b) where b ∈ R. Then Lb = 0 so b ∈ r(L) = 0.This means that L = l(b) = l(0) = R, as required.The converse fails (even if R is left Kasch) by the following example.If F is a field and V is an F -space of dimension 2, write R = a v a a ∈ F, v ∈ V .Then R is a commutative, local, artinian, Kasch ring, but R is not pseudo-morphic.Proof.R is clearly commutative, local and artinian, and it is Kasch because J = J(R) = 0 V 0 is the only maximal ideal and l(J) = J = 0. Let 0 = v ∈ V and consider α = 0 v 0 .Then Rα = 0 Rv 0 , and we claim Rα = l(β) is impossible for β ∈ R. Indeed, such a β is not a unit and β = 0, so β = 0 v 0 , 0 = v ∈ V.But then l(β) = J = Rα.Every left nonsingular, left finitely Kasch ring R is semisimple.