S-Cohn-Jordan Extensions

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1 S-Cohn-Jordan Extensions Jerzy Matczuk Institute of Mathematics, Warsaw University, Banacha 2, Warsaw, Poland e.mail: Abstract Let a monoid S act on a ring R by injective endomorphisms and A(R; S) denote the S-Cohn-Jordan extension of R. A series of results relating properties of R and that of A(R; S) are presented. In particular it is shown that: (1) A(R; S) is semiprime (prime) iff R is semiprime (prime), provided R is left noetherian; (2) if R is semiprime left Goldie ring, then so is A(R; S), Q(A(R; S)) = A(Q(R); S) and udimr = udima(r; S); (3) A(R; S) is semisimple iff R is semisimple, provided R is left artinian. Some applications to the skew semigroup ring R#S are given. Introduction Throughout the paper R stands for an associative ring with unity and φ denotes an action of a multiplicative monoid S on R by injective endomorphisms. By this we mean that a homomorphism φ: S End(R) is given, such that φ(s) is an injective endomorphism of R, for any s S. It is assumed that all endomorphisms of R preserve unity. A classical result of Cohn (see Theorem [3]) says that if the monoid S possesses a group S 1 S = G of left quotients, then there exists an over-ring A(R; S) of R such that: 1. the action φ of S on R can be extended to an action φ: G Aut(A(R; S)) of the group G on the ring A(R; S) 2. every element a A(R; S) is of the form a = φ(s) 1 (b), for some suitable b R and s S. The ring A(R; S) is uniquely determined up to an R-isomorphism. The above mentioned theorem of Cohn was originally formulated for Ω-algebras and the construction of A(R; S) was given as a limit of a suitable directed system. The research was supported by Polish KBN grant No. 1 P03A

2 2 J.MATCZUK Picavet ([16]) gave a new construction of A(R; S), in the case R is a monoid acted by a monoid S which has a group of left quotients. His construction was very similar to that of the classical localization of a ring with respect to a multiplicatively closed set containing zero divisors. He did not assumed that S acts by injective endomorphisms. Thus, in general, the resulted monoid S 1 R = A(R; S) was not an over-monoid of R but an over-monoid of some homomorphic image of R. Yet another construction of A(R; S) was given by Jordan ([6]), in the case when S = x. He recognized A(R; x ) as a subring of the left localization of the skew polynomial ring R[x; φ(x)] with respect to the set of powers of the indeterminate x. With the help of this presentation he began systematic studies of the ring extension R A(R; x ). Due to works of Cohn and Jordan, A(R; S) will be called a S-Cohn-Jordan extension of R. Such name was also used in [11]. The possibility of enlarging an object and replacing the action by endomorphisms by the action of automorphisms should be a powerful tool, similarly as a localization in the ring theory. This is indeed the case. One can see [9], [8], [12], [16], [13], [14] for examples of such applications in various algebraic contexts. The aim of the paper is to compare some of algebraic properties of R and that of A(R; S). The obtained results not only extend results of Jordan ([6]) from a single endomorphism case to the case of an action of a monoid, but some of them are new even for the action of a single endomorphisms. In Section 1 we work under the assumption that the S-Cohn-Jordan extension A(R; S) of R exists. Theorem 1.20 is the main result of this section. It states that if the ring R is left noetherian, then A(R; S) is prime (semiprime) if and only if R is prime (semiprime). An example showing necessity of the finiteness assumption is provided. This section contains also some other results. In particular Theorem 1.17 offers a characterization of semiprimeness of S-Cohn-Jordan extensions A(R; S) in terms of properties of R and S. Proposition1.14 describes primitive central idempotents of A(R; S), in the case there is a finite common bound for the cardinality of sets of orthogonal idempotents of R. In Sections 2 and 3 we assume that the multiplicative monoid S possesses a group of left quotients. Our approach in Section 2 owes much to that of [6] and most of presented results are generalizations and extensions of the one obtained by Jordan in the case S = x. We begin with identifying A(R; S) as a subring of the localization S 1 (R# φ S) of the skew semigroup ring R# φ S. The most interesting results are Theorems 2.19, 2.22 and The first one gives necessary and sufficient conditions for A(R; S) to be left artinian. The second one implies that if R is a semisimple ring, then A(R; S) is also semisimple and block decompositions of R and A(R; S) into simple components are of the same type. The third one says that if R is semiprime left Goldie ring, then so is A(R; S), the classical left quotient ring Q(A(R; S)) of A(R; S) is equal to A(Q(R); S) and left uniform dimensions of R and A(R; S) are the same. The localization S 1 (R# φ S) is naturally isomorphic to A(R; S)# φ S 1 S. This observation and results of Section 2 enable us to reduce some problems concerning the skew

3 S-COHN-JORDAN EXTENSIONS 3 semigroup ring R# φ S to the case of A(R; S)# φ G where the ring R is replaced by A(R; S) and the action of the monoid S by injective endomorphisms is replaced by the action of its quotient group G by automorphisms. Using this approach we show in Section 3 that for certain monoids S the skew semigroup ring R# φ S is a semiprime left Goldie ring, provided R is such. Then also both rings have the same left uniform dimension. 1 Elementary Properties Let R be a unital ring, S a multiplicative monoid and φ an action of S on R by injective endomorphisms. For any s S, the endomorphism φ(s) End(R) will be denoted by φ s. Definition 1.1. We say that an over-ring A(R; S) of R is an S-Cohn-Jordan extension of R if: 1. the action of S on R extends to an action of S (also denoted by φ) on A(R; S) by automorphisms. 2. for any a A(R; S), there exists s S such that φ s (a) R. Henceforth, as in the above definition, φ s will denote the automorphism φ(s) of A(R; S), where s S. Let us remark that it is far from being clear what are the necessary and sufficient conditions for the existence of A(R; S). Notice that if S End(R) and A(R; S) exists, then S can be embedded in a group. In particular S has to be a cancellative monoid in this case. On the other hand, if T is an arbitrary set of injective endomorphisms of a ring R and A is an over ring of R such that every endomorphism from T can be extended to an automorphism of A and the condition (2) of Definition 1.1 is satisfied, then the free monoid F on the set T acts in a natural way on R and B becomes an F-Cohn- Jordan extension of R. In the same time, it may happen that the submonoid S of End(R) generated by T is not cancellative, so the S-Cohn-Jordan extension of R does not exist. The following example presents endomorphisms σ, τ of a ring R such that the monoid σ, τ is not cancellative but for any over-ring A of R such that σ and τ can be extended to automorphisms σ, τ of A, the monoid σ, τ is free. Example 1.2. Let K be a field and R = K[X] be the polynomial ring in the set X = {x i } i=1 of commuting indeterminates. Define K-linear endomorphisms σ, τ of R by setting { x2i if i is a power of 2 σ(x i ) = else x i { x3i if i is a power of 3 and τ(x i ) = else x i Let A be an over-ring of R and σ, τ automorphisms of A such that σ R = σ and τ R = τ. Notice that if w, v F, where F = σ, τ Aut(A) is the monoid generated by σ, τ, then wσ (x 1 ) = x 2 k and vτ (x 1 ) = x 3 l for some k, l > 0. Using this and an inductive argument it is easy to see that F is a free monoid.

4 4 J.MATCZUK Such an over-ring A exists. For example one can take A = K[x i i Z] and extend σ and τ to A by setting σ (x i ) = τ (x i ) = x i+1, for all i 0. In this way A becomes F-Cohn-Jordan extension of R. Observe that στ τσ but στσ = τσ 2 in End(R), i.e. S = σ, τ is not cancellative and S-Cohn-Jordan extension of R does not exist. Notice that in the above example the automorphisms σ and τ of A are not uniquely determined by σ and τ. For example one could define σ (x 0 ) = x 0 and σ (x i ) = x i+2, for i 1 and A would stay F-Cohn-Jordan extension of R. Such situation can not happen (see the proof of Lemma 2.4) when the monoid S has a group of left quotients. Moreover, as it was mentioned in the introduction, A(R; S) always exists and is uniquely defined up to an R-isomorphism in this case. In general it is also not clear what are the necessary and sufficient conditions for uniqueness of A(R; S), provided A(R; S) exists. Let us recall that a monoid S possesses a group of left quotients if and only if S is left and right cancellative and satisfies the left Ore condition, that is, for any s 1, s 2 S, there exist t 1, t 2 S such that t 1 s 1 = t 2 s 2. Suppose A(R; S) exists and let W be a submonoid of S. It is also unclear, when A(R; W ) exists and, if so, when A(R; W ) can be naturally embedded in A(R; S). Notice that one can always restrict the action of S on A(R; S), to the action of W on A(R; S). If additionally B = {a A(R; S) φ t (a) R for some t W } is a subring of A(R; S), then A(R; W ) exists and is equal to B. The above remarks give immediately first two statements of the following: Proposition 1.3. Suppose that A(R; S) exists. Then: 1. Let n 1 and W = s n s S S be a submonoid of S generated by n-th powers of elements of S. Then A(R; W ) exists and A(R; W ) = A(R; S). 2. Let W be a submonoid of S such that A(R; W ) = A(R; S). Then, for any submonoid W Ŵ S, A(R; Ŵ ) exists and A(R; Ŵ ) = A(R; S). 3. Let W be a submonoid of S containing one-sided ideal of S, then A(R; W ) exists and A(R; W ) = A(R; S). Proof. (3) It is enough to show that, for any a A(R; S), there exists t W such that φ t (a) R. Let I be a one-sided ideal of S contained in the submonoid W of S. Without loosing generality, we may assume that I is equal either to ks or Sk, for some k W. Assume I = ks. Let a A(R; S) and s S be such that φ s (a) R. Then, for t = ks W we have φ t (a) R. Assume I = Sk. Let a A(R; S) and s S be such that φ s (φ k (a)) R. Then, for t = sk W we have φ t (a) R. Remark 1.4. Let W be a submonoid of S containing one-sided ideal of S, say W contains ks, for some k W. Suppose that A(R; W ) exists. It is easy to see that the action of

5 S-COHN-JORDAN EXTENSIONS 5 every element s S on R extends to an action of s on A(R; W ) by an automorphism, by setting φ s (a) := (φ k ) 1 (φ ks )(a), for a A(R; W ). One can easily check that the above defines an action of S on A(R; W ) if and only if φ ksh (a) = φ ks (φ k ) 1 φ kh (a), for any s, h S and a A(R; W ). Notice that such equality clearly holds for a R. Corollary 1.5. Suppose that a submonoid W of S contains an element k such that ks = Sk W. Then the following conditions are equivalent: 1. The W -Cohn-Jordan extension A(R; W ) of R exists. 2. The S-Cohn-Jordan extension A(R; S) of R exists. Moreover, one can take A(R; S) = A(R; W ), whenever one of the extensions exists. Proof. (1) (2) Let A = A(R; W ). We claim that A = A(R; S). In virtue of the Remark 1.4, it is enough to show that the equality φ k φ ksh = φ k φ ks (φ k ) 1 φ kh holds in Aut(A), for any s, h S. Let s, h S. Since ks = Sk, there exists t S such that ks = tk. Therefore φ k φ ks (φ k ) 1 φ kh = φ kt φ kh = φ ktkh = φ k 2 sh = φ k φ ksh. This completes the proof of the implication. The implication (2) (1) is a direct consequence of Proposition 1.3(3). From now on we assume that the S-Cohn-Jordan extension A(R; S) exists. We will frequently make use of the following: Lemma 1.6. Let X be a finite subset of A(R; S). Then there exists s S such that φ s (X) R. Proof. Let a X and Y = X\{a}. If t, v S are such that φ v (a) R and φ t (φ v (Y )) R. Then φ s (X) R, where s = tv. This yields the proof. The element s from the above lemma resembles a common left denominator of a finite set of elements in a localization of a ring. The proof of the following proposition is left as an exercise. Proposition 1.7. For any element a A(R; S), we have: 1. a is regular in A(R; S) if and only if φ s (a) is regular in R, for any s S such that φ s (a) R. 2. a is invertible in A(R; S) if and only if there exists s S such that φ s (a) R is invertible in R. It is easy to see, using Lemma 1.6, that many properties of rings lift from R to A(R; S). In particular we have: Proposition Let T denote one of the following classes of rings: a class of all division, simple, von Neumann regular, prime, semiprime rings. If R T then A(R; S) T.

6 6 J.MATCZUK 2. Let P denote one of the following classes of rings: a class of all domains, reduced rings, n n matrix rings, commutative or, more generally, rings satisfying a fixed polynomial identity. Then A(R; S) P if and only if R P. Proof. We left the easy proof for the reader. As an exemplification of the use of Lemma 1.6 we show only that R is a n n matrix ring, provided A(R; S) is such. Let us recall that a ring T is an n n matrix ring if and only if T contains the set {e ij 1 i, j n, e ij e kl = δ jk e il, eii = 1} of n n matrix units. Suppose now that E = {e ij } 1 i,j n A(R; S) is a set of matrix units. By Lemma 1.6, there is s S such that φ s (E) R. Clearly φ s (E) is the set of matrix units of R, i.e. R is an n n matrix ring. The following two examples show that A(R; S) can be in the class T although R does not belong to T. The first one, which is a minor reformulation of Example 3.4.[6], shows that A(R; S) can be a division ring with R being not simple. Example 1.9. Let A = K(x i i Z) be the field of rational functions over a field K in commuting indeterminates {x i } i Z and S = σ, where σ is the K-automorphism of A given by σ(x i ) = x i+1, for i Z. Let us set R = K(x i i 1)[x j j 0] A and R 0 = K(x i i 1)[x 0 ] R. Then S acts in a natural way both on R 0 and R and for any a A, there exists n 1 such that σ n (a) K(x i i 1) R 0. This means A = A(R; S) = A(R 0 ; S). Remark that R 0 is noetherian while R is not. The following example shows that there exists a not semiprime ring R acted by S = σ, such that A(R; S) is a prime ring. For elements a, b of a ring T, [a, b] will denote the commutator ab ba. Example Let K be a field and B denote the free product B = K[x i x 2 i = 0; i Z] K K y i i Z of the commutative K-algebra K[x i x 2 i = 0; i Z] and the free K-algebra K y i i Z. Define A = B/I, where I is the ideal of B generated by all commutators [x i, y j ], where i, j Z and i j. The natural images of x i s and y i s in A will be denoted by the same letters. Notice that the algebra B possesses a K-automorphism σ given by σ(x i ) = x i+1 and σ(y i ) = y i+1, for all i Z. Since σ(i) = I, σ induces an automorphism of A, which also will be denoted by σ. Let R = K[x i, y j i Z, j 0] A. Then σ(r) R and i=0 σ i (R) = A, so A(R; S) = A, where S = σ. Notice that, for every i 0, the element x i is central in R and x 2 i = 0. Thus R is not semiprime. In fact, one can check that B(R) = i 0 x ir is the prime radical of R. The algebra A is prime. Indeed, for any v, w A \ {0}, we can pick an index m Z such that, for any x i appearing in either v or w, one has m < i. Then vy m w 0, i.e. vaw 0.

7 S-COHN-JORDAN EXTENSIONS 7 For any ring T, Z(T ) will stand for the center of T. Notice that, in the above example, Z(R) = K[x i i 0] R but Z(A(R; S)) = K. The following proposition offers a characterization of central elements in A(R; S). Proposition equivalent: 1. For any element a A(R; S), the following conditions are (a) a is central in A(R; S); (b) φ s (a) Z(R), for any s S such that φ s (a) R; (c) For any b A(R; S), there exists s S such that φ s (a), φ s (b) R and [φ s (a), φ s (b)] = The following conditions are equivalent: (a) Z(R) Z(A(R; S)); (b) φ s (Z(R)) Z(R), for any s S. If one of the above equivalent conditions holds, then A(Z(R), S) = Z(A(R; S)). Proof. We left the proof of (1) as an easy exarcise. The statement (2) is a direct consequence of (1). The last statement of the lemma is clear, as always R Z(A(R; S)) Z(R). The following lemma is a direct consequence of Lemma 1.1 and Theorem 1.3 of [2] (see also Proposition 2.3 [8]). Lemma Suppose that there is a finite bound on the cardinality of sets of orthogonal idempotents of R. Thus R = n i=1 R i is a finite product of indecomposable rings. Then: 1. For any injective endomorphism σ of R, there exists a permutation ρ of the index set {1,..., n} such that σ(r i ) R ρ(i). 2. For any injective endomorphism σ of R, σ n! (e i ) = e i, where e i denotes the identity of R i, 1 i n. Remark Making use of Lemma 1.6, it is easy to see that, for any natural number k, the following conditions are equivalent: 1. k is the upper bound of the cardinality of sets of orthogonal idempotents of R. 2. k is the upper bound of the cardinality of sets of orthogonal idempotents of A(R; S). Now we are ready to formulate the following: Proposition Suppose that there is a finite bound on the cardinality of sets of orthogonal idempotents of R. Then:

8 8 J.MATCZUK 1. Let e A(R; S) be a central idempotent of A(R; S), then e R. 2. Let e R be a central idempotent of R. Then e is central in A(R; S). Moreover e is a central primitive idempotent of R if and only if e is a central primitive primitive idempotent of A(R; S). 3. If e A(R; S) is a central idempotent then er = M n (B) for some ring B if and only if ea(r; S) = M n (C) for some ring C. Proof. (1). Suppose e A(R; S) is a central idempotent of A(R; S). Remark 1.13 guarantees that there is a finite bound on the cardinality of sets of orthogonal idempotents of A(R; S). In particular e can be decompose into a finite sum of central primitive idempotents and Lemma 1.12 applied to A(R; S) yields that there exists m 1 such that φ s m(e) = e, for any s S. This implies that e R. (2). Suppose that e R is a central idempotent of R. With the help of Lemma 1.12 we can find, similarly as in the proof of (1), an m 1 such that φ s m(e) = e, for any s S. Let a A(R; S) and s S be such that φ s (a) R. Then [φ s m(e), φ s m(a)] = 0 and Proposition 1.11 (1) implies that e is central in A(R; S). Now, using (1), it is easy to complete the proof of (2). (3) Let e A(R; S) be a central idempotent. Then, by (1), e R. Suppose ea(r; S) = M m (C). In particular, ea(r; S) contains a set E = {e ij 1 i, j m, e ij e kl = δ jk e il, eii = e} of m m matrix units. Using Lemmas 1.12 and 1.6, we can find s S, such that φ s (e) = e and φ s (E) R. Thus er contains the set φ s (E) of m m matrix units. This implies that er is m m matrix ring. The reverse implication is obvious as any over-ring of a unital m m matrix ring is an m m matrix ring. The following example is a modification of Example It shows that the assumption imposed on idempotents in Proposition 1.14 and Lemma 1.12 is essential. Example Let K be a field and B the free product B = ( i Z K i) K K y j j Z, where K i = K, for all i Z. Let e i B stand for the identity of K i B, i Z. Define A = B/I, where I is the ideal of B generated by the set {[e i, y j ] i, j Z, i j}. The natural images of e i s and y i s in A will be denoted by the same letters. Notice that the algebra A possesses a K-automorphism σ such that σ(e i ) = e i+1 and σ(y i ) = y i+1, for all i Z. Let R = K[e i, y j i, j Z, j 1] A. Then e i s, for i 0, are central, primitive idempotents of R but A(R; S) = A does not have nontrivial central idempotents. Notice also that R is not prime but A is prime. Lemma Suppose A(R; S) exists. Let C be a left Ore set of regular elements of R such that φ s (C) C, for all s S. Then: 1. The action of S on R extends to the action of S on C 1 R.

9 S-COHN-JORDAN EXTENSIONS 9 2. If either S satisfies the left Ore condition or C is the set of all regular elements of R, then C = s S (φ s) 1 (C) is a left Ore set of regular elements of A(R; S) and C 1 A(R; S) = A(C 1 R; S). 3. If R has a left artinian quotient ring (for example R is a semiprime left Goldie ring) and C is the set of all regular elements of R, then φ s (C) C, for all s S. Proof. (1) The assumption that φ s (C) C, for any s S, implies that φ s can be extended to an endomorphism of C 1 R by setting φ s (c 1 r) = φ s (c) 1 φ s (r). Then, it is standard to check that Φ: S End(C 1 R) given by Φ(s) = φ s is a homomorphism of monoids. (2) Let a A(R; S) and s S. Notice that if φ s (a) C, then a is regular in A(R; S). Indeed, suppose that ab = 0, for some b A. Let t S be such that φ t (φ s (b)) R. Then φ ts (a)φ ts (b) = 0 is an equation in R and b = 0 follows, as φ ts (a) C and φ ts is monic. A similar argument shows that ba = 0 implies b = 0. Thus, by the above, C = s S (φ s) 1 (C) consists of regular elements of A(R; S). The additional assumptions imposed either on S or on C in (2) yield that C is multiplicatively closed. If C is the set of all regular elements of R, then the first part of the proof implies that C is the set of all regular elements of A(R; S), so C is multiplicatively closed in this case. Suppose S satisfies the left Ore condition and let a (φ s1 ) 1 (C), b (φ s2 ) 1 (C). We can pick t 1, t 2 S such that t 1 s 1 = t 2 s 2 := v. Then φ v (a), φ v (b) C. This implies ab C and shows that C is multiplicatively closed. Now we verify that C satisfies the left Ore condition. Let a A and d C. Then we can pick t S such that φ t (a) R and φ t (d) C. Since C satisfies the left Ore condition, we can find c C and r R such that cφ t (a) = rφ t (d). Thus (φ t ) 1 (c)a = (φ t ) 1 (r)d with (φ t ) 1 (c) C. This shows that C is indeed a left Ore set in A. Now it is easy to check that C 1 A(R; S) contains C 1 R as a subring, the action of S on A(R; S) extends to C 1 A(R; S) and for any c 1 a C 1 A(R; S) one can pick t S such that φ t (c 1 a) C 1 R, i.e. C 1 A(R; S) = A(C 1 R; S) (3) The statement (3) is exactly Proposition 2.4 [5]. We have seen in Example 1.10 that A(R; S) can be semiprime although the ring R is not semiprime. The following theorem offers a characterization of semiprimeness of A(R; S) in terms of properties of R. Theorem Suppose A(R; S) exists. The following conditions are equivalent: 1. A(R; S) is semiprime. 2. for any nonzero left ideal I of R, there exists s S such that (Rφ s (I)) 2 0. Proof. Suppose A(R; S) is not semiprime. Then we can pick a nonzero left ideal L of A(R; S) such that L 2 = 0. Eventually replacing L by φ t (L), for a suitable t S, we may assume that I = L R 0. Then, for any s S, we have Rφ s (I) A(R; S)φ s (L) φ s (L) and, consequently, (Rφ s (I)) 2 = 0. This shows that (2) (1).

10 10 J.MATCZUK (1) (2) Suppose A(R; S) is semiprime. Let I be a left ideal of R such that (Rφ s (I)) 2 = 0, for any s S. Then, for any a 1, a 2 A(R; S) and r 1, r 2 I there is s S such that φ s (a i ) R, i = 1, 2. Thus φ s (a 1 r 1 )φ s (a 2 r 2 ) (Rφ s (I)) 2 = 0. This implies that (A(R; S)I) 2 = 0 and I = 0 follows. A left ideal I of R is called S-stable if φ s (I) I, for all s S. As a direct application of the above theorem we obtain: Corollary If A(R; S) is semiprime then R does not contain nonzero nilpotent S- stable left ideals. The following proposition offers a characterization of nilpotent ideals of A(R; S). Proposition Let n N, L be a left ideal of A(R; S) and I s = R φ s (L), s S. Then: 1. L n = 0 if and only if (I s ) n = 0, for any s S 2. L is nil (nil of bounded index n) if and only if I s is nil (nil of bounded index n), for any s S. Proof. (1) Suppose that (I s ) n = 0, for every s S and let a 1,..., a n L. By Lemma 1.6, we may pick t S such that φ t (a i ) R φ t (L) = I t, 1 i n. Hence φ t (a 1 )φ t (a 2 )... φ t (a n ) = 0 and a 1 a 2... a n = 0 follows, as φ t is injective. This shows that L n = 0. The reverse implication is clear as I s φ s (L), for any s S. The proof of the statement (2) is similar to that of (1). The family {I s = R φ s (L)} s S of left ideals of R, which has been associated to the left ideal L of A(R; S) in the above proposition, will play an important role in the next section. We close this section with the following: Theorem Suppose R is left noetherian and A(R; S) exists. The following conditions are equivalent: 1. R is prime (semiprime). 2. A(R; S) is prime (semiprime). Proof. The implication (1) (2) is given by Proposition 1.8. (2) (1) Suppose A(R; S) is semiprime. A result of Mushrub (Cf. [11]) says that for any injective endomorphism σ of a left noetherian ring R, its prime radical B(R) of R is σ-stable. Therefore, B(R) is a nilpotent S-stable ideal of R and Corollary 1.18 implies that B(R) = 0, i.e. R is semiprime. Suppose now that A(R; S) is prime. Then, by the above, R is semiprime. Let C denote the set of all regular elements of R. By Goldie s theorem, we know that R has a classical

11 S-COHN-JORDAN EXTENSIONS 11 ring of quotients Q = C 1 R which is a semisimple ring. Therefore, by Lemma 1.16, φ s (C) C, for all s S, the action of S on R can be extended to Q, A(Q, S) exists and is equal to C 1 A(R; S), where C = s S (φ s) 1 (C). The ring C 1 A(R; S) is prime, as A(R; S) is such. Thus C 1 A(R; S) = A(Q; S) does not contain non-trivial central idempotents. Therefore, by Proposition 1.14, the semisimple ring Q also does not contain non-trivial central idempotents, i.e. Q is simple and Goldie s theorem implies that R is prime. 2 A(R; S) for S being an Ore monoid Henceforth we will assume that the monoid S has a group of left quotients S 1 S = G. Recall that this holds exactly when the semigroup S is left and right cancellative and satisfies the left Ore condition, that is, for any s 1, s 2 S, there exist t 1, t 2 S such that t 1 s 1 = t 2 s 2. As in the previous section, φ: S End(R) stands for an action of S on a ring R by injective endomorphisms. It was mentioned in the introduction, that if such action of S on R is given, then the corresponding S-Cohn-Jordan extension A(R; S) exists and is uniquely determined up to an R-isomorphism. At the beginning of this section we present a new, very short, proof of existence of A(R; S). The proof is based on extending arguments used by Jordan (Cf.[6]) for proving the existence of A(R; S) in the case S = σ, where σ is an injective endomorphism of R. R#S will stand for the skew semigroup ring R# φ S. Since R is a unital ring and S is a monoid, both R and S are naturally embedded in R#S. Elements of R#S will be denoted as s S a ss, where a s = 0 for almost all s S. Recall that multiplication in R#S is given by multiplications in R and S together with the condition sa = φ s (a)s, for a R, s S. Keeping the above notation, we have: Lemma Elements of S are regular in R#S and S is a left Ore set in R#S. In particular, we can consider the left localization S 1 (R#S) of R#S. 2. In S 1 (R#S) we have: s 1 Rs (ts) 1 Rts, for all s, t S. 3. Let A(R) = s S s 1 Rs S 1 (R#S). Then: (a) A(R) is a subring of S 1 (R#S); (b) sa(r)s 1 = A(R), for any s S; (c) The action φ of S on R extends to an action φ: G = S 1 S Aut(A(R)) of G on A(R) by setting φ(t)(a) = φ t (a) = tat 1, for t G and a A(R). Proof. (1) Using the facts that the monoid S is cancellative and S acts on R by injective endomorphisms, one can directly check that elements of S are regular in R#S.

12 12 J.MATCZUK Let 0 a = n i=1 a is i R#S and s S. Since S satisfies the left Ore condition, elements s i s 1 S 1 S, 1 i n, can be presented with common left denominator, i.e. there exist elements t, w i S such that ts i = w i s, for any i. Then 0 t n i=1 a is i = ( n i=1 φ t(a i )w i )s. This shows Sa Rs 0, for any 0 a R#S and s S, i.e. S is a left Ore set in R#S. (2) Let a R and s, t S. Then s 1 as = s 1 t 1 tat 1 ts = (ts) 1 φ t (a)ts (ts) 1 Rts, i.e. (2) holds. (3)(a) Let s 1, s 2 S. Then, by the left Goldie condition, there exist t 1, t 2 S such that t 1 s 1 = t 2 s 2 := u. Then, by (2), s 1 i Rs i u 1 Ru, for 1 i 2. This easily yields the statement (a). (b) Let s S. By construction of A(R), s 1 A(R)s A(R). Thus, for proving (b), it is enough to show that sa(r)s 1 A(R). To this end, let t S. Then st 1 = t1 1 s 1, for some s 1, t 1 S and s(t 1 Rt)s 1 = t 1 1 (s 1 Rs 1 1 )t 1 = t 1 1 φ s1 (R)t 1 t 1 1 Rt 1 A(R). This implies that sa(r)s 1 A(R) and (b) follows. The statement (c) is an easy consequence of (b). The statement (c) from the above lemma and the definition of A(R) give immediately the following: Corollary 2.2. The ring A(R) S 1 (R#S) is the S-Cohn-Jordan extension of R. Remark 2.3. Notice that, by the construction of A(R), G = S 1 S and A(R) are naturally embedded in S 1 (R#S). Moreover, for any t S and s S a ss R#S we have t 1 s S a ss = s S (t 1 a s t)(t 1 s) A(R)G. This shows that A(R)G = S 1 (R#S). Using this, it is easy to check that A(R)#G is isomorphic to S 1 (R#S). The following lemma is known, we present its proof for completeness. Lemma 2.4. Let R B 1 and R B 2 be S-Cohn-Jordan extensions of R and φ: S Aut(B 1 ), ψ : S Aut(B 2 ) be appropriate extensions of the action φ of S on R to B 1 and B 2, respectively. Then B 1 and B 2 are R-isomorphic. Proof. Let a B 1 and s 1, s 2 S be such that such φ si (a) R, i = 1, 2. Let t 1, t 2 S be such that t 1 s 1 = t 2 s 2 = u. It is easy to check that ψs 1 1 (φ s1 (a)) = ψu 1 (φ u (a)) = ψs 1 2 (φ s2 (a)) This means that the map f : B 1 B 2 defined by setting f(a) = ψs 1 (φ s (a)), where s S is such that φ s (a) R, is well-defined. Using the fact that for a, b B 1 there exist s S such that φ s (a), φ s (b) R, it is easy to check that f is a homomorphism. f is in fact an isomorphism with the inverse g : B 2 B 1 given by g(b) = φ 1 s (ψ s (b)), where s S is such that ψ s (b) R. Corollary 2.5. The construction of Cohn from Theorem 7.3.4[3] and that of Picavet [16] lead to the same ring extension of R which is R-isomorphic to A(R; S) = A(R). Due to Lemma 2.1, the action φ of S on R extends to the action (also denoted by φ) of the group G = S 1 S on A(R; S). Thus, for s S, we have (φ s ) 1 = φ s 1 and if g = k 1 s G, then φ g = φ k 1 s = φ k 1φ s.

13 S-COHN-JORDAN EXTENSIONS 13 Definition 2.6. The set of left ideals {I s } s S k, s S, we have R φ s 1(I sk ) = I k. of R is called S-admissible if, for any The following lemma explains the reason why we are interested in S-admissible sets of left ideals. Lemma 2.7. Let L be a left ideal of A(R; S) and I s = φ s (L) R, for s S. Then {I s } s S is an S-admissible set of left ideals of R. Proof. Let s, k S. Notice that I s is an intersection of a left ideal of A(R; S) and R, so it is a left ideal of R. Moreover, R φ s 1(I sk ) = R φ s 1(φ sk (L) R) = R φ k (L) φ s 1(R) = R φ k (L) = I k, as R φ s 1(R). This shows that {I s } s S is an S-admissible set of left ideals of R. Let us record the following technical lemma. Lemma 2.8. Let {I s } s S be an S-admissible set of left ideals of R. Then, for any k, s, l S, we have: 1. Rφ s (I k ) I sk 2. R φ s 1(Rφ s (I k )) I k 3. φ k 1(I k ) φ (sk) 1(I sk ) 4. φ l (I k ) φ ls 1(I sk ) Proof. Since {I s } s S is an S-admissible set of left ideals of R, I k = R φ s 1(I sk ), for any k, s S. Let us fix s, k, l S. Then Rφ s (I k ) = Rφ s (R φ s 1(I sk )) R(φ s (R) I sk ) I sk, as I sk is a left ideal of R. This proves the statement (1). Using (1), we get R φ s 1(Rφ s (I k )) R φ s 1(I sk ) = I k, i.e. (2) holds. Observe that φ k 1(I k ) = φ k 1(R φ s 1(I sk )) φ (sk) 1(I sk ), as φ k 1φ s 1 = φ (sk) 1. This gives (3). Finally, applying φ lk to the formula (3), we obtain (4). Lemma 2.9. Let {I s } s S be an S-admissible set of left ideals of R. Then L = s S φ s 1(I s) is a left ideal of A(R; S). Proof. Let us fix s, t S. Since S satisfies the left Ore condition, there are k 1, k 2 S such that k 1 s = k 2 t = u. By Lemma 2.8(3), φ s 1(I s ) φ (k1 s) 1(I k 1 s) = φ u 1(I u ). Similarly φ t 1(I t ) φ u 1(I u ). The above shows that, for any s, t S, there exists u S such that φ s 1(I s ), φ t 1(I t ) φ u 1(I u ). This implies, in particular, that L is an additive subgroup of A(R; S). Let a A(R; S) and b φ t 1(I t ). Thus there are s S, r R and c I t such that a = φ s 1(r) and b = φ t 1(c). Since S satisfies the left Ore condition, there exist k 1, k 2 S

14 14 J.MATCZUK such that k 1 s = k 2 t = u. Then also s 1 = u 1 k 1 and t 1 = u 1 k 2. Therefore, making use of Lemma 2.8(1), we obtain: ab = φ s 1(r)φ t 1(c) = φ u 1(φ k1 (r)φ k2 (c)) φ u 1(Rφ k2 (I t )) φ u 1(I k2 t) = φ u 1(I u ) L. by The set of all S-admissible sets of left ideals of R has a natural partial ordering given {I s } s S {J s } s S if and only if I s J s, for all s S. Theorem There is an order-preserving one-to-one correspondence between the set L of all left ideals of A(R; S) ordered by inclusion and the partially ordered set R of all S-admissible sets of left ideals of R. The correspondence is given by the map Γ: L R defined by the formula Γ(L) = {R φ s (L)} s S. The inverse bijection : R L is given by ({I s } s S ) = s S φ s 1(I s), for {I s } s S R. Proof. By Lemmas 2.7 and 2.9, the maps Γ, are well-defined and clearly preserve ordering. Every element of A(R; S) is of the form φ s 1(a), for some s S and a R. Thus it is clear that Γ(L) = s S φ s 1(φ s(l) R) = L, for any left ideal L of A(R; S), i.e. Γ = id L. Let {I k } k S be an S-admissible set of left ideals of R. Then obviously (2.I) {I k } k S Γ ({I k } k S ) = {J k } k S, where J k = R s S φ ks 1(I s ). Let a J k. Then there are s S and b I s such that a = φ ks 1(b). Since S satisfies the left Ore condition, we can pick t, l S such that tk = ls. Hence a = φ t 1 l(b) and φ t (a) = φ l (b) φ l (I s ) I ls = I tk, where the last inclusion is given by Lemma 2.8(1). Therefore we obtain a R φ t 1(I tk ) = I k, as {I s } s S is an S-admissible set. This shows that J k I k, for any k S. This together with (2.I) yield that {I k } k S = Γ ({I k } k S ) and complete the proof of the theorem. The above theorem is a generalization of Theorem 4.7[6], where a similar bijection was constructed in the case S = σ. The idea of relating a left ideal L of A(R; S) with its orbit {φ s (L) R} s S comes from [6]. However our definition of an S-admissible set is not a direct generalization of the definition of α-sequence introduced by Jordan in [6]. For any left ideal I of R we will associate an S-admissible set of left ideals of R. For doing this some preparation is needed. Lemma For a left ideal I of R the following conditions are equivalent: 1. A(R; S)I R I (then equality holds). 2. R φ s 1(Rφ s (I)) I, for any s S.

15 S-COHN-JORDAN EXTENSIONS R φ s 1(Rφ s (I)) = I, for any s S. Proof. Notice that always I R φ s 1(Rφ s (I)), for s S. Thus the equivalence (2) (3) is clear. (2) (1) Let r A(R; S)I R, i.e. r = n i=1 a ir i R, where a i A(R; S) and r i I. By Lemma 1.6, we can find s S such that φ s (a i ) R, for 1 i n. Then φ s (r) Rφ s (I) and, by (1), r R φ s 1(Rφ s (I)) I follows. (1) (2) Suppose (1) holds. Let a R and s S be such that φ s (a) Rφ s (I). Since Rφ s (I) A(R; S)φ s (I) = φ s (A(R; S)I), we obtain a R A(R; S)I I, i.e. (2) holds. Definition We say that a left ideal I of R is S-closed if I satisfies one of the equivalent conditions of the above lemma. Remark Notice that if {I k } k S is an S-admissible set of left ideals of R then, by Lemma 2.8(2), I k is S-closed, for any k S. Corollary Let s S. If I is an S-closed left ideal of R, then R φ s 1(I) is also an S-closed left ideal of R. Proof. Let s S. Clearly J = R φ s 1(I) is a left ideal of R. Observe that φ s (A(R; S)J R) = φ s (A(R; S)(R φ s 1(I)) R) A(R; S)I φ s (R) A(R; S)I R I, as I is S-closed. This implies that A(R; S)J R φ s 1(I) R = J, i.e. J is S-closed. Definition For a left ideal I of R and k S, we set c k (I) = s S φ s 1(Rφ sk(i)) R. Notice that c k (I) always contains the left ideal Rφ k (I) of R. Lemma Let I be a left ideal of R. Then: 1. For any k, s, t S, we have φ s 1(Rφ sk (I)) φ (ts) 1(Rφ tsk (I)); 2. For any k, s, t S, there exists v S such that φ s 1(Rφ sk (I)) φ (vt) 1(Rφ (vt)k (I)); 3. For any k, t S, c k (I) = R s S φ (ts) 1(Rφ (ts)k(i)) = R s S φ (st) 1(Rφ (st)k(i)); 4. {c k (I)} k S is an S-admissible set of left ideals of R. In particular, for every k S, c k (I) is an S-closed left ideal of R. Proof. (1) Let k, t, s S. Since R φ t 1(R), we have φ s 1(R) φ (ts) 1(R). Now it is easy to complete the proof of (1). (2) Let k, t, s S. Since, S satisfies the left Goldie condition, we can pick v, u S such that vt = us = z. Making use of (1), we have φ s 1(Rφ sk (I)) φ (us) 1(Rφ usk (I)) = φ z 1(Rφ zk (I)) = φ (vt) 1(Rφ (vt)k (I)), i.e. (2) holds. The statement (3) is an easy consequence of statements (1) and (2).

16 16 J.MATCZUK (4) Making use of (1) and the fact for any s 1, s 2 S we can find t 1, t 2 S such that t 1 s 1 = t 2 s 2, one can check that c k (I) = s S φ s 1(Rφ sk(i)) R is an additive subgroup of R, for any k S. For any r R and a φ s 1(Rφ sk (I)) R we have φ s (ra) Rφ sk (I), i.e. ra φ s 1(Rφ sk (I)) R. The above implies that c k (I) is a left ideal of R, for any k S. Since R φ t 1(R), for t S, we have: R φ t 1(c tk (I)) = R s S φ t 1(φ s 1(Rφ stk (I)) R) = R s S φ (st) 1(Rφ (st)k (I)) = c k (I), where the last equality is given by (2). This shows that {c k (I)} k S satisfies the condition from Definition 2.6, i.e. {c k (I)} k S is an S-admissible set of left ideals of R. Now Remark 2.13 completes the proof of (4). Lemma Let k S and I, J be left ideals of R such that I J. If I is S-closed, then c k (I) c k (J). Proof. Suppose I is S-closed and let a J \ I. Clearly c k (I) c k (J). In case c k (I) = c k (J), we would have φ k (a) φ k (J) c k (J) = c k (I). Then a φ k 1(c k (I)) R = s S φ (sk) 1(Rφ sk(i)) R. I is S-closed so, by Lemma 2.11, every term of the last sum is contained in I. Thus a I would follow, which is impossible. Therefore c k (I) c k (J). This yields the lemma. Lemma Let L, M be left ideals of A(R; S) such that L M. Then: 1. There exists s S such that φ s (L) R φ s (M) R. 2. If φ s (L) R φ s (M) R, then φ ts (L) R φ ts (M) R, for any t S. Proof. Let a M \ L and s S be such that φ s (a) R. Clearly φ s (L) R φ s (M) R and φ s (a) φ s (L) This means that φ s (a) (φ s (M) R) \ (φ s (L) R), i.e. (1) holds. (2) Suppose φ s (L) R φ s (M) R. Let t S and b (φ s (M) R) \ φ s (L). Then φ t (b) φ ts (L) and φ t (b) φ ts (M) R, i.e. (2) holds. Now we are ready to prove the following theorem: Theorem The following conditions are equivalent: 1. The ring A(R; S) is left artinian. 2. There exists a finite bound on lengths of chains of S-closed left ideal of R. Moreover, if one of the above conditions holds, then the length of A(R; S) as a left A(R; S)- module is equal to the length of the longest chain of S-closed left ideals of R.

17 S-COHN-JORDAN EXTENSIONS 17 Proof. (1) (2) Suppose A(R; S) is left artinian. Then A(R; S) has a finite length, say m, as a left A(R; S)-module. Let 0 I 1 I 2... I n be a chain of S-closed left ideals of R and I i = {c k (I i )} k S, for 1 i n. Then, by Lemmas 2.16 and 2.17, 0 I 1 I 2... I n is a strictly increasing sequence of S-admissible sets of left ideals of R. Thus, by Theorem 2.10, (I 1 ) (I 2 )... (I n ) is a strictly increasing sequence of nonzero left ideals of A(R; S). Therefore n m. (2) (1) Let 0 L 1 L 2... L m be a chain of left ideals of A(R; S). We claim, by making use of induction on m, that (2.II) there exists s S such that 0 φ s (L 1 ) R φ s (L 2 ) R... φ s (L m ) R. For m = 1, the statement is clear. Suppose that m 2 and s 1 S is such that 0 φ s1 (L 1 ) R φ s1 (L 2 ) R... φ s1 (L m 1 ) R. Applying Lemma 2.18(1) to L m 1 L m, we can pick s 2 S such that φ s2 (L m 1 ) R φ s2 (L m ) R. Since S satisfies the left Ore condition, there exist t 1, t 2 S, such that t 1 s 1 = t 2 s 2 and Lemma 2.18(2) guarantees that the statement (2.II) holds for s = t 1 s 1. Notice that, by Remark 2.13, all left ideals (φ s (L i ) R), 1 i m, of R appearing in (2.II) are S-closed. This shows that m n, where n denotes the common boundary of lengths of strictly increasing chains of S-closed left ideals of R. This completes the proof of the theorem. The above theorem gives immediately the following: Corollary If the ring R is left artinian, then so is A(R; S). Theorem 2.19 is a generalization of Theorem 5.2 from [6]. The idea of the proof of the implication (1) (2) is the same as the one presented in [6]. The proof of the reverse implication in Theorem 5.2[6] was based on the use of an appropriate counterpart of Theorem One could adopt the methods of [6] to complete the proof of Theorem However we have presented another, elementary arguments based on Lemma Notice that Corollary 2.20 is a direct consequence of (2.II). Thus, this approach gives a very short, direct proof of Corollary The following observation is a direct consequence of Lemma 1.16 and Corollary 2.20 applied to the ring C 1 R. Corollary Suppose that R has a left artinian quotient ring C 1 R. Then A(R; S) has a left artinian quotient ring C 1 A(R; S) = A(C 1 R; S), where C = s S φ s 1(C). Theorem Suppose the ring R is left artinian. Then: 1. R is a semisimple ring if and only if A(R; S) is a semisimple ring. 2. If R = k i=1 e ir, with e i R = M ni (B i ), is a block decomposition of the semisimple ring R, then A(R; S) = k i=1 e ia(r; S) is a block decomposition of A(R; S) and e i A(R; S) = M ni (D i ) for some division ring. Moreover, for 1 i k, the division ring D i is an extension of B i.

18 18 J.MATCZUK Proof. By Corollary 2.20, A(R; S) is a left artinian ring. Now, the statement (1) is a direct consequence of Theorem 1.20 and the fact that every left artinian ring is left noetherian. (2) The first part of the statement is a direct consequence of (1) and Proposition Let E denote the set of matrix units of the extension of simple artinian rings e i R e i A(R; S). Then B i and D i can be recovered as a centralizers of E in e i R and e i A(R; S), respectively. This completes the proof of the theorem. Since a semisimple ring is always artinian, the above theorem gives immediately: Corollary If R is semisimple (simple artinian), then A(R; S) is semisimple (simple artinian). Example 1.9 shows that the converse implication does not hold when one only assume that R is noetherian. Namely, in this example, A(R 0 ; S) is a field while R 0 is a polynomial ring in one indeterminate over a field. Theorem 2.22 can be viewed as an extension of some results obtained in [6] and [8]. Namely, when S = σ is a monoid generated by a single endomorphism σ of R, then Corollary 2.23 was obtained by Jordan in [6]. For such S = σ the statement (2) of Theorem 2.22 was obtained independently in [8]. The methods of proving are different from those used in this paper but both proofs have made use of properties of injective endomorphisms of semisimple rings obtained in [2]. The left uniform dimension of a ring T will be denoted by udimt. It is easy to see, by applying Lemma 1.6, that if L 1... L n is a direct sum of nonzero left ideals of A(R; S) then there exists t S such that φ t (L i ) R 0, for all 1 i n. Consequently φ t (L 1 ) R... φ t (L n ) R is a direct sum of nonzero left ideals of R. This means that always udima(r; S) udimr. It is known (Cf. [12]) that the inequality can be strict. In fact, Mushrub presented in [12] an example in which udimr = and udima(r; S) = 1. Our Example 2.25 is of similar nature but we believe that it is simpler and much shorter than the one from [12]. From now on Q(R) will denote the classical left quotient ring of a semiprime left Goldie ring R. Theorem Let R be a semiprime left Goldie ring. Then: 1. A(R; S) is also a semiprime left Goldie ring and Q(A(R; S)) = A(Q(R); S). 2. udimr = udima(r; S). Proof. (1) R is a semiprime ring. Hence, by Proposition 1.8, A(R; S) is also semiprime. Now, making use of Corollary 2.21 and Goldie s Theorem, it is easy to complete the proof of the statement (1). (2) When R is semisimple, then the equality udimr = udima(r; S) is given by Theorem 2.22(2). Recall that for any semiprime left Goldie ring B, its ring of quotients Q(B) have the same uniform dimension as B. Now, the statement (1) together with Goldie s Theorem yield that, for a semiprime left Goldie ring R, we have udima(r; S) = udimq(a(r; S)) = udima(q(r); S) = udimq(r) = udimr.

19 S-COHN-JORDAN EXTENSIONS 19 Example Let B = K F({x i } i=1) denote the Mal cev-neumann division ring of series, where F({x i } i=1) is the free group on the set {x i } i=1. Let σ be the K-endomorphism of B defined by σ(x i ) = x i+2 and set R = K x 1, x 2 K K F({x i } i=3) B. Then σ(r) K F({x i } i=3) R. Thus the semigroup S = σ acts on R. Notice that every nonzero element from σ(r) is invertible in R, so Proposition 1.7 implies that A(R; S) is a division ring. Therefore we have udimr = and udima(r; S) = 1. 3 An Application It is well-known that if R is a semiprime left Goldie ring then so is the polynomial ring R[x] and udimr[x] = udimr. On the other hand, the easiest examples of a left but not right Ore domains can be constructed as Ore extensions of the form R[x; σ], where R is a field and σ is an injective endomorphism of R which is not onto. The aim of this section is to extend the above results to the case of the skew semigroup ring R#S for a certain class of monoids S acting by injective endomorphisms (see Theorem 3.8). For doing this we will need a similar result for strongly Z-graded rings. It seems that Proposition 3.3 is not recorded in the literature even though there are many generalizations of Shock s result (Cf. [17]) to various ring extensions (see for example [1], [4], [8], [10]). For completeness, we present its proof, although it is very close to the proof of the classical Shock s counterpart for polynomial rings (Cf. Theorem 6.65 and Exercise of [7]). Let R = i Z R i be a strongly Z-graded ring. For a R, a k R k will denote the k-th homogenous component of a. If a = m k n a k, with nonzero a m and a n, then we say that a is of degree deg a = n; the length l(a) of a is defined by l(a) = n m + 1. We set deg 0 = and l(0) = 0. Let I = i Z I i be a homogenous left ideal of R. Recall that, since R is strongly Z-graded, I = RI 0 (indeed: I n R 0 I n = R n (R n I n ) RI 0 ). We will also use the fact that ar n 0 R n a, for any nonzero homogenous element a R and n Z. For a R, lann R (a) will stand for the left annihilator of a in R, i.e. lann R (a) = {r R ra = 0}. We say that 0 a = m i n a i R is nice if lann R (a) = lann R (a n ) (Shock used good polynomials i.e. elements for which all nonzero homogenous components have the same left annihilator in R). Notice that lann R (a n ) is a homogenous left ideal, so lann R (a n ) = Rlann R0 (a n ). The easy proof of the following lemma is left for the reader. Lemma Any nonzero homogenous element is nice. 2. If a R is a nice element and deg a = n then, for any homogenous element r R s \ lann R (a), ra is nice of degree n + s. 3. If 0 b R, then there is r R 0 such that rb is nice. For a left ideal L of R we will write L < e R (L < u R) if L is an essential (uniform) left ideal of R.

20 20 J.MATCZUK The following lemma is a generalization of Lemma 6.68 [7], its proof is an easy adaptation of the arguments of Shock. Lemma 3.2. Let K be a left ideal of R 0. Then: 1. If K < u R 0, then RK < u R. 2. If K < e R 0, then RK < e R. Proof. (1) Suppose that K is a uniform left ideal of R 0. Assume that RK is not uniform. Thus there exist nonzero elements a, b RK such that Ra Rb = 0. Among such elements choose a and b such that l(a) + l(b) = s is as small as possible. By Lemma 3.1, we may assume that both a and b are nice and deg a = deg b = 0. We may also assume that l(a) l(b). Since a 0, b 0 K and K is uniform, there exist x, y R 0 such that xa 0 = yb 0 0. Let us consider the element c = xa yb RK. The choice of x, y yields l(c) < l(a). Notice that c 0, as Ra Rb = 0. Since l(c) + l(b) < s, Rc Rb 0. Therefore, there are w, v R such that (3.I) 0 wb = vc = vxa vyb. By Lemma 3.1(2) xa and yb are nice elements of the same degree 0. Thus lann R (xa) = lann R (yb), as yb 0 = xa 0. Observe that the equation (3.I) and the assumption Ra Rb = 0 imply that vxa = 0. Then also vyb = 0. This means that the right hand side of (3.I) is 0. The obtained contradiction shows that RK is a uniform right ideal of R. (2) Suppose K is essential in R 0. Assume that Ra RK = 0, for some 0 a R. Choose a as above of the minimal possible length. Since K is essential in R 0, we can find with a help of Lemma 3.1, a homogenous element r R such that ra is nice, deg ra = 0 and (ra) 0 K. Thus, eventually replacing a by ra, we may assume that deg a = 0 and a 0 K. Then l(a a 0 ) < l(a) and a a 0 0, as RK Ra = 0. Thus, by the choice of a, there is v R such that 0 v(a a 0 ) RK. Since RK Ra = 0, va = 0 follows. Hence also va 0 = 0, a contradiction. Z(R) will denote the left singular ideal of R, that is Z(R) = {a R lann R (a) < e R}. Proposition 3.3. Suppose R is strongly Z-graded. Then: 1. udimr = udimr Z(R) = Z(R 0 )R = {a R lann R0 (a) < e R 0 }. Proof. The statement (1) is a direct consequence of Lemma 3.2 and the fact that direct sums of left ideals of R 0 lift, by multiplying on the left by R, to direct sums of left ideals of R. (2) Let Y = {a R lann R0 (a) < e R 0 }. First observe that Z(R 0 )R = Y. Indeed, let 0 a Y be a homogenous element, say a R n. Then lann R0 (a)ar n = 0. This means that a (ar n )R n Z(R 0 )R and yields Y Z(R 0 )R. The reverse implication is clear.

21 S-COHN-JORDAN EXTENSIONS 21 Lemma 3.2(2) implies that Y = Z(R 0 )R Z(R). Suppose there exists 0 a Z(R) \ Y and choose such a of minimal length. Let n = deg a. The choice of a implies that a n / Y, that is, L = lann R0 (a n ) is not essential in R 0. Let K 0 be a left ideal of R 0 such that L K = 0. Then RK RL = 0. Let 0 b lann R (a) with deg b = m. Then b m a n = 0, i.e. b m lann R (a n ) = RL. This implies that RK lann R (a) = 0 and contradicts the assumption that a / Y. This completes the proof of (2). In the previous sections we have used to denote the free product of two algebras. Henceforth we also use for denoting the crossed product B G of a group G over a ring B (Cf.[15]). Corollary 3.4. Let B be a ring acted by a poly-infinite cyclic group G by automorphisms. Then: 1. udim(b G) = udimb and Z(B G) = Z(B)(B G). 2. If B is semiprime (prime) then the crossed-product B G is a semiprime (prime) left Goldie ring if and only if B is left Goldie. Proof. Suppose first that G is an infinite cyclic group. Then B G is a strongly G = Z- graded ring. Then, it is known and easy to see that if B is a semiprime (prime) ring then so is B G. Recall that a semiprime ring T is left Goldie if and only if udimt < and Z(T ) = 0. Therefore Proposition 3.3 yields the statements (1) and (2) in the case G is infinite cyclic. When G is an arbitrary poly-infinite cyclic group, then it is known that R G can be presented as iterated crossed product R G 1... G s, where all G i s are infinite cyclic groups. Then, an easy inductive argument completes the proof. Lemma 3.5. Let S be a left Ore set of regular elements of R. If S 1 R is a semiprime (prime) left Goldie ring, then R is a semiprime (prime) left Goldie ring and Q(R) = Q(S 1 R). Proof. We present the proof for R being semiprime. Let r R be a regular element. It is easy to see that lann S 1 R(a) = 0, i.e. r is a left regular element of S 1 R. By assumption, S 1 R is semiprime left Goldie ring. Thus, the left hand version of Proposition [7] implies that r is regular in S 1 R. This means that every regular element of R is invertible in Q = Q(S 1 R). Let q = a 1 b Q, where a, b S 1 R. Then there are s 1, s 2 S and a 0, b 0 R such that a = s 1 1 a 0, b = s 1 2 b 0. Since S is a left Ore set in R, there exist c R and s S such that ts 1 = cs 2. Then q = a 1 b = (ta 0 ) 1 (cb 0 ). This shows that every element q Q can be written in the form q = (ã) 1 b, for some suitable ã, b R. Therefore the semisimple ring Q is the left quotient ring of R and Goldie s Theorem yields the thesis.