- Email: [email protected]
On prime ideals and radicals of polynomial rings and graded rings
Journal of Pure and Applied Algebra 218 (2014) 323–332
Contents lists available at ScienceDirect
Journal of Pure and Applied Algebra journal homepage: www.elsevier.com/locate/jpaa
On prime ideals and radicals of polynomial rings and graded rings P.-H. Lee a,∗ , E.R. Puczyłowski b a
Department of Mathematics, National Taiwan University, Taipei 106, Taiwan
b
Institute of Mathematics, University of Warsaw, Warsaw, Poland
article
info
Article history: Received 21 February 2013 Received in revised form 12 May 2013 Available online 24 June 2013 Communicated by C.A. Weibel MSC: 16W50; 16N60; 16N40; 16D25
abstract We extend some known results on radicals and prime ideals from polynomial rings and Laurent polynomial rings to Z-graded rings, i.e, rings graded by the additive group of integers. The main of them concerns the Brown–McCoy radical G and the radical S, which for a given ring A is defined as the intersection of prime ideals I of A such that A/I is a ring with a large center. The studies are related to some open problems on the radicals G and S of polynomial rings and situated in the context of Koethe’s problem. © 2013 Elsevier B.V. All rights reserved.
1. Introduction One of the equivalent formulations of Koethe’s problem is [6] whether the Jacobson radical J (A[x]) of the polynomial ring A[x] in one indeterminate over an arbitrary ring A contains Nil(A)[x], where Nil(A) denotes the nil radical of A. Recall that for a given ring A the Brown–McCoy radical G(A) of A is defined as the intersection of all ideals I of A such that A/I is a simple ring with unity. It is evident that J (A) ⊆ G(A). Denote by P the class of non-zero prime rings A with a large center, i.e., such that every non-zero ideal I of A contains a non-zero central element of A. For a given ring A we define S (A) = {I | I is a prime ideal of A such that A/I ∈ P }. In [13] the following theorem was proved. Theorem 1.1. (i) [13, Theorem 1] The polynomial ring A[x] is Brown–McCoy radical if and only if A cannot be mapped homomorphically onto a ring in P ; (ii) [13, Corollary 4] G(A[x]) = S (A)[x] for every ring A. It is clear that Nil(A) ⊆ S (A) for every ring A, so the above theorem gives an approximation of a positive solution of Koethe’s problem. This approximation was later considerably improved (cf. [2–4,14,15]). The strongest among these results is the one [15] which shows that if A is a nil ring and A[x] contains primitive ideals (i.e., Koethe’s problem has a negative solution), then they are of the form I [x] for some ideals I of A. Thus the Brown–McCoy radical of polynomial rings in one indeterminate is described quite well and this description is of importance in the context of Koethe’s problem. Much less is known about the Brown–McCoy radical of polynomial rings in several indeterminates. Applying Theorem 1.1 one gets that for two commuting indeterminates G(A[x, y]) = S (A[x])[y], so the problem reduces to a description of S (A[x]). It is known that S (A[x]) = (A ∩ S (A[x]))[x] for every ring A, but A ∩ S (A[x]) is not described. In [4, Question 2] the following problem was raised. Problem 1.2 (Ferrero and Wisbauer). Is S (A[x]) = S (A)[x] for every ring A?
∗
Corresponding author. E-mail addresses: [email protected] (P.-H. Lee), [email protected] (E.R. Puczyłowski).
0022-4049/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.jpaa.2013.06.004
324
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
This problem is still open. If it had a positive solution, then for every set X of commuting indeterminates, the Brown– McCoy radical G(A[X ]) of the polynomial ring A[X ] would be equal to S (A)[X ]. However presently it is even not known, whether the following problem has a positive solution? Problem 1.3. Is Nil(A)[x] ⊆ S (A[x]) for every ring A? In [3] it was shown that this problem has a positive solution for algebras of positive characteristic and in [14], generally but assuming that Koethe’s has a positive solution. Thus a positive solution of this problem would give a new approximation of a positive solution of Koethe’s problem. Following [13] we call an epimorphism f : A[x] → P, where P is a prime ring, proper if f (xJ [x]) ̸= 0 for every non-zero ideal J of A. Theorem 1.1 is a consequence of the following proposition. Proposition 1.4 ([13, Proposition 1]). If there exists a proper epimorphism of A[x] onto a simple ring with unity, then the center Z (A) of A is non-zero. In that context, one can ask the following problem. Problem 1.5. Suppose that there is a proper epimorphism of A[x] onto a ring P ∈ P . Is then Z (A) ̸= 0? It is not hard to see (c.f., Corollary 5.5) that a positive solution of Problem 1.5 gives a positive solution of Problem 1.2. Recently it was shown [16] that there are positively graded rings (i.e. rings graded by the additive semigroup of positive integers), which are graded-nil, i.e. whose homogeneous elements are nilpotent, but are not Jacobson radical. Hence the counterpart of Koethe’s problem for positively graded rings has a negative solution. On the other hand, it was proved [14] that primitive ideals of graded-nil positively graded rings are homogeneous, which extends an analogous result [15] obtained earlier for polynomial rings over nil rings. In this context it is interesting to study graded versions of other results and open problems on polynomial rings, which are related to Koethe’s problem. In this paper we study counterparts of the above quoted results and problems for, generally speaking, Z-graded rings, i.e., rings graded by the additive group of integers. The paper is organized as follows. In the preliminary section we recall and introduce some notions on Z-graded rings and on rings with a large center. We also collect here a number of results, which will be needed later. In Section 3 we extend the notion of proper epimorphisms from polynomial rings to Z-graded rings and study a graded counterpart of Problem 1.5 and answer it in the positive in some particular cases. Many properties of polynomial rings A[x] and Laurent polynomial rings A[x, x−1 ] are strictly connected. This is so because on the one hand, if A is a ring with unity, A[x, x−1 ] is the localization of A[x] with respect to the set of positive powers of x and, on the other hand, the subring A[x + x−1 ] of A[x, x−1 ] is isomorphic to A[x] and is close to A[x, x−1 ]. In Section 4 we show that these and the so-called diagonal embedding of a Z-graded ring R into R[x, x−1 ] can be applied to show that Problem 1.5 is equivalent to its counterpart, i.e. Problem 3.1(i), asked for Z-graded rings. In the last section we obtain some relationships between the Brown–McCoy and the S radicals of Z-graded rings. We also prove that all graded-nil Z-graded rings are Brown–McCoy radical and, in case of algebras of a positive characteristic, S-radical. 2. Preliminaries All rings in this paper are associative but not necessarily with unities. For a given ring A, we denote by A1 the ring obtained by adjoining a unity to A. Taking an ideal M of A1 maximal with respect to A ∩ M = 0, one obtains the ring A∗ = A1 /M with unity such that A, identified with (A + M )/M, is an essential ideal of A∗ (i.e., I ∩ A ̸= 0 for every non-zero ideal I of A∗ ) and A∗ /A is a homomorphic image (possibly equal to 0, which is precisely the case when A is a ring with unity) of the ring of integers Z. In what follows, A∗ denotes a ring obtained in this way from a given ring A. The nil, Jacobson and Brown–McCoy radicals are denoted by Nil, J and G, respectively, and Z (A) denotes the center of A for every ring A. We will apply several times the following lemma. Lemma 2.1. For every ring A, G(A) ∩ Z (A) ⊆ J (Z (A)). Proof. Take any a ∈ G(A) ∩ Z (A). Since A is an ideal of A∗ , a ∈ G(A) ⊆ G(A∗ ). Hence no maximal ideal of A∗ contains 1 − a. This implies that (1 − a)A∗ = A∗ . Consequently there is t ∈ A∗ such that (1 − a)(1 − t ) = 1. Since A is an ideal of A∗ and a ∈ Z (A), this implies that t ∈ Z (A). Consequently every element of G(A) ∩ Z (A) is quasi-invertible in Z (A) and the result follows. Now we will recall some notions and basic results concerning graded rings and then describe some properties of rings in P .
Z-graded rings A ring R is called Z-graded if R is the direct sum of its additive subgroups Ri , where i runs over integers, called homogeneous components of R, and Ri Rj ⊆ Ri+j for arbitrary i, j. Elements belonging to H (R) = i∈Z Ri are called homogeneous elements
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
325
of R. Every element r ∈ R can be presented in the form r = rm + rm+1 + · · · + rn , where ri ∈ Ri is called the i-th homogeneous component of r. This presentation is unique modulo zero homogeneous components and uniquely determines the set supp(r ) = {i | ri ̸= 0}. Writing an element of R in this form we always mean such its presentation. If rn ̸= 0, then rn is called the leading component of r and n will be denoted by δ(r ) and called the degree of r. If, moreover, rm ̸= 0, then the length l(r ) of r is defined as n − m. Clearly a non-zero element of R is of length 0 if and only if it is homogeneous. If R is a non-zero Z-graded ring, then R∗ (or R1 ) can also be regarded as a Z-graded ring in such a way that, if R0 ̸= 0, the 0-component of R∗ (or R1 ) is identified with R∗0 (or R10 ) and all the other homogeneous components of R∗ (or R1 ) and R coincide. If R0 = 0, then the 0-component of R∗ (or R1 ) is identified with Z/{n ∈ Z | nR = 0} (or Z). A Z-graded ring R is called non-negatively graded if Ri = 0 for all negative i. If, in addition, R0 = 0, then we say that R is positively graded. A Z-graded ring R is said to be trivially graded if Ri = 0 for all i ̸= 0. Note that every ring A can be made trivially graded by putting A0 = A and Ai = 0 for all 0 ̸= i ∈ Z. Note that if a Z-graded ring R has a unity, then the unity belongs to R0 . So, if R is positively graded, it cannot have unity and we can get an extension of the grading from R to R∗ (or R1 ) in the way mentioned above. A subring (ideal) T of R is called homogeneous if, together with an element, T contains all its homogeneous components, i.e., T = i∈Z (T ∩ Ri ). Clearly Z (R) is a homogeneous subring of R and the subring (ideal) of R generated by homogeneous elements is a homogeneous subring (ideal). If I is a homogeneous ideal of a Z-graded ring R, then R/I is also Z-graded with components (R/I )i = (Ri + I )/I. It is clear that every homogeneous ideal of R/I has the form J /I for some homogeneous ideal J of R containing I. For an arbitrary ideal I of R we denote by Ih the largest homogeneous ideal of R contained in I, i.e., Ih = i∈Z (I ∩ Ri ). We conclude with some results, which will be used several times in the paper. They are known but we include their simple proofs for completeness. Proposition 2.2. Let R be a Z-graded ring and I a homogeneous ideal of R. Then the following are equivalent: (i) I is a prime ideal, (ii) AB ⊆ I implies A ⊆ I or B ⊆ I for homogeneous ideals A, B of R, (iii) aRb ⊆ I implies a ∈ I or b ∈ I for a, b ∈ H (R). Proof. (i) ⇒ (ii). Clear. (ii) ⇒ (iii). Let (a) and (b) be the ideals of R generated by a and b. Then (a) and (b) are homogeneous and (a)R(b) ⊆ I. Hence (a) ⊆ I or (b) ⊆ I, and so a ∈ I or b ∈ I. (iii) ⇒ (i). Suppose that A, B are ideals of R such that AB ⊆ I. Assume that A ̸⊆ I and B ̸⊆ I. Choose a ∈ A\I and b ∈ B\I. We may assume that an and bm are the leading components of a and b respectively and that an ̸∈ I and bm ̸∈ I. Then for every r ∈ Rk , the n + k + m-component of arb is equal to an rbm . Since arb ∈ AB ⊆ I, we have an rbm ∈ I for all r ∈ H (R). Consequently an Rbm ⊆ I, a contradiction. Corollary 2.3. Let R be a Z-graded ring and I an ideal of R. If I is a prime ideal of R, so is Ih . Proof. If A, B are homogeneous ideals of R such that AB ⊆ Ih ⊆ I, then A ⊆ I or B ⊆ I, so A ⊆ Ih or B ⊆ Ih . Thus Ih is a prime ideal of R by Proposition 2.2. For an arbitrary ring A the Laurent polynomial ring R = A[x, x−1 ] is Z-graded in the standard way (i.e., Ri = Axi ) and the polynomial ring A[x] is non-negatively graded with respect to that grading. Proposition 2.4. (i) Let I be a prime ideal of A[x]. Then I = I0 + xA[x] provided that xA[x] ⊆ I, and Ih = I0 [x], where I0 = A ∩ I is a prime ideal I0 of A. (ii) If I is a prime ideal of A[x, x−1 ], then Ih = I0 [x, x−1 ] where I0 = A ∩ I is a prime ideal I0 of A. Proof. (i) If xA[x] ⊆ I, it is clear that I = I0 + xA[x] where I0 = A ∩ I and A[x]/I = (A + I )/I ≃ A/I0 so I0 is a prime ideal n n n of A. Assume that xA[x] ̸⊆ I. Write Ih = ( I ∩ Ax ) = n ≥0 n≥0 In x where In = {a ∈ A | ax ∈ I } are ideals of A and ∗ I0 = A ∩ I. Since I is a prime ideal of A[x], it is an ideal of A [x]. Applying this one gets that I0 ⊆ I1 ⊆ I2 ⊆ · · ·. Obviously In [x](xA[x])n ⊆ In xn A[x] ⊆ I. Since xA[x] ̸⊆ I and I is a prime ideal, we get that In [x] ⊆ I, so In = I0 . Consequently, Ih = I0 [x] and A[x]/Ih ≃ (A/I0 )[x] so I0 is a prime ideal ofn A by Corollary 2.3. n (ii) As in (i) we may write Ih = n∈Z In x where In = {r ∈ R | rx ∈ I } are ideals of A. Since I is a prime ideal of A[x, x−1 ], it is an ideal of A∗ [x, x−1 ]. This implies that all In are equal to I0 = A ∩ I and, consequently, Ih = I0 [x, x−1 ]. Clearly A[x, x−1 ]/Ih ≃ (A/I0 )[x, x−1 ] and so I0 is a prime ideal of A by Corollary 2.3. Applying Proposition 2.4(i) one easily gets the following corollary. An ideal I of A[x] or A[x, x−1 ] is called A-disjoint if I ∩ A = 0. Corollary 2.5. Let A be a ring, P a non-zero prime ring and f : A[x] → P a ring epimorphism. Then the following are equivalent: (i) f is proper; (ii) Ker f is an A-disjoint ideal of A[x] and xA[x] ̸⊆ Kerf ; (iii) (Ker f )h = 0 in A[x] regarded as a Z-graded ring in the standard way.
326
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
Prime rings with a large center In this part we state some results on rings in P , which we will need later. We denote by P 1 the class of rings in P , which have unities. These rings were studied by many authors in various contexts. For more information concerning them we refer the reader to [1] where they were called ‘‘rings intrinsic over its center’’. The following proposition collects some easy-to-check properties of the class P . For a prime ring A and a multiplicative subset T of non-zero elements from Z = Z (A) we denote by AT −1 the localization of A at T . In case T = Z \0 we denote A(Z \0)−1 simply by AZ −1 . The subring of AT −1 generated by a subring U of A will be denoted by UT −1 . Proposition 2.6. (i) If A is a prime ring with Z = Z (A) ̸= 0, then A ∈ P if and only if AZ −1 is a simple ring with unity. Moreover, if A ∈ P , then AT −1 ∈ P 1 for every multiplicative subset T of non-zero elements from Z . (ii) If A ∈ P and I is a non-zero ideal of A, then I ∈ P . (iii) If I ∈ P and I is an essential ideal of A, then A ∈ P . Note that from Proposition 2.6(ii) and (iii) it follows immediately that A ∈ P if and only if A∗ ∈ P . This allows us to reduce some studies to rings with unity. Proposition 2.7. For a ring A, the following are equivalent: (i) A ∈ P ; (ii) A[x] ∈ P ; (iii) A[x, x−1 ] ∈ P . Proof. (i) ⇒ (ii). Since A is a prime ring with Z = Z (A) ̸= 0, so R = A[x] is also a prime ring with Z (R) = Z [x] ̸= 0. From A ∈ P it follows that Q = AZ −1 is a simple ring with center C = ZZ −1 by Proposition 2.6(i). Now RZ −1 = Q [x] ≃ Q C C [x] −1 −1 −1 −1 and Z (R)Z −1 = C [x] ≃ C C [ x ] , so RZ ( R ) = RZ ( Z ( R ) Z ) ≃ Q C ( x ) is a simple ring with unity. Hence R ∈P C C by Proposition 2.6(i) again. (ii) ⇒ (i). Obviously A is a prime ring if A[x] is. Let I be a non-zero ideal of A; then I [x] is a non-zero ideal of A[x] and so I [x] ∩ Z (A[x]) ̸= 0. Hence I ∩ Z (A) ̸= 0. (i) ⇔ (iii). Similar to (i) ⇔ (ii). Note that the quotient field of C [x, x−1 ] is C (x). 3. On proper epimorphisms of Z-graded rings Corollary 2.5(ii) motivates the following extension of the notion of proper epimorphisms from polynomial rings to Z-graded rings. Let R be a Z-graded ring and P a non-zero prime ring. A ring epimorphism f : R → P is called proper if (Kerf )h = 0, i.e., f (a) ̸= 0 for any non-zero a ∈ H (R). Note that if f : R → P is a proper epimorphism of a Z-graded ring R onto a non-zero prime ring P, then R is also a prime ring. Moreover, if I is a non-zero homogeneous ideal of R, then f (I ) is a non-zero ideal of P and hence is itself a prime ring, and the restriction of f to I is also a proper epimorphism of I onto f (I ). Applying Proposition 2.4(ii) one verifies readily that a ring epimorphism f of A[x, x−1 ], regarded as Z-graded ring, onto a prime ring P is proper if and only if Ker f is an A-disjoint ideal of A[x, x−1 ]. The following are some variants of Problem 1.5 asked for Z-graded rings. Problem 3.1. Let R be a Z-graded ring. (i) Suppose that there exists a proper epimorphism of R onto some P ∈ P . Must R contain a non-zero central element? (ii) Suppose that there exists a proper epimorphism of R onto some P ∈ P 1 . Must R contain a non-zero central element? (iii) Suppose that there exists a proper epimorphism of R onto a simple ring P with unity. Must R contain a non-zero central element? Note that if f : R → P is a proper epimorphism of a Z-graded ring R onto some P ∈ P , then R is also in P provided that Problem 3.1(i) has a positive solution. We will get some partial answers to these questions. Let R be a Z-graded ring and let f : R → P be a proper epimorphism of R onto a ring P ∈ P . Note that if I is a nonzero homogeneous ideal of R, then f (I ) is a non-zero ideal of P, so it contains a non-zero central element of P. Define lI = min{l(r ) | r ∈ I , 0 ̸= f (r ) ∈ Z (P )}. First we answer Problem 3.1(i) in the affirmative under some additional condition. Theorem 3.2. Let R be a Z-graded ring and f : R → P be a proper epimorphism of R onto a ring P ∈ P . If there is a upper bound for all lI , where I is a non-zero homogeneous ideals I of R, then R contains a non-zero central element.
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
327
Proof. Let M be a homogeneous ideal of R with maximal lM . Then f induces a proper epimorphism of M onto a ring f (M ) which is in P by Proposition 2.6(ii). Thus lJ is defined for non-zero homogeneous ideals J of M. Note that lJ remains unchanged when a non-zero homogeneous ideal J of R contained in M is considered as an ideal of M. We claim that lJ = lM for every non-zero homogeneous ideal J of M. Indeed let J¯ = R∗ JR∗ be the ideal of R generated by J which is clearly homogeneous. Since R is a prime ring, J¯3 is also a non-zero homogeneous ideal of R and is contained in J. Thus lJ¯3 ≥ lJ ≥ lJ¯ ≥ lM and the maximality of lM proves the claim. Hence, we may replace R by M and assume that lI = k for every non-zero homogeneous ideal I of R. Let r be an element of R with l(r ) = k such that f (r ) is a non-zero central element of P. Write r = rm + · · · + rn , where rm ̸= 0 ̸= rn and n − m = k. We claim that rn is central. If not, then there is a homogeneous element a ∈ R such that b = rn a − arn ̸= 0. Let I be the ideal of R generated by b. It contains an element u with l(u) = k such that f (u) is a non-zero central element of P. Write u = ut + · · · + us , where s − t = k = n − m. Then u = xi byi for some homogeneous elements s xi , yi from R∗ with δ(xi ) + δ(yi ) = s − δ(b). Then v = xi (ra − ar )yi = vt + · · · + vs , where vj = xi (rj+n−s a − arj+n−s )yi and in particular vs = us . Since f (r ) is central, we get f (v) = 0 and so f (u − v) = f (u) is a non-zero central element of P but l(u − v) < l(u) = k, a contradiction. Hence, rn is a non-zero central element of R. Next we answer to Problem 3.1(ii) in the affirmative for positively graded rings. Theorem 3.3. Let R be a positively graded ring and f : R → P be a proper epimorphism of R onto a ring P ∈ P 1 . Then R contains a non-zero central element. Proof. By Theorem 3.2 it suffices to show that there is a common upper bound for all lI , where I is a non-zero homogeneous ideal of R. Let r = r1 + r2 + · · · + rn be an element of degree n such that f (r ) = 1. Suppose that I = i≥1 Ii is a non-zero homogeneous ideal of R. Let u = uk + uk+1 + · · · + um be an element of I with length m − k = lI such that f (u) is a non-zero central element of P. Let u∗ = uk r + uk+1 + · · · + um . Clearly u∗ ∈ Ik+1 + Ik+2 + · · · and f (u∗ ) = f (u). Consequently l(u∗ ) ≥ lI = m − k. However δ(u∗ ) ≤ max{m, k + n}, so we have m − k ≤ l(u∗ ) ≤ max{m − k − 1, n − 1}, which implies that lI ≤ l(u∗ ) ≤ n − 1 and we are done. Corollary 3.4. Let A be a ring. If there is a proper epimorphism of xA[x] onto a ring in P 1 , then A is a prime ring containing a non-zero central element. 4. Diagonal embedding and its application to problems on proper epimorphisms Let B be a ring with unity and A its subring sharing the unity. The extension A ⊆ B is called centralizing if there are central elements b1 = 1, b2 , . . . , bn of B such that B = Ab1 + Ab2 + · · · + Abn . Lemma 4.1. Let A ⊆ B be a centralizing extension. (i) [9, Theorem 10.2.4] If Q is a prime ideal of B, then Q ∩ A is a prime ideal of A. (ii) [9, Theorem 10.2.9] For every prime ideal P of A there is a prime ideal Q of B such that P = Q ∩ A. Note that for t = x + x−1 ∈ A[x, x−1 ], we have x2 − tx + 1 = 0, which shows that if A is a ring with unity, then A[t ] ⊆ A[x, x−1 ] is a centralizing extension. Moreover, A[t ] ≃ A[x] in the canonical way. Consequently Lemma 4.1 can be applied to extend some results on prime ideals or radicals from polynomial rings to Laurent polynomial rings. For the Jacobson and Brown–McCoy radicals it was done in [6,7]. Unfortunately such a nice method cannot be applied to extend results known for positively (or non-negatively) graded rings to Z-graded rings. There is however a method, which allows one to apply the above idea sometimes. Namely one can embed a given Z-graded ring R into R[x, x−1 ], translate the studied properties of R to properties of R[x, x−1 ] and then apply the above described method to reduce the studied property to the polynomial ring R[x] and next relate it back to R. Using that method we will show in this section that Problems 1.5 and 3.1(i) are equivalent. i Let R be a Z-graded ring. The map f : R → R[x, x−1 ] defined by f ( ri ) = ri x is a ring embedding. It will be called the diagonal embedding. Obviously, if R is non-negatively (positively) graded, then f is an embedding of R into R[x] (into xR[x]). Lemma 4.2. Let R be a Z-graded ring with unity, f : R → R[x, x−1 ] the diagonal embedding, I an ideal of R such that Ih = 0 and I the ideal of R[x, x−1 ] generated by f (I ). Then (i) I = f (I )xi ; (ii) R ∩ I = 0; (iii) f (R) ∩ I = f (I ); (iv) R[x, x−1 ] = f (R)xi and (R/I )[x, x−1 ] ≃ R[x, x−1 ]/I . Proof. The properties (i) and (ii) are particular cases of [12, Lemma 3]. k j (iii) The inclusion ⊇ is clear. Suppose that i ri xi ∈ f (R) belongs to I, where ri ∈ Ri . Then, by (i), i ri xi = j( k ajk x )x ,
i i( j+k=i ajk )x , so j+k=i ajk = ri ∈ Ri for each i, implying i ajk = 0 unless k = i and j = 0. Consequently, i ri xi = a x = f ( a ) ∈ f ( I ) . i 0i i 0i (iv) For aj ∈ Rj we have aj xk = aj xj xk−j = f (aj )xk−j ∈ f (R)xi and so R[x, x−1 ] = f (R)xi . Thus the map ψ : −1 −1 i i R[x, x ] → R[x, x ] defined by the rule ψ( i ri x )= so is π ◦ ψ : R[x, x−1 ] → i f (ri )x is a ring epimorphism and −1 −1 −1 R[x, x ]/I, where π is the canonical epimorphism of R[x, x ] onto R[x, x ]/I. Suppose that i ri xi ∈ Ker (π ◦ ψ). Then
where ajk ∈ Rk such that
k
ajk ∈ I for each j. Now
i ri x
i
=
328
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
j i j i j aij , si = j bij where aij , bij ∈ Rj . Then i( j aij x )x = i( j bij x )x , k k or equivalently, k( j ak−j,j )x = k( j bk−j,j )x , so j ak−j,j = j bk−j,j for all k, implying j aij = j bij for all i, j. Hence, ri = si ∈ I for all i. This implies that Ker (π ◦ ψ) ⊆ I [x, x−1 ]. The opposite inclusion is clear. Consequently R[x, x−1 ]/I ≃ R[x, x−1 ]/I [x, x−1 ] ≃ (R/I )[x, x−1 ] and we are done.
i
f (ri )xi =
i
f (si )xi for some si ∈ I. Write ri =
Now we will prove that Problems 1.5 and 3.1(i) are equivalent. It is a consequence of the following result. Theorem 4.3. Let R be a Z-graded ring. If R contains an ideal I with Ih = 0 such that R/I ∈ P , then the polynomial ring R[t ] contains an R-disjoint ideal M such that R[t ]/M ∈ P . Proof. Replacing R by R∗ and I by an ideal of R∗ maximal with respect to I ′ ∩ R = I, we may assume that R is a ring with unity, because Ih′ = 0 and R∗ /I ′ ∈ P by Proposition 2.6(iii) since R∗ /I ′ contains an essential ideal (R + I ′ )/I ′ ≃ R/I ∈ P . Moreover, if we obtain an ideal M ′ in R∗ [t ] such that M ′ ∩ R∗ = 0 and R∗ [t ]/M ′ ∈ P , then M = M ′ ∩ R[t ] will satisfy the desired properties: M ∩ R = M ′ ∩ R[t ] ∩ R = M ′ ∩ R = 0 and R[t ]/M = R[t ]/(M ′ ∩ R[t ]) ≃ (R[t ] + M ′ )/M ′ ∈ P by Proposition 2.6(ii) since (R[t ] + M ′ )/M ′ is a non-zero ideal of R∗ [t ]/M ′ . Thus assume that R is a ring with unity. Let f be the diagonal embedding of R into R[x, x−1 ] and I the ideal of R[x, x−1 ] generated by f (I ). Since R/I is a prime ring, so is (R/I )[x, x−1 ]. Hence it follows from Lemma 4.2(iv) that I is a prime ideal of R[x, x−1 ]. Note that we may assume that t = x + x−1 ; then R[t ] ⊆ R[x, x−1 ] is a centralizing extension. Thus M = I ∩ R[t ] is a prime ideal of R[t ] by Lemma 4.1(i) and M ∩ R = I ∩ R[t ] ∩ R = I ∩ R = 0 by Lemma 4.2(ii). Then A = (R[t ] + I )/I ≃ R[t ]/M is a prime ring and A ⊆ B = R[x, x−1 ]/I is a centralizing extension. By Lemma 4.2(iv) there is an isomorphism (R/I )[x, x−1 ] ≃ B and so B ∈ P by Proposition 2.7. Moreover, this isomorphism maps Z (R/I )[x, x−1 ] and Z (R/I )[x + x−1 ] onto Z (B) and Z = Z (A) respectively. Since Z (R/I )[x + x−1 ] ⊆ Z (R/I )[x, x−1 ] is an integral extension, so is the extension Z (A) ⊆ Z (B). Consider the localizations AZ −1 and BZ −1 of A and B at Z \ 0. Then AZ −1 ⊆ BZ −1 is a centralizing extension with Z (AZ −1 ) = ZZ −1 and Z (BZ −1 ) = Z (B)Z −1 . Since Z ⊆ Z (B) is an integral extension, so is the extension ZZ −1 ⊆ Z (B)Z −1 . However, ZZ −1 is a field, so Z (B)Z −1 is also a field and hence BZ −1 = BZ (B)−1 which is a simple ring with unity by Proposition 2.6(i). If A ∈ / P , then A would contain a non-zero ideal P such that P ∩ Z = 0. Taking P to be an ideal of R which is maximal with respect to P ∩ Z = 0, we may assume that P is a prime ideal of A. By Lemma 4.1(ii), there exists a prime ideal Q of B such that Q ∩ A = P. From the simplicity of BZ −1 it follows that QZ −1 = BZ −1 and so P ∩ Z ̸= 0, a contradiction. Therefore, R[t ]/M ≃ A ∈ P and so M is as required. Corollary 4.4. Problems 1.5 and 3.1(i) are equivalent. Proof. It is clear that if Problem 3.1(i) has a positive solution, then Problem 1.5 also has a positive solution. We will show the converse. Let R be a Z-graded ring and f : R → P be a proper epimorphism for some P ∈ P . Applying Theorem 4.3 we get that R[x] contains an R-disjoint ideal M such that R[x]/M ∈ P . If xR[x] ⊆ M, then R[x]/M = (R + M )/M ≃ R, so R ∈ P and Z (R) ̸= 0. If xR[x] ̸⊆ M, then by Corollary 2.5 the canonical epimorphism of R[x] onto R[x]/M is proper. Hence a positive solution of Problem 1.5 gives that Z (R) ̸= 0, so Problem 3.1(i) has a positive solution. 5. On the Brown–McCoy and S radicals of Z-graded rings We start with introducing a Z-graded analog of the radical S and obtaining some relations among this radical, the Brown– McCoy radical and the radical S for Z-graded rings. Denote by Ph the class of non-zero Z-graded prime rings P such that I ∩ Z (P ) ̸= 0 for every non-zero homogeneous ideal I of P, and for a given Z-graded ring R define Sh (R) as the intersection of all homogeneous ideals I of R such that R/I ∈ Ph . We will examine these notions for polynomial rings and Laurent polynomial rings regarded as Z-graded rings in the standard way. Proposition 5.1. (i) Let I be a homogeneous ideal of A[x] such that xA[x] ̸⊆ I. Then A[x]/I ∈ Ph if and only if I = I0 [x] for some ideal I0 of A such that A/I0 ∈ P . (ii) Let I be a homogeneous ideal of A[x, x−1 ]. Then A[x, x−1 ]/I ∈ Ph if and only if I = I0 [x, x−1 ] for an ideal I0 of A such that A/I0 ∈ P . Proof. (i) Assume first that A[x]/I ∈ Ph . Applying Proposition 2.4(i) we get that I = I0 [x] for some prime ideal I0 of A. For every ideal K of A which contains I0 strictly, K [x]/I0 [x] is a non-zero homogeneous ideal of A[x]/I0 [x], so it contains a non-zero central element kxn + I0 [x] of A[x]/I0 [x]. It is clear that k + I0 is a non-zero element in K /I0 belonging to the center of A/I0 . Hence A/I0 ∈ P . Conversely, assume that I = I0 [x] and A/I0 ∈ P . Then I0 [x] is a homogeneous prime ideal of A[x]. Now, if J is a homogeneous ideal of A[x] which contains I0 [x] strictly, then there is an n such that Jn = {a ∈ A | axn ∈ J } is an ideal of A which contains I0 strictly. Since A/I0 ∈ P , we get that for some k ∈ Jn , 0 ̸= k + I0 belongs to the center of A/I0 . It is clear that 0 ̸= kxn + I0 [x] ∈ J /I0 [x] belongs to the center of J /I0 [x]. Hence A[x]/I0 [x] ∈ Ph . (ii) The proof is analogous to that of (i). Corollary 5.2. Sh (A[x]) = S (A)[x] and Sh (A[x, x−1 ]) = S (A)[x, x−1 ] for every ring A.
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
329
Proof. If I is a homogeneous ideal I of A[x] such that xA[x] ⊆ I, then I = I0 + xA[x] for some ideal I0 of A. It is clear that A[x]/I ∈ Ph if and only if A/I0 ∈ P and, by Proposition 5.1(i), if and only if A[x]/I0 [x] ∈ Ph . Now the result is a consequence of Proposition 5.1(i). The result for A[x, x−1 ] follows directly from Proposition 5.1(ii). This corollary shows that Sh can be treated, for Z-graded rings, as a natural counterpart of the radical S. From Corollary 5.2 it follows that Problem 1.2 can be restated as whether S (A[x]) = Sh (A[x]) for every ring A. So one can also ask the following extension of Problem 1.2. Problem 5.3. Is S (R) = Sh (R) for every Z-graded ring R? The inclusion S (R) ⊆ Sh (R) is verified in the next proposition. Proposition 5.4. Let R be a Z-graded ring. Then (i) R ̸= Sh (R) if and only if there is a homogeneous ideal I of R such that R/I contains a non-nilpotent central element. (ii) S (R) is a homogeneous ideal of R and S (R) ⊆ Sh (R). (iii) If Problem 3.1(i) has a positive solution, then S (R) = Sh (R), i.e., Problem 5.3 also has a positive solution. Proof. (i) The ‘‘only if’’ implication is an immediate consequence of the definition of Sh (R). To get the ‘‘if’’ implication suppose that I is a homogeneous ideal of R and R/I contains a non-nilpotent central element a + I. Applying Zorn’s Lemma one can find a homogeneous ideal M of R which is maximal among the homogeneous ideals of R containing I such that an ̸∈ M for all n = 1, 2, . . .. Then R/M ∈ Ph . (ii) Let R = i∈Z Ri ; then R[x] = i∈Z Ri [x] gives a Z-grading of R[x]. By Theorem 1.1 we have S (R)[x] = G(R[x]). From [5, Corollary 6] it follows that G ( R [ x ]) is homogeneous with respect to the above gradation. Consequently S (R)[x] = i∈Z (Ri [x] ∩ S (R)[x]) and so S (R) = i∈Z (Ri ∩ S (R)). This shows that S (R) is homogeneous. Let I be a homogeneous ideal of R such that R/I ∈ Ph . Then R/I is a Z-graded prime ring. Take any non-zero ideal J /I of R/I. Let r = rm + · · · + rn be a non-zero element of J /I of the smallest length, where ri ∈ (R/I )i and rn ̸= 0. Since R/I ∈ Ph , the ideal of R/I generated by rn contains a non-zero homogeneous central element c = at rn bt , where at , bt are homogeneous elements of R∗ /I. Clearly r ′ = at rbt ∈ J /I. The length of r ′ is equal to that of r and its leading component is equal to c. For every homogeneous element a ∈ R, we have ar ′ − r ′ a ∈ J /I. Since ac = ca, the length of ar ′ − r ′ a is smaller than that of r. Consequently ar ′ − r ′ a = 0, so r ′ is a non-zero element of J /I belonging to the center of R/I. Hence R/I ∈ P which implies that S (R) ⊆ Sh (R). (iii) Let I be an ideal of R such that R/I ∈ P . The canonical epimorphism f of the Z-graded ring R/Ih onto R/I is proper. Hence for every non-zero homogeneous ideal K of R/Ih the restriction of f to K is a proper epimorphism of the Z-graded ring K onto f (K ) ∈ P . Assuming that Problem 3.1(i) has a positive solution we get that 0 ̸= Z (K ) ⊆ Z (R/Ih ). Consequently R/Ih ∈ Ph , so Sh (R) ⊆ Ih ⊆ I. Hence Sh (R) ⊆ S (R) and so Sh (R) = S (R) by (ii). That is, Problem 5.3 has a positive solution. Corollary 5.5. (i) S (A[x]) ⊆ S (A)[x] for every ring A. (ii) If Problem 1.5 has a positive solution, then S (A[x]) = S (A)[x] for every ring A, i.e., Problem 1.2 has a positive solution. Proof. (i) An immediate consequence of Corollary 5.2 and Proposition 5.4(ii). (ii) Suppose that I is a prime ideal of A[x] such that A[x]/I ∈ P . If xA[x] ⊆ I, then I = I0 + xA[x] for an ideal I0 of A and A/I0 ∈ P . If xA[x] ̸⊆ I, then Ih = I0 [x] for some ideal I0 of A by Proposition 2.4(i). By Corollary 2.5, the canonical epimorphism of (A/I0 )[x] onto A[x]/I is proper. Hence assuming a positive solution of Problem 1.5 (and so of Problem 3.1 (i) by Corollary 4.4) we get that (A/I0 )[x] ∈ P and so A/I0 ∈ P by Proposition 2.7. In either case, we have always that S (A)[x] ⊆ I0 [x] ⊆ I. Consequently S (A)[x] ⊆ S (A[x]). Theorem 5.6. Let R be a Z-graded ring. (i) If R ∈ Ph and G(R) ̸= 0, then Z (G(R)) ⊆ R0 and S (R0 ) ̸= R0 . (ii) If S (R0 ) = R0 , then G(R) ⊆ Sh (R). (iii) If R is positively graded, then G(R) = Sh (R). Proof. (i) By [5, Corollary 6] G(R) is a homogeneous ideal of R. Since R ∈ Ph and G(R) ̸= 0, Z (G(R)) is a non-zero Z-graded ring and it contains no non-zero nilpotent elements. By Lemma 2.1, Z (G(R)) is Jacobson radical, so its i-components are nil for all i ̸= 0 by [10, Theorem 22.6]. Consequently 0 ̸= Z (G(R)) ⊆ R0 . Hence R0 contains non-nilpotent central elements, so R0 is not S-radical. (ii) Let I be a homogeneous ideal of R such that R/I ∈ Ph . Since R0 is S-radical, so is (R/I )0 = (R0 + I )/I ≃ R0 /(R0 ∩ I ). Applying (i) we get that G(R/I ) = 0 and so G(R) ⊆ I. Hence G(R) ⊆ Sh (R). (iii) Applying (ii) we get that G(R) ⊆ Sh (R) since R0 = 0. Conversely let M be a maximal ideal of R such that R/M has a unity. Then R/Mh is a prime ring by Corollary 2.3 and the canonical epimorphism f of R/Mh onto R/M is proper. Since R/M is simple, the restriction of f to any homogeneous ideal K of R/Mh onto R/M is also proper. Thus K has a non-zero center by Theorem 3.3. This implies that R/Mh ∈ Ph . Hence Sh (R) ⊆ Mh ⊆ M. Consequently Sh (R) ⊆ G(R). From Corollary 5.2 it follows that Theorem 1.1(ii) can be restated as G(A[x]) = Sh (A[x]) for every ring A. Hence Theorem 5.6 can be treated as a generalization of Theorem 1.1(ii) to positively graded rings. Now we will give an example showing that it does not extend to non-negatively graded rings, i.e., that there is a non-negatively graded ring R such that G(R) ̸= Sh (R).
330
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
Example. Let K be a commutative local domain with non-zero Jacobson radical J and F the field of quotients of K . Let A be a simple F -algebra with unity, which is a domain but not a division algebra. As a specific example one can take K = ZS −1 , the localization of Z at the set S of odd integers so that F is the field Q of rational numbers and take A to be the Weyl algebra over Q. Let a be a non-invertible element of A. It is easy to see that B = Aa is a simple F -algebra without unity. Applying Corollary 3.4 we get that xB[x] is Brown–McCoy radical. Set R = J + xB[x]; then xB[x] is an ideal of R such that R/(xB[x]) ≃ J is a commutative Jacobson radical ring and so a Brown–McCoy radical ring. Hence R is itself a Brown–McCoy radical ring. Obviously R is a homogeneous subring of A[x] graded in the standard way, so R is non-negatively graded. Now xB[x] is a homogeneous ideal of R and R/(xB[x]) ∈ Ph , so Sh (R) ⊆ xB[x] ̸= R = G(R). Note that if a Z-graded ring R is graded-nil, i.e., H (R) consists of nilpotent elements, then Z (R) is a nil ring. Hence, R/I ̸∈ Ph for any homogeneous ideal I of R, that is, R = Sh (R). Moreover, if R is positively graded, then R = G(R) by Theorem 5.6(iii). Now we extend the latter result to Z-graded rings. Theorem 5.7. Every Z-graded ring which is graded-nil is Brown–McCoy radical. Proof. Assume that the result does not hold. Then there exist a Z-graded ring R, which is graded nil, and a ring epimorphism f : R → P, where P is a simple ring with unity 1. Let r ∈ R and f (r ) = 1. We claim that supp(r ) contains a negative integer. Indeed, if not, then f ( k≥0 Rk ) is a ring with unity, so replacing R by k≥0 Rk we can assume that R is non-negatively graded. Since R+ = k≥1 Rk is an ideal of R, R0 is nil and P is a simple ring with unity, we obtain that f (R+ ) = P. Now R+ /(R+ ∩ (Ker f )h ) is a graded-nil positively graded ring and f induces an epimorphism f : R+ /(R+ ∩ (Ker f )h ) → P defined by f (r ) = f (r ) for r = r + R+ ∩ (Kerf )h . Clearly, f is a proper epimorphism and so R+ /(R+ ∩ (Ker f )h ) contains a non-zero central element by Theorem 3.3. Hence R+ /(R+ ∩ (Ker f )h ) is a prime ring containing a non-zero homogeneous central element which is nilpotent, a contradiction. Consequently for every r ∈ R such that f (r ) = 1, the minimal integer m in supp(r ) is negative. Let us choose R, P, f and r such that m is the largest possible and the index of nilpotency of rm is the smallest possible. Note that supp(r ) contains an odd integer, as otherwise r ∈ R = k∈Z R2k and R is Z-graded with components Rk = R2k and in that grading the minimal integer in supp(r ) is m/2, which contradicts the maximality of m. Let r = a + b where a is the sum of components of r with odd degrees and b is the sum of other components. Then f (a)+ f (b) = 1 and f (a)2 = 1 − 2f (b)+ f (b)2 . Consequently f (t ) = 1 where t = a2 + 2b − b2 . Note that t ∈ R and the m-th component tm of t 2 in the grading of R considered above is equal to rm and all the components of t with degrees < m are equal to 0. Hence by the minimality of the nilpotency index of rm , we get that tm = 0. Thus the minimal integer in supp(t ) is > m, a contradiction. The converse of Theorem 5.7 does not hold. Indeed from Corollary 3.4 it follows that if A is a simple domain without unity, then the ring xA[x] is Brown–McCoy radical. Note also that the homogeneous subrings of a Brown–McCoy radical Z-graded ring R need not be necessarily Brown–McCoy radical. For instance, let B be the subring of the above-mentioned A generated by a non-zero element. Since B is not nil, xB[x] is not Jacobson radical. But xB[x] is a commutative ring, so it is not Brown– McCoy radical either. However the following slightly modified version of Theorem 5.7 gives already a characterization of graded-nil Z-graded rings. Proposition 5.8. Every homogeneous subring of a Z-graded ring R is Brown–McCoy radical if and only if R is graded-nil. Proof. The ‘‘if’’ part is an immediate consequence of Theorem 5.7. Now we will prove the opposite implication. Let r ∈ Ri , i ̸= 0; then the commutative homogeneous subring ⟨r ⟩ of R generated by r is Brown–McCoy radical by assumption and so is Jacobson radical. By [10, Theorem 22.6] each component of ⟨r ⟩ is nil. In particular, r is nilpotent. Assume now that r ∈ R0 . If r is not nilpotent, then there exists a prime ideal P of ⟨r ⟩ such that r n ̸∈ P for all n > 0. Then U = ⟨r ⟩/P is a commutative domain and so is U ∗ . By assumption, all the subrings of ⟨r ⟩ ⊆ R0 are Brown–McCoy radical since they are homogeneous. Hence every subring of U is also Brown–McCoy radical and so Jacobson radical. Consider the unital epimorphism f : Z[x] → U ∗ such that f (x) = r + P. Then f (Z) is either a finite field or a ring isomorphic to Z since it is a domain. Note that Ker f is not zero; otherwise xZ[x] ≃ U would be Jacobson radical, a contradiction. Take a non-zero polynomial w(x) ∈ Ker f and write w(x) = xn v(x), where v(x) = a0 + a1 x + · · · + am xm ∈ Z[x] with a0 ̸= 0. Since Ker f is a prime ideal of Z[x] and xn ̸∈ Ker f , we have v(x) ∈ Ker f . Consequently a0 ∈ K = Z ∩ (xZ[x] + Ker f ) ̸= 0. Since f (K ) is a subring of f (xZ[x]) = U, it is Jacobson radical. On the other hand, f (K ) is an ideal of f (Z), a ring containing no non-zero Jacobson radical ideals, so f (K ) = 0. From K = K ∩ Ker f = Z ∩ Ker f it follows that Z/K ≃ f (Z). Thus the proper homomorphic image Z/K of Z is a domain and so is a finite field. Now K [x] ⊆ Ker f , so (Z/K )[x] ≃ Z[x]/K [x] can be mapped homomorphically onto U ∗ . This implies that either U ∗ ≃ F [x] or U ∗ ≃ F for some finite field F . In either case, no non-zero subring of U ∗ = U can be Jacobson radical, a contradiction. As we have seen above, a Z-graded ring R which is Brown–McCoy radical need not be graded-nil. However, from the Proof of Proposition 5.8, if R is Jacobson radical and, in addition, R0 is nil, then R is graded-nil. Thus all homogeneous subrings of R are Brown–McCoy radical. It is natural to ask whether they must also be Jacobson radical. We do not know the answer yet in the general case. However, it is true at least for non-negatively graded rings. Since R0 is nil it suffices to prove that for positively graded rings. Theorem 5.9. If T is a homogeneous subring of a positively graded ring R, then J (R) ∩ T ⊆ J (T ). In particular, if R is Jacobson radical, then so is T .
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
331
Proof. It is enough to show that if r ∈ T and in R1 , then (1 − r )−1 ∈ T 1 . Let f : R1 → R1 [x] be 1 − r is invertible 1 −1 −1 −1 the diagonal embedding. Write (1 − r ) = ri where ri ∈ Ri ; then f (1 − r ) = f ((1 − r ) ) = ri xi and it suf−1 1 1 fices to show that f (1 − r ) ∈ T [x]. Note that f (r ) ∈ T [x] since T is a homogeneous subring of R, so f (1 − r )−1 = (1 − f (r ))−1 = 1 + f (r ) + f (r )2 + · · · ∈ T 1 [[x]]. On the other hand, f (1 − r )−1 = f ((1 − r )−1 ) ∈ R1 [x], so f (1 − r )−1 ∈ R1 [x] ∩ T 1 [[x]] = T 1 [x] and we are done. Applying Proposition 5.8 and Theorem 5.9 one gets the following characterization of non-negatively graded rings in which all homogeneous subrings are Jacobson radical. Corollary 5.10. Every homogeneous subring of a non-negatively graded ring R is Jacobson radical if and only if R0 is a nil ring and R+ = i≥1 Ri is a Jacobson radical ring. Remark 5.11. One could ask whether the conclusion of Theorem 5.9 holds for all subrings instead of only homogeneous subrings. This is not the case. In [18] it was shown that there is a Jacobson radical polynomial ring R[x], which is not nil. If the conclusion of Theorem 5.9 held for all subrings of R[x], then all of them would be Jacobson radical. Regarding R[x] as a trivially Z-graded ring and applying Corollary 5.10 we would get that R[x] is nil, a contradiction. In the Proof of Theorem 5.9 we applied the diagonal embedding but one could get it following the idea from the Proof of [11, Proposition 2]. However none of these methods applies to Z-graded rings. The following observations concern particular cases. Proposition 5.12. Suppose that A[x, x−1 ] is Jacobson radical and B is a subring of A. Then B[x, x−1 ] Jacobson radical. Proof. Since the Laurent polynomial ring A[x, x−1 ] is Jacobson radical, it follows from [8, Lemma 2.2] that the polynomial ring A[x] is Jacobson radical. In view of [6, Theorem 1] we have that the matrix rings Mn (A) are nil for all n ≥ 1. Hence Mn (B) are nil for all n ≥ 1, and by [6, Theorem 1] again B[x] is Jacobson radical. Applying [8, Lemma 2.2] once more we have that B[x, x−1 ] is Jacobson radical. Modifying slightly arguments from [3] we will show that graded-nil Z-graded algebras of a positive characteristic are S-radical. We need the following lemma proved in [3]. Lemma 5.13. Let A be an algebra of characteristic p > 0 and a0 , a1 , . . . , an ∈ A. If a = a0 + a1 + · · · + an is a central element of A, then ap is the sum of elements of the form ai1 · · · aip with i1 + · · · + ip ≡ 0 (mod p). Theorem 5.14. Every Z-graded algebra of characteristic p > 0, which is graded-nil, is S-radical. Proof. Assume on the contrary that there are Z-graded algebras of characteristic p, which are graded-nil but not S-radical. Then for each such an algebra R there is an epimorphism f : R → P with P ∈ P . Let r ∈ R be such an element that f (r ) is a non-zero central element of P. We can assume that the algebra R, the epimorphism f and the element r, are chosen in such a way leading component rn of that the length l(r ) of r is the smallest possible and among those the nilpotency index of the p p r = m≤i≤n ri is the smallest possible. Since f (r ) is central, it follows from Lemma 5.13 that f (r ) = f (r ) = f (r ) for some
r =
ri1 · · · rip =
i1 +···+ip ≡0 (mod p)
n
r i,
i=m
· · · rip ∈ Rip . Thus r ∈ k Rkp . The ring R = k Rkp is Z-graded with components Rk = Rkp . p The restriction of f to R maps it onto f (R) ∈ P and f (r ) = f (r ) is a non-zero central element of f (R). The length l(r ) of p r in this grading is ≤ n − m = l(r ) and the n-component r n of r in this grading is rn . The minimality of l(r ) gives that p l(r ) = l(r ). However, r n = rn would then be the leading component of r with nilpotency index smaller than that of rn , a
where r i =
i1 +···+ip =ip ri1
contradiction. We do not know whether Theorem 5.14 can be extended to all graded-nil Z-graded rings, so the following problem is open. Problem 5.15. Is S (R) = R for every graded-nil Z-graded ring R? In solving this problem the following result might be of some use. Proposition 5.16. (i) If R is a graded-nil Z-graded ring, which is not S-radical, then R contains a finitely generated homogeneous subring, which can be mapped homomorphically onto a ring P ∈ P such that Z (P ) is Jacobson radical. (ii) If R is a graded-nil positively graded ring, which is not S-radical, then R contains a homogeneous subring, which can be mapped homomorphically onto a Jacobson radical ring in P . Proof. Suppose that R is a graded-nil Z-graded ring, which is not S-radical; then there exist a P ′ ∈ P and an epimorphism f ′ : R → P ′ . Let r ∈ R be such that f ′ (r ) is a non-zero central element of P ′ and R′ be the subring of R generated by the
332
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
homogeneous components of r. Then R′ is a finitely generated, graded-nil, homogeneous subring of R. Since f ′ (R′ ) contains the non-nilpotent central element f ′ (r ), there exists an ideal M which is maximal with respect to not containing any power of f ′ (r ). Thus P = f ′ (R′ )/M ∈ P , so there is an epimorphism f : R′ → P. (i) By Theorem 5.7 we get that R′ is Brown–McCoy radical. Hence P is also Brown–McCoy radical, so Z (P ) is Jacobson radical by Lemma 2.1. Thus R′ is a subring of R as required. (ii) We are done if P is Jacobson radical. Thus suppose that P contains primitive ideals. If I is a primitive ideal of P, then J = f −1 (I ) is a primitive ideal of R′ since R′ /J ≃ P /I. By [16, Theorem 1] we get that J is a homogeneous ideal of R′ . Since R′ /J is a primitive ring and all homogeneous elements of R′ /J are nilpotent, we have Z (R′ /J ) = 0. Consequently Z (P /I ) = 0 and Z (P ) ⊆ I. Thus the Jacobson radical J (P ) of P contains Z (P ) and so J (P ) ̸= 0. Hence J (P ) ∈ P by Proposition 2.6(ii). Now f −1 (J (P )) = {f −1 (I ) | I primitive ideal of P } and all f −1 (I ) are homogeneous ideals of R. Consequently f −1 (J (P )) is a homogeneous subring of R. Thus f −1 (J (P )) is a subring of R as required. Acknowledgments We are grateful to the referee for informing us about a recent paper [17] by A. Smoktunowicz. The Proposition 0.1 in [17] coincides with our Theorem 5.9. This research was supported by the Polish National Center of Science Grant No DEC-2011/03/B/ST1/04893. References [1] E.P. Armendariz, G.F. Birkenmeier, J.K. Park, Ideal intrinsic extensions with connections to PI-rings, J. Pure Appl. Algebra 213 (2009) 1756–1776. [2] K.I. Beidar, Y. Fong, E.R. Puczyłowski, Polynomial rings over nil rings cannot be homomorphically mapped onto rings with nonzero idempotents, J. Algebra 238 (2001) 389–399. [3] M. Chebotar, W.-F. Ke, P.-H. Lee, E.R. Puczyłowski, A note on polynomial rings over nil rings, in: Modules and Comodules, in: Trends Math., Birkhäuser Verlag, Basel, 2008, pp. 169–172. [4] M. Ferrero, R. Wisbauer, Unitary strongly prime rings and related radicals, J. Pure Appl. Algebra 181 (2003) 209–226. [5] E. Jespers, E.R. Puczyłowski, The Jacobson and Brown–McCoy radicals of rings graded by free groups, Comm. Algebra 19 (1991) 551–558. [6] J. Krempa, Logical connections between some open problems concerning nil rings, Fund. Math. 76 (1972) 121–130. [7] J. Krempa, On radical properties of polynomial rings, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 20 (1972) 545–548. [8] J. Krempa, A. Sierpinska, The Jacobson radical of certain group and semigroup rings, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 26 (1978) 963–967. 1979. [9] J.C. McConnell, J.C. Robson, Noncommutative Noetherian Rings, John Wiley & Sons, Chichester, 1987. [10] D. Passman, Infinite Crossed Products, Academic Press, San Diego, 1989. [11] E.R. Puczyłowski, Radicals of polynomial rings in non-commutative indeterminates, Math. Z. 182 (1983) 63–67. [12] E.R. Puczyłowski, A note on graded algebras, Proc. Amer. Math. Soc. 113 (1991) 1–3. [13] E.R. Puczyłowski, A. Smoktunowicz, On maximal ideals and the Brown–McCoy radical of polynomial rings, Comm. Algebra 26 (1998) 2473–2482. [14] A. Smoktunowicz, R[x, y] is Brown–McCoy radical if R[x] is Jacobson radical, in: Proceedings of the Third International Algebra Conference, Tainan, 2002, Kluwer Acad. Publ., Dordrecht, 2003, pp. 235–240. [15] A. Smoktunowicz, On primitive ideals in polynomial rings over nil rings, Algebra Represent. Theory 8 (2005) 69–73. [16] A. Smoktunowicz, On primitive ideals in graded rings, Canad. Math. Bull. 51 (2008) 460–466. [17] A. Smoktunowicz, A note on nil and Jacobson radicals in graded rings. http://arxiv.org/abs/1301.2835. [18] A. Smoktunowicz, E.R. Puczyłowski, A polynomial ring that is Jacobson radical and not nil, Israel J. Math. 124 (2001) 317–325.
Contents lists available at ScienceDirect
Journal of Pure and Applied Algebra journal homepage: www.elsevier.com/locate/jpaa
On prime ideals and radicals of polynomial rings and graded rings P.-H. Lee a,∗ , E.R. Puczyłowski b a
Department of Mathematics, National Taiwan University, Taipei 106, Taiwan
b
Institute of Mathematics, University of Warsaw, Warsaw, Poland
article
info
Article history: Received 21 February 2013 Received in revised form 12 May 2013 Available online 24 June 2013 Communicated by C.A. Weibel MSC: 16W50; 16N60; 16N40; 16D25
abstract We extend some known results on radicals and prime ideals from polynomial rings and Laurent polynomial rings to Z-graded rings, i.e, rings graded by the additive group of integers. The main of them concerns the Brown–McCoy radical G and the radical S, which for a given ring A is defined as the intersection of prime ideals I of A such that A/I is a ring with a large center. The studies are related to some open problems on the radicals G and S of polynomial rings and situated in the context of Koethe’s problem. © 2013 Elsevier B.V. All rights reserved.
1. Introduction One of the equivalent formulations of Koethe’s problem is [6] whether the Jacobson radical J (A[x]) of the polynomial ring A[x] in one indeterminate over an arbitrary ring A contains Nil(A)[x], where Nil(A) denotes the nil radical of A. Recall that for a given ring A the Brown–McCoy radical G(A) of A is defined as the intersection of all ideals I of A such that A/I is a simple ring with unity. It is evident that J (A) ⊆ G(A). Denote by P the class of non-zero prime rings A with a large center, i.e., such that every non-zero ideal I of A contains a non-zero central element of A. For a given ring A we define S (A) = {I | I is a prime ideal of A such that A/I ∈ P }. In [13] the following theorem was proved. Theorem 1.1. (i) [13, Theorem 1] The polynomial ring A[x] is Brown–McCoy radical if and only if A cannot be mapped homomorphically onto a ring in P ; (ii) [13, Corollary 4] G(A[x]) = S (A)[x] for every ring A. It is clear that Nil(A) ⊆ S (A) for every ring A, so the above theorem gives an approximation of a positive solution of Koethe’s problem. This approximation was later considerably improved (cf. [2–4,14,15]). The strongest among these results is the one [15] which shows that if A is a nil ring and A[x] contains primitive ideals (i.e., Koethe’s problem has a negative solution), then they are of the form I [x] for some ideals I of A. Thus the Brown–McCoy radical of polynomial rings in one indeterminate is described quite well and this description is of importance in the context of Koethe’s problem. Much less is known about the Brown–McCoy radical of polynomial rings in several indeterminates. Applying Theorem 1.1 one gets that for two commuting indeterminates G(A[x, y]) = S (A[x])[y], so the problem reduces to a description of S (A[x]). It is known that S (A[x]) = (A ∩ S (A[x]))[x] for every ring A, but A ∩ S (A[x]) is not described. In [4, Question 2] the following problem was raised. Problem 1.2 (Ferrero and Wisbauer). Is S (A[x]) = S (A)[x] for every ring A?
∗
Corresponding author. E-mail addresses: [email protected] (P.-H. Lee), [email protected] (E.R. Puczyłowski).
0022-4049/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.jpaa.2013.06.004
324
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
This problem is still open. If it had a positive solution, then for every set X of commuting indeterminates, the Brown– McCoy radical G(A[X ]) of the polynomial ring A[X ] would be equal to S (A)[X ]. However presently it is even not known, whether the following problem has a positive solution? Problem 1.3. Is Nil(A)[x] ⊆ S (A[x]) for every ring A? In [3] it was shown that this problem has a positive solution for algebras of positive characteristic and in [14], generally but assuming that Koethe’s has a positive solution. Thus a positive solution of this problem would give a new approximation of a positive solution of Koethe’s problem. Following [13] we call an epimorphism f : A[x] → P, where P is a prime ring, proper if f (xJ [x]) ̸= 0 for every non-zero ideal J of A. Theorem 1.1 is a consequence of the following proposition. Proposition 1.4 ([13, Proposition 1]). If there exists a proper epimorphism of A[x] onto a simple ring with unity, then the center Z (A) of A is non-zero. In that context, one can ask the following problem. Problem 1.5. Suppose that there is a proper epimorphism of A[x] onto a ring P ∈ P . Is then Z (A) ̸= 0? It is not hard to see (c.f., Corollary 5.5) that a positive solution of Problem 1.5 gives a positive solution of Problem 1.2. Recently it was shown [16] that there are positively graded rings (i.e. rings graded by the additive semigroup of positive integers), which are graded-nil, i.e. whose homogeneous elements are nilpotent, but are not Jacobson radical. Hence the counterpart of Koethe’s problem for positively graded rings has a negative solution. On the other hand, it was proved [14] that primitive ideals of graded-nil positively graded rings are homogeneous, which extends an analogous result [15] obtained earlier for polynomial rings over nil rings. In this context it is interesting to study graded versions of other results and open problems on polynomial rings, which are related to Koethe’s problem. In this paper we study counterparts of the above quoted results and problems for, generally speaking, Z-graded rings, i.e., rings graded by the additive group of integers. The paper is organized as follows. In the preliminary section we recall and introduce some notions on Z-graded rings and on rings with a large center. We also collect here a number of results, which will be needed later. In Section 3 we extend the notion of proper epimorphisms from polynomial rings to Z-graded rings and study a graded counterpart of Problem 1.5 and answer it in the positive in some particular cases. Many properties of polynomial rings A[x] and Laurent polynomial rings A[x, x−1 ] are strictly connected. This is so because on the one hand, if A is a ring with unity, A[x, x−1 ] is the localization of A[x] with respect to the set of positive powers of x and, on the other hand, the subring A[x + x−1 ] of A[x, x−1 ] is isomorphic to A[x] and is close to A[x, x−1 ]. In Section 4 we show that these and the so-called diagonal embedding of a Z-graded ring R into R[x, x−1 ] can be applied to show that Problem 1.5 is equivalent to its counterpart, i.e. Problem 3.1(i), asked for Z-graded rings. In the last section we obtain some relationships between the Brown–McCoy and the S radicals of Z-graded rings. We also prove that all graded-nil Z-graded rings are Brown–McCoy radical and, in case of algebras of a positive characteristic, S-radical. 2. Preliminaries All rings in this paper are associative but not necessarily with unities. For a given ring A, we denote by A1 the ring obtained by adjoining a unity to A. Taking an ideal M of A1 maximal with respect to A ∩ M = 0, one obtains the ring A∗ = A1 /M with unity such that A, identified with (A + M )/M, is an essential ideal of A∗ (i.e., I ∩ A ̸= 0 for every non-zero ideal I of A∗ ) and A∗ /A is a homomorphic image (possibly equal to 0, which is precisely the case when A is a ring with unity) of the ring of integers Z. In what follows, A∗ denotes a ring obtained in this way from a given ring A. The nil, Jacobson and Brown–McCoy radicals are denoted by Nil, J and G, respectively, and Z (A) denotes the center of A for every ring A. We will apply several times the following lemma. Lemma 2.1. For every ring A, G(A) ∩ Z (A) ⊆ J (Z (A)). Proof. Take any a ∈ G(A) ∩ Z (A). Since A is an ideal of A∗ , a ∈ G(A) ⊆ G(A∗ ). Hence no maximal ideal of A∗ contains 1 − a. This implies that (1 − a)A∗ = A∗ . Consequently there is t ∈ A∗ such that (1 − a)(1 − t ) = 1. Since A is an ideal of A∗ and a ∈ Z (A), this implies that t ∈ Z (A). Consequently every element of G(A) ∩ Z (A) is quasi-invertible in Z (A) and the result follows. Now we will recall some notions and basic results concerning graded rings and then describe some properties of rings in P .
Z-graded rings A ring R is called Z-graded if R is the direct sum of its additive subgroups Ri , where i runs over integers, called homogeneous components of R, and Ri Rj ⊆ Ri+j for arbitrary i, j. Elements belonging to H (R) = i∈Z Ri are called homogeneous elements
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
325
of R. Every element r ∈ R can be presented in the form r = rm + rm+1 + · · · + rn , where ri ∈ Ri is called the i-th homogeneous component of r. This presentation is unique modulo zero homogeneous components and uniquely determines the set supp(r ) = {i | ri ̸= 0}. Writing an element of R in this form we always mean such its presentation. If rn ̸= 0, then rn is called the leading component of r and n will be denoted by δ(r ) and called the degree of r. If, moreover, rm ̸= 0, then the length l(r ) of r is defined as n − m. Clearly a non-zero element of R is of length 0 if and only if it is homogeneous. If R is a non-zero Z-graded ring, then R∗ (or R1 ) can also be regarded as a Z-graded ring in such a way that, if R0 ̸= 0, the 0-component of R∗ (or R1 ) is identified with R∗0 (or R10 ) and all the other homogeneous components of R∗ (or R1 ) and R coincide. If R0 = 0, then the 0-component of R∗ (or R1 ) is identified with Z/{n ∈ Z | nR = 0} (or Z). A Z-graded ring R is called non-negatively graded if Ri = 0 for all negative i. If, in addition, R0 = 0, then we say that R is positively graded. A Z-graded ring R is said to be trivially graded if Ri = 0 for all i ̸= 0. Note that every ring A can be made trivially graded by putting A0 = A and Ai = 0 for all 0 ̸= i ∈ Z. Note that if a Z-graded ring R has a unity, then the unity belongs to R0 . So, if R is positively graded, it cannot have unity and we can get an extension of the grading from R to R∗ (or R1 ) in the way mentioned above. A subring (ideal) T of R is called homogeneous if, together with an element, T contains all its homogeneous components, i.e., T = i∈Z (T ∩ Ri ). Clearly Z (R) is a homogeneous subring of R and the subring (ideal) of R generated by homogeneous elements is a homogeneous subring (ideal). If I is a homogeneous ideal of a Z-graded ring R, then R/I is also Z-graded with components (R/I )i = (Ri + I )/I. It is clear that every homogeneous ideal of R/I has the form J /I for some homogeneous ideal J of R containing I. For an arbitrary ideal I of R we denote by Ih the largest homogeneous ideal of R contained in I, i.e., Ih = i∈Z (I ∩ Ri ). We conclude with some results, which will be used several times in the paper. They are known but we include their simple proofs for completeness. Proposition 2.2. Let R be a Z-graded ring and I a homogeneous ideal of R. Then the following are equivalent: (i) I is a prime ideal, (ii) AB ⊆ I implies A ⊆ I or B ⊆ I for homogeneous ideals A, B of R, (iii) aRb ⊆ I implies a ∈ I or b ∈ I for a, b ∈ H (R). Proof. (i) ⇒ (ii). Clear. (ii) ⇒ (iii). Let (a) and (b) be the ideals of R generated by a and b. Then (a) and (b) are homogeneous and (a)R(b) ⊆ I. Hence (a) ⊆ I or (b) ⊆ I, and so a ∈ I or b ∈ I. (iii) ⇒ (i). Suppose that A, B are ideals of R such that AB ⊆ I. Assume that A ̸⊆ I and B ̸⊆ I. Choose a ∈ A\I and b ∈ B\I. We may assume that an and bm are the leading components of a and b respectively and that an ̸∈ I and bm ̸∈ I. Then for every r ∈ Rk , the n + k + m-component of arb is equal to an rbm . Since arb ∈ AB ⊆ I, we have an rbm ∈ I for all r ∈ H (R). Consequently an Rbm ⊆ I, a contradiction. Corollary 2.3. Let R be a Z-graded ring and I an ideal of R. If I is a prime ideal of R, so is Ih . Proof. If A, B are homogeneous ideals of R such that AB ⊆ Ih ⊆ I, then A ⊆ I or B ⊆ I, so A ⊆ Ih or B ⊆ Ih . Thus Ih is a prime ideal of R by Proposition 2.2. For an arbitrary ring A the Laurent polynomial ring R = A[x, x−1 ] is Z-graded in the standard way (i.e., Ri = Axi ) and the polynomial ring A[x] is non-negatively graded with respect to that grading. Proposition 2.4. (i) Let I be a prime ideal of A[x]. Then I = I0 + xA[x] provided that xA[x] ⊆ I, and Ih = I0 [x], where I0 = A ∩ I is a prime ideal I0 of A. (ii) If I is a prime ideal of A[x, x−1 ], then Ih = I0 [x, x−1 ] where I0 = A ∩ I is a prime ideal I0 of A. Proof. (i) If xA[x] ⊆ I, it is clear that I = I0 + xA[x] where I0 = A ∩ I and A[x]/I = (A + I )/I ≃ A/I0 so I0 is a prime ideal n n n of A. Assume that xA[x] ̸⊆ I. Write Ih = ( I ∩ Ax ) = n ≥0 n≥0 In x where In = {a ∈ A | ax ∈ I } are ideals of A and ∗ I0 = A ∩ I. Since I is a prime ideal of A[x], it is an ideal of A [x]. Applying this one gets that I0 ⊆ I1 ⊆ I2 ⊆ · · ·. Obviously In [x](xA[x])n ⊆ In xn A[x] ⊆ I. Since xA[x] ̸⊆ I and I is a prime ideal, we get that In [x] ⊆ I, so In = I0 . Consequently, Ih = I0 [x] and A[x]/Ih ≃ (A/I0 )[x] so I0 is a prime ideal ofn A by Corollary 2.3. n (ii) As in (i) we may write Ih = n∈Z In x where In = {r ∈ R | rx ∈ I } are ideals of A. Since I is a prime ideal of A[x, x−1 ], it is an ideal of A∗ [x, x−1 ]. This implies that all In are equal to I0 = A ∩ I and, consequently, Ih = I0 [x, x−1 ]. Clearly A[x, x−1 ]/Ih ≃ (A/I0 )[x, x−1 ] and so I0 is a prime ideal of A by Corollary 2.3. Applying Proposition 2.4(i) one easily gets the following corollary. An ideal I of A[x] or A[x, x−1 ] is called A-disjoint if I ∩ A = 0. Corollary 2.5. Let A be a ring, P a non-zero prime ring and f : A[x] → P a ring epimorphism. Then the following are equivalent: (i) f is proper; (ii) Ker f is an A-disjoint ideal of A[x] and xA[x] ̸⊆ Kerf ; (iii) (Ker f )h = 0 in A[x] regarded as a Z-graded ring in the standard way.
326
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
Prime rings with a large center In this part we state some results on rings in P , which we will need later. We denote by P 1 the class of rings in P , which have unities. These rings were studied by many authors in various contexts. For more information concerning them we refer the reader to [1] where they were called ‘‘rings intrinsic over its center’’. The following proposition collects some easy-to-check properties of the class P . For a prime ring A and a multiplicative subset T of non-zero elements from Z = Z (A) we denote by AT −1 the localization of A at T . In case T = Z \0 we denote A(Z \0)−1 simply by AZ −1 . The subring of AT −1 generated by a subring U of A will be denoted by UT −1 . Proposition 2.6. (i) If A is a prime ring with Z = Z (A) ̸= 0, then A ∈ P if and only if AZ −1 is a simple ring with unity. Moreover, if A ∈ P , then AT −1 ∈ P 1 for every multiplicative subset T of non-zero elements from Z . (ii) If A ∈ P and I is a non-zero ideal of A, then I ∈ P . (iii) If I ∈ P and I is an essential ideal of A, then A ∈ P . Note that from Proposition 2.6(ii) and (iii) it follows immediately that A ∈ P if and only if A∗ ∈ P . This allows us to reduce some studies to rings with unity. Proposition 2.7. For a ring A, the following are equivalent: (i) A ∈ P ; (ii) A[x] ∈ P ; (iii) A[x, x−1 ] ∈ P . Proof. (i) ⇒ (ii). Since A is a prime ring with Z = Z (A) ̸= 0, so R = A[x] is also a prime ring with Z (R) = Z [x] ̸= 0. From A ∈ P it follows that Q = AZ −1 is a simple ring with center C = ZZ −1 by Proposition 2.6(i). Now RZ −1 = Q [x] ≃ Q C C [x] −1 −1 −1 −1 and Z (R)Z −1 = C [x] ≃ C C [ x ] , so RZ ( R ) = RZ ( Z ( R ) Z ) ≃ Q C ( x ) is a simple ring with unity. Hence R ∈P C C by Proposition 2.6(i) again. (ii) ⇒ (i). Obviously A is a prime ring if A[x] is. Let I be a non-zero ideal of A; then I [x] is a non-zero ideal of A[x] and so I [x] ∩ Z (A[x]) ̸= 0. Hence I ∩ Z (A) ̸= 0. (i) ⇔ (iii). Similar to (i) ⇔ (ii). Note that the quotient field of C [x, x−1 ] is C (x). 3. On proper epimorphisms of Z-graded rings Corollary 2.5(ii) motivates the following extension of the notion of proper epimorphisms from polynomial rings to Z-graded rings. Let R be a Z-graded ring and P a non-zero prime ring. A ring epimorphism f : R → P is called proper if (Kerf )h = 0, i.e., f (a) ̸= 0 for any non-zero a ∈ H (R). Note that if f : R → P is a proper epimorphism of a Z-graded ring R onto a non-zero prime ring P, then R is also a prime ring. Moreover, if I is a non-zero homogeneous ideal of R, then f (I ) is a non-zero ideal of P and hence is itself a prime ring, and the restriction of f to I is also a proper epimorphism of I onto f (I ). Applying Proposition 2.4(ii) one verifies readily that a ring epimorphism f of A[x, x−1 ], regarded as Z-graded ring, onto a prime ring P is proper if and only if Ker f is an A-disjoint ideal of A[x, x−1 ]. The following are some variants of Problem 1.5 asked for Z-graded rings. Problem 3.1. Let R be a Z-graded ring. (i) Suppose that there exists a proper epimorphism of R onto some P ∈ P . Must R contain a non-zero central element? (ii) Suppose that there exists a proper epimorphism of R onto some P ∈ P 1 . Must R contain a non-zero central element? (iii) Suppose that there exists a proper epimorphism of R onto a simple ring P with unity. Must R contain a non-zero central element? Note that if f : R → P is a proper epimorphism of a Z-graded ring R onto some P ∈ P , then R is also in P provided that Problem 3.1(i) has a positive solution. We will get some partial answers to these questions. Let R be a Z-graded ring and let f : R → P be a proper epimorphism of R onto a ring P ∈ P . Note that if I is a nonzero homogeneous ideal of R, then f (I ) is a non-zero ideal of P, so it contains a non-zero central element of P. Define lI = min{l(r ) | r ∈ I , 0 ̸= f (r ) ∈ Z (P )}. First we answer Problem 3.1(i) in the affirmative under some additional condition. Theorem 3.2. Let R be a Z-graded ring and f : R → P be a proper epimorphism of R onto a ring P ∈ P . If there is a upper bound for all lI , where I is a non-zero homogeneous ideals I of R, then R contains a non-zero central element.
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
327
Proof. Let M be a homogeneous ideal of R with maximal lM . Then f induces a proper epimorphism of M onto a ring f (M ) which is in P by Proposition 2.6(ii). Thus lJ is defined for non-zero homogeneous ideals J of M. Note that lJ remains unchanged when a non-zero homogeneous ideal J of R contained in M is considered as an ideal of M. We claim that lJ = lM for every non-zero homogeneous ideal J of M. Indeed let J¯ = R∗ JR∗ be the ideal of R generated by J which is clearly homogeneous. Since R is a prime ring, J¯3 is also a non-zero homogeneous ideal of R and is contained in J. Thus lJ¯3 ≥ lJ ≥ lJ¯ ≥ lM and the maximality of lM proves the claim. Hence, we may replace R by M and assume that lI = k for every non-zero homogeneous ideal I of R. Let r be an element of R with l(r ) = k such that f (r ) is a non-zero central element of P. Write r = rm + · · · + rn , where rm ̸= 0 ̸= rn and n − m = k. We claim that rn is central. If not, then there is a homogeneous element a ∈ R such that b = rn a − arn ̸= 0. Let I be the ideal of R generated by b. It contains an element u with l(u) = k such that f (u) is a non-zero central element of P. Write u = ut + · · · + us , where s − t = k = n − m. Then u = xi byi for some homogeneous elements s xi , yi from R∗ with δ(xi ) + δ(yi ) = s − δ(b). Then v = xi (ra − ar )yi = vt + · · · + vs , where vj = xi (rj+n−s a − arj+n−s )yi and in particular vs = us . Since f (r ) is central, we get f (v) = 0 and so f (u − v) = f (u) is a non-zero central element of P but l(u − v) < l(u) = k, a contradiction. Hence, rn is a non-zero central element of R. Next we answer to Problem 3.1(ii) in the affirmative for positively graded rings. Theorem 3.3. Let R be a positively graded ring and f : R → P be a proper epimorphism of R onto a ring P ∈ P 1 . Then R contains a non-zero central element. Proof. By Theorem 3.2 it suffices to show that there is a common upper bound for all lI , where I is a non-zero homogeneous ideal of R. Let r = r1 + r2 + · · · + rn be an element of degree n such that f (r ) = 1. Suppose that I = i≥1 Ii is a non-zero homogeneous ideal of R. Let u = uk + uk+1 + · · · + um be an element of I with length m − k = lI such that f (u) is a non-zero central element of P. Let u∗ = uk r + uk+1 + · · · + um . Clearly u∗ ∈ Ik+1 + Ik+2 + · · · and f (u∗ ) = f (u). Consequently l(u∗ ) ≥ lI = m − k. However δ(u∗ ) ≤ max{m, k + n}, so we have m − k ≤ l(u∗ ) ≤ max{m − k − 1, n − 1}, which implies that lI ≤ l(u∗ ) ≤ n − 1 and we are done. Corollary 3.4. Let A be a ring. If there is a proper epimorphism of xA[x] onto a ring in P 1 , then A is a prime ring containing a non-zero central element. 4. Diagonal embedding and its application to problems on proper epimorphisms Let B be a ring with unity and A its subring sharing the unity. The extension A ⊆ B is called centralizing if there are central elements b1 = 1, b2 , . . . , bn of B such that B = Ab1 + Ab2 + · · · + Abn . Lemma 4.1. Let A ⊆ B be a centralizing extension. (i) [9, Theorem 10.2.4] If Q is a prime ideal of B, then Q ∩ A is a prime ideal of A. (ii) [9, Theorem 10.2.9] For every prime ideal P of A there is a prime ideal Q of B such that P = Q ∩ A. Note that for t = x + x−1 ∈ A[x, x−1 ], we have x2 − tx + 1 = 0, which shows that if A is a ring with unity, then A[t ] ⊆ A[x, x−1 ] is a centralizing extension. Moreover, A[t ] ≃ A[x] in the canonical way. Consequently Lemma 4.1 can be applied to extend some results on prime ideals or radicals from polynomial rings to Laurent polynomial rings. For the Jacobson and Brown–McCoy radicals it was done in [6,7]. Unfortunately such a nice method cannot be applied to extend results known for positively (or non-negatively) graded rings to Z-graded rings. There is however a method, which allows one to apply the above idea sometimes. Namely one can embed a given Z-graded ring R into R[x, x−1 ], translate the studied properties of R to properties of R[x, x−1 ] and then apply the above described method to reduce the studied property to the polynomial ring R[x] and next relate it back to R. Using that method we will show in this section that Problems 1.5 and 3.1(i) are equivalent. i Let R be a Z-graded ring. The map f : R → R[x, x−1 ] defined by f ( ri ) = ri x is a ring embedding. It will be called the diagonal embedding. Obviously, if R is non-negatively (positively) graded, then f is an embedding of R into R[x] (into xR[x]). Lemma 4.2. Let R be a Z-graded ring with unity, f : R → R[x, x−1 ] the diagonal embedding, I an ideal of R such that Ih = 0 and I the ideal of R[x, x−1 ] generated by f (I ). Then (i) I = f (I )xi ; (ii) R ∩ I = 0; (iii) f (R) ∩ I = f (I ); (iv) R[x, x−1 ] = f (R)xi and (R/I )[x, x−1 ] ≃ R[x, x−1 ]/I . Proof. The properties (i) and (ii) are particular cases of [12, Lemma 3]. k j (iii) The inclusion ⊇ is clear. Suppose that i ri xi ∈ f (R) belongs to I, where ri ∈ Ri . Then, by (i), i ri xi = j( k ajk x )x ,
i i( j+k=i ajk )x , so j+k=i ajk = ri ∈ Ri for each i, implying i ajk = 0 unless k = i and j = 0. Consequently, i ri xi = a x = f ( a ) ∈ f ( I ) . i 0i i 0i (iv) For aj ∈ Rj we have aj xk = aj xj xk−j = f (aj )xk−j ∈ f (R)xi and so R[x, x−1 ] = f (R)xi . Thus the map ψ : −1 −1 i i R[x, x ] → R[x, x ] defined by the rule ψ( i ri x )= so is π ◦ ψ : R[x, x−1 ] → i f (ri )x is a ring epimorphism and −1 −1 −1 R[x, x ]/I, where π is the canonical epimorphism of R[x, x ] onto R[x, x ]/I. Suppose that i ri xi ∈ Ker (π ◦ ψ). Then
where ajk ∈ Rk such that
k
ajk ∈ I for each j. Now
i ri x
i
=
328
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
j i j i j aij , si = j bij where aij , bij ∈ Rj . Then i( j aij x )x = i( j bij x )x , k k or equivalently, k( j ak−j,j )x = k( j bk−j,j )x , so j ak−j,j = j bk−j,j for all k, implying j aij = j bij for all i, j. Hence, ri = si ∈ I for all i. This implies that Ker (π ◦ ψ) ⊆ I [x, x−1 ]. The opposite inclusion is clear. Consequently R[x, x−1 ]/I ≃ R[x, x−1 ]/I [x, x−1 ] ≃ (R/I )[x, x−1 ] and we are done.
i
f (ri )xi =
i
f (si )xi for some si ∈ I. Write ri =
Now we will prove that Problems 1.5 and 3.1(i) are equivalent. It is a consequence of the following result. Theorem 4.3. Let R be a Z-graded ring. If R contains an ideal I with Ih = 0 such that R/I ∈ P , then the polynomial ring R[t ] contains an R-disjoint ideal M such that R[t ]/M ∈ P . Proof. Replacing R by R∗ and I by an ideal of R∗ maximal with respect to I ′ ∩ R = I, we may assume that R is a ring with unity, because Ih′ = 0 and R∗ /I ′ ∈ P by Proposition 2.6(iii) since R∗ /I ′ contains an essential ideal (R + I ′ )/I ′ ≃ R/I ∈ P . Moreover, if we obtain an ideal M ′ in R∗ [t ] such that M ′ ∩ R∗ = 0 and R∗ [t ]/M ′ ∈ P , then M = M ′ ∩ R[t ] will satisfy the desired properties: M ∩ R = M ′ ∩ R[t ] ∩ R = M ′ ∩ R = 0 and R[t ]/M = R[t ]/(M ′ ∩ R[t ]) ≃ (R[t ] + M ′ )/M ′ ∈ P by Proposition 2.6(ii) since (R[t ] + M ′ )/M ′ is a non-zero ideal of R∗ [t ]/M ′ . Thus assume that R is a ring with unity. Let f be the diagonal embedding of R into R[x, x−1 ] and I the ideal of R[x, x−1 ] generated by f (I ). Since R/I is a prime ring, so is (R/I )[x, x−1 ]. Hence it follows from Lemma 4.2(iv) that I is a prime ideal of R[x, x−1 ]. Note that we may assume that t = x + x−1 ; then R[t ] ⊆ R[x, x−1 ] is a centralizing extension. Thus M = I ∩ R[t ] is a prime ideal of R[t ] by Lemma 4.1(i) and M ∩ R = I ∩ R[t ] ∩ R = I ∩ R = 0 by Lemma 4.2(ii). Then A = (R[t ] + I )/I ≃ R[t ]/M is a prime ring and A ⊆ B = R[x, x−1 ]/I is a centralizing extension. By Lemma 4.2(iv) there is an isomorphism (R/I )[x, x−1 ] ≃ B and so B ∈ P by Proposition 2.7. Moreover, this isomorphism maps Z (R/I )[x, x−1 ] and Z (R/I )[x + x−1 ] onto Z (B) and Z = Z (A) respectively. Since Z (R/I )[x + x−1 ] ⊆ Z (R/I )[x, x−1 ] is an integral extension, so is the extension Z (A) ⊆ Z (B). Consider the localizations AZ −1 and BZ −1 of A and B at Z \ 0. Then AZ −1 ⊆ BZ −1 is a centralizing extension with Z (AZ −1 ) = ZZ −1 and Z (BZ −1 ) = Z (B)Z −1 . Since Z ⊆ Z (B) is an integral extension, so is the extension ZZ −1 ⊆ Z (B)Z −1 . However, ZZ −1 is a field, so Z (B)Z −1 is also a field and hence BZ −1 = BZ (B)−1 which is a simple ring with unity by Proposition 2.6(i). If A ∈ / P , then A would contain a non-zero ideal P such that P ∩ Z = 0. Taking P to be an ideal of R which is maximal with respect to P ∩ Z = 0, we may assume that P is a prime ideal of A. By Lemma 4.1(ii), there exists a prime ideal Q of B such that Q ∩ A = P. From the simplicity of BZ −1 it follows that QZ −1 = BZ −1 and so P ∩ Z ̸= 0, a contradiction. Therefore, R[t ]/M ≃ A ∈ P and so M is as required. Corollary 4.4. Problems 1.5 and 3.1(i) are equivalent. Proof. It is clear that if Problem 3.1(i) has a positive solution, then Problem 1.5 also has a positive solution. We will show the converse. Let R be a Z-graded ring and f : R → P be a proper epimorphism for some P ∈ P . Applying Theorem 4.3 we get that R[x] contains an R-disjoint ideal M such that R[x]/M ∈ P . If xR[x] ⊆ M, then R[x]/M = (R + M )/M ≃ R, so R ∈ P and Z (R) ̸= 0. If xR[x] ̸⊆ M, then by Corollary 2.5 the canonical epimorphism of R[x] onto R[x]/M is proper. Hence a positive solution of Problem 1.5 gives that Z (R) ̸= 0, so Problem 3.1(i) has a positive solution. 5. On the Brown–McCoy and S radicals of Z-graded rings We start with introducing a Z-graded analog of the radical S and obtaining some relations among this radical, the Brown– McCoy radical and the radical S for Z-graded rings. Denote by Ph the class of non-zero Z-graded prime rings P such that I ∩ Z (P ) ̸= 0 for every non-zero homogeneous ideal I of P, and for a given Z-graded ring R define Sh (R) as the intersection of all homogeneous ideals I of R such that R/I ∈ Ph . We will examine these notions for polynomial rings and Laurent polynomial rings regarded as Z-graded rings in the standard way. Proposition 5.1. (i) Let I be a homogeneous ideal of A[x] such that xA[x] ̸⊆ I. Then A[x]/I ∈ Ph if and only if I = I0 [x] for some ideal I0 of A such that A/I0 ∈ P . (ii) Let I be a homogeneous ideal of A[x, x−1 ]. Then A[x, x−1 ]/I ∈ Ph if and only if I = I0 [x, x−1 ] for an ideal I0 of A such that A/I0 ∈ P . Proof. (i) Assume first that A[x]/I ∈ Ph . Applying Proposition 2.4(i) we get that I = I0 [x] for some prime ideal I0 of A. For every ideal K of A which contains I0 strictly, K [x]/I0 [x] is a non-zero homogeneous ideal of A[x]/I0 [x], so it contains a non-zero central element kxn + I0 [x] of A[x]/I0 [x]. It is clear that k + I0 is a non-zero element in K /I0 belonging to the center of A/I0 . Hence A/I0 ∈ P . Conversely, assume that I = I0 [x] and A/I0 ∈ P . Then I0 [x] is a homogeneous prime ideal of A[x]. Now, if J is a homogeneous ideal of A[x] which contains I0 [x] strictly, then there is an n such that Jn = {a ∈ A | axn ∈ J } is an ideal of A which contains I0 strictly. Since A/I0 ∈ P , we get that for some k ∈ Jn , 0 ̸= k + I0 belongs to the center of A/I0 . It is clear that 0 ̸= kxn + I0 [x] ∈ J /I0 [x] belongs to the center of J /I0 [x]. Hence A[x]/I0 [x] ∈ Ph . (ii) The proof is analogous to that of (i). Corollary 5.2. Sh (A[x]) = S (A)[x] and Sh (A[x, x−1 ]) = S (A)[x, x−1 ] for every ring A.
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
329
Proof. If I is a homogeneous ideal I of A[x] such that xA[x] ⊆ I, then I = I0 + xA[x] for some ideal I0 of A. It is clear that A[x]/I ∈ Ph if and only if A/I0 ∈ P and, by Proposition 5.1(i), if and only if A[x]/I0 [x] ∈ Ph . Now the result is a consequence of Proposition 5.1(i). The result for A[x, x−1 ] follows directly from Proposition 5.1(ii). This corollary shows that Sh can be treated, for Z-graded rings, as a natural counterpart of the radical S. From Corollary 5.2 it follows that Problem 1.2 can be restated as whether S (A[x]) = Sh (A[x]) for every ring A. So one can also ask the following extension of Problem 1.2. Problem 5.3. Is S (R) = Sh (R) for every Z-graded ring R? The inclusion S (R) ⊆ Sh (R) is verified in the next proposition. Proposition 5.4. Let R be a Z-graded ring. Then (i) R ̸= Sh (R) if and only if there is a homogeneous ideal I of R such that R/I contains a non-nilpotent central element. (ii) S (R) is a homogeneous ideal of R and S (R) ⊆ Sh (R). (iii) If Problem 3.1(i) has a positive solution, then S (R) = Sh (R), i.e., Problem 5.3 also has a positive solution. Proof. (i) The ‘‘only if’’ implication is an immediate consequence of the definition of Sh (R). To get the ‘‘if’’ implication suppose that I is a homogeneous ideal of R and R/I contains a non-nilpotent central element a + I. Applying Zorn’s Lemma one can find a homogeneous ideal M of R which is maximal among the homogeneous ideals of R containing I such that an ̸∈ M for all n = 1, 2, . . .. Then R/M ∈ Ph . (ii) Let R = i∈Z Ri ; then R[x] = i∈Z Ri [x] gives a Z-grading of R[x]. By Theorem 1.1 we have S (R)[x] = G(R[x]). From [5, Corollary 6] it follows that G ( R [ x ]) is homogeneous with respect to the above gradation. Consequently S (R)[x] = i∈Z (Ri [x] ∩ S (R)[x]) and so S (R) = i∈Z (Ri ∩ S (R)). This shows that S (R) is homogeneous. Let I be a homogeneous ideal of R such that R/I ∈ Ph . Then R/I is a Z-graded prime ring. Take any non-zero ideal J /I of R/I. Let r = rm + · · · + rn be a non-zero element of J /I of the smallest length, where ri ∈ (R/I )i and rn ̸= 0. Since R/I ∈ Ph , the ideal of R/I generated by rn contains a non-zero homogeneous central element c = at rn bt , where at , bt are homogeneous elements of R∗ /I. Clearly r ′ = at rbt ∈ J /I. The length of r ′ is equal to that of r and its leading component is equal to c. For every homogeneous element a ∈ R, we have ar ′ − r ′ a ∈ J /I. Since ac = ca, the length of ar ′ − r ′ a is smaller than that of r. Consequently ar ′ − r ′ a = 0, so r ′ is a non-zero element of J /I belonging to the center of R/I. Hence R/I ∈ P which implies that S (R) ⊆ Sh (R). (iii) Let I be an ideal of R such that R/I ∈ P . The canonical epimorphism f of the Z-graded ring R/Ih onto R/I is proper. Hence for every non-zero homogeneous ideal K of R/Ih the restriction of f to K is a proper epimorphism of the Z-graded ring K onto f (K ) ∈ P . Assuming that Problem 3.1(i) has a positive solution we get that 0 ̸= Z (K ) ⊆ Z (R/Ih ). Consequently R/Ih ∈ Ph , so Sh (R) ⊆ Ih ⊆ I. Hence Sh (R) ⊆ S (R) and so Sh (R) = S (R) by (ii). That is, Problem 5.3 has a positive solution. Corollary 5.5. (i) S (A[x]) ⊆ S (A)[x] for every ring A. (ii) If Problem 1.5 has a positive solution, then S (A[x]) = S (A)[x] for every ring A, i.e., Problem 1.2 has a positive solution. Proof. (i) An immediate consequence of Corollary 5.2 and Proposition 5.4(ii). (ii) Suppose that I is a prime ideal of A[x] such that A[x]/I ∈ P . If xA[x] ⊆ I, then I = I0 + xA[x] for an ideal I0 of A and A/I0 ∈ P . If xA[x] ̸⊆ I, then Ih = I0 [x] for some ideal I0 of A by Proposition 2.4(i). By Corollary 2.5, the canonical epimorphism of (A/I0 )[x] onto A[x]/I is proper. Hence assuming a positive solution of Problem 1.5 (and so of Problem 3.1 (i) by Corollary 4.4) we get that (A/I0 )[x] ∈ P and so A/I0 ∈ P by Proposition 2.7. In either case, we have always that S (A)[x] ⊆ I0 [x] ⊆ I. Consequently S (A)[x] ⊆ S (A[x]). Theorem 5.6. Let R be a Z-graded ring. (i) If R ∈ Ph and G(R) ̸= 0, then Z (G(R)) ⊆ R0 and S (R0 ) ̸= R0 . (ii) If S (R0 ) = R0 , then G(R) ⊆ Sh (R). (iii) If R is positively graded, then G(R) = Sh (R). Proof. (i) By [5, Corollary 6] G(R) is a homogeneous ideal of R. Since R ∈ Ph and G(R) ̸= 0, Z (G(R)) is a non-zero Z-graded ring and it contains no non-zero nilpotent elements. By Lemma 2.1, Z (G(R)) is Jacobson radical, so its i-components are nil for all i ̸= 0 by [10, Theorem 22.6]. Consequently 0 ̸= Z (G(R)) ⊆ R0 . Hence R0 contains non-nilpotent central elements, so R0 is not S-radical. (ii) Let I be a homogeneous ideal of R such that R/I ∈ Ph . Since R0 is S-radical, so is (R/I )0 = (R0 + I )/I ≃ R0 /(R0 ∩ I ). Applying (i) we get that G(R/I ) = 0 and so G(R) ⊆ I. Hence G(R) ⊆ Sh (R). (iii) Applying (ii) we get that G(R) ⊆ Sh (R) since R0 = 0. Conversely let M be a maximal ideal of R such that R/M has a unity. Then R/Mh is a prime ring by Corollary 2.3 and the canonical epimorphism f of R/Mh onto R/M is proper. Since R/M is simple, the restriction of f to any homogeneous ideal K of R/Mh onto R/M is also proper. Thus K has a non-zero center by Theorem 3.3. This implies that R/Mh ∈ Ph . Hence Sh (R) ⊆ Mh ⊆ M. Consequently Sh (R) ⊆ G(R). From Corollary 5.2 it follows that Theorem 1.1(ii) can be restated as G(A[x]) = Sh (A[x]) for every ring A. Hence Theorem 5.6 can be treated as a generalization of Theorem 1.1(ii) to positively graded rings. Now we will give an example showing that it does not extend to non-negatively graded rings, i.e., that there is a non-negatively graded ring R such that G(R) ̸= Sh (R).
330
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
Example. Let K be a commutative local domain with non-zero Jacobson radical J and F the field of quotients of K . Let A be a simple F -algebra with unity, which is a domain but not a division algebra. As a specific example one can take K = ZS −1 , the localization of Z at the set S of odd integers so that F is the field Q of rational numbers and take A to be the Weyl algebra over Q. Let a be a non-invertible element of A. It is easy to see that B = Aa is a simple F -algebra without unity. Applying Corollary 3.4 we get that xB[x] is Brown–McCoy radical. Set R = J + xB[x]; then xB[x] is an ideal of R such that R/(xB[x]) ≃ J is a commutative Jacobson radical ring and so a Brown–McCoy radical ring. Hence R is itself a Brown–McCoy radical ring. Obviously R is a homogeneous subring of A[x] graded in the standard way, so R is non-negatively graded. Now xB[x] is a homogeneous ideal of R and R/(xB[x]) ∈ Ph , so Sh (R) ⊆ xB[x] ̸= R = G(R). Note that if a Z-graded ring R is graded-nil, i.e., H (R) consists of nilpotent elements, then Z (R) is a nil ring. Hence, R/I ̸∈ Ph for any homogeneous ideal I of R, that is, R = Sh (R). Moreover, if R is positively graded, then R = G(R) by Theorem 5.6(iii). Now we extend the latter result to Z-graded rings. Theorem 5.7. Every Z-graded ring which is graded-nil is Brown–McCoy radical. Proof. Assume that the result does not hold. Then there exist a Z-graded ring R, which is graded nil, and a ring epimorphism f : R → P, where P is a simple ring with unity 1. Let r ∈ R and f (r ) = 1. We claim that supp(r ) contains a negative integer. Indeed, if not, then f ( k≥0 Rk ) is a ring with unity, so replacing R by k≥0 Rk we can assume that R is non-negatively graded. Since R+ = k≥1 Rk is an ideal of R, R0 is nil and P is a simple ring with unity, we obtain that f (R+ ) = P. Now R+ /(R+ ∩ (Ker f )h ) is a graded-nil positively graded ring and f induces an epimorphism f : R+ /(R+ ∩ (Ker f )h ) → P defined by f (r ) = f (r ) for r = r + R+ ∩ (Kerf )h . Clearly, f is a proper epimorphism and so R+ /(R+ ∩ (Ker f )h ) contains a non-zero central element by Theorem 3.3. Hence R+ /(R+ ∩ (Ker f )h ) is a prime ring containing a non-zero homogeneous central element which is nilpotent, a contradiction. Consequently for every r ∈ R such that f (r ) = 1, the minimal integer m in supp(r ) is negative. Let us choose R, P, f and r such that m is the largest possible and the index of nilpotency of rm is the smallest possible. Note that supp(r ) contains an odd integer, as otherwise r ∈ R = k∈Z R2k and R is Z-graded with components Rk = R2k and in that grading the minimal integer in supp(r ) is m/2, which contradicts the maximality of m. Let r = a + b where a is the sum of components of r with odd degrees and b is the sum of other components. Then f (a)+ f (b) = 1 and f (a)2 = 1 − 2f (b)+ f (b)2 . Consequently f (t ) = 1 where t = a2 + 2b − b2 . Note that t ∈ R and the m-th component tm of t 2 in the grading of R considered above is equal to rm and all the components of t with degrees < m are equal to 0. Hence by the minimality of the nilpotency index of rm , we get that tm = 0. Thus the minimal integer in supp(t ) is > m, a contradiction. The converse of Theorem 5.7 does not hold. Indeed from Corollary 3.4 it follows that if A is a simple domain without unity, then the ring xA[x] is Brown–McCoy radical. Note also that the homogeneous subrings of a Brown–McCoy radical Z-graded ring R need not be necessarily Brown–McCoy radical. For instance, let B be the subring of the above-mentioned A generated by a non-zero element. Since B is not nil, xB[x] is not Jacobson radical. But xB[x] is a commutative ring, so it is not Brown– McCoy radical either. However the following slightly modified version of Theorem 5.7 gives already a characterization of graded-nil Z-graded rings. Proposition 5.8. Every homogeneous subring of a Z-graded ring R is Brown–McCoy radical if and only if R is graded-nil. Proof. The ‘‘if’’ part is an immediate consequence of Theorem 5.7. Now we will prove the opposite implication. Let r ∈ Ri , i ̸= 0; then the commutative homogeneous subring ⟨r ⟩ of R generated by r is Brown–McCoy radical by assumption and so is Jacobson radical. By [10, Theorem 22.6] each component of ⟨r ⟩ is nil. In particular, r is nilpotent. Assume now that r ∈ R0 . If r is not nilpotent, then there exists a prime ideal P of ⟨r ⟩ such that r n ̸∈ P for all n > 0. Then U = ⟨r ⟩/P is a commutative domain and so is U ∗ . By assumption, all the subrings of ⟨r ⟩ ⊆ R0 are Brown–McCoy radical since they are homogeneous. Hence every subring of U is also Brown–McCoy radical and so Jacobson radical. Consider the unital epimorphism f : Z[x] → U ∗ such that f (x) = r + P. Then f (Z) is either a finite field or a ring isomorphic to Z since it is a domain. Note that Ker f is not zero; otherwise xZ[x] ≃ U would be Jacobson radical, a contradiction. Take a non-zero polynomial w(x) ∈ Ker f and write w(x) = xn v(x), where v(x) = a0 + a1 x + · · · + am xm ∈ Z[x] with a0 ̸= 0. Since Ker f is a prime ideal of Z[x] and xn ̸∈ Ker f , we have v(x) ∈ Ker f . Consequently a0 ∈ K = Z ∩ (xZ[x] + Ker f ) ̸= 0. Since f (K ) is a subring of f (xZ[x]) = U, it is Jacobson radical. On the other hand, f (K ) is an ideal of f (Z), a ring containing no non-zero Jacobson radical ideals, so f (K ) = 0. From K = K ∩ Ker f = Z ∩ Ker f it follows that Z/K ≃ f (Z). Thus the proper homomorphic image Z/K of Z is a domain and so is a finite field. Now K [x] ⊆ Ker f , so (Z/K )[x] ≃ Z[x]/K [x] can be mapped homomorphically onto U ∗ . This implies that either U ∗ ≃ F [x] or U ∗ ≃ F for some finite field F . In either case, no non-zero subring of U ∗ = U can be Jacobson radical, a contradiction. As we have seen above, a Z-graded ring R which is Brown–McCoy radical need not be graded-nil. However, from the Proof of Proposition 5.8, if R is Jacobson radical and, in addition, R0 is nil, then R is graded-nil. Thus all homogeneous subrings of R are Brown–McCoy radical. It is natural to ask whether they must also be Jacobson radical. We do not know the answer yet in the general case. However, it is true at least for non-negatively graded rings. Since R0 is nil it suffices to prove that for positively graded rings. Theorem 5.9. If T is a homogeneous subring of a positively graded ring R, then J (R) ∩ T ⊆ J (T ). In particular, if R is Jacobson radical, then so is T .
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
331
Proof. It is enough to show that if r ∈ T and in R1 , then (1 − r )−1 ∈ T 1 . Let f : R1 → R1 [x] be 1 − r is invertible 1 −1 −1 −1 the diagonal embedding. Write (1 − r ) = ri where ri ∈ Ri ; then f (1 − r ) = f ((1 − r ) ) = ri xi and it suf−1 1 1 fices to show that f (1 − r ) ∈ T [x]. Note that f (r ) ∈ T [x] since T is a homogeneous subring of R, so f (1 − r )−1 = (1 − f (r ))−1 = 1 + f (r ) + f (r )2 + · · · ∈ T 1 [[x]]. On the other hand, f (1 − r )−1 = f ((1 − r )−1 ) ∈ R1 [x], so f (1 − r )−1 ∈ R1 [x] ∩ T 1 [[x]] = T 1 [x] and we are done. Applying Proposition 5.8 and Theorem 5.9 one gets the following characterization of non-negatively graded rings in which all homogeneous subrings are Jacobson radical. Corollary 5.10. Every homogeneous subring of a non-negatively graded ring R is Jacobson radical if and only if R0 is a nil ring and R+ = i≥1 Ri is a Jacobson radical ring. Remark 5.11. One could ask whether the conclusion of Theorem 5.9 holds for all subrings instead of only homogeneous subrings. This is not the case. In [18] it was shown that there is a Jacobson radical polynomial ring R[x], which is not nil. If the conclusion of Theorem 5.9 held for all subrings of R[x], then all of them would be Jacobson radical. Regarding R[x] as a trivially Z-graded ring and applying Corollary 5.10 we would get that R[x] is nil, a contradiction. In the Proof of Theorem 5.9 we applied the diagonal embedding but one could get it following the idea from the Proof of [11, Proposition 2]. However none of these methods applies to Z-graded rings. The following observations concern particular cases. Proposition 5.12. Suppose that A[x, x−1 ] is Jacobson radical and B is a subring of A. Then B[x, x−1 ] Jacobson radical. Proof. Since the Laurent polynomial ring A[x, x−1 ] is Jacobson radical, it follows from [8, Lemma 2.2] that the polynomial ring A[x] is Jacobson radical. In view of [6, Theorem 1] we have that the matrix rings Mn (A) are nil for all n ≥ 1. Hence Mn (B) are nil for all n ≥ 1, and by [6, Theorem 1] again B[x] is Jacobson radical. Applying [8, Lemma 2.2] once more we have that B[x, x−1 ] is Jacobson radical. Modifying slightly arguments from [3] we will show that graded-nil Z-graded algebras of a positive characteristic are S-radical. We need the following lemma proved in [3]. Lemma 5.13. Let A be an algebra of characteristic p > 0 and a0 , a1 , . . . , an ∈ A. If a = a0 + a1 + · · · + an is a central element of A, then ap is the sum of elements of the form ai1 · · · aip with i1 + · · · + ip ≡ 0 (mod p). Theorem 5.14. Every Z-graded algebra of characteristic p > 0, which is graded-nil, is S-radical. Proof. Assume on the contrary that there are Z-graded algebras of characteristic p, which are graded-nil but not S-radical. Then for each such an algebra R there is an epimorphism f : R → P with P ∈ P . Let r ∈ R be such an element that f (r ) is a non-zero central element of P. We can assume that the algebra R, the epimorphism f and the element r, are chosen in such a way leading component rn of that the length l(r ) of r is the smallest possible and among those the nilpotency index of the p p r = m≤i≤n ri is the smallest possible. Since f (r ) is central, it follows from Lemma 5.13 that f (r ) = f (r ) = f (r ) for some
r =
ri1 · · · rip =
i1 +···+ip ≡0 (mod p)
n
r i,
i=m
· · · rip ∈ Rip . Thus r ∈ k Rkp . The ring R = k Rkp is Z-graded with components Rk = Rkp . p The restriction of f to R maps it onto f (R) ∈ P and f (r ) = f (r ) is a non-zero central element of f (R). The length l(r ) of p r in this grading is ≤ n − m = l(r ) and the n-component r n of r in this grading is rn . The minimality of l(r ) gives that p l(r ) = l(r ). However, r n = rn would then be the leading component of r with nilpotency index smaller than that of rn , a
where r i =
i1 +···+ip =ip ri1
contradiction. We do not know whether Theorem 5.14 can be extended to all graded-nil Z-graded rings, so the following problem is open. Problem 5.15. Is S (R) = R for every graded-nil Z-graded ring R? In solving this problem the following result might be of some use. Proposition 5.16. (i) If R is a graded-nil Z-graded ring, which is not S-radical, then R contains a finitely generated homogeneous subring, which can be mapped homomorphically onto a ring P ∈ P such that Z (P ) is Jacobson radical. (ii) If R is a graded-nil positively graded ring, which is not S-radical, then R contains a homogeneous subring, which can be mapped homomorphically onto a Jacobson radical ring in P . Proof. Suppose that R is a graded-nil Z-graded ring, which is not S-radical; then there exist a P ′ ∈ P and an epimorphism f ′ : R → P ′ . Let r ∈ R be such that f ′ (r ) is a non-zero central element of P ′ and R′ be the subring of R generated by the
332
P.-H. Lee, E.R. Puczyłowski / Journal of Pure and Applied Algebra 218 (2014) 323–332
homogeneous components of r. Then R′ is a finitely generated, graded-nil, homogeneous subring of R. Since f ′ (R′ ) contains the non-nilpotent central element f ′ (r ), there exists an ideal M which is maximal with respect to not containing any power of f ′ (r ). Thus P = f ′ (R′ )/M ∈ P , so there is an epimorphism f : R′ → P. (i) By Theorem 5.7 we get that R′ is Brown–McCoy radical. Hence P is also Brown–McCoy radical, so Z (P ) is Jacobson radical by Lemma 2.1. Thus R′ is a subring of R as required. (ii) We are done if P is Jacobson radical. Thus suppose that P contains primitive ideals. If I is a primitive ideal of P, then J = f −1 (I ) is a primitive ideal of R′ since R′ /J ≃ P /I. By [16, Theorem 1] we get that J is a homogeneous ideal of R′ . Since R′ /J is a primitive ring and all homogeneous elements of R′ /J are nilpotent, we have Z (R′ /J ) = 0. Consequently Z (P /I ) = 0 and Z (P ) ⊆ I. Thus the Jacobson radical J (P ) of P contains Z (P ) and so J (P ) ̸= 0. Hence J (P ) ∈ P by Proposition 2.6(ii). Now f −1 (J (P )) = {f −1 (I ) | I primitive ideal of P } and all f −1 (I ) are homogeneous ideals of R. Consequently f −1 (J (P )) is a homogeneous subring of R. Thus f −1 (J (P )) is a subring of R as required. Acknowledgments We are grateful to the referee for informing us about a recent paper [17] by A. Smoktunowicz. The Proposition 0.1 in [17] coincides with our Theorem 5.9. This research was supported by the Polish National Center of Science Grant No DEC-2011/03/B/ST1/04893. References [1] E.P. Armendariz, G.F. Birkenmeier, J.K. Park, Ideal intrinsic extensions with connections to PI-rings, J. Pure Appl. Algebra 213 (2009) 1756–1776. [2] K.I. Beidar, Y. Fong, E.R. Puczyłowski, Polynomial rings over nil rings cannot be homomorphically mapped onto rings with nonzero idempotents, J. Algebra 238 (2001) 389–399. [3] M. Chebotar, W.-F. Ke, P.-H. Lee, E.R. Puczyłowski, A note on polynomial rings over nil rings, in: Modules and Comodules, in: Trends Math., Birkhäuser Verlag, Basel, 2008, pp. 169–172. [4] M. Ferrero, R. Wisbauer, Unitary strongly prime rings and related radicals, J. Pure Appl. Algebra 181 (2003) 209–226. [5] E. Jespers, E.R. Puczyłowski, The Jacobson and Brown–McCoy radicals of rings graded by free groups, Comm. Algebra 19 (1991) 551–558. [6] J. Krempa, Logical connections between some open problems concerning nil rings, Fund. Math. 76 (1972) 121–130. [7] J. Krempa, On radical properties of polynomial rings, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 20 (1972) 545–548. [8] J. Krempa, A. Sierpinska, The Jacobson radical of certain group and semigroup rings, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 26 (1978) 963–967. 1979. [9] J.C. McConnell, J.C. Robson, Noncommutative Noetherian Rings, John Wiley & Sons, Chichester, 1987. [10] D. Passman, Infinite Crossed Products, Academic Press, San Diego, 1989. [11] E.R. Puczyłowski, Radicals of polynomial rings in non-commutative indeterminates, Math. Z. 182 (1983) 63–67. [12] E.R. Puczyłowski, A note on graded algebras, Proc. Amer. Math. Soc. 113 (1991) 1–3. [13] E.R. Puczyłowski, A. Smoktunowicz, On maximal ideals and the Brown–McCoy radical of polynomial rings, Comm. Algebra 26 (1998) 2473–2482. [14] A. Smoktunowicz, R[x, y] is Brown–McCoy radical if R[x] is Jacobson radical, in: Proceedings of the Third International Algebra Conference, Tainan, 2002, Kluwer Acad. Publ., Dordrecht, 2003, pp. 235–240. [15] A. Smoktunowicz, On primitive ideals in polynomial rings over nil rings, Algebra Represent. Theory 8 (2005) 69–73. [16] A. Smoktunowicz, On primitive ideals in graded rings, Canad. Math. Bull. 51 (2008) 460–466. [17] A. Smoktunowicz, A note on nil and Jacobson radicals in graded rings. http://arxiv.org/abs/1301.2835. [18] A. Smoktunowicz, E.R. Puczyłowski, A polynomial ring that is Jacobson radical and not nil, Israel J. Math. 124 (2001) 317–325.