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О ПОЛНОМ РАДИКАЛЕ МОНОИДНЫХ КОЛЕЦ
Л. М. Мартынов
Омский государственный педагогический университет, г. Омск, Россия
Информация о статье
Дата поступления 31.03.2017
Дата принятия в печать 04.04.2017
Аннотация. Изучается задача нахождения полного радикала C(AM) полугруппового кольца AM, где A - ассоциативное кольцо, M - моноид. Доказано равенство C(AM) = = C(A)M для случая, когда моноид M имеет нетривиальный идеал и квадрат любого такого идеала отличен от самого идеала. В частности, отсюда следует, что полный радикал обладает сильным свойством Амицура С^х]) = С^^х].
Дата онлайн-размещения 15.07.2017
Ключевые слова
Ассоциативное кольцо, полный радикал, полугрупповое кольцо, полиномиальное кольцо
Автор выражает благодарность проф. В.А. Романькову за полезные советы при подготовке английской версии статьи
ON THE COMPLETE RADICAL OF A MONOID RING
L. M. Martynov
Omsk State Pedagogical University, Omsk, Russia
Article info Abstract. For an associative ring A and monoid M, we study the problem of finding the
Received complete radical C(AM) of the semigroup ring AM. For the case when M has a non-trivial
31.03.2017 ideal I, and I2 * I for any such I, we prove that C(AM) has the strong Amitsur property,
namely, C(A[xj) = C(A)[xj.
Accepted 04.04.2017
Available online 15.07.2017
Keywords
Associative ring, complete radical, semigroup ring, polynomial ring
Acknowledgements
The author is grateful to Professor V. A. Roman'kov for useful advices in preparing the English version of the article.
In the theory of Abelian groups, the concepts of a complete (divisible) and the reduced groups play an important role. In [1], we define the analogues of these
concepts for arbitrary algebras and formulate the main problems in their study. Remind that an algebra of a variety of algebras is called complete if it has no homo-
morphisms onto non-trivial algebra that belongs to a minimal subvariety of this variety. Algebra is called reduced if it does not contain a non-trivial complete subalgebra. The concepts of completeness and reducibility were studied for universal algebras, modules, groups, semigroups, monoassociative algebras (in particular, associative algebras and rings), monounary algebras and lattices. The survey of these researches can be found in [2].
Recall that according to Propositions 1 and 2 in [1], the class C of all complete algebras of any variety V is closed under homomorphic images and extensions in V. By Proposition 5 in [1], the class S of all reduced algebras of V is closed under subalgebras, Cartesian products, and extensions in V. As has been shown in [3], any algebra A from a variety V, that satisfies to the condition of transverbality on minimal subvarieties (see [4]), contains the greatest complete scatter C(i.e., disjoint family of subalgebras), which admits such congruence Pc that the factor algebra A/pc is reduced. Thus, the strict radical in the sense of Kurosh [5] can be defined for V. Here the class C is the radical class and the class S is the semisimple class.
In [3], it was noted that every variety of modules, groups and, in view of Remark to Theorem 2 in [6], associative, alternative, Jordan (%eA) and Lie algebras over arbitrary associative and commutative ring A with unit, is transverbal on minimal subvarieties. Thus, for each algebra A of the mentioned varieties, C(A) is generated by all complete subalgebras of A and coincides with the complete radical of A. This subalgebra, as has already been mentioned, is congruence-admitted and therefore in the case of groups it is a normal subgroup, and in the case of associative rings, it is two-sided ideal.
In the associative case, the complete radicals of some group or semigroup rings were found in [7] and [8]. The complete radical of the ring of all square matrices of given size over an associative ring is characterized in [9]. Note that this matrix ring is also semigroup ring over the corresponding semigroup of matrix units with zero.
In this paper, we study a problem of finding the complete radical C(AM) of the semigroup ring AM for an associative ring A and monoid M. Such semigroup rings we call further as monoid rings. We will call M as monoid without idempotent ideals, if M has a proper non-trivial two-sided ideal I and I2* I for every such I. Let we note that the polynomial ring A[x] is the monoid ring in the case when of the free monogenic monoid [x] = {1, x, x2, ...}, which is, obviously, monoid without idempotent ideals. Note that a free (commutative) mo-
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noid with any set of free generators and the monoid obtained by adding the external unit to a non-zero nilpotent semigroup are monoids without idempotent ideals. We can conclude that the considered problem belongs to the actual radicals theory of associative rings. Namely, studying of properties of various radicals of semigroup rings, in particular, polynomial rings, is one of actual directions in the associative algebras theory.
The main aim of this work is to prove equality C(AM) = C(A)M for each associative ring A and every monoid M without idempotent ideals. In particular, it will follow that the complete radical has the strong Amitsur property C(A[x]) = C(A)[x].
In the further of the paper "ring" means "associative ring" (not necessarily with unity element) and "ideal" means "two-sided ideal". Recall that radical R is called strict if the radical R(A) of a ring A contains every R-radical subring B (i.e., subring with the property R(B) = B) of A. As already mentioned above, the complete radical C is strict.
Note that existence of a strict radical R and a ring A for which R(A[x])) * R(A)[x] is shown in [10]. In addition, it follows from [7, Theorem 3] that similar inequality holds for the complete radical of a monoid ring when the monoid is a group.
We remind that the polynomial property of a radical R means that the radical of a polynomial ring is again a polynomial ring. i.e., R(A[x]) = B[x] for any ring A and some ring B. It is known (see, eg., [11, Proposition 1.1]) that the polynomial property of a radical R is equivalent to the equality R(A[x]) = (R(A[x]) n A)[x] for any ring A. Apparently, the first similar equality was proved byAmitsur [12, Theorem 1] for the Jacobson radical J. Therefore the polynomial property is called the Amitsur property. If for any ring A and radical R, the equality R(A[x]) = R(A)[x] is true, one say that R has the strong Amitsur property. Note that the Jacobson radical J is not strict radical (along with many other well-known radicals of associative rings) and does not satisfy to the strong Amitsur property.
Recall the definitions of some concepts applied to associative rings and let us agree on some notations. A ring is called complete if it has no homomorphisms on nonzero rings from atoms of the lattice ¿(Ass) of subvarieties of the variety Ass of all rings. If a ring has no complete non-zero subrings then it is called reduced. Note that only the zero ring is both complete and reduced. It is clear that any ring belonging to an atom of the lattice ¿(Ass) is reduced and any simple ring which does not belong to any atom of ¿(Ass) is complete.
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The additive group of a ring A is denoted as A+. The ring obtained from the additive group A+ of A by introducing zero multiplication is denoted by A0. The ring of residues of integers modulo n is denoted as Zn. The finite field of q elements is denoted as Fq. Recall that the orders of the finite fields can only be primery numbers of the form q = pn where p is a prime and n > 0.
For a system E of ring identities, varE denotes the corresponding variety. It is well known that the atoms of the lattice ¿(Ass) are given by the varieties Fp = var{px = 0, xp = x} and = var{px = 0, xy = 0}for all primes p. It is obvious that the varieties Fp and Zp are generated by the rings Fp and Zp0 , respectively.
For a variety V of rings, V(A) denotes V-verbal of the ring A, i.e., the smallest ideal of A such that the quotient of A by this ideal belongs to V. For varieties Zp, Fp and an arbitrary ring A, we indicate the following formulas to calculate the coresponding verbals: Zp(A) = pA + A2, Fp(A) = pA + Ap, where Ap is the ideal of A generated by all elements of the form an - a (a e A). It is easy to see that a ring A is complete if and only if Zp(A) = A and Fp(A) = A for any prime p. From the formulas above immediately follows that if A+ is a complete (divisible) group then the ring A is complete.
We remind that transverbility of the variety Ass on minimal subvarieties means that for any ring A, ideal I of A and atom A of the lattice ¿(Ass), the A-verbal A(I) is ideal of A.
It is clear that the complete radical C(A) of a ring A is equal to the sum of all complete subrings of A. A ring A is complete if and only if C(A) = A. In particular, the complete radical C(A) of any ring A is complete ideal of A. A ring A is reduced if and only if C(A) = О.
Note that the complete radical is, in some sense, antipode to the well-known Jacobson radical. For example, any nilpotent ring with reduced additive group is semisimple in our sense, but it is the radical ring in sense of Jacobson. On the other hand, any field which is not finite prime is radical ring in our sense, but it is semisimple in sense of Jacobson.
Following [13], we say that class K of rings is poly-nomially extensible if A £ K implies A[x] £ K. It follows from [13, Proposition 3.4] that if a radical R has the Amitsur property then the corresponding semisimple class S is polynomially extensible. As shown in [13, Corollary 3.8 (iii)], the Amitsur property of a radical in the general case does not depend on the property of polynomial extensibility.
Definitions of other commonly used concepts of the theory of radicals of rings and well-known charac-
terizations of radical and semisimple classes can be found in monographs [14; 15].
Recall that any nonzero element x of a monoid ring AM definitely planning in the form of a formal sum x = a01 + a1x1 + a2x2 +...+ akxk for some elements ai e A and xi e M (i = 0, 1, 2,..., k), where 1 is the unit of M, and if a0 = 0 then ai ^ 0 and xi^ 1 for i ^ 0. The elements a0, a1, a2, ..., ak of A are called coefficients of the x. The set A1 = {a1| a e A} is subring of AM, and A1 is isomorphic to A. Therefore, as usual, we assume that A is a subring of AM. Elements of A will be called constants for AM. The homomorphism ct: AM ^ A, a(a0 + + + ... + akxk) = a0 for any element x = a0 + a1x1+...+ akxk from AM, we call constant endomorphism of the ring AM.
The main result of this paper will follow from the next auxiliary assertions.
Lemma 1. For any ring A, the complete radical C(A) admits all endomorphism of A.
Proof. Let 9 be an endomorphism of A. Since C(A) is the complete ideal of A, <p(C(A)) is complete subring of A as homomorphic image of the complete ring C(A). But the complete radical contains all complete subrings of A and therefore <p(C(A)) c C(A).
Remark 1. A natural analogue of Lemma 1 is true for any strict radical.
Lemma 2. For any ring A and monoid M, the monoid ring AM is complete if and only if A is complete ring. In particular, the class C of all complete rings is closed with respect to construction of monoid rings.
Proof. If A is a complete ring then A is complete subring of AM and therefore A c C(AM). Since C(A) is ideal of AM, the inclusion AM c C(AM) and hence the equality AM = C(AM) are true.
Inversely, if AM is a complete ring then the ring A is complete as the endomorphic image of the complete ring AM under the constant endomorphism ct: AM ^ A.
Corollary 1. The class C of all complete rings is pol-ynomially extensible.
Note that the analogue to Corollary 1 is true for any strict radical [10, Proposition 3.1].
Lemma 3. For any ring A and monoid M the following two properties are equivalent:
(1) C(AM) = (C(AM) n A)M;
(2) C(AM) = BM for some ring B, i.e., the complete radical of a monoid ring is a monoid ring over the same monoid.
Proof. (1) ^ (2) obviously as one can take B = = C(AM) n A.
(2) ^ (1). Let C(AM) = BM for some ring B. Then the constants of C(AM) and BM coincide and therefore C(AM) n A = B. Hence, C(AM) = (C(AM) n A)M.
Note that an analogue of Lemma 3 for any polynomial ring A[x] and a radical R is proved in [11, Proposition 1.1].
A useful criterion for the polynomial property of radical R was obtained in [16] (see also [15], Theorem 4.9.22, p. 267): the radical R had the Amitsur property if and only if R(A[x]) n A = O implies R(A[x]) = O for any ring A. For a-Amitsur property, the analogous criterion is also true [17, Theorem 2.4]. Below we give a similar criterion for the complete radical of a monoid ring by adding two more assumptions.
The following statement is one of the main in the paper.
Proposition 1. For any ring A and monoid M, the following properties are equivalent:
(a) C(AM) = (C(AM) n A)M;
(b) C(AM) n A = O implies C(AM) = O;
(c) C(A) = O implies C(AM) = O;
(d) C(AM) = C(A)M.
Proof. (a) ^ (c). Suppose that C(A) = O, i.e., A is reduced ring. From (a) in view of completeness of the radical C(AM), we conclude the completeness of the ring (C(AM) n A)M. By Lemma 2 C(AM) n A is complete a ring. On the other hand, being a subring of the reduced ring A, C(AM) n A is reduced ring. This is possible only in the case when C(AM) n A = O. Hence in view of
(a), we obtain C(AM) = O. This means that (c) holds.
(c) ^ (b). Suppose that C(AM) n A = O. Since C(A) c C(AM) n A, we have C(A) = O. According to the condition (c), we have C(AM) = O. Thus, the condition
(b) is holds.
(b) ^ (a). First, we prove the inclusion
(C(AM)nA)M c C(AM). (1)
Denote I = C(AM) n A. In view of Lemma 1, endo-morphic image ct(C(AM), where ct: AM ^ A is constant endomorphism of the ring AM, coincides with I and therefore I is complete subring of A. By Lemma 2, IM is a complete subring of AM and therefore IM c C(AM), i.e., (1) holds.
Further, it is easy to understand that
C(AM)/IM = C(AM/IM) = C(BM), (2)
where B = A/I. Further, using, in particular (2), we have: C(BM) n B = (C(AM)/IM) n ((A + IM)/IM) =
= (C(AM) n (A + IM))/IM = = ((C(AM) n A)+IM)/IM = (I + IM)/ IM = O. (3)
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From relations (3), we obtain C(BM) n B = O. Therefore, in view of (b) C(BM) = O. Given (2), we obtain equality C(AM/IM) = C(AM)/IM = O. This implies the inclusion IM 3 C(AM), then
(C(AM) n A)M 3 C(AM). (4)
Now from (1) and (4) we get the required equality C(AM) = (C(AM) n A)M, i.e., (a) holds.
So, we have proved the equivalence of conditions (a), (b) and (c). It remains to prove the equivalence of (a) and (d).
(d) ^ (a) follows from Lemma 3.
(a) ^ (d). According to (a) C(AM) = = (A n C(AM))M and therefore C(AM) e C implies (A n C(AM))M e C. Hence by Lemma 2 (A n C(AM)) e C. As A n C(AM) is a complete ideal of the ring A, we have the inclusion
A n C(AM) ç C(A). (5)
On the other hand, in view of the obvious inclusions C(A) ç A and C(A) ç C(AM), we get the inclusion
C(A) ç A n C(AM). (6)
From the inclusions (5) and (6), we obtain now the equality A n C(AM) = C(A). Hence in view of the condition (a) C(AM) = (A n C(AM))M, we get the required equality C(AM) = C(A)M, i.e. the condition (d) holds.
Proposition 1 is proved.
Remark 2. Proposition 1 is valid for any strict radical.
Lemma 4. Additive group A+ of any reduced ring A is reduced.
Proof. Let A be a reduced ring. If to suppose that the group A+ is not reduced then it has non-zero complete (divisible) subgroup B. Then pB = B for any prime p. Consider the subring H of the ring A generated by B. Obviously, Zp(H) = pH + H2 3 B and Fp(H) = pH + Hp 3 B. As Zp(H) and Fp(H) are ideals of H, Zp(H) 3 H and Fp(H) 3 H. Hence Zp(H) = H and Fp(H) = H for any prime p. It means that H is nonzero complete subring of A. However, this contradicts to the reducibility of the ring A. Thus, the group A+ is reduced.
The following statement is of independent interest.
Proposition 2. If A is a reduced ring then AM is a reduced ring for any monoid M without idempotent ideals, i.e., the class S of all reduced rings is closed with respect to such monogenic rings.
Proof. Let A be a reduced ring. Suppose that the ring AM is not reduced. Then C = C(AM) is a nonzero complete ideal of the ring AM. Endomorphic image ct(C) = H of the ring C at the constant endomorphism
BecTHUK OMCKoro yHUBepckrreTa 2017. № 2(84). C. 8-13
ct: AM ^ A is a complete subring of the reduced ring A. Therefore, H is a zero subring of A. Thus, A n C = O.
Then the set T of all elements of the monoid M occurring in the recording elements of the ideal С does not contain the unit 1 of the monoid M and is therefore non-trivial ideal of the monoid M. Hence T2 ^ T and consequently C2 ^ C. We denote by N the set of all elements c from C, which is in the form of a linear combination c = ax + a2x2 +...+ akxk, there is an element t of T\ T2. It is easy to understand that the set I of all coefficients at t in the representations of the elements of the set N together with the zero of the ring A is an ideal of A. Since by Lemma 4 A+ is the reduced group, I + is its reduced non-zero subgroup. But then there exists such prime p that pI + ^ I + and therefore pI ^ I. Choose the element a e I \ pI and let x be an element of N with the coefficient a at t in the entry of x.
Consider now the ideal pC + C2 of C and show that an element x of C does not belong to pC + C2. In fact, if we assume that x = pg + h for some g e C and h e C2, it is easy to understand that the summand at in the representation of the element x must be equal to term of the form pyt from the representation as a linear combination of the element pg. Then the equation a = py is true for some ye I. This contradicts the choice of the element a. Hence pC + C2 ^ C and the factor ring C/(pC + C2) is a nonzero ring of the atom Zp of the lattice ¿(As). However, the latter means that the complete radical С is not a complete ideal of a ring AM, which is contradictory.
Thus, the assumption С = C(AM) is a nonzero ideal of AM led to contradiction. Therefore, C(AM) = O, i.e., AM is reduced ring.
Proposition 2 is proved.
Corollary 2. The class S of all reduced rings is pol-ynomially extensible.
The main result of this article is the following statement.
Theorem. For an associative ring A and monoid M without idempotent ideals, the equality C(AM) = C(A)M is true.
Proof. Sufficiently to note that according to Proposition 2, the property (c) of Proposition 1 holds and therefore the equivalent property (d) of Proposition 1 is true.
In the case AM = A[x] we get the following important statement.
Corollary 3. The complete radical has the strong Amitsur property.
Remark 3. Restrictions on the monoid M in the theorem are essential. If to refuse the requirement about the existence of non-trivial ideal in the monoid M then from [7, Theorem 3] it follows that the group ring FpG, where G is a non-Abelian finite p-group G, is not reduced while the prime finite field Fp is reduced. It is easy to see that a similar situation will be for the monoid M = G u 1, obtained from a group G attaching the external unit other than the unit of the group G. In this monoid M, the group G is a non-trivial ideal of M and G2 = G.
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ИНФОРМАЦИЯ ОБ АВТОРЕ
Мартынов Леонид Матвеевич - доктор физико-математических наук, профессор, профессор кафедры математики и методики обучения математике, Омский государственный педагогический университет, 644099, Россия, г. Омск, наб. им. Тухачевского, 14; e-mail: [email protected].
ДЛЯ ЦИТИРОВАНИЯ
Мартынов Л. М. О полном радикале моноидных колец // Вестн. Ом. ун-та. 2017. № 2 (84). С. 8-13. (На англ. яз.).
INFORMATION ABOUT THE AUTHOR
Martynov Leonid Matveevich - Doctor of Physical and Mathematical Sciences, Professor, Professor of the Chair of Mathematics and of Methods of Teaching mathematics, Omsk State Pedagogical University, 14, Naberezhnaya Tukchachevskogo, Omsk, 644099, Russia; e-mail: [email protected].
FOR CITATIONS
Martynov L.M. On the complete radical of a monoid ring. Vestnik Omskogo universiteta = Herald of Omsk University, 2017, no. 2 (84), pp. 8-13.
