In this section, we show that n-Amitsur rings do indeed exist and we also provide some characterizations. Moreover, we shall introduce new radicals, associated with these rings, and study some of their properties.
We shall require the following known result:
Proposition 2.1. ([6, 13])
If γ is a radical, then for any ring A.
For any radical γ, consider the radical class
defined by Tumurbat and Wisbauer in [13]. It is clear that, for any ring A and any radical if and only if for every radical γ.
Using the method of proof of ([4] , Proposition 4.9.18), we may show the following:
Proposition 2.2.
If γ is a hereditary radical, then for any ring A.
Proposition 2.3.
If A is an n-Amitsur (respectively, hereditary n-Amitsur) ring, then
for any radical (respectively, hereditary radical) γ.
Proof. We prove the result for the case when A is an n-Amitsur ring. Let γ be an arbitrary radical. We have . Notice that since , it is clear that . Therefore .
Now
Hence and so . Thus , whence .
The assertion now follows, taking into account that .
Let I be an ideal of a ring A. If there exists a radical (respectively, hereditary radical) γ such that , then I is called a radical ideal (respectively, h-radical ideal) of A. From Proposition 2.3, it is clear that if A is an n-Amitsur (respectively, hereditary n-Amitsur) ring, then is a radical ideal (respectively, h-radical ideal) of A, for any radical (respectively, hereditary radical) γ.
Proposition 2.4.
Let A be an n-Amitsur (respectively, hereditary n-Amitsur) ring. Then implies that , for any radical (respectively, hereditary radical) γ
Proof. Assume that A is an n-Amitsur ring and let γ be an arbitrary radical Suppose that . Since
it is clear that . Thus . Hence ; that is . The proof is similar when A is a hereditary n-Amitsur ring.
In what follows, for a radical γ and a ring A, we put
Lemma 2.5.
, for any ring A and any radical γ.
Proof. Suppose that there exists a radical γ and a ring A such that . Then there is a nonzero element . Hence there exists , which is a pre-image of . By Proposition 2.1, . Taking into account that radical classes are closed under extensions, we have . Consequently, and , which is a contradiction.
Theorem 2.6.
For a ring A, the following conditions are equivalent:
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(i) A is a n-Amitsur ring;
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(ii) for every radical γ;
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(iii) , where and H is any ideal of ;
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(iv) every radical ideal I of is a polynomial ring;
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(v) for every radical γ.
Proof. (ii) implies (i). If γ is a radical, then we know that . Since radical classes are closed under extensions, . Hence, if , then and so . Therefore A is a n-Amitsur ring.
(iii) implies (ii). Let γ be an arbitrary radical. Since , we take and . Then . Thus, if (iii) holds, then .
(i) implies (iii). This is clear.
(i) implies (iv). Let I be radical ideal of A[Xn]. Then for some radical γ. If A is a n-Amitsur ring, then and so I is a polynomial ring.
(iv) implies (i). Let γ be a radical such that , that is, I is a radical ideal of the ring A[Xn]. Then, by condition
(iv), I is a polynomial ring. Hence there exists such that . Now it is easy to see that
(i) implies (v). If A is an n-Amitsur ring, then (v) follows from Proposition 2.3.
(v) implies (i). Suppose that for any radical γ. Then
Taking into account Proposition 2.1, we have the desired result.
The next theorem may be proved in a similar way to the theorem above, and so we omit its proof. Recall that a subring I of a ring A is called an accessible subring of A if there exists a finite sequence of subrings such that .
Theorem 2.7.
For a ring A, the following statements are equivalent:
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(i) A is a hereditary n-Amitsur ring;
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(ii) for every hereditary radical γ;
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(iii) , where and H is any accessible subring of ;
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(iv) every h-radical ideal I of is a polynomial ring.
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(v) for every hereditary radical γ.
Proposition 2.8.
If F is a finite field, then F is not a hereditary n-Amitsur ring, for every natural number n.
Proof. If F is a finite field, then is a special radical. Now , where pm is the number of elements in F. Therefore and hence F is not a hereditary n-Amitsur ring.
Following the reasoning in the proof of Proposition 2.6 of [13], we may prove the next result.
Lemma 2.9.
Let Y be any subset of Xn of cardinality 1, where n>1. If A is a ring such that is an Amitsur ring (respectively, hereditary Amitsur ring), then implies that , for any radical (respectively, hereditary radical) γ.
Proof. Let γ be an arbitray radical and A be a ring such that is a an Amitsur ring. Assume that . Since is an Amitsur ring,
For any , there exist elements and such that
where, for each there exists an exponent such that
or a ∈ A. The number of nonzero summands of a is called the length of a and is denoted by ℓ (a).
Suppose that . For each , . Choose with minimal. If a depends on only one indeterminate, then the coefficients of a belong to and so a=0, which is a contradiction. Therefore a depends on at least two indeterminates, that is, . We can write
where and . If , take instead of a. The number of indeterminates in is less than the number of indeterminates in a. Continuing with this procedure, we can find such that either for some subset Y of Xn of cardinality 1, or . In the first case, , which is a contradiction. The second case contradicts the minimality of . Thus we conclude that , as desired.
Let n be a natural number. We recall that a class α of rings said to be n-polynomially extensible if for all . If n=1, the class α of rings is said to be polynomially extensible. It is clear that if a class α of rings is polynomially extensible, then it is n-polynomially extensible, for any natural number n.
For any ring A, let A0 be the zero-ring on the additive group of A, that is, A0 has the additive group of A and multiplication defined by ab=0 for all . Clearly, the class of all zero-rings is n-polynomially extensible, for any natural number n. Moreover, in [9], it was proved that every zero-ring is an Amitsur ring. We are now in a position to prove the following proposition.
Proposition 2.10.
For any ring A, the ring A0 is an n-Amitsur ring.
Proof. Let γ be an arbitrary radical. From Proposition 2.1,
Suppose that
Then . Hence, by the previous lemma, , which is a contradiction with Lemma 2.5.
The Baer radical class is the upper radical determined by the class of all prime rings and also coincides with the upper radical determined by the class of all semiprime rings. Rings belonging to the Baer radical class are called Baer radical rings. It is well known that the Baer radical class is n-polynomially extensible, for any natural number n. In addition, it was proved in [9] that every Baer radical ring is a hereditary Amitsur ring. Hence we have:
Proposition 2.11.
If A is a Baer radical ring, then A is a hereditary n-Amitsur ring.
Proof. is a Baer radical ring and therefore is also a Baer radical ring. If , then, by the previous lemma, . This is a contradiction with Lemma 2.5. Thus . Hence, by Theorem 2.7, A is a hereditary n-Amitsur ring.
The following lemma is required in the proof of the next proposition.
Lemma 2.12. ([8])
Let A be an infinite integral domain, and let be the polynomial ring over A with Λ commuting indeterminates. For every radical class γ, if and only if .
Proposition 2.13.
If A is an infinite integral domain whose every proper homomorphic image is a Baer radical ring, then A is a hereditary n-Amitsur ring.
Proof. Let A be an infinite integral domain without proper prime homomorphic images. Let γ be a hereditary radical and suppose that . By the lemma above, . We also know that . If , then we
have . By assumption, is a Baer
radical ring and hence and are Baer radical rings. Then is not a semiprime ring and so it has a nonzero ideal such that . Let be the ideal of generated by the
coefficients of . Therefore
since γ is hereditary Then, from the lemma above, , where . However, from Lemma 2.5, we have , which is a contradiction.
Example 2.14.([4])
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(i) The ring
is an infinite integral domain and every proper homomorphic image of W is a Baer radical ring. Therefore W is a hereditary n-Amitsur ring.
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(ii) Taking into account Lemma 2.12, it can be easily seen that any infinite field is an n-Amitsur ring.
Next, we introduce the following classes of rings, which rely on the concept of n-Amitsur ring:
is a ring whose nonzero prime homomorphic images are not hereditary n-Amitsur rings
is a ring whose prime homomorphic images have no nonzero ideals that are hereditary n-Amitsur rings,
It is easy to see that each of the classes and is homomorphically closed and that . Moreover, we will show that they are radical classes. For this purpose, we require the following class of rings:
,
Lemma 2.15.
τ n is a hereditary class of rings.
Proof. Suppose that . Let γ be a hereditary radical. Since A is a prime ring, is also a prime ring. So, if , then . Moreover, all the coefficients of any are in $I$ and also in and therefore . Taking into account Proposition 2.1, , as desired.
Let us recall that the essential cover of a class α of rings, denoted by ϵα, is the class of all rings that contain an essential ideal in α.
Lemma 2.16.
The essential cover is a special class of rings.
Proof. Since is a hereditary class of prime rings, is also a hereditary class of prime rings. It remains to show that the class is essentially closed. Let and . Then there exists a nonzero ideal I of A such that and . Since I is a prime ring, A is a prime ring and hence B is also a prime ring, because the class of prime rings is essentially closed. Let IB be the ideal of B generated by I. By Andrunakievich's Lemma, , where . By Lemma 2.15, is a hereditary n-Amitsur ring. Thus .
Corollary 2.17.
is a prime ring having a nonzero ideal I that is a hereditary n-Amitsur ring.
Theorem 2.18.
is a radical class, and is a special radical class.
Proof. We claim that . If , then every nonzero prime homomorphic image of $A$ is not a hereditary n-Amitsur ring, that is, . Therefore . Let . Since , there is a nonzero prime homomorphic image of A which is a hereditary n-Amitsur ring. Hence and so , which is a contradiction. Therefore . Lemma 2.15, now yields that is a radical class.
Next, we show that . Suppose that . Then there exists such that . Since , there is a homomorphic image of A such that . Therefore , which is a contradiction. Hence . Now suppose that . Then there exists such that . Since , there is a nonzero prime homomorphic image of A such that . Since is homomorphically closed, . On the other hand, every homomorphic image of A is not in , which is a contradiction. Thus . By Lemma 2.16, is a special class of rings and so is a special radical.
Theorem 2.19.
The semisimple class is n-polynomially extensible.
Proof. Let . By Theorem 2.18, is a special radical. Therefore there exist ideals Hi of $A$ such that, for each , and . We claim that , for any . If for some , then
Thus . Since is a special class of rings, is a prime ring. On the other hand, there exists an ideal Ji of Bi such that and Ji is a prime hereditary n-Amitsur ring. We have, for any
Thus , where . But , which is contradiction. Thus and consequently , for any . If , then there exists . But all the coefficients ai of of are in . Hence and so , that is, .
We recall from [5] that a radical γ is called small if for each proper radical , where denotes the class of all associative rings. Dually, a nonzero radical γ is large if for each proper radical .
Let denote the collection of all strongly hereditary (that is, closed under subrings) and large radicals, and let denote the collection of all radicals γ such that for every . In [10], S. Tumurbat et al. proved that is a complete sublattice in the lattice of all radicals. Now we have the following:
Proposition 2.20.
Each of the radicals and belongs to
Proof. The radical contains all fields and all rings of prime order p (where denotes the ring with the additive group of and with multiplication defined by ab=0 for all ). In [10], it was shown that every nonzero strongly hereditary radical contains a prime field or a simple zero ring of prime order. Hence for every . Thus and also .
Definition 2.21.
We call a radical γ an n-bad radical if, for every prime ring , there exists a hereditary radical such that .
In what follows, and denote, respectively, the class of all n-bad radicals and the class of all special radicals that are n-bad radicals.
Proposition 2.22.
The classes and satisfy the following conditions:
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(i) If for each , then ;
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(ii) If for each , then .
Proof. (i) Let A be a prime ring in . For each , there exists a hereditary radical such that
Then we may choose any one of , .
Statement (ii) may be proved in a similar way.
Theorem 2.23.
Both and are complete sublattices in the lattice of all radicals.
Proof. We have for each n-bad radical . Therefore and . By Proposition 2.22, . Thus is a complete sublattice in the lattice of all radicals. The proof that is a complete sublattice in the lattice of all radicals is similar.
In [3], B.J. Gardner, J. Krempa and R. Wiegandt posed the question as to whether there exists a (hereditary) radical γ with polynomially extensible semisimple class such that γ does not have the Amitsur property. If does not have the Amitsur property, then we have a positive answer to this question.