International Journal of Mathematics and Mathematical Sciences
Volume 2016 (2016), Article ID 7159180, 5 pages
http://dx.doi.org/10.1155/2016/7159180

Research Article

A Note on Primitivity of Ideals in Skew Polynomial Rings of Automorphism Type

Edilson Soares Miranda

Departamento de Ciências, Centro de Ciências Exatas, Universidade Estadual de Maringá, 87360-000 Goioerê, PR, Brazil

Received 4 March 2016; Accepted 11 May 2016

Academic Editor: Kaiming Zhao

Copyright © 2016 Edilson Soares Miranda. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We extend results about primitive ideals in polynomial rings over nil rings originally proved by Smoktunowicz (2005) for -primitive ideals in skew polynomial rings of automorphism type.

1. Introduction

Throughout this paper denotes an associative ring but does not necessarily have an identity element and an automorphism of , unless otherwise stated. We denote by the skew polynomial rings of automorphism type whose elements are polynomials , , for every , with usual addition and the following multiplication: for all .

A ring is said to be a Jacobson ring if every prime ideal of is an intersection of (either left or right) primitive ideals of . In [1], Smoktunowicz proved that if is a nil ring and an ideal of , then is Jacobson radical if and only if is Jacobson radical, where is the ideal of generated by coefficients of polynomial from . Also if is a nil ring and is a primitive ideal of , then for some ideal of and affirmative answer to this question is equivalent to the Köthe conjecture. Our main results state that if is a nil ring and an ideal of , then is -Jacobson radical if and only if is -Jacobson radical, where is the ideal of generated by coefficients of polynomial from . Also if is a nil ring and is a -primitive ideal of , then for some ideal of . This result includes, as particular cases, all the above results.

Now we recall some terminology and results; see [2–4]. A right ideal of a ring is called modular in if and only if there exists an element such that for every . An ideal of a ring is said to be a -invariant if and only if . An ideal of is said to be a right -primitive in if and only if there exists a modular maximal right ideal -invariant of such that is the maximal ideal contained in . For , denotes the degree of and the leading coefficient of .

2. Results

We begin with the following results that extend ([1, Lemma  1]) and the proof is also similar to the one in the paper.

Lemma 1. Let be a ring, a right ideal of , , a right ideal of , and such that for every . If , then, for every , there are such that and .

Proof. We proceed by induction on . If , we put . Suppose the lemma holds for some . Let

with and . Consider

Since , then , . Thus

We denote by the usual extension of to a ring with identity and by again the natural extension of to .

The next lemma extends ([1, Lemma  2]).

Lemma 2. Let be an ideal of with and a right ideal of with . Consider , , and

(i)If , , and , then there exists such that and .(ii)Let be a right ideal of , such that for every , and . If with , , and , where , then, for every , there exists such that and .

Proof. (i) Let , , and . We can write

Then

Hence

with , , , and . Put

Therefore

Since and , then and .
(ii) By Lemma 1, for every , there exists such that

Consider

For every denote

Note that ; thus for every

Hence

Because and , then . We have that, for every , there exists such that . If , then is the required. If , by first part of this lemma, there exists

such that and . Thus for all . If , then is the required. If , using similar arguments as above, we can find such that

with and for every . Hence is the required.

Let , a right ideal of , , and such that for all . Following [1] we have the following. We say that is a “good number for ,” if, for all sufficiently large , there are such that with . Let ; we denote

Lemma 3. Let be a right ideal of maximal in the set of all right ideals -invariants with such that for all . Suppose with for some . If there is no right ideal of with , , and , then there exists a positive integer and such that if with , , and , then , , and is a good number for all .

Proof. Let be minimal positive integer such that there exists and with and . It is clear that . If , put and

Thus is a right ideal of with and . By assumption , then , where and . Put

Comparing the leading coefficients of and , we have that

which contradicts the minimality of . Therefore ; consequently .
Suppose that for some . Put ; using similar arguments as above we can have a contradiction. Hence .
If there exists with , , and , then using similar arguments as above we can show that and . Moreover, if , put ; we have that is a right ideal of with and .
By assumption . Thus , where , , and . Therefore . Consequently is a good number for all .

Lemma 4. Let be a right ideal with , , and such that for all sufficiently large there are such that and , where is a right ideal of and such that for every . Then there exists a positive integer and such that if with , , and one has that , , and is a good number for all .

Proof. Let be minimal positive integer such that for all sufficiently large there are such that and . Put

with , , and . By Lemma 1 and minimality of we have that . Using the same ideas of Lemma 3, we have that and . Since , we have that the first part of lemma is satisfied.
Let ; we denote by the right ideal of :

For sufficiently large there are such that and . Put

For every we have that , where and . Consequently

Since , we can write

Put ; thus . Therefore is a good number for all .

Lemma 5. Let be a right ideal of , , such that for all and is good number for all , where . Assume that for every with , , and one has that and . If there are and with

then is a good number for .

Proof. Since is a good number for and , then for every sufficiently large there are and such that

with , . Consider

Since , then

a contradiction.
Thus there exists sufficiently large such that ; hence is a good number for . Then for all sufficiently large there are such that and . We denote

Consider

Since , then . Moreover

Consequently is a good number for .

The following theorem extends ([1, Theorem  1]).

Theorem 6. Let be a nil ring and let be a -primitive ideal in . Then , where is an ideal -invariant of .

Proof. Assume by contradiction that there are with

Since is a -primitive ideal in , there is a right ideal of with and such that for all . Moreover is a maximal in the set of right ideals -invariants and is the maximal ideal contained in . We have that ; otherwise , which is impossible because is a nil ring. By definition of it follows that .
If for some with , then . In fact, if , let be the minimal positive integer with respect to . Thus . Then ; hence . Consequently , a contradiction.
Let be a right ideal of with and . We have that . There exists such that . Consider

Since is an ideal -prime and , then . Consequently , because is the maximal ideal contained in . Then . There exists such that . By Lemma 2, for every , there are such that and . Lemmas 3 and 4 imply that there are and such that if with , , and , then and . Moreover is a good number for all . Let be minimal such that is a good number for all . We have that . Let . Since is a good number for , then for sufficiently large there are , such that

Consider , then and . For some , there are , , such that , , and . Put

where and . Since , then . Moreover,

Since is a nil ring, consider , where is a minimal with respect to the condition . Thus for all . We have that

Put . Thus,

for every . Since , if is not a good number for , then Lemma 5 implies that

In this case, there exists such that . Consequently , , and . Then . Therefore is a good number for . Then for sufficiently large there are , such that

Let

Since , , and , then . Thus ; hence for every .
Let

We have that . Thus . Put

We can write as

Thus for all sufficient large

Then is a good number for all . This contradicts the minimality of .

Recall that the -Jacobson radical of a ring is defined as the intersection of all -primitive ideals of . A ring is a -Jacobson radical if .

Theorem 7. Let be a nil ring and let be an ideal of . Consider the ideal of generated by coefficients of polynomial from . Then is -Jacobson radical if and only if is -Jacobson radical.

Proof. Assume by contradiction that is not -Jacobson radical. Then there is a -primitive ideal of such that . We have that there is an ideal of such that . Therefore is a -primitive ideal of . By Theorem 6, there is an ideal of such that . It is clear that . Since

then is a -primitive ideal, a contradiction. Using the fact that , the converse follows.

Corollary 8. If is a nil ring, then the polynomial ring of type automorphism can not be homomorphically mapped onto a -simple -primitive ring.

Competing Interests

The author declares that they have no competing interests.

References

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T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics, Springer, New York, NY, USA, 1991.