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

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