On 2-primal Ore extensions
Let R be a ring, be an automorphism of R and δ be a σ-derivation of R. We define a δ property on R. We say that R is a δ-ring if aδ(a) ∊ P(R) implies a ∊ P(R), where P(R) denotes the prime radical of R. We ultimately show the following. Let R be a Noetherian δ-ring, which is also an algebra over Q...
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Bhat, V.K. 2017-09-28T13:38:53Z 2017-09-28T13:38:53Z 2007 On 2-primal Ore extensions / V.K. Bhat // Український математичний вісник. — 2007. — Т. 4, № 2. — С. 173-179. — Бібліогр.: 15 назв. — англ. 1810-3200 2000 MSC. 16XX, 16N40, 16P40, 16W20, 16W25. https://nasplib.isofts.kiev.ua/handle/123456789/124514 Let R be a ring, be an automorphism of R and δ be a σ-derivation of R. We define a δ property on R. We say that R is a δ-ring if aδ(a) ∊ P(R) implies a ∊ P(R), where P(R) denotes the prime radical of R. We ultimately show the following. Let R be a Noetherian δ-ring, which is also an algebra over Q, σ and δ be as usual such that σ(δ(a)) = δ(σ(a)), for all a ∊ R and σ(P) = P, P any minimal prime ideal of R. Then R[x, σ(, δ] is a 2-primal Noetherian ring. en Інститут прикладної математики і механіки НАН України Український математичний вісник On 2-primal Ore extensions Article published earlier |
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Let R be a ring, be an automorphism of R and δ be a σ-derivation of R. We define a δ property on R. We say that R is a δ-ring if aδ(a) ∊ P(R) implies a ∊ P(R), where P(R) denotes the prime radical of R. We ultimately show the following. Let R be a Noetherian δ-ring, which is also an algebra over Q, σ and δ be as usual such that σ(δ(a)) = δ(σ(a)), for all a ∊ R and σ(P) = P, P any minimal prime ideal of R. Then R[x, σ(, δ] is a 2-primal Noetherian ring.
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On 2-primal Ore extensions / V.K. Bhat // Український математичний вісник. — 2007. — Т. 4, № 2. — С. 173-179. — Бібліогр.: 15 назв. — англ. |
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Український математичний вiсник
Том 4 (2007), № 2, 173 – 179
On 2-primal Ore extensions
Vijay K. Bhat
(Presented by I. V. Protasov)
Abstract. Let R be a ring, σ be an automorphism of R and δ be a
σ-derivation of R. We define a δ property on R. We say that R is a δ-ring
if aδ(a) ∈ P (R) implies a ∈ P (R), where P(R) denotes the prime radical
of R. We ultimately show the following.
Let R be a Noetherian δ-ring, which is also an algebra over Q, σ and
δ be as usual such that σ(δ(a)) = δ(σ(a)), for all a ∈ R and σ(P ) = P ,
P any minimal prime ideal of R. Then R[x, σ, δ] is a 2-primal Noetherian
ring.
2000 MSC. 16XX, 16N40, 16P40, 16W20, 16W25.
Key words and phrases. 2-primal, Minimal prime, prime radical, nil
radical, automorphism, derivation.
1. Introduction
A ring R always means an associative ring. Q denotes the field of ra-
tional numbers. Spec(R) denotes the set of prime ideals of R. MinSpec(R)
denotes the sets of minimal prime ideals of R. P(R) and N(R) denote the
prime radical and the set of nilpotent elements of R respectively. Let I
and J be any two ideals of a ring R. Then I ⊂ J means that I is strictly
contained in J.
This article concerns the study of ore extensions in terms of 2-primal
rings. 2-primal rings have been studied in recent years and the 2-primal
property is being studied for various types of rings. In [13], Greg Marks
discusses the 2-primal property of R[x, σ, δ], where R is a local ring, σ is
an automorphism of R and δ is a σ-derivation of R.
Recall that a σ-derivation of R is an additive map δ : R → R such
that δ(ab) = δ(a)σ(b) + aδ(b), for all a, b ∈ R. In case σ is the identity
map, δ is called just a derivation of R. For example for any endomorphism
τ of a ring R and for any a ∈ R, ̺ : R → R defined as ̺(r) = ra − aτ(r)
Received 14.11.2006
Sincere thanks to referee for some suggestions to give the manuscript a modified shape.
ISSN 1810 – 3200. c© Iнститут математики НАН України
174 On 2-primal Ore extensions
is a τ -derivation of R. Also let R = K[x], K a field. Then the formal
derivative d/dx is a derivation of R.
Minimal prime ideals of 2-primal rings have been discussed by Kim
and Kwak in [10]. 2-primal near rings have been discussed by Argac
and Groenewald in [2]. Recall that a ring R is 2-primal if and only
if nil radical and prime radical of R are same if and only if the prime
radical is a completely semiprime ideal. An ideal I of a ring R is called
completely semiprime if a2 ∈ I implies a ∈ I, where a ∈ R. We also
note that a reduced is 2-primal and a commutative ring is also 2-primal.
For further details on 2-primal rings, we refer the reader to [7, 9, 10, 14].
Before proving the main result, we find a relation between the minimal
prime ideals of R and those of the Ore extension R[x, σ, δ], where R is a
Noetherian Q-algebra, σ is an automorphism of R and δ is a σ-derivation
of R. This is proved in Theorem (2.1). Recall that R[x, σ, δ] is the usual
polynomial ring with coefficients in R in which multiplication is subject to
the relation ax = xσ(a)+δ(a) for all a ∈ R. We take any f(x) ∈ R[x, σ, δ]
to be of the form f(x) =
∑n
i=0
xiai. We denote R[x, σ, δ] by O(R).
Ore-extensions including skew-polynomial rings and differential oper-
ator rings have been of interest to many authors. See [1, 3, 4, 8, 11, 12].
Recall that in [11], a ring R is called σ-rigid if there exists an endomor-
phism σ of R with the property that aσ(a) = 0 implies a = 0 for a ∈ R.
In [12], Kwak defines a σ(∗)-ring R to be a ring if aσ(a) ∈ P (R) implies
a ∈ P (R) for a ∈ R and establishes a relation between a 2-primal ring
and a σ(∗)-ring. The property is also extended to the skew-polynomial
ring R[x, σ].
Let R be a ring, σ be an automorphism of R and δ be a σ-derivation
of R. We introduce a property on R and say that R is a δ-ring if aδ(a) ∈
P (R) implies a ∈ P (R), where P(R) denotes the prime radical of R. We
note that a ring with identity is not a δ-ring.
Now let R be a Noetherian δ-ring, which is also an algebra over Q such
that σ(δ(a)) = δ(σ(a)), for all a ∈ R; σ(P ) = P for all P ∈ MinSpec(R)
and δ(P (R)) ⊆ P (R). Then R[x, σ, δ] is 2-primal. This is proved in
Theorem (2.4).
2. Ore extensions
We begin with the following definition:
Definition 2.1. Let R be a ring. Let σ be an automorphism of R and
δ be a σ-derivation of R. We say that R is a δ-ring if a δ(a) ∈ P (R)
implies a ∈ P (R).
V. K. Bhat 175
Recall that an ideal I of a ring R is called σ-invariant if σ(I) = I
and is called δ-invariant if δ(I) ⊆ I. If an ideal I of R is σ-invariant
and δ-invariant, then I[x, σ, δ] is an ideal of R[x, σ, δ]. Also I is called
completely prime if ab ∈ I implies a ∈ I or b ∈ I for a, b ∈ R.
Gabriel proved in Lemma (3.4) of [5] that if R is a Noetherian Q-
algebra and δ is a derivation of R, then δ(P ) ⊆ P , for all P ∈MinSpec(R).
We generalize this for σ-derivation δ of R and give a structure of minimal
prime ideals of O(R) in the following Theorem.
Theorem 2.1. Let R be a Noetherian Q-algebra. Let σ be an automor-
phism of R and δ be a σ-derivation of R such that σ(δ(a)) = δ(σ(a)),
for a ∈ R. Then P ∈ MinSpec(O(R)) such that σ(P ∩ R) = P ∩ R im-
plies P ∩R ∈ MinSpec(R) and P1 ∈ MinSpec(R) such that σ(P1) = P1
implies O(P1) ∈ MinSpec(O(R)).
Proof. Let P1 ∈ MinSpec(R) with σ(P1) = P1. Let T = R[[t, σ]], the
skew power series ring. Now it can be seen that etδ is an automor-
phism of T and P1T ∈ MinSpec(T ). We also know that (etδ)k(P1T ) ∈
MinSpec(T ) for all integers k ≥ 1. Now T is Noetherian by Exer-
cise (1ZA(c)) of [6], and therefore Theorem (2.4) of [6] implies that
MinSpec(T) is finite. So exists an integer an integer n ≥ 1 such that
(etδ)n(P1T ) = P1T ; i.e. (entδ)(P1T ) = P1T . But R is a Q-algebra,
therefore, etδ(P1T ) = P1T . Now for any a ∈ P1, a ∈ P1T also, and so
etδ(a) ∈ P1T ; i.e. a+ tδ(a)+(t2/2!)δ2(a)+ · · · ∈ P1T , which implies that
δ(a) ∈ P1. Therefore δ(P1) ⊆ P1.
Now it can be easily seen that O(P1) ∈ Spec(O(R)). Suppose that
O(P1) /∈ MinSpec(O(R)), and P2 ⊂ O(P1) is a minimal prime ideal
of O(R). Then we have P2 = O(P2 ∩ R) ⊂ O(P1) ∈ MinSpec(O(R)).
Therefore P2 ∩ R ⊂ P1, which is a contradiction as P2 ∩ R ∈ Spec(R).
Hence O(P1) ∈ MinSpec(O(R)).
Conversely let P ∈ MinSpec(O(R)) with σ(P ∩R) = P ∩R. Then it
can be easily seen that P ∩ R ∈ Spec(R) and O(P ∩ R) ∈ Spec(O(R)).
Therefore O(P ∩ R) = P . We now show that P ∩ R ∈ MinSpec(R).
Suppose that P3 ⊂ P ∩ R, and P3 ∈ MinSpec(R). Then O(P3) ⊂
O(P ∩ R) = P . But O(P3) ∈ Spec(O(R)) and, O(P3) ⊂ P , which is not
possible. Thus we have P ∩ R ∈ MinSpec(R).
Proposition 2.1. Let R be a 2-primal ring. Let σ and δ be as usual
such that δ(P (R)) ⊆ P (R). If P ∈ MinSpec(R) is such that σ(P ) = P ,
then δ(P ) ⊆ P .
Proof. Let P ∈ MinSpec(R). Now for any a ∈ P , there exists b /∈ P such
that ab ∈ P (R) by Corollary (1.10) of [14]. Now δ(P (R)) ⊆ P (R), and
176 On 2-primal Ore extensions
therefore δ(ab) ∈ P (R); i.e. δ(a)σ(b)+aδ(b) ∈ P (R) ⊆ P . Now aδ(b) ∈ P
implies that δ(a)σ(b) ∈ P . Also σ(P ) = P and by Proposition (1.11)
of [14], P is completely prime, we have δ(a) ∈ P . Hence δ(P ) ⊆ P .
Theorem 2.2. Let R be a δ-ring. Let σ and δ be as above such that
δ(P (R)) ⊆ P (R). Then R is 2-primal.
Proof. Define a map ρ : R/P (R) → R/P (R) by ρ(a + P (R)) = δ(a) +
P (R) for a ∈ R and τ : R/P (R) → R/P (R) a map by τ(a + P (R)) =
σ(a) + P (R) for a ∈ R, then it can be seen that τ is an automorphism
of R/P(R) and ρ is a τ -derivation of R/P(R). Now aδ(a) ∈ P (R) if and
only if (a+P (R))ρ(a+P (R)) = P (R) in R/P(R). Thus as in Proposition
(5) of [8], R is a reduced ring and, therefore R is 2-primal.
Proposition 2.2. Let R be a ring. Let σ and δ be as usual. Then:
1. For any completely prime ideal P of R with δ(P ) ⊆ P , P [x, σ, δ] is
a completely prime ideal of R[x, σ, δ].
2. For any completely prime ideal U of R[x, σ, δ], U∩R is a completely
prime ideal of R.
Proof. (1) Let P be a completely prime ideal of R. Now let f(x) =∑n
i=0
xiai ∈ R[x, σ, δ] and g(x) =
∑m
j=0
xjbj ∈ R[x, σ, δ] be such that
f(x)g(x) ∈ P [x, σ, δ]. Suppose f(x) /∈ P [x, σ, δ]. We will show that
g(x) ∈ P [x, σ, δ]. We use induction on n and m. For n = m = 1, the
verification is easy. We check for n = 2 and m = 1. Let f(x) = x2a+xb+c
and g(x) = xu + v. Now f(x)g(x) ∈ P [x, σ, δ] with f(x) /∈ P [x, σ, δ].
The possibilities are a /∈ P or b /∈ P or c /∈ P or any two out of these
three do not belong to P or all of them do not belong to P. We verify
case by case.
Let a /∈ P . Since x3σ(a)u + x2(δ(a)u + σ(b)u + av) + x(δ(b)u +
σ(c)u + bv) + δ(c)u + cv ∈ P [x, σ, δ], we have σ(a)u ∈ P , and so u ∈ P .
Now δ(a)u + σ(b)u + av ∈ P implies av ∈ P , and so v ∈ P . Therefore
g(x) ∈ P [x, σ, δ].
Let b /∈ P . Now σ(a)u ∈ P . Suppose u /∈ P , then σ(a) ∈ P and
therefore a, δ(a) ∈ P . Now δ(a)u+σ(b)u+av ∈ P implies that σ(b)u ∈ P
which in turn implies that b ∈ P , which is not the case. Therefore we
have u ∈ P . Now δ(b)u + σ(c)u + bv ∈ P implies that bv ∈ P and
therefore v ∈ P . Thus we have g(x) ∈ P [x, σ, δ].
Let c /∈ P . Now σ(a)u ∈ P . Suppose u /∈ P , then as above a,
δ(a) ∈ P . Now δ(a)u + σ(b)u + av ∈ P implies that σ(b)u ∈ P . Now
u /∈ P implies that σ(b) ∈ P ; i.e. b, δ(b) ∈ P . Also δ(b)u+σ(c)u+bv ∈ P
implies σ(c)u ∈ P and therefore σ(c) ∈ P which is not the case. Thus
V. K. Bhat 177
we have u ∈ P . Now δ(c)u + cv ∈ P implies cv ∈ P , and so v ∈ P .
Therefore g(x) ∈ P [x, σ, δ].
Now suppose the result is true for k, n = k > 2 and m = 1. We will
prove for n = k+1. Let f(x) = xk+1ak+1 +xkak + · · ·+xa1 +a0, and g(x)
= xb1 + b0 be such that f(x)g(x) ∈ P [x, σ, δ], but f(x) /∈ P [x, σ, δ]. We
will show that g(x) ∈ P [x, σ, δ]. If ak+1 /∈ P , then equating coefficients
of xk+2, we get σ(ak+1)b1 ∈ P , which implies that b1 ∈ P . Now equating
coefficients of xk+1, we get σ(ak)b1 + ak+1b0 ∈ P , which implies that
ak+1b0 ∈ P , and therefore b0 ∈ P . Hence g(x) ∈ P [x, σ, δ].
If aj /∈ P , 0 ≤ j ≤ k, then using induction hypothesis, we get that
g(x) ∈ P [x, σ, δ]. Therefore the statement is true for all n. Now using
the same process, it can be easily seen that the statement is true for all
m also. The details are left to the reader.
(2) Let U be a completely prime ideal of R[x, σ, δ]. Suppose a, b ∈ R
are such that ab ∈ U ∩ R with a /∈ U ∩ R. This means that a /∈ U as
a ∈ R. Thus we have ab ∈ U ∩ R ⊆ U , with a /∈ U . Therefore we have
b ∈ U , and thus b ∈ U ∩ R.
Corollary 2.1. Let R be a δ-ring, where σ and δ as usual such that
δ(P (R)) ⊆ P (R). Let P ∈ MinSpec(R) be such that σ(P ) = P . Then
P [x, σ, δ] is a completely prime ideal of R[x, σ, δ].
Proof. R is 2-primal by Theorem (2.2), and so by Proposition (2.1)
δ(P ) ⊆ P . Further more P is a completely prime ideal of R by Proposi-
tion (1.11) of [10]. Now use Proposition (2.2).
We now prove the following Theorem, which is crucial in proving
Theorem 2.4.
Theorem 2.3. Let R be a δ-ring, where σ and δ as usual such that
δ(P (R)) ⊆ P (R) and σ(P ) = P for all P ∈ MinSpec(R). Then R[x, σ, δ]
is 2-primal if and only if P (R)[x, σ, δ] = P (R[x, σ, δ]).
Proof. Let R[x, σ, δ] be 2-primal. Now by Corollary (2.1) P (R[x, σ, δ]) ⊆
P (R)[x, σ, δ]. Let f(x) =
∑n
j=0
xjaj ∈ P (R)[x, σ, δ]. Now R is a 2-
primal subring of R[x, σ, δ] by Theorem (2.2), which implies that aj is
nilpotent and thus aj ∈ N(R[x, σ, δ]) = P (R[x, σ, δ]), and so we have
xjaj ∈ P (R[x, σ, δ]) for each j, 0 ≤ j ≤ n, which implies that f(x) ∈
P (R[x, σ, δ]). Hence P (R)[x, σ, δ] = P (R[x, σ, δ]).
Conversely suppose P (R)[x, σ, δ] = P (R[x, σ, δ]). We will show that
R[x, σ, δ] is 2-primal. Let g(x) =
∑n
i=0
xibi ∈ R[x, σ, δ], bn 6= 0, be such
that (g(x))2 ∈ P (R[x, σ, δ]) = P (R)[x, σ, δ]. We will show that g(x) ∈
P (R[x, σ, δ]). Now leading coefficient σ2n−1(an)an ∈ P (R) ⊆ P , for all
178 On 2-primal Ore extensions
P ∈ MinSpec(R). Now σ(P ) = P and P is completely prime by Propo-
sition (1.11) of [10]. Therefore we have an ∈ P , for all P ∈ MinSpec(R);
i.e. an ∈ P (R). Now since δ(P (R)) ⊆ P (R) and σ(P ) = P for all
P ∈ MinSpec(R), we get (
∑n−1
i=0
xibi)
2 ∈ P (R[x, σ, δ]) = P (R)[x, σ, δ]
and as above we get an−1 ∈ P (R). With the same process in a finite
number of steps we get ai ∈ P (R) for all i, 0 ≤ i ≤ n. Thus we have
g(x) ∈ P (R)[x, σ, δ]; i.e. g(x) ∈ P (R[x, σ, δ]). Therefore P (R[x, σ, δ]) is
completely semiprime. Hence R[x, σ, δ] is 2-primal.
Theorem 2.4. Let R be a Noetherian δ-ring, which is also an algebra
over Q such that σ(δ(a)) = δ(σ(a)), for all a ∈ R; σ(P ) = P for all
P ∈ MinSpec(R) and δ(P (R)) ⊆ P (R), where σ and δ are as usual.
Then R[x, σ, δ] is 2-primal.
Proof. We use Theorem (2.1) to get that P (R)[x, σ, δ] = P (R[x, σ, δ]),
and now the result is obvious by using Theorem (2.3).
Corollary 2.2. Let R be a commutative Noetherian δ-ring, which is also
an algebra over Q such that σ(δ(a)) = δ(σ(a)), for all a ∈ R; σ(P ) = P
for all P ∈MinSpec(R), where σ and δ are as usual. Then R[x, σ, δ] is
2-primal.
Proof. Using Theorem (1) of [15] we get δ(P (R)) ⊆ P (R). Now rest is
obvious.
The above gives rise to the following questions:
If R is a Noetherian Q-algebra (even commutative), σ is an automor-
phism of R and δ is a σ-derivation of R. Is R[x, σ, δ] 2-primal? The main
problem is to get Theorem (2.3) satisfied.
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Contact information
Vijay Kumar Bhat School of Applied Physics and Mathematics,
SMVD University,
P/o Kakryal, Udhampur, J and K,
India 182121
E-Mail: vijaykumarbhat2000@yahoo.com
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