Abstract Let R be a σ-prime ring with characteristic not two and d be a nonzero derivation of R commuting with σ. The purpose of this paper is to give suitable conditions under which R must be commutative. Mathematics Subject Classification: 16W25, 16W20, 16U80

L Oukhtite

S Salhi

CiteSeerX

Int. J. Contemp. Math. Sci., Vol. 1, 2006, no. 9, 439-448
Derivations and commutativity
of σ−prime rings
L. Oukhtite and S. Salhi
Université Moulay Ismäıl, Faculté des Sciences et Techniques
Département de Mathématiques, Groupe d’Algèbre et Applications
B. P. 509 Boutalamine, Errachidia; Maroc.
oukhtite@math.net
Abstract
Let R be a σ-prime ring with characteristic not two and d be a
nonzero derivation of R commuting with σ. The purpose of this paper
is to give suitable conditions under which R must be commutative.
Mathematics Subject Classification: 16W25, 16W20, 16U80.
Keywords: σ-prime rings, σ-ideals, derivations.
1 Introduction
Throughout the present paper all rings will be associative. A ring R equipped
with an involution σ is said to be σ−prime if aRb = aRσ(b) = 0 implies that
a = 0 or b = 0. Recall that a ring R is prime if aRb = 0 implies that a = 0
or b = 0. Obviously, every prime ring with involution σ is σ-prime but the
converse is in general not true. An ideal I of R is a σ-ideal if I is invariant under
σ, i.e, σ(I) = I. In all that follows Saσ(R) will denote the set of symmetric and
skew symmetric elements of R, i.e, Saσ(R) = {x ∈ R/σ(x) = ±x}. As usual
the commutator xy − yx will denoted by [x, y] = xy − yx. We shall use basic
commutator identities: [xy, z] = x[y, z]+[x, z]y and [x, yz] = y[x, z]+[x, y]z.
An additive mapping d of R to itself is a derivation if d(xy) = d(x)y + xd(y)
holds for all pairs x, y ∈ R. A mapping F : R → R is said to be centralizing
on a subset S of R if [F (s), s] ∈ Z(R) for all s in S, where Z(R) denotes the
center of R; in the special case where [F (s), s] = 0 for all s in S, the mapping
F is said to be commuting on S.
The history of commuting and centralizing mappings goes back to 1955
when Divinsky [2] proved that a simple artinien ring is commutative if it has a
440 L. Oukhtite and S. Salhi
commuting non trivial automorphism. Two years later, Posner [6] have proved
that the existence of a nonzero centralizing derivation on a prime ring forces
the ring to be commutative (Posner’s second theorem). This work was the
motivation for our first result. More precisely, we are interested to know wether
an analogous result can be proved for σ−prime rings with the restriction that
the derivation is centralizing not on R but on a σ−ideal of R. The following
theorem treats the case of σ-prime rings with characteristic not two.
Theorem 1 Let R be a σ-prime ring with characteristic not two, I be a
nonzero σ−ideal of R and d be a nonzero derivation of R which commutes
with σ. If [d(x), x] ∈ Z(R) for all x in I, then R is commutative.
In the special case of a generalized inner derivation f of a ring R, i.e,
f(x) = ax +xb with a, b ∈ R, Brešar [1] remarked that the condition that f is
centralizing on a subset S of R can be written in the form d(x)x−xg(x) ∈ Z(R)
for all x ∈ S, where d (resp. g) is the inner derivation induced by a (resp.
b), that is d(x) = [a, x] and g(x) = [x, b]. Motivating by this observation,
Brešar introduced a more general concept than centralizing derivations by
considering the situation when derivations f and g of a prime ring R satisfy
f(x)x−xg(x) ∈ Z(R) for all x in some subset S of R. He proved that a prime
ring R, equipped with derivations 0 = f and g satisfying f(x)x−xg(x) ∈ Z(R)
for all x in a nonzero left ideal U of R, must be commutative. Our aim
in the following theorem is to prove a similar result for σ−prime rings with
characteristic not two.
Theorem 2 Let R be a σ-prime ring with characteristic not two and I be a
nonzero σ−ideal of R. Suppose there exist derivations d1 and d2 which commute
with σ such that d1(x)x − xd2(x) ∈ Z(R) for all x ∈ I. If d2 = 0, then R is
commutative.
In [7], Kyoo-Hong Parck and Yong-Soo Sung state that if a prime ring
R with characteristic different from two has a nonzero derivation d such that
[ad(x), x] = 0 for all x in R, then a = 0 or R is commutative. The main purpose
of the following theorem is to prove analogous result for σ-prime rings. The
result we shall prove is:
Theorem 3 Let R be a σ-prime ring with characteristic not two, I be a
nonzero σ−ideal of R and d = 0 be a derivation of R which commutes with σ.
If [ad(x), x] = 0 for all x in I, then a = 0 or R is commutative.
P.H. Lee and T.L Lee [3] have shown that if a prime ring of characteristic
different from two has a nonzero derivation d satisfying [d(R), d(R)] ⊂ Z(R),
then R is commutative. Our next goal is to generalize the result of [3] for
σ-prime rings.
Derivations and commutativity 441
Theorem 4 Let R be a σ-prime ring with characteristic not two, I be a
nonzero σ−ideal of R and d a nonzero derivation such that σd = dσ. If
d(I) ⊂ I and [d(I), d(I)] ⊂ Z(R), then R is commutative.
2 Proof of the main results
In all that follows, R will represent a σ-prime ring with characteristic not two.
We will make frequent and important use of the following lemmas.
Lemma 1 ([5], 3) of Theorem 1) Let I be a nonzero σ-ideal of R. If a, b in R
are such that aIb = 0 = aIσ(b), then a = 0 or b = 0.
Proof. Suppose a = 0, there exists some x ∈ I such that ax = 0. Indeed,
otherwise
aRx = 0 and aRσ(x) = 0 for all x ∈ I
and therefore a = 0. Since aIRb = 0 and aIRσ(b) = 0, we then obtain
axRb = axRσ(b) = 0.
In view of the σ-primeness of R this yields b = 0.
Lemma 2 Let I be a nonzero σ-ideal of R and 0 = d be a derivation on R
which commutes with σ. If [x,R]Id(x) = 0 for all x ∈ I, then R is commutative.
Proof. Let x ∈ I. Since t = x − σ(x) ∈ I, then [t, r]Id(t) = 0 for all r ∈ R.
As t ∈ Saσ(R), we then get
[t, r]Id(t) = σ([t, r])Id(t) = 0, for all r ∈ R.
According to Lemma 1 we obtain
d(t) = 0 or [r, t] = 0, for all r ∈ R.
If d(t) = 0, then d(x) = d(σ(x)) = σ(d(x)). Therefore
0 = [x, r]Id(x) = [x, r]Iσ(d(x))
and thus d(x) = 0 or [r, x] = 0 for all r ∈ R, by Lemma 1. Consequently,
either d(x) = 0 or x ∈ Z(R).
If [r, t] = 0 for all r ∈ R, then t ∈ Z(R) and thus [x, r] = [σ(x), r] for all r ∈ R.
Hence
[x, r]Id(x) = 0 = σ([x, r])Id(x).
Again using Lemma 1, we get d(x) = 0 or x ∈ Z(R).
In conclusion, for each x ∈ I either d(x) = 0 or x in Z(R). Let us consider
442 L. Oukhtite and S. Salhi
G1 = {x ∈ I/d(x) = 0} and G2 = {x ∈ I / x in Z(R) }. It is clear that G1
and G2 are additive subgroups of I such that I = G1 ∪ G2. But a group can
not be a union of two its proper subgroups and hence I = G1 or I = G2.
If I = G1, then d(x) = 0 for all x ∈ I. For any s ∈ R, replace x by xs to get
xd(s) = 0, for all x ∈ I so that Id(s) = 0 for all s ∈ R. In particular
0 = 1Id(s) = σ(1)Id(s) for all s in R
and Lemma 1 gives d = 0, a contradiction. Hence, I = G2 so that I ⊆ Z(R).
Let r, s ∈ R and x ∈ I. From rsx = rxs = srx we conclude that [r, s]I = 0
and then
[r, s]I1 = [r, s]Iσ(1) = 0.
As 0 = 1, in view of Lemma 1, we then get [r, s] = 0 for all r, s ∈ R. Therefore,
R is a commutative ring.
Lemma 3 Let I be a nonzero σ-ideal of R and 0 = d be a derivation on R
which commutes with σ. If [d(x), x] = 0 for all x ∈ I, then R is commutative.
Proof. Let x, y ∈ I; we linearize [d(x), x] = 0 to get
[d(x), y] + [d(y), x] = 0 for all x, y ∈ I. (1)
If in (1) we replace y by yx where y ∈ I we obtain
[d(x), y]x + [d(y), x]x + [y, x]d(x) = 0 for all x, y ∈ I.
Hence (1) yields
[x, y]d(x) = 0 for all x, y ∈ I. (2)
Thus, for any r ∈ R, we see that 0 = [x, ry]d(x) = [x, r]yd(x) = 0 and therefore
[x, r]Id(x) = 0 for all r ∈ R, x ∈ I. We then conclude, by Lemma 2, that R
is commutative.
Lemma 4 Let d be a derivation of R satisfying dσ = ±σd and let I be a
nonzero σ-ideal of R. If d2(I) = 0 then d = 0.
Proof. For any x ∈ I, we have d2(x) = 0. Replacing x by xy, we obtain
d2(x)y + 2d(x)d(y) + xd2(y) = 0 for all x, y ∈ I.
The fact that d2(I) = 0 together with char(R) = 2, give d(x)d(y) = 0. If we
replace x by xz where z ∈ I, in the last equality, we get d(x)zd(y) = 0 for
all x, y, z ∈ I, so that d(x)Id(y) = 0. Since d commutes with σ we have, by
Lemma 1, that
d(x) = 0 for all x ∈ I. (3)
In this replace x by xr where r ∈ R, to obtain xd(r) = 0 and therefore
IRd(r) = 0 for all r ∈ R. Since I is a nonzero σ-ideal and R is σ-prime this
last relation yields d(r) = 0 for all r ∈ R and consequently d = 0.
Derivations and commutativity 443
Lemma 5 Let d1 and d2 be derivations of R such that d1σ = ±σd1 and
d2σ = ±σd2. If I = 0 is a σ-ideal of R such that d2(I) ⊂ I and d1 d2(I) = 0,
then d1 = 0 or d2 = 0.
Proof. Let u, v ∈ I; then
0 = d1d2(uv) = d2(u)d1(v) + d1(u)d2(v). (4)
Replacing u by d2(u) in (4) we get
d22(u)d1(v) = 0 for all u, v ∈ I. (5)
If we replace v by vw where w ∈ I, in (5), we obtain
d22(u)vd1(w) = 0 for all u, v, w ∈ I.
It then follows that d22(u)Id1(w) = 0 for all u, w ∈ I. The fact that σ(I) = I
combined with d1σ = ±σd1 give
d22(u)Id1(w) = d
2
2(u)Iσ(d1(w)) = 0
and we would conclude by Lemma 1 that either d1(w) = 0 for all w ∈ I so
that d1 = 0 or d
2
2(u) = 0 for all u ∈ I. If d22(u) = 0 for all u ∈ I, we obtain
from Lemma 4 that d2 = 0.
Lemma 6 Let d be a nonzero derivation of R which commutes with σ and I
be a nonzero σ-ideal of R. If a ∈ I∩Saσ(R) is such that [d(I), a] ⊂ Z(R), then
a ∈ Z(R).
Proof. As [d(a2), a] = 2a[d(a), a], we have a[d(a), a] ∈ Z(R) because char(R) =
2. This implies [d(a), a]R[a, r] = 0 for all r ∈ R. Since a ∈ Saσ(R), then
[d(a), a]R[a, r] = [d(a), a]Rσ([a, r]) = 0
and so a ∈ Z(R) or [d(a), a] = 0. Let us suppose that [d(a), a] = 0. By our hy-
pothesis we see that [d[a, u], a] = [[d(a), u], a] which leads us to [[d(a), u], a] ∈
Z(R) for all u ∈ I. If we replace u by au in the last relation, we obtain
a[[d(a), u], a] ∈ Z(R) for all u ∈ I. Hence, for r ∈ R, we have
ra[[d(a), u], a] = a[[d(a), u], a]r = ar[[d(a), u], a]
so that [r, a][[d(a), u], a] = 0. Let t ∈ R, from [rt, a][[d(a), u], a] = 0 we obtain
[r, a]t[[d(a), u], a] = 0 and thus [r, a]R[[d(a), u], a] = 0. As a ∈ Saσ(R), then
[r, a]R[[d(a), u], a] = σ([r, a])R[[d(a), u], a] = 0
444 L. Oukhtite and S. Salhi
and the σ−primeness of R yields a ∈ Z(R) or [[d(a), u], a] = 0 for all u ∈ I. If
[[d(a), u], a] = 0, then [d(a), [u, a]] = 0 for all u ∈ I, that is dd(a)da(u) = 0 for
all u ∈ I where dz is the inner derivation induced by z, i.e, dz(x) = [x, z]. In
view of Lemma 5, the fact that dd(a)(I) ⊂ I combined with σda = ± daσ and
σdd(a) = ± dd(a) yield da = 0 or dd(a) = 0 so that a ∈ Z(R) or d(a) ∈ Z(R).
Let us now suppose that d(a) ∈ Z(R); the fact that
[d(au), a] = d(a)[u, a] + a[d(u), a] for all u ∈ I
forces d(a)[u, a] + a[d(u), a] ∈ Z(R) for all u ∈ I. Then
0 = [d(a)[u, a] + a[d(u), a], a] = d(a)[[u, a], a] for all u ∈ I
which gives d(a)R[[u, a], a] = 0 for all u ∈ I. Since a ∈ Saσ(R) and d commutes
with σ, then d(a)R[[u, a], a] = σ(d(a))R[[u, a], a] = 0 and the σ-primeness of
R yields d(a) = 0 or [[u, a], a] = 0 for all u ∈ I. If [[u, a], a] = 0 for all u ∈ I,
then the inner derivation da satisfies d
2
a(I) = 0. Since daσ = ±σda, according
to Lemma 4, we conclude that da = 0 which proves a ∈ Z(R). If d(a) = 0,
then a[d(u), a] = [d(au), a] ∈ Z(R) which would force [a, r]R[d(u), a] = 0 for
all u ∈ I, r ∈ R. We then conclude a ∈ Z(R) or [d(u), a] = 0 for all u ∈ I.
Suppose that [d(u), a] = 0 for all u ∈ I, then
0 = [d(u[v, a]), a] = [d(u)[v, a], a] = d(u)[[v, a], a] for all u, v ∈ I. (6)
Replace u by uw where w ∈ I, in (6), to get
d(u)w[[v, a], a] = 0, for all u, v, w ∈ I,
and whence
d(u)I [[v, a], a] = 0 = σ(d(u))I [[v, a], a] for all u, v ∈ I.
As d = 0, according to Lemma 1, [[v, a], a] = 0 so that d2a(v) = 0 for all v ∈ I
and Lemma 4 yields da = 0; hence a ∈ Z(R).
Proof of Theorem 1. Suppose that [d(x), x] ∈ Z(R) for all x ∈ I, then
[d(x), y] + [d(y), x] ∈ Z(R) for all x, y ∈ I.
In particular [d(x), x2] + [d(x)x + xd(x), x] ∈ Z(R) for all x ∈ I, so that
x[d(x), x] + [d(x), x]x + [d(x), x]x + x[d(x), x] ∈ Z(R) for all x ∈ I.
Therefore x[d(x), x] ∈ Z(R) for all x ∈ I. Thus, for any r ∈ R, we have that
rx[d(x), x] = x[d(x), x]r = xr[d(x), x]
Derivations and commutativity 445
and so
[r, x][d(x), x] = 0, for all x ∈ I.
In particular, we obtain [d(x), x]2 = 0 for all x ∈ I, so as [d(x), x] ∈ Z(R) we
then have
[d(x), x]R[d(x), x]σ([d(x), x]) for all x ∈ I.
The fact that [d(x), x]σ([d(x), x]) ∈ Saσ(R) combined with the σ-primeness of
R yield
[d(x), x] = 0 or [d(x), x]σ([d(x), x]) = 0.
Suppose that [d(x), x]σ([d(x), x]) = 0, then
[d(x), x]Rσ([d(x), x]) = [d(x), x]R[d(x), x].
Again using the σ-primeness of R we get [d(x), x] = 0. In both cases we have
[d(x), x] = 0, for all x ∈ I.
Applying Lemma 3, R is a commutative ring.
Proof of Theorem 2. If I ∩Z(R) = 0; as d1(x)x−xd2(x) ∈ I ∩Z(R), then
d1(x)x = xd2(x) for all x ∈ I. Linearizing this equality we get
d1(x)y + d1(y)x = xd2(y) + yd2(x) for all x, y ∈ I. (7)
Replace y by yx in (7), the fact that d1(x)x = xd2(x) combined with (7) assure
that
[x, yd2(x)] = 0 for all x, y ∈ I. (8)
In this replace y by ry, where r ∈ R, to get [x, r]yd2(x) = 0 in such a way
that
[x,R]Id2(x) = 0 for all x, z ∈ I.
As d2 = 0, from Lemma 2 we conclude that R is commutative.
Now, assume that I ∩ Z(R) = 0. Let c ∈ I ∩ Z(R); we can suppose that
σ(c) = ±c. Indeed, if σ(c) = c we consider t = c− σ(c), then t ∈ I ∩Z(R) and
σ(t) = −t. We linearize d1(x)x − xd2(x) ∈ Z(R) to get
d1(x)y + d1(y)x − xd2(y) − yd2(x) ∈ Z(R) for all x, y ∈ I. (9)
Replace y by c in (9) and using d2(c) ∈ Z(R), to get
c(d1(x) − d2(x)) + (d1(c) − d2(c))x ∈ Z(R) for all x ∈ I. (10)
Taking y = c2 in (9), in view of (10), we are forced to conclude that
c(d1(c) − d2(c))x ∈ Z(R). Hence, as I is a σ−ideal, we then obtain
c(d1(c) − d2(c))R[x, r] = 0 = c(d1(c) − d2(c))Rσ([x, r]) for all x ∈ I, r ∈ R
446 L. Oukhtite and S. Salhi
which leads us to I ⊂ Z(R) and R is commutative by Lemma 3 or d1(c) = d2(c).
If d1(c) = d2(c), then in view of (10) we get c(d1(x) − d2(x)) ∈ Z(R) and so
d1(x)− d2(x) ∈ Z(R) for all x ∈ I. Therefore, d(I) ⊂ Z(R) where d = d1 − d2.
If d = 0, then it follows from Lemma 3 that R is commutative.
If d1 = d2 then
d1(x)x − xd1(x) = [d1(x), x] ∈ Z(R) for all x ∈ I.
and in view of Theorem 1 we are forced to conclude that R is commutative.
Proof of Theorem 3. Let x, y in I; from [ad(x + y), x + y] = 0 it follows
that
[ad(x), y] + [ad(y), x] = 0 for all x, y ∈ I. (11)
In this we replace y by yx, to obtain
[ad(x), yx] + [ad(y)x, x] + [ayd(x), x] = 0 for all x, y ∈ I,
hence
[ad(x), y]x + [ad(y), x]x + ay[d(x), x] + [ay, x]d(x) = 0 for all x, y ∈ I.
By (11) we conclude that
ay[d(x), x] + a[y, x]d(x) + [a, x]yd(x) = 0 for all x, y ∈ I. (12)
Replacing y by ay in (12), we get
a2y[d(x), x] + a2[y, x]d(x) + a[a, x]yd(x) + [a, x]ayd(x) = 0 for all x, y ∈ I.
We have, using (12), that [a, x]ayd(x) = 0 for all y ∈ I and therefore
[a, x]aId(x) = 0 for all x ∈ I. (13)
If x ∈ I∩Saσ(R), then [a, x]aId(x) = [a, x]aIσ(d(x)) = 0 and Lemma 1 gives
d(x) = 0 or [a, x]a = 0. For any y ∈ I; as y + σ(y) ∈ I ∩ Saσ(R) we have that
d(y + σ(y)) = 0 or [a, y + σ(y)]a = 0.
If d(y+σ(y)) = 0 then d(y) = −σ(d(y)) and so d(y) = 0 or [a, y]a = 0, by (13).
We now suppose [a, y + σ(y)]a = 0; similarly as y − σ(y) ∈ I ∩ Saσ(R) either
d(y − σ(y)) = 0 or [a, y − σ(y)]a = 0. If d(y − σ(y)) = 0 then d(y) = σ(d(y))
and, using (13), we see that d(y) = 0 or [a, y]a = 0.
If [a, y − σ(y)]a = 0 then 0 = [a, y − σ(y)]a + [a, y + σ(y)]a = 2[a, y]a, and
since char(R) = 2 we have that [a, y]a = 0. Therefore, the additive group
I is union of two subgroups A and B where A = {x ∈ I/d(x) = 0} and
B = {x ∈ I/[a, x]a = 0}. But an additive group can’t be the union of two its
proper subgroups. Thus I = A or I = B. If I = A, that is d(x) = 0 for all
Derivations and commutativity 447
x ∈ I, then for any r ∈ R, 0 = d(rx) = d(r)x. This implies that d(r)I = 0 and
Lemma 1 gives d = 0, a contradiction. Consequently, I = B and so [a, x]a = 0
for all x ∈ I. In this replace x by xy where y ∈ I, to get [a, x]ya = 0 so
that [a, x]Ia = 0 = [a, x]Iσ(a). According to Lemma 1, we obtain a = 0 or
a ∈ Z(R). If 0 = a ∈ Z(R), we obtain [ad(x), x] = a[d(x), x] = 0 for all x ∈ I
and thus
aR[d(x), x] = 0 = σ(a)R[d(x), x] for all x ∈ I.
But R is σ−prime so we are forced to [d(x), x] = 0 for all x ∈ I. Applying
Lemma 3, R is then commutative.
Proof of Theorem 4. Let I = 0 be a σ-ideal of R with [d(I), d(I)] ⊂ Z(R).
Applying Lemma 6, this yields d(I) ∩ Saσ(R) ⊂ Z(R). Hence
d(u + σ(u)) ∈ Z(R) and d(u − σ(u)) ∈ Z(R) for all u ∈ I.
It follows that 2d(u) ∈ Z(R) for all u ∈ I. Thus d(I) ⊂ Z(R) since char(R) =
2. In particular [d(x), x] = 0 for all x ∈ I and using Lemma 3, we conclude
that R is commutative.
Remark. As an immediate consequence of Theorem 4, if d is a nonzero
derivation of R which commutes with σ and satisfying 0 = d2(I) ⊂ Z(R),
for a nonzero σ-ideal I of R, then R is commutative. Indeed, let u, v ∈ I;
then d2([u, v]) = 2[d(u), d(v)] ∈ Z(R) and as char(R) = 2, we are forced to
[d(I), d(I)] ⊂ Z(R). Accordingly, R is commutative by Theorem 4.
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448 L. Oukhtite and S. Salhi
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Received: December 15, 2005


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