Periodic Alternative Rings With Commutative Nilpotent Elements

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International Journal of Scientific and Innovative Mathematical Research (IJSIMR)
Volume 3, Special Issue 1, July 2015, PP 76-78, ISSN 2347-307X (Print)
www.arcjournals.org
© ARC Page | 76
PERIODIC ALTERNATIVE RINGS WITH COMMUTATIVE
NILPOTENT ELEMENTS
K.Madhusudhan Reddy
Assistant Professor,
VITUniversity,Vellore,Tamil Nadu,India
C.Jayasubba Reddy
Assistant Professor, S.V.University,
Tirupathi,Andharapradesh,India.
Abstract: Bell [1] studied some results for periodic alternative rings. In 2006 Yaqub [7] studied the
structure of weakly periodic like rings with some conditions on N. He proved that if R is an associative ring
with every noncentral element x is of the form a + b, a is nilpotent and b is potent, and N is commutative,
then R is commutative. In this section we prove some results on the commutativity of a periodic alternative
ring with noncentral elements expressed as sum of nilpotent and potent elements.
Throughout this paperR represents a periodic alternative ring with noncentral elements expressed
as sum of a nilpotent element and a potent element. A ring R is called periodic if for every x in R, there
exists positive integers m and n such that xm = xn. A ring R is called weakly periodic if every x in R can be
written in the form x = a + b for some nilpotent element a and some potent element b. A ring R is weakly
periodic like if every x in R which is not in the center of R can be written in the form x = a + b, where a is
nilpotent and b is potent.
We use the following Lemmas proved in [1, 5, 4, 3].
Lemma 1 : Suppose that R is a ring with center C and N is the set of nilpotents of R. such that
(i) N is commutative,
(ii) for all a in N and x in R, [a, x] commutes with x,
(iii) for all x in R\C, xn – x ∈N for some integer n = n(x) > 1. Then R is commutative.
Lemma 2 :Suppose R is a ring such that for every x in R, these exists an integer n = n(x) > 1 such that x
– xnis in the center of R. Then R is commutative.
Lemma 3 : Suppose R is a ring and the set N of nilpotents of R is commutative. Then the product of any
potent element and any nilpotent element is nilpotent.
Lemma 4 :Suppose R is a ring in which every element is central or potent. Then R is commutative.Now
we prove the properties of periodic alternative rings.
Lemma 5 : A periodic alternative ring R has each of the following properties:
(i) For each x∈R, there exists some positive integer k = k(x) such that xk is idempotent.
(ii) For each x∈R, there exists a positive integer n = n(x) > 1 for which x – xn is nilpotent.
(iii) If A is an ideal of R and y + A is a nonzero nilpotent element of R/A,thenR contains a nilpotent
element a such that x ≡ a(mod A).
Proof : (i) If xn = xmfor n>m, then xj+k(n - m) = xj for each positive integer k and each j≥m. Thus am(n - m) is
idempotent.
(ii) Let xn = xm, n>m> 1.Then xm-1(x – xn-m+1) = 0 = xm - 2x(x - xn-m+1) = xm- 2xn-m+1 (x – xn-m+1).
Therefore xm-2 (x – xn-m+1)2 = 0 and the results follows by the obvious induction.
(iii) If y + A is a nonzero nilpotent element of R/A, then there is a positive integer k such that yq∈A
for each q ≥k. By the proofs of (i) and (ii), R contains a nilpotent element a = y – yq with q ≥k.
Obviouslya≡y (mod A).
Theorem 1 : Let R be a periodic alternative ring and the nilpotent elements of which commute with each
other. Then the commutator ideal C(R) is nil, and the nilpotent elements form an ideal.

PERIODIC ALTERNATIVE RINGS WITH COMMUTATIVE NILPOTENT ELEMENTS
International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page | 77
Proof : We shall use without explicit mention Artin’s result that any two elements of an alternative ring
generate an associative subring.
Let N denote the set of nilpotent elements, and use the standard argument for the commutative case to show
that a1 – a2∈Nwhenever a1, a2∈N. We prove by induction on k that if ak = 0 and r∈R, it is true that (ar)k =
(ra)k = 0.
Consider first the case k = 2. Let a be any element of R such that a2 = 0 and j a positive integer for which
(ar) j = e is idempotent (possibly zero). Then re – ere is nilpotent and hence commutes with a, that is,
(ar) (ar)j – a(ar)j r (ar)j = r(ar)ja – (ar)jr(ar)ja (1)
and multiplying on the right by a yields (ar)(ar)ja= 0, so that (ar)j+2 = (ra)j+2 = 0. It follows that a
commutes with both arand ra, so that (ra)2 = (ar)2 = 0.
Now suppose the result holds for all z with zn= 0, m<k and suppose ak = 0, k≥ 3. Determining j as above and
multiplying 1 by r on the left and a on the right, we get
(ra)j+2 = ra2s + ta2, (2)
wheres and t are elements of the subring generated by r and a. Since (a2)k-1 = 0, the inductive hypothesis
implies both ra2s and ta2 are nilpotent. Therefore 2 shows that ra and ar are nilpotent. Again, invoking the
fact that a must commute with ar and ra, we see that (ra)k = (ar)k = 0. Hence N is an ideal. We apply
Lemma 5(ii) to show that R/N has the “bn = b property”. By Jacobson’s well known theorem for the
associative case [21] and Artin’s theorem, R\N is commutative and the proof of the theorem is complete.
Theorem 2: Let R be a periodic alternative ring with noncentral element expressed as a sum of a
nilpotent element and a potent element
(i) If N is an ideal, then for each x in R, either x∈C or x – xn∈N for some integer n> 1.
(ii) Every ideal A of R is periodic alternative ring with noncentral element expressed as a sum of a
nilpotent element and a potent element.
Proof :(i) Let x∈R\C. Then x = a + b for some nilpotent element a and some potent element b, with bn =
b, n> 1. Hence x – a = b = bn= (x – a)n
Since N is an ideal, it follows that x – xn∈N.
(ii) Suppose i ∈A, i∉C. Then i = a + b, a nilpotent and bn = b potent for some n> 1. Since a is a nilpotent,
am = 0 for some integer m, and hence
Aaibbb
mm
nnn
∈−=== )(
(since A is an ideal and
0=
m
n
a
). Thus b∈A and hence a = i – b∈A, which proves (ii).
Theorem 3 : If R is a periodic alternative ring with noncentral element expressed as a sum of a nilpotent
element and a potent element and N⊆C, then R is commutative.
Proof : This follows from Theorem 2 (i) and Lemma 2.
Theorem 4: Let R be a periodic alternative ring with noncentral element expressed as a sum of a
nilpotent element and a potent element and N is commutative. Then N is an ideal and R/N is commutative.
Proof : Let a∈ N with ak = 0 and x∈R. If x∈C, then (ax)k = akxk = 0, hence ax∈N. On the other hand, if
x∉C, then x = b + u, b potent and u nilpotent. By Lemma 3, ab∈N. Since N is commutative, au∈N and
ax = ab + au ∈N. Similarly xa∈N. Since N is commutative, the difference of any two elements in N is also
in N and hence N is an ideal. By Theorem 2(i) every element of R/N is central or potent and hence by
Lemma 4, R\N is commutative.
This proves the theorem.
Theorem 5: Let R be a periodic alternative ring with noncentral element expressed as a sum of a
nilpotent element and a potent element and if N is commutative, then R is commutative.
Proof : First, we show that all idempotents are central. Suppose e is an idempotent and x∈R. If e + ex –
exe ∈C, then [e, ex – exe] = 0 and hence ex = exe. On the other hand, if e + ex – exe∉C, then e + (ex –
exe) = (e + ex – exe) + 0
are representations as a sum of a potent element and a nilpotent element and hence ex – exe = 0 by
uniqueness assumption. Similarly xe – exe = 0 and hence e∈C. We now distinguish two cases
Case 1:R has an identity 1. We claim that N⊆C (3)
Suppose not. Let w∈N\C. Then 1 + w ∉C. Since we cannot have both 2(1 + w) and 3(1 + w) in C, assume
without loss of generality, that 2(1 + w) ∉C. By Theorem 4, N is an ideal. Since 1 + w∉C and 2(1 + w) ∉C,
it follows by Theorem 2 (i), that for some integers m> 1 and n> 1,(1 + w) – (1 + w)m∈N;
2(1 + w) – (2(1 + w))n∈N.

K.Madhusudhan Reddy & C.Jayasubba Reddy
International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page | 78
Thus, computing in R/N, we have
nm
wwww ))1(2()1(2;)1(1 +=++=+
, (4)
whereq = (m – 1)(n – 1) + 1. Then q> 1, since m> 1 and n> 1. It is readily verified that 4 implies
qq
wwww )
)1(2()1(2;)1(1 +=++=+
; sinceq = (m – 1) (n – 1) + 1 > 1.
Therefore,
0)
1
)(2
2( =
+− w
q
, (q> 1).
Since w∈N,
w+1
is a unit in R/N, and hence the above equation implies that
0
1
)
22
(=
−
q
. Thus (2q
– 2)k 1 = 0 for some positive integer k. Therefore (R, +) is a torsion-group. More over, since (1 + w)q – (1 +
w)∈N, the subring 〈1 + w〉 generated by 1 + w is finite. Furthermore, since 1 + w is a unit in R, 1 + w is
potent. Now we have
1 + (w) = (1 + w) + 0,
and hence by the uniqueness assumption, applied to the noncentral element 1 + w, it follows that w = 0, a
contradiction. This contradiction proves 3. Since N⊆C, R is commutative, by Theorem 3.
Case 2 : 1 ∉R.
Let a be nonzero potent element with an = a, n> 1, and let e = an-1. Then e is a nonzero idempotent which
(as shown above) must be central. Therefore eR is a ring with identity. Moreover, by Theorem 2(ii), eR is a
periodic alternative ring with noncentral element expressed as a sum of a nilpotent element and a potent
element. Also, eR inherits the uniqueness assumption made on the ground ring R, and thus eR satisfies all
the hypothesis of the theorem. Since eR has an identity, it follows, by case 1, thateR is commutative.
Hence, e[x, y] = 0 for all x, y in R. In particular, e[a, y] = 0 for all y in R, and thus (since e = an-1) an-1[a, y]
= 0. Therefore (since an = a), ay = yafor all y in R. SoAll potent elements of R are central.
(5)
To complete the proof, suppose x, y∈R, and suppose x∉C, y∉C.
Then by hypothesis,
x = a + b, y = a′ + b′; a, a′ nilpotent; b, b′ potent. Therefore by 5 and hypothesis that N is commutative.
[x, y] = [a + b, a′ + b′] = [a, a′] = 0.
Thus R is commutative.
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