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EDN: UNTTNT УДК 512.6
Identities Related to Homo-derivation on Ideal in Prime Rings
Nadeem ur Rehman* Hafedh Alnoghashi^
Department of Mathematics Aligarh Muslim University Aligarh, India
Received 20.11.2022, received in revised form 25.12.2022, accepted 20.02.2023 Abstract. This study aims to investigate the commutativity of a prime ring R with a non-zero ideal I and a homo-derivation ё that satisfies certain algebraic identities. We also provided some examples of why our results hypothesis is essential.
Keywords: prime ring, homo-derivation.
Citation: N. ur Rehman, H. Alnoghashi, Identities Related to Homo-derivation on Ideal in Prime Rings, J. Sib. Fed. Univ. Math. Phys., 2023, 16(3), 370-384. EDN: UNTTNT.
1. Introduction
Throughout this article, R denotes a ring with centre Z. A ring R is said to be a prime ring if V a,b £ R, aRb = {0} implies a = 0 or b = 0. For a,b £ R, the symbol [a, b] (resp. a o b) denotes the (resp. anti-) commutator ab — ba (resp. ab + ba). An additive mapping $ : R — R is said to be a homo-derivation if $(ab) = $(a)$(b) + $(a)b + a$(b) V a,b £ R, [14]. The only additive map which is both derivation and homo-derivation on prime ring is the zero map. If D C R, then a mapping $ : R — R preserves D if $(D) C D. A map $ : R — R is called zero-power valued on D if $ preserves D and if for all a £ D, there is a positive integer m(a) > 1 such that $m(a) = 0, El Sofy [14].
In 2001, Ashraf and Rehman [13] showed that if R is a prime ring, I a non-zero ideal of R and $ : R — R is a derivation of R, then R is commutative if R satisfies any one of the following: $(ab) ± ab £ Z, $(ab) ± ba £ Z, $(a)$(b) ± ab £ Z, $(a)$(b) ± ba £ Z, for all a,b £ I.
In 2016, Asmaa Melaibari et. al. [2] showed that if R is a prime ring, I a non-zero ideal of R, and $ : R — R is a non-zero homo-derivation of R, then R is commutative if $([a, b]) = 0 for all a,b £ I. Moreover, if char(R)= 2, $ is zero-power valued on R, and $([a, b]) £ Z for all a, b £ R, then R is commutative.
In 2018, Alharfie and Muthana [7] showed that if R is a prime ring of characteristic = 2, I a non-zero left ideal of R, and $ a homo-derivation of R, which is a zero-power valued on I, then R is commutative if R satisfies any one of the following: a$(b) ± ab £ Z, a$(b) ± ba £ Z, a$(b) ± [a, b] £ Z, $(b)a ± [a, b] £ Z, $(a), b] ± ab £ Z, [$(a), b] ± ba £ Z, for all a,b £ I.
*rehman100@gmail.com, nu.rehman.mm@amu.ac.in https://orcid.org/0000-0003-3955-7941 t halnoghashi@gmail.com https://orcid.org/0000-0003-0253-6573 © Siberian Federal University. All rights reserved
In 2019, they showed that if R is a prime ring, I a non-zero ideal of R and $ : R ^ R is a zero-power valued homo-derivation on I, then R is commutative if $(ab) — ab € Z for all a,b € I [6, Theorem 3.1], or $(ab) + ab € Z for all a,b € I [6, Theorem 3.2]. In fact, we will prove the same result in [6, Theorem 3.1] by replacing the condition "$ is zero-power valued" by "char(R)= 2," as in Theorem 1.2(ii). For ( [6, Theorem 3.2]), they made a mistake in the proof because they replaced $ by —$, given that —$ is homo-derivation, but this is not true in general, take $(a) = —a of any ring of characteristic = 2. So, we will prove the previous result as in Theorem 1.2(i) again. Additional references may be found at [1,3-5,9,10,12,15-19].
Motivated by these results, we will investigate the commutativity of prime ring R with homoderivation $ and a non-zero ideal of I that fulfills specific algebraic identities. We also provided some examples of why our results hypothesis is essential.
Theorem 1.1. Let R be a prime ring, I a non-zero ideal of R and $ a homo-derivation of R preserves I. If any one of the following holds, then R is commutative.
(i) char(R)= 2 and $(a)$(b) + ab € Z V a,b € I;
(i) $(a) = —a and $(a)$(b) — ab € Z V a,b € I;
(ii) char(R)= 2 and $(a)$(b) + ba € Z V a,b € I;
(iii) $(a)$(b) — ba € Z V a,b € I.
Theorem 1.2. Let R be a prime ring, I a non-zero ideal of R and $ a homo-derivation of R preserves I. If any one of the following holds, then R is commutative.
(i) $ is zero-power valued on I and $(ab) + ab € Z V a,b € I;
(ii) char(R)= 2 and $(ab) — ab € Z V a,b € I;
(iii) $(ab) + ba € Z V a,b € I;
(iv) $(ab) — ba € Z V a,b € I.
The next example demonstrates that the hypothesis $ is zero-power valued on I in Theorem 1.2(i) is essential.
Example 1.1. Let Z be the ring of all integers, R = M2(Z), I = 2R, and define $ : R ^ R by $(a) = —a. Then it is easy to see that $ is a homo-derivation such that $(ab) + ab € Z V a,b € I, but $ is not zero-power valued on I and R is not commutative.
The next example shows that char(R)= 2 cannot be omitted in the hypothesis of Theorem 1.2(ii).
Example 1.2. Let R = M2(Z2), I = R, and define $ : R ^ R by $(a) = —a. Moreover, $ is a homo-derivation such that $(ab) — ab € Z V a,b € I, but char(R)= 2 and R is not commutative.
Theorem 1.3. Let R be a prime ring, I a non-zero ideal of R and $ a non-zero homo-derivation of R preserves I. If any one of the following holds, then R is commutative.
(i) $(a) = —a and $(ab) + $(a)$(b) € Z V a,b € I;
(ii) char(R)= 2 and $(ab) — $(a)$(b) € Z V a,b € I;
(iii) char(R)= 2 and $(ab) + $(b)$(a) € Z V a,b € I;
(iv) char(R)= 2 and $(ab) — $(b)$(a) € Z V a,b € 1.
The next example demonstrates that Theorem 1.3(ii) cannot be satisfied without char(R)= 2.
Example 1.3. As in Example 1.2, we see that $(ab)—$(a)$(b) € Z, R is prime and char(R )= 2, but R is not commutative.
The next example demonstrates how essential primeness is to our results.
define ïï : R ^ R by ïï(a) = —a. Then it is easy to see that ïï is a homo-derivation on R, which satisfies the next conditions ïï(a)ïï(b) + ab G Z, â(a)â(b) + ba G Z, ïï(ab) — ab G Z, ïï(ab) + ba G Z, ïï(ab) — ba G Z, ïï(ab) — ïï(a)ïï(b) G Z, ïï(ab) + ïï(b)ïï(a) G Z, and ïï(ab) — ïï(b)ïï(a) G Z, V a, bl, but R is not commutative.
2. Preliminaries
The following fundamental identities that satisfy V a,b,c G R :
will be applied without being mentioned.
To achieve our aim, we will use the next lemmas.
Lemma 2.1 ([8], Lemma 4). Let R a prime ring and {b, ab} C Z. Then a € Z or b = 0.
Lemma 2.2 ([13], Lemma 2.2). Let R be a prime ring. If R contains a non-zero commutative right ideal, then R is a commutative ring.
Lemma 2.3 ([11], Lemma 2.5). Let R be a prime ring and I a non-zero ideal of R. If (i) [a,b] € Z; (ii) a o b € Z V a, b € I. Then R is commutative.
Lemma 2.4. Let R be a prime ring and I a non-zero ideal of R. For a,b € R, if al b = {0}, then a = 0 or b = 0.
Proof. Let a,b € R and alb = {0}. Then alRb = {0}, and so b = 0 or al = {0}. Now, if al = {0}, then a,RJ = {0}. Hence, a = 0 or I = {0}. Since I = {0}, we get a = 0. □
Remark 2.1. Let R be a prime ring, I an ideal of R and $ a homo-derivation of R. If $(a) = a V a € I, then char(R)= 2 or I = {0}.
Proof. Suppose that char(R)= 2. From definition of $, we infer that $(ab) = $(a)$(b) + $(a)b + a$(b) V a,b € I and so ab = ab + ab + ab and hence 2ab = 0 thus ab = 0, that is, 12 = {0} and since R is prime, we see that 1 = {0}. □
[ab, c] = a[b, c] + [a, c]b;
[a, bc] = b[a, c] + [a, b]c; a o bc = (a o b)c — b[a, c] = b(a o c) + [a, b]c; ab o c = a(b o c) — [a, c]b = (a o c)b + a[b, c],
3. The main result
Lemma 3.1. If char(R)= 2 and ê(a)ê(b) + ab G Z V a,b G I, then R is commutative. Proof. Assume that
ê(a)ê(b) + ab G Z Va, b G I. (1)
Suppose that I n Z = {0}.
ê(a)ê(b) + ab = 0. (2)
Replacing b by bt in (2) and using it, where t G I, we have
ê(a)ê(b)ê(t) + ê(a)bê(t) = 0. (3)
Adding ±ê(a)bt in (3) and applying (2), we get ê(a)b(ê(t) —1) = 0, that is, ê(a)1 (ê(t) —1) = {0} and by Lemma 2.4, we infer that ê(a) = 0 or ê(t) — t = 0. If ê(t) — t = 0, then ê(t) = t and by Remark 2.1 a contradiction. Now, in case ê(a) = 0. Using the last expression in (2), we infer that ab = 0, and so I = {0}, a contradiction.
So I n Z = {0}. Substituting bz for b in (1) and applying it, where 0 = z G I n Z, we obtain
ê(a)ê(b)ê(z) + ê(a)bê(z) G Z. (4)
Adding ±û(a)bz in (4), we conclude that ê(a)(ê(b)ê(z) + bz) + ê(a)b(ê(z) — z) G Z and by using (1) in the last relation, we see that [d(a)b(û(z) — z),â(a)] = 0. That is,
[&(a),ê(a)]b(ê(z) — z) + ê(a)[b(ê(z) — z),ê(a)] = 0.
Hence,
ê(a)[b(ê(z) — z),tf(a)]=0. (5)
Writing z instead of a and b by z' in (1), respectively, where z,z' G I n Z, we obtain
ê(z)ê(z') G Z. (6)
Taking z by z2 in (6) and applying it, we get ê(z)ê(z)ê(z') G Z and by using (6) in the previous expression, we conclude ê(z)ê(z') =0 or ê(z) G Z. If ê(z) G Z, then from (5), we see that (ê(z) — z)â(a)[b,â(a)] = 0, that is, (â(z) — z)1ê(a)[b,ê(a)] = {0} and so ê(z) — z = 0 or $(a)[b, $(a)] = 0. If ê(z) — z = 0, then
ê(z)= z = 0. (7)
Now, from definition of ê, we have ê(za) = ê(z)ê(a) + ê(z)a + zê(a). Using (7) in the last relation, we get ê(za) = 2zê(a) + za. Taking a by za in (1) and using the previous relation, we infer that 2zê(a)ê(b) + zaê(b) + zab G Z. Applying (1) in the last expression, we find that zê(a)ê(b) + zaê(b) G Z. Thus,
(ê(a) + a)ê(b) G Z. (8)
Taking b = z in (8) and using (7), we see that ê(a) + a G Z. Using the previous relation and Lemma 2.1 in (8), we arrive at ê(a) + a = 0 or ê(b) G Z. In case ê(b) G Z. Applying the previous relation in (1), we obtain ab G Z. Putting b = z in the last relation, we have a G Z. Hence, I Ç R, and by Lemma 2.2, R is commutative. Now, if ê(a) + a = 0, then ê(a) = —a. Taking a = z in the last relation and by (7), we get a contradiction.
Suppose that
$(a)[b,$(a)] =0. (9)
Replacing b by sb in (9) and applying it, where s € 1, we get $(a)s[b, $(a)] = 0, that is, $(a)1 [b,$(a)] = {0} and so $(a) = 0 or [b,$(a)] = 0. Suppose that [b,$(a)] = 0. Substituting at for a in the last expression and using it, where t € 1, we conclude $(a)[b,t] + [b,a]$(t) = 0. Taking b = a in the previous expression, we see that $(a)[a, t] = 0. Writing ba instead of t in the last relation and applying it, where b € 1, we find that $(a)b[a,t] = 0, that is, $(a)1 [a,t] = {0} and so $(a) =0 or [a, t] = 0. If [a, t] = 0, then 1 C Z, and by Lemma 2.2, we get R is commutative. Now, suppose that
$(a) = 0. (10)
Applying (10) in (1), we infer that ab € Z. Putting 0 = b = z € 1 n Z and by Lemma 2.1, we get a € Z, and by Lemma 2.2, we obtain R is commutative. Now, suppose that $(z)$(z') = 0. Replacing z' by z in the last expression, we get
$(z)$(z) = 0. (11)
Substituting b$(z) for b in (5) and using (11), we conclude $(a)[—b$(z)z,$(a)] = 0, that is, $(a)[b$(z), $(a)] =0 by applying the last relation in (5), we obtain $(a)[—bz,$(a)] = 0, that is, $(a)[b, $(a)] = 0. As in (9), we have R is commutative. □
Lemma 3.2. If $(a)$(b) — ab € Z V a,b € 1, then R is commutative or any homo-derivation $ is of the form $(a) = —a.
Proof. Assume that
$(a)$(b) — ab € Z Va,b € 1. (12)
Suppose that 1 n Z = {0}.
$(a)$(b) — ab = 0. (13)
Writing bt instead of b in (13) and using it, where t € 1, we see that $(a)$(b)$(t)+$(a)b$(t) = 0. Adding ±$(a)bt in the previous expression and applying (13), we get $(a)b($(t) +1) = 0, that is, $(a)1 ($(t) +1) = {0} and so $(a) =0 or $(t) +1 = 0. In case $(a) = 0 and using it in (13), we obtain ab = 0 and so 1 = {0}, a contradiction. If $(t) +1 = 0, then $(t) = —t.
Now, in case 1 n Z = {0}. Replacing b by bz in (12) and applying it, where 0 = z € 1 n Z, we have
$(a)$(b)$(z) + $(a)b$(z) € Z. (14)
Adding ±$(a)bz in (14), we conclude $(a)($(b)$(z) — bz) + $(a)b($(z) + z) € Z. Using (12) in the last relation, we obtain
$(a)[b($(z)+ z),$(a)]=0. (15)
Now as Lemma 3.1 in Eq. (5), we have R is commutative or $(z) + z = 0. If $(z) + z = 0, then $(z) = —z = 0 and by applying the previous expression in (14), we get
$(a)($(b) + b) € Z. (16)
Putting 0 = a = z € 1 n Z in (16), this gives $(z)($(b) + b) € Z, that is, —z($(b) + b) € Z, and hence $(b) + b € Z by Lemma 2.1. Using the previous relation in (16) and by Lemma 2.1, we see that $(a) € Z or $(b) + b = 0. If $(b) + b = 0, then $(b) = —b. In case $(a) € Z and by (12), we infer that ab € Z. Putting 0 = b = z € 1 n Z and by Lemma 2.1, we conclude that a € Z, and by Lemma 2.2, we get R is commutative. □
Lemma 3.3. If char(R)= 2 and ê(a)ê(b) + ba G Z V a,b G I, then R is commutative. Proof. Assume that
ê(a)ê(b) + ba G Z Va, b G I. (17)
Suppose that I n Z = {0}.
ê(a)ê(b) + ba = 0. (18)
Substituting bt for b in (18) and applying it, where t G I, we conclude
ê(a)ê(b)ê(t) + ê(a)ê(b)t + ê(a)bê(t) + bta = 0.
Adding ±bat in the last expression and using (18), we get
ê(a)ê(b)ê(t) + ê(a)bê(t) + b[t, a] = 0.
Adding ±ê(a)tb in the above relation and applying (18), we obtain
ê(a)bê(t) — ê(a)tb + b[t,a]=0. (19)
Writing ê(a)b instead of b in (19), this gives
ê(a)2bê(t) — ê(a)tê(a)b + ê(a)b[t, a] = 0. (20)
Left multiplying (19) by ê(a), we see that
ê(a)2bê(t) — ê(a)2tb + ê(a)b[t, a] = 0. (21)
Comparing (20) and (21), this gives ê(a)[ê(a),t]b = 0, that is, ê(a)[ê(a),t]1 = 0 and since I = 0, we infer that ê(a)[ê(a),t] = 0. Now as in Lemma 3.1 in (9), we have ê(a) =0 or R is commutative. If R is commutative, then I n R = {0}, and so I = {0}, a contradiction. In case ê(a) = 0 and from (18), we get ba = 0 and so I = {0}, contradiction.
So, I n Z = {0}. Replacing b by bz in (17) and using it, where 0 = z G I n Z, we conclude
ê(a)ê(b)ê(z) + ê(a)bê(z) G Z.
Adding ±ê(a)zb in the previous expression, we get
ê(a)(ê(b)ê(z) + zb) + ê(a)(bê(z) — zb) G Z.
Using (17) in the last relation, we see that ê(a)[b(ê(z) — z), ê(a)] = 0. Now, as in Lemma 3.1 in Eq. (5), we have R is commutative. □
Lemma 3.4. If ê(a)ê(b) — ba G Z V a,b G I, then R is commutative. Proof. Assume that
ê(a)ê(b) — ba G Z Va, b G I. (22)
In case I n Z = {0} as Lemma 3.3, we have a contradiction. So, I n Z = {0}. Substituting bz for b in (22) and applying it, where 0 = z G I n Z, we conclude
ê(a)ê(b)ê(z) + ê(a)bê(z) G Z. (23)
Adding ±ê(a)zb in (23), we get
ê(a)(ê(b)ê(z) — zb) + ê(a)b(ê(z) + z) G Z.
Using (22) in the above expression, we obtain ê(a)[b(ê(z)+z), ê(a)] = 0. Now, as in Lemma 3.2 in Eq. (15), we have R is commutative or ê(a) = —a. Now, in case ê(a) = —a. Applying the previous relation in (22), we get [a, b] G Z and by using Lemma 2.3, we obtain R is commutative. □
Lemma 3.5. If $ is zero-power valued on 1 and $(ab) + ab € Z V a,b € 1, then R is commutative.
Proof. Assume that
$(ab) + ab € Z Va,b € 1. (24)
Writing bt instead of b in (24), where t € 1, we see that ($(ab) + ab)($(t) +1) € Z and by applying (24) in the last relation and by Lemma 2.1, we get $(ab) + ab = 0 or $(t) +1 € Z. If $(ab) + ab = 0, then
($(a) + a)($(b)+ b)=0. (25)
Replacing b by $n-1(b) in (25), where $n(b) = 0, we conclude
($(a) + a)$n-1(b)=0. (26)
Substituting $n-2(b) for b in (25) and using (26), we see that
($(a)+ a)$n-2(b) =0
and by repeating the previous steps, we conclude that ($(a) + a)b = 0. Now as in Eq. (25), we have ab = 0 and so 1 = {0}, contradiction. So $(t) +1 € Z. As in Eq. (25), we get t € Z and by Lemma 2.2, R is commutative. □
Lemma 3.6. If char(R)= 2 and $(ab) — ab € Z V a,b € 1, then R is commutative.
Proof. Assume that
$(ab) — ab € Z Va,b € 1. (27)
Writing bt instead of b in (27), where t € 1, we conclude
($(ab) — ab)($(t) + t) + 2ab$(t) € Z. (28)
Thus 2[ab$(t), $(t)+t] = 0, that is, [ab$(t), $(t)+t] = 0. Replacing a by sa in the previous expression and applying it, where s € 1, we get [s, $(t) + t]ab$(t) = 0, that is, [s, $(t) +1]1 b$(t) = {0} and so [s,$(t) +1] =0 or b$(t) = 0. If b$(t) = 0, then 1$(t) = {0} and hence $(t) = 0. Using the last relation in (27), we obtain
ab € Z. (29)
Replacing a by b and b by a in (29), we see that ba € Z. From the last expression and (29), we have [a,b] € Z and by Lemma 2.3(i), R is commutative. Suppose that [s,$(t) +1] =0. Substituting sr for s in the previous relation and applying it, where r € R, we get s[r, $(t) +1] = 0, that is, 1 [r,$(t)+1] = {0} and so [r,$(t)+1] = 0 thus $(t)+1 € Z. Again writing ab instead of t in the last expression, we see that $(ab) + ab € Z. Comparing the last relation and (27), we conclude a o b € Z, and by Lemma 2.3(ii), R is commutative. □
Lemma 3.7. If $(ab) + ba € Z V a,b € 1, then R is commutative. Proof. Assume that
$(ab) + ba € Z Va,b € 1. (30)
Suppose that 1 n Z = {0}.
${ab) + ba = 0. (31)
Replacing b by bt in (31), where t G 1, we conclude {${ab) + ab)${t) + b[t, a] = 0. Adding ±ba${t) in the previous expression and using (31), we get b([t,a] — a${t)) = 0 and since 1 = {0}, we infer that
[t, a] — a${t) = 0. (32)
Substituting sa for a in (32), where s G 1, we obtain
s[t, a] + [t, s]a — sa${t) = 0. (33)
Left multiplying (32) by s, we see that
s[t,a] — sa${t)=0. (34)
Comparing (33) and (34), this gives [t, s]a = 0, that is, [t,s]1 = {0} and since 1 = {0}, we infer that [t, s] = 0 and so 1 C Z and by Lemma 2.2, R is commutative.
Suppose that 1 n Z = {0}. Writing bz instead of b in (30) and applying it, where 0 = z G 1 n Z, we conclude
{${ab) + ab)${z) G Z. (35)
Putting a = b = z in (30), where 0 = z G 1 n Z, we get
${z2) G Z. (36)
Taking z by z2 in (35), we have {${ab) + ab)${z2) G Z and by using (36) in the last relation, we see that ${ab) + ab G Z or ß(z2) = 0. If ß(ab) + ab G Z and by applying (30), then [a, b] G Z and so R is commutative. Suppose that
ß(z2)=0. (37)
Replacing b by bt in (30), we conclude ß(ab)ß(t) + ß(ab)t + abd(t) + bta G Z, where t G 1. Putting a = b = z in the last expression and using (37), where 0 = z G 1 n Z, this gives d{t) +1 G Z. Taking t by ab in the previous relation where a,b G 1, we get d{ab) + ab G Z. Using the last expression in (30), we conclude [a, b] G Z and so R is commutative. □
Lemma 3.8. If ${ab) — ba G Z V a,b G 1, then R is commutative.
Proof. Let 1 n Z = {0}. As in Lemma 3.7. Suppose that 1 n Z = {0}. As in Lemma 3.7 and applying Lemma 2.3(ii). □
Lemma 3.9. If $ = 0 and ${ab) + ß{a)ß{b) G Z V a,b G 1, then R is commutative or any homo-derivation $ is of the form ${a) = —a.
Proof. Assume that
${ab) + ${a)${b) G Z Va,b G 1. (38)
Suppose that 1 n Z = {0}.
${ab)+ ${a)${b) = 0. (39)
Substituting bt for b in (39) and using it, where t € 1, we see that ($(a) + a)b$(t) = 0, that is, ($(a) + a)1 $(t) = {0} and so $(a) + a = 0 or $(t) = 0. In case
$(t) = 0 Vt € 1. (40)
Writing rt instead of t in the last relation and applying it, where r € R, we get $(r)t = 0, that is $(r)1 = {0} and so $(r) = 0 V r € R, a contradiction. In case $(a) + a = 0, we infer that $(a) = — a.
Suppose that 1 n Z = {0}. Replacing b by bz in (38) and using it, where 0 = z € 1 n Z, we conclude
($(ab) + $(a)$(b))$(z) + (a + $(a))b$(z) € Z. (41)
Using (38) in (41), we obtain
[(a + $(a))b, $(z)]$(z) = 0. (42)
Substituting (z + $(z))b for b in (42) and applying it, where 0 = z € 1 n Z, we conclude
[a + $(a), $(z)]zb$(z) = 0 implies that [a + $(a), $(z)]b$(z) = 0, that is, [a + $(a), $(z)]1 $(z) = {0} and so [a + $(a), $(z)] =0 or $(z) = 0. In case
$(z) = 0. (43)
Putting b = z in (38) and using (43), where 0 = z € 1 n Z, we see that $(a)z € Z and hence
$(a) € Z. (44)
Writing ab instead of a in (44) and applying it, where b € 1, we get $(a)b + a$(b) € Z and by using (44) in the previous expression, we have $(a)[b, a] = 0. Replacing b by sb in the last relation and applying it, where s € 1, we obtain $(a)s[b, a] = 0, that is $(a)1 [b, a] = {0} and so $(a) = 0 or [b, a] = 0. If [b, a] = 0, then 1 C Z and so R is commutative. In case $(a) = 0 the same as in (40), we infer that $ = 0, a contradiction. Now, suppose that $(z) = 0 and [a + $(a), $(z)] = 0. By using the last expression in (42), this gives (a + $(a))[b, $(z)]$(z) = 0. Substituting bs for b in the previous relation and applying it, where s € 1, we conclude (a + $(a))[b, $(z)]s$(z) = 0, that is, (a + $(a))[b, $(z)]1$(z) = {0} and so (a + $(a))[b, $(z)] =0 or $(z) = 0. But $(z) = 0 and hence (a+$(a))[b, $(z)] = 0. Writing sb instead of b in the last expression and using it, where s € 1, we get (a + $(a))s[b, $(z)] = 0, that is, (a + $(a))1 [b,$(z)] = {0} and so a + $(a) =0 or [b,$(z)] = 0. If a + $(a) = 0, then $(a) = —a. Now, suppose that [b,$(z)] = 0. Replacing b by br in the last relation and applying it, where r € R, we have b[r, $(z)] = 0, that is, 1 [r, $(z)] = {0} and so [r, $(z)] = 0 this implies that
$(z) € Z. (45)
Using (45) and (38) in (41), we see that (a + $(a))b$(z) € Z and by applying (45) in the previous expression, we get (a + $(a))b € Z or $(z) = 0. In case $(z) =0 as (43). Now, suppose that (a + $(a))b € Z. Substituting br for b in the last relation and using it, where r € R, we obtain (a + $(a))b = 0 or r € Z. If r € Z, then R is commutative. If (a + $(a))b = 0, then (a + $(a))1 = {0} and so a + $(a) = 0, and hence $(a) = —a. □
Lemma 3.10. If char(R)= 2 and $(ab) — $(a)$(b) € Z V a,b € 1, then R is commutative.
Proof. Assume that
$(ab) — $(a)$(b) € Z Va,b € 1. (46)
$(a)b + a$(b) £ Z. (47)
Writing bt instead of b in (47), where t £ I, we conclude
($(a)b + a$(b))t + a($(b) + b)$(t) £ Z
and by applying (47) in the last expression, we get [a($(b) + b)$(t),t] = 0. Replacing a by sa in the previous relation and using it, where s £ I, we obtain [s,t]a($(b) + b)$(t) = 0, that is, [s,t]1 ($(b) + b)$(t) = {0} and so [s,t] = 0 or ($(b) + b)$(t) = 0. If [s,t] = 0, then I C Z and hence R is commutative. Suppose that ($(b) + b)$(t) = 0. Substituting at for t in the last expression and applying it, where a £ I, we see that ($(b) + b)a$(t) = 0, that is, ($(b) + b)1$(t) = {0} and so $(b) + b = 0 or $(t) = 0. In case $(t) = 0, the same as in (40), we infer that $ = 0, a contradiction. If $(b) + b = 0, then $(b) = —b and by using the last relation in (46), we conclude 2ab £ Z and so ab £ Z. Now as Lemma 3.6 in Eq.(29) we get R is commutative. □
Lemma 3.11. If char(R)= 2 and $(ab) + $(b)$(a) £ Z V a,b £ I, then R is commutative.
Proof. Assume that
$(ab) + $(b)$(a) £ Z Va,b £ I. (48)
Suppose that I n Z = {0}. Then
$(ab)+ $(b)$(a) = 0, (49)
that is,
$(a) o $(b) + $(a)b + a$(b) = 0. (50)
Writing ab instead of b in (50), we conclude $(a) o $(ab) + $(a)ab + a$(ab) = 0. Using (49) in the previous expression, we get
— ($(a) o $(b)$(a)) + $(a)ab — a$(b)$(a) = 0.
That is,
— ($(a) o $(b))$(a) + $(a)ab — a$(b)$(a) = 0. (51) Right multiplying (50) by $(a), this gives
($(a) o $(b))$(a) + $(a)b$(a) + a$(b)$(a) = 0. (52)
Comparing (51) and (52), we obtain
$(a)b$(a)+ $(a)ab = 0. (53)
Replacing b by bt in (51), where t £ I, we see that
$(a)bt$(a) + $(a)abt = 0. (54)
Right multiplying (53) by t, where t £ I, we conclude
$(a)b$(a)t + $(a)abt = 0. (55)
Comparing (54) and (55), we get
ê(a)b[ê(a),t]=0, (56)
that is, ê(a)1 [ê(a),t] = {0} and so ê(a) =0 or [ê(a),t] = 0. Assume that ê(a) = 0, the same as in (40), we infer that ê = 0, a contradiction. In case [ê(a),t] = 0. Substituting ab for a in the last relation and using it, where b G I, we obtain ê(a)[b,t] + [a,t]ê(b) = 0. Putting t = b in the last expression, this gives [a, b]ê(b) = 0. Writing as instead of a in the previous relation and applying it, where s G I, we see that [a, b]sê(b) = 0, that is, [a, b]1ê(b) = {0} and so [a, b] = 0 or ê(b) = 0. If ê(b) = 0, then ê = 0, a contradiction. In case [a, b] = 0, we infer that R is commutative.
Now, suppose that I n Z = {0}. We can write Eq. (48) as
(ê(a) o ê(b)) + ê(a)b + aê(b) G Z. (57)
Replacing b by bz in (57) and using it, where 0 = z G I n Z, we get
(ê(a) o ê(b)ê(z)) + (ê(a) o bê(z)) + aê(b)ê(z) + abê(z) G Z. (58)
Putting a = z and adding ±ê(z)bê(z) in (58), we have
{(ê(z) o ê(b)) + zê(b) + ê(z)b}ê(z) + b{ê(z)2 + zê(z)} G Z. (59)
Taking a = b = z in (57) and since char(R)= 2, we infer that
ê(z)2 + zê(z) G Z. (60)
Using (57) and (60) in (59), we obtain {(ê(z) o ê(b)) + zê(b) + ê(z)b}[ê(z),b] = 0. Applying (57) in the last expression, we get {(ê(z) o ê(b)) + zê(b) + ê(z)b}1 [ê(z), b] = {0} and so (ê(z) o ê(b)) + zê(b) + ê(z)b = 0 or [ê(z), b] = 0. Suppose that
[ê(z),b]=0. (61)
Substituting rb for b in (61) and using it, where r G R, we conclude [ê(z),r]b = 0, that is, [ê(z),r]1 = {0} and so [ê(z),r] = 0 and hence
ê(z) G Z. (62)
Adding ±ê(a)bê(z) in (58) and applying (57) and (62), we see that (bê(a) + ab)ê(z) G Z and by using (62) in the last relation, we get bê(a) + ab G Z or ê(z) = 0. In case
ê(z) = 0. (63)
Putting b = z in (57), where 0 = z G I n Z and applying the previous expression, we conclude zê(a) G Z and so
ê(a) G Z. (64)
Writing ab instead of a in (64) and using it, where b G I, we obtain ê(a)b + aê(b) G Z and by applying (64) in the last relation, we get [ê(a)b + aê(b),r] = 0, where r G R, by using (64) in the last expression, we see that ê(a)[b, r] + [a, r]ê(b) = 0. Taking r = b in the previous relation, this gives [a, b]ê(b) = 0. Replacing a by as in the last expression, where s G I, we conclude [a,b]sê(b) = 0, that is, [a,b]1ê(b) = {0} and so [a,b] =0 or ê(b) = 0. If [a,b] = 0, then R is commutative. If ê(b) = 0, then ê = 0, a contradiction. Now, suppose that
bê(a) + ab G Z. (65)
Putting 0 = b = z G I n Z in (65), we get
ê(a) + a G Z.
(66)
Adding ±ba in (65), this gives b($(a) + a) + [a, b] £ Z. Using 66) in the last relation, we obtain [[a, b],b] = 0. Substituting ta for a in the previous expression and applying it, where t £ I, we have 2[t, b][a, b] = 0 and hence [t, b][a, b] = 0. Writing sa instead of a in the last relation and using it, where s £ I, we see that [t, b]s[a, b] = 0, that is, [t, b]1 [a, b] = {0} and so [t, b] = 0 or [a, b] = 0. In two cases R is commutative.
Now, in case ($(z) o $(b)) + z$(b) + $(z)b = 0 as in Eq.(50), we conclude as Eq.(56), $(z)b[$(z),t] = 0, that is, $(z)1 [$(z),t] = {0} and so $(z) = 0 or [$(z),t] = 0. In case [$(z),t] = 0, as in (61). In case $(z) = 0 as in (63). □
Lemma 3.12. If char(R)= 2 and $(ab) — $(b)$(a) £ Z V a,b £ I, then R is commutative.
Proof. Assume that
ê(ab) — ê(b)ê(a) G Z Va, b G I.
Suppose that I n Z = {0}. Then
That is,
ê(ab) — ê(b)ê(a) = 0.
[ê(a), ê(b)] + ê(a)b + aê(b) = 0.
(67)
(68)
(69)
Replacing b by ab in (69), we have [$(a),$(ab)] + $(a)ab + a$(ab) = 0. Using (68) in the last relation, we get [$(a), $(b)$(a)] + $(a)ab + a$(b)$(a) = 0. Thus,
[$(a), $(b)]$(a) + $(a)ab + a$(b)$(a) = 0. Right multiplying (69) by $(a) this gives
[$(a),$(b)]$(a) + $(a)b$(a) + a$(b)$(a) = 0. Comparing (70) and (71), we obtain
$(a)b$(a) — $(a)ab = 0. Substituting bt for b in (72), where t £ I, we see that
$(a)bt$(a) — $(a)abt = 0. Right multiplying (72) by t, where t £ I, we conclude
$(a)b$(a)t — $(a)abt = 0.
(70)
(71)
(72)
(73)
(74)
Comparing (73) and (74), we get ê(a)b[ê(a),t] = 0, that is, ê(a)1 [ê(a),t] = {0} and so ê(a) = 0 or [ê(a),t] = 0. If ê(a) = 0, then ê = 0, a contradiction. In case [ê(a),t] = 0. Writing ab instead of a in the previous relation and using it, where b G I, we obtain ê(a)[b,t] + [a,t]ê(b) = 0. Putting t = b in the last expression, this gives [a, b]ê(b) = 0. Replacing a by as in the last relation and applying it, where s G I, we see that [a, b]sê(b) = 0, that is, [a, b]1ê(b) = {0} and so [a, b] = 0 or ê(b) = 0. If ê(b) = 0, then ê = 0, a contradiction. In case [a, b] = 0, we infer that R is commutative.
Now, suppose that 1 n Z = {0}. Putting a = b = z in (67), where 0 = z € 1 n Z, we conclude
$(z) € Z. (75)
Substituting bz for b in (67) and using it and (75), where 0 = z € 1 n Z, we get
[$(a), $(b)]$(z) + [$(a), b]$(z) + a$(b)$(z) + ab$(z) € Z.
Adding ±$(a)b$(z) in the previous expression and suing (67) and (75), we obtain
([$(a),b] + ab — $(a)b)$(z) € Z
and so [$(a), b] + ab — $(a)b € Z or $(z) = 0. In case
$(z) = 0. (76)
Taking b = z in (67) and applying the above relation, where 0 = z € 1 nZ, we see that $(a)z € Z and so
$(a) € Z. (77)
By using (77) in (67), we have $(a)b + a$(b) € Z and so $(a)[b, a] = 0. Writing sb instead of b in the last expression, where s € 1, we get $(a)s[b, a] = 0, that is, $(a)1 [b,a] = {0} and so $(a) = 0 or [b, a] = 0. If [b, a] = 0, then R is commutative. If $(a) = 0, then $ = 0, a contradiction. Now, suppose that
$(a),b] + ab — $(a)b € Z. (78)
Putting b = z in (78), where 0 = z € 1 n Z, we see that
a — $(a) € Z. (79)
Replacing b by $(b) in (78), this gives
[$(a), $(b)] + a$(b) — $(a)$(b) € Z. (80)
Comparing (80) and (67), we conclude $(a)($(b) + b) € Z. Taking a = z in the previous relation, where 0 = z € 1 nZ, we get $(z)($(b)+ b) € Z and so $(z) = 0 or $(b) + b € Z. In case $(z) = 0 as Eq. (76). In case $(b) + b € Z. Putting b = a in the last expression, where a € 1, we obtain
$(a) + a € Z. (81)
Comparing (80) and (79), we see that 2a € Z and so a € Z, and hence R is commutative. □
By using Lemmas 3.1-3.12, we get the proof of Theorems 1.1-1.3.
The authors are greatly indebted to the referee for his/her valuable suggestions, which have immensely improved the paper.
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Тождества, связанные с гомообразованием на идеале в первичных кольцах
Надим Рехман Хафед Альногаши
Кафедра математики Алигархский мусульманский университет Алигарх, Индия
Аннотация. Целью данной работы является исследование коммутативности первичного кольца М с ненулевым идеалом I и гомодифференцированием д, удовлетворяющим некоторым алгебраическим тождествам. Мы также привели несколько примеров того, почему наша гипотеза о результатах важна.
Ключевые слова: первичное кольцо, гомопроисхождение.
