Chinese medical knowledge-based question answering (cMed-KBQA) is a vital component of the intelligence question-answering assignment. Its purpose is to enable the model to comprehend questions and then deduce the proper answer from the knowledge base. Previous methods solely considered how questions and knowledge base paths were represented, disregarding their significance. Due to entity and path sparsity, the performance of question and answer cannot be effectively enhanced. To address this challenge, this paper presents a structured methodology for the cMed-KBQA based on the cognitive science dual systems theory by synchronizing an observation stage (System 1) and an expressive reasoning stage (System 2). System 1 learns the question's representation and queries the associated simple path. Then System 2 retrieves complicated paths for the question from the knowledge base by using the simple path provided by System 1. Specifically, System 1 is implemented by the entity extraction module, entity linking module, simple path retrieval module, and simple path-matching model. Meanwhile, System 2 is performed by using the complex path retrieval module and complex path-matching model. The public CKBQA2019 and CKBQA2020 datasets were extensively studied to evaluate the suggested technique. Using the metric average F1-score, our model achieved 78.12% on CKBQA2019 and 86.60% on CKBQA2020.
Citation: Meiling Wang, Xiaohai He, Zhao Zhang, Luping Liu, Linbo Qing, Yan Liu. Dual-process system based on mixed semantic fusion for Chinese medical knowledge-based question answering[J]. Mathematical Biosciences and Engineering, 2023, 20(3): 4912-4939. doi: 10.3934/mbe.2023228
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Chinese medical knowledge-based question answering (cMed-KBQA) is a vital component of the intelligence question-answering assignment. Its purpose is to enable the model to comprehend questions and then deduce the proper answer from the knowledge base. Previous methods solely considered how questions and knowledge base paths were represented, disregarding their significance. Due to entity and path sparsity, the performance of question and answer cannot be effectively enhanced. To address this challenge, this paper presents a structured methodology for the cMed-KBQA based on the cognitive science dual systems theory by synchronizing an observation stage (System 1) and an expressive reasoning stage (System 2). System 1 learns the question's representation and queries the associated simple path. Then System 2 retrieves complicated paths for the question from the knowledge base by using the simple path provided by System 1. Specifically, System 1 is implemented by the entity extraction module, entity linking module, simple path retrieval module, and simple path-matching model. Meanwhile, System 2 is performed by using the complex path retrieval module and complex path-matching model. The public CKBQA2019 and CKBQA2020 datasets were extensively studied to evaluate the suggested technique. Using the metric average F1-score, our model achieved 78.12% on CKBQA2019 and 86.60% on CKBQA2020.
Semirings have significant applications in theory of automata, optimization theory, and in theoretical computer sciences (see [1,2,3]). A group of Russian mathematicians was able to create novel probability theory based on additive inverse semirings, called idempotent analysis (see[4,5]) having interesting applications in quantum physics. Javed et al. [6] identified a proper subclass of semirings known as MA-Semirings. The development of commutator identities and Lie type theory of semirings [6,7,8,9,10] and derivations [6,7,8,11,12] make this class quite interesting for researchers. To investigate commuting conditions for rings through certain differential identities and certain ideals are still interesting problems for researchers in ring theory (see for example [13,14,15,16,17,18,19]) and some of them are generalized in semirings (see [6,8,9,10,11,20]). In this paper we investigate commuting conditions of prime MA-semirings through certain differential identities and Jordan ideals (Theorems 2.5–2.8) and also study differential identities with the help of Jordan ideals (Theorem 2.3, Theorem 2.4, Theorem 2.10). In this connection we are able to generalize a few results of Oukhtite [21] in the setting of semirings. Now we present some necessary definitions and preliminaries which will be very useful for the sequel. By a semiring S, we mean a semiring with absorbing zero '0' in which addition is commutative. A semiring S is said to be additive inverse semiring if for each s∈S there is a unique s′∈S such that s+s′+s=s and s′+s+s′=s′, where s′ denotes the pseudo inverse of s. An additive inverse semiring S is said to be an MA-semiring if it satisfies s+s′∈Z(S),∀s∈S, where Z(S) is the center of S. The class of MA-semirings properly contains the class of distributive lattices and the class of rings, we refer [6,8,11,22] for examples. Throughout the paper by semiring S we mean an MA-semiring unless stated otherwise. A semiring S is prime if aSb={0} implies that a=0 or b=0 and semiprime if aSa={0} implies that a=0. S is 2-torsion free if for s∈S,2s=0 implies s=0. An additive mapping d:S⟶S is a derivation if d(st)=d(s)t+sd(t). The commutator is defined as [s,t]=st+t′s. By Jordan product, we mean s∘t=st+ts for all s,t∈S. The notion of Jordan ideals was introduced by Herstein [23] in rings which is further extended canonically by Sara [20] for semirings. An additive subsemigroup G of S is called the Jordan ideal if s∘j∈G for all s∈S,j∈G. A mapping f:S⟶S is commuting if [f(s),s]=0, ∀s∈S. A mapping f:S⟶S is centralizing if [[f(s),s],r]=0, ∀s,r∈S. Next we include some well established identities of MA-semirings which will be very useful in the sequel. If s,t,z∈S and d is a derivation of S, then [s,st]=s[s,t], [st,z]=s[t,z]+[s,z]t, [s,tz]=[s,t]z+t[s,z], [s,t]+[t,s]=t(s+s′)=s(t+t′), (st)′=s′t=st′, [s,t]′=[s,t′]=[s′,t], s∘(t+z)=s∘t+s∘z, d(s′)=(d(s))′. To see more, we refer [6,7].
From the literature we recall a few results of MA-semirings required to establish the main results.
Lemma 1. [11] Let G be a Jordan ideal of an MA-semiring S. Then for all j∈G (a). 2[S,S]G⊆G (b). 2G[S,S]⊆G (c). 4j2S⊆G (d). 4Sj2⊆G (e). 4jSj⊆G.
Lemma 2. [11] Let S be a 2-torsion free prime MA-semiring and G a Jordan ideal of S. If aGb={0} then a=0 or b=0.
In view of Lemma 1 and Lemma 2, we give some very useful remarks.
Remark 1. [11]
a). If r,s,t∈S,u∈G, then 2[r,st]u∈G.
b). If aG={0} or Ga={0}, then a=0.
Lemma 3. [12] Let G be a nonzero Jordan ideal and d be a derivation of a 2-torsion free prime MA-semiring S such that for all u∈G, d(u2)=0. Then d=0.
Lemma 4. Let G be a nonzero Jordan ideal of a 2-torsion free prime MA-semiring S. If a∈S such that for all g∈G, [a,g2]=0. Then [a,s]=0,∀s∈S and hence a∈Z(S).
Proof. Define a function da:S⟶S by da(s)=[a,s], which is an inner derivation. As every inner derivation is derivation, therefore in view of hypothesis da is derivation satisfying da(g2)=[a,g2]=0,∀g∈G. By Lemma 3, da=0, which implies that da(s)=[a,s]=0, for all s∈S. Hence a∈Z(S).
Lemma 5. Let S be a 2-torsion free prime MA-semiring and G a nonzero Jordan ideal of S. If S is noncommutative such that for all u,v∈G and r∈S
a[r,uv]b=0, | (2.1) |
then a=0 or b=0.
Proof. In (2.1) replacing r by ar and using MA-semiring identities, we obtain
aa[r,uv]b+a[a,uv]rb=0 | (2.2) |
Using (2.1) again, we get a[a,uv]Sb=0. By the primeness of S, we have either b=0 or a[a,uv]=0. Suppose that
a[a,uv]=0 | (2.3) |
In view of Lemma 1, replacing v by 2v[s,t] in (2.3) and using 2-torsion freeness of S, we get 0=a[a,uv[s,t]]=auv[a,[s,t]]+a[a,uv][s,t]. Using (2.3) again auv[a,[s,t]]=0 and therefore auG[a,[s,t]]={0}. By the Lemma 2, we have either aG={0} or [a,[s,t]]=0. By Remark 1, aG={0} implies a=0. Suppose that
[a,[s,t]]=0 | (2.4) |
In (2.4) replacing s by sa, we get [a,s[a,t]]+[a,[s,t]a]=0 and therefore [a,s[a,t]]+[a,[s,t]]a=0. Using (2.4) again, we get [a,s][a,t]=0. By the primeness of S, [a,s]=0 and therefore a∈Z(S). Hence from (2.2), we can write aS[r,uv]b={0}. By the primeness of S, we obtain a=0 or
[r,uv]b=0 | (2.5) |
In (2.5) replacing r by rs and using (2.5) again, we get [r,uv]Sb={0}. By the primeness of S, we have either b=0 or [r,uv]=0. Suppose that
[r,uv]=0 | (2.6) |
In (2.6) replacing y by 2v[s,t] and using (2.6) again, we obtain 2[r,uv[s,t]]=0. As S is 2-torsion free, [r,uv[s,t]]=0 which further gives uG[r,[s,t]]={0}. As G≠{0}, by Lemma 2 [r,[s,t]]=0 which shows that S is commutative, a contradiction. Hence we conclude that a=0 or b=0.
Theorem 1. Let S be a 2-torsion free prime MA-semiring and G a nonzero Jordan ideal of S. If d1 and d2 are derivations of S such that for all u∈G,
d1d2(u)=0 | (2.7) |
then either d1=0 or d2=0.
Proof. Suppose that d2≠0. We will show that d1=0. In view of Lemma 1, replacing u by 4u2v,v∈G in (2.7), we obtain d1d2(4u2v)=0 and by the 2-torsion freeness of S, we have d1d2(u2v)=0. Using (2.7) again, we obtain
d2(u2)d1(v)+d1(u2)d2(v)=0 | (2.8) |
By lemma 1, replacing v by 2[r,jk]v,j,k∈G in (2.8), we get
d2(u2)d1(2[r,jk]v)+d1(u2)d2(2[r,jk]v)=0 |
and
2d2(u2)[r,jk]d1(v)+2d2(u2)d1([r,jk])v+2d1(u2)[r,jk]d2(v)+2d1(u2)d2([r,jk])v=0 |
Using (2.8) again and hence by the 2-torsion freeness of S, we obtain
d2(u2)[r,jk]d1(v)+d1(u2)[r,jk]d2(v)=0 | (2.9) |
In (2.9), replacing v by 4v2t,t∈S and using (2.9) again, we obtain
4d2(u2)[r,jk]v2d1(t)+4d1(u2)[r,jk]v2d2(t)=0 |
As S is 2-torsion free, therefore
d2(u2)[r,jk]v2d1(t)+d1(u2)[r,jk]v2d2(t)=0 | (2.10) |
In (2.10), taking t=d2(g),g∈G and using (2.7), we obtain
d1(u2)[r,jk]v2d2(d2(g))=0 | (2.11) |
In (2.11) writing a for d1(u2) and b for v2d2(d2(g)), we have a[r,jk]b=0,∀r∈S,j,k∈G.
Firstly suppose that S is not commutative. By Lemma 5, we have a=0 or b=0. If d1(u2)=a=0, then by Lemma 3, d1=0. Secondly suppose that S is commutative. In (2.7) replacing u by 2u2, we obtain 0=d1d2(2u2)=2d1d2(u2)=4d1(ud2(u))=4(d1(u)d2(u)+ud1d2(u)). Using (2.7) and the 2-torsion freeness of S, we obtain d1(u)d2(u)=0. By our assumption S is commutative, therefore d1(u)Sd2(u)={0}. By the primeness of S, we have either d1(G)={0} or d2(G)={0}. By Theorem 2.4 of [11], we have d1=0 or d2=0. But d2≠0. Hence d1=0 which completes the proof.
Theorem 2. Let S be a 2-torsion free prime MA-semiring and G a nonzero Jordan ideal of S. If d1 and d2 are derivations of S such that for all u∈G
d1(d2(u)+u′)=0, | (2.12) |
then d1=0.
Proof. Firstly suppose that S is commutative. Replacing u by 2u2 in (2.12) and using (2.12) again, we obtain d1(u)d2(u)=0 which further implies d1(u)Sd2(u)={0}. In view of Theorem 2.4 of [11], by the primeness of S we have d1=0 or d2=0. If d2=0, then from (2.12), we obtain d1(u)=0,∀u∈G and hence by Lemma 3, we conclude d1=0. Secondly suppose that S is noncommutative. Further suppose that d2≠0. We will show that d1=0. In (2.12) replacing u by 4u2v,v∈G, and using (2.12) again, we obtain 2(d2(u2)d1(v)+d1(u2)d2(v))=0. As S is 2-torsion free, therefore
d2(u2)d1(v)+d1(u2)d2(v)=0 | (2.13) |
In (2.13) replacing v by 2[r,jk]v,r∈S,j,k,v∈G, we obtain
d2(u2)d1(2[r,jk])v+2d2(u2)[r,jk]d1(v)+d1(u2)d2(2[r,jk])v+2d1(u2)[r,jk]d2(v)=0 |
As by MA-semiring identities, 2[r,jk]=2j[r,k]+2[r,j]k, by Lemma 1 2[r,jk]∈G. Therefore using (2.13) again and the 2-torsion freeness of S, we obtain
d2(u2)[r,jk]d1(v)+d1(u2)[r,jk]d2(v)=0 | (2.14) |
In (2.14) replacing v by 4v2t,t∈S and using (2.14) again, we get
d2(u2)[r,jk]v2d1(t)+d1(u2)[r,jk]v2d2(t)=0 | (2.15) |
In (2.15) taking t=t(d2(w)+w′),w∈G, we get
d2(u2)[r,jk]v2d1(t(d2(w)+w′))+d1(u2)[r,jk]v2d2(t(d2(w)+w′))=0 |
and therefore
d2(u2)[r,jk]v2d1(t)(d2(w)+w′)+d2(u2)[r,jk]v2td1((d2(w)+w′))
+d1(u2)[r,jk]v2d2(t)(d2(w)+w′)+d1(u2)[r,jk]v2td2(d2(w)+w′)=0 |
Using (2.12) and (2.15) in the last expression, we obtain
(d1(u2))[r,jk](v2td2(d2(w)+w′))=0 | (2.16) |
Applying Lemma 5 on (2.15), we get either d1(u2)=0 or v2td2(d2(w)+w′)=0. If d1(u2)=0 then by Lemma 3, d1=0. If v2Sd2(d2(w)+w′)={0}, the by the primeness of S, we have v2=0 or d2(d2(w)+w′)=0. If v2=0,∀v∈G, then G={0}, a contradiction. Suppose that for all w∈G
d2(d2(w)+w′)=0 | (2.17) |
In (2.17)replacing w by 4z2u,z,u∈G, and using (2.17) again, we obtain
d2(z2)d2(u)=0 | (2.18) |
In (2.18), replacing u by 4xz2,x∈G and using (2.18) again, we obtain d2(z2)Gd2(z2)={0}. By Lemma 2, d2(z2)=0 and hence by Lemma 3, we conclude that d2=0. Taking d2=0 in the hypothesis to obtain d1(u)=0 and hence by Theorem 2.4 of [11], we have d1=0.
Theorem 3. Let G be a nonzero Jordan ideal of a 2-torsion free prime MA-semiring S and d1 and d2 be derivations of S such that for all u,v∈G
[d1(u),d2(v)]+[u,v]′=0 | (2.19) |
Then S is commutative.
Proof. If d1=0 or d2=0, then from (2.19), we obtain [G,G]={0}. By Theorem 2.3 of [11] S is commutative. We assume that both d1 and d2 are nonzero. In (2.19) replacing u by 4uw2 and using MA-semiring identities and 2-torsion freeness of S, we get
d1(u)[2w2,d2(v)]+([d1(u),d2(v)]+[u,v]′)2w2+u([d1(2w2),d2(v)]
+[2w2,v]′)+[u,d2(v)]d1(2w2)=0 |
Using (2.19) again, we get
d1(u)[2w2,d2(v)]+[u,d2(v)]d1(2w2)=0 |
and by the 2-torsion freeness of S, we have
d1(u)[w2,d2(v)]+[u,d2(v)]d1(w2)=0 | (2.20) |
Replacing u by 2u[r,jk] in (2.20) and using it again, we obtain
d1(u)[r,jk][w2,d2(v)]+[u,d2(v)][r,jk]d1(w2)=0 | (2.21) |
In (2.21) replacing u by 4su2 and using (2.21) again, we obtain
d1(s)u2[r,jk][w2,d2(v)]+[s,d2(v)]u2[r,jk]d1(w2)=0 | (2.22) |
In (2.22) replacing s by d2(v)s and then using commutator identities, we get
d1d2(v)su2[r,jk][w2,d2(v)]=0 | (2.23) |
Therefore d1d2(v)Su2[r,jk][w2,d2(v)]={0}. By the primeness of S, we obtain either d1d2(v)=0 or u2[r,jk][w2,d2(v)]=0. Consider the sets
G1={v∈G:d1d2(v)=0} |
and
G2={v∈G:u2[r,jk][w2,d2(v)=0} |
We observe that G=G1∪G2. We will show that either G=G1 or G=G2. Suppose that v1∈G1∖G2 and v2∈G2∖G1. Then v1+v2∈G1+G2⊆G1∪G2=G. We now see that 0=d1d2(v1+v2)=d1d2(v2), which shows that v2∈G1, a contradiction. On the other hand 0=u2[r,jk][w2,d2(v1+v2)]=u2[r,jk][w2,d2(v1)], which shows that v1∈G2, a contradiction. Therefore either G1⊆G2 or G2⊆G1, which respectively show that either G=G1 or G=G2. Therefore we conclude that for all v∈G, d1d2(v)=0 or u2[r,jk][w2,d2(v)]=0. Suppose that d1d2(v)=0,v∈G. then by Lemma 2.1, d1=0 or d2=0. Secondly suppose that u2[r,jk][w2,d2(v)]=0,u,v,w,j,k∈G,r∈S. By Lemma 5, we have either u2=0 or [w2,d2(v)]=0. But u2=0 leads to G={0} which is not possible. Therefore [w2,d2(v)]=0 and employing Lemma 4, [d2(v),s]=0,s∈S. Hence by Theorem 2.2 of [22], S is commutative.
Theorem 4. Let G be a nonzero Jordan ideal of a 2-torsion free prime MA-semiring S and d1 and d2 be derivations of S such that for all u,v∈G
d1(u)d2(v)+[u,v]′=0 | (2.24) |
Then d1=0 or d2=0 and thus S is commutative.
Proof. It is quite clear that if at least one of d1 and d2 is zero, then we obtain [G,G]={0}. By Theorem 2.3 of [11] and Theorem 2.1 of [22], S is commutative. So we only show that at least one of the derivations is zero. Suppose that d2≠0. In (2.24), replacing v by 4vw2,w∈G, we obtain d1(u)d2(4vw2)+[u,4vw2]′=0 and therefore using MA-semirings identities, we can write
4d1(u)vd2(w2)+4d1(u)d2(v)w2+4v[u,w2]′+4[u,v]′w2=0 |
In view of Lemma 1 as 2w2∈G, using (2.24) and the 2-torsion freeness of S, we obtain
d1(u)vd2(w2)+v[u,w2]′=0 | (2.25) |
In (2.25) replacing v by 2[s,t]v, s,t∈S and hence by the 2-torsion freeness of S, we get
d1(u)[s,t]vd2(w2)+[s,t]v[u,w2]′=0 | (2.26) |
Multiplying (2.25) by [s,t] from the left, we get
[s,t]d1(u)vd2(w2)+[s,t]v[u,w2]′=0 |
and since S is an MA-semiring, therefore
[s,t]d1(u)vd2(w2)=[s,t]v[u,w2] | (2.27) |
Using (2.27) into (2.26), we obtain d1(u)[s,t]vd2(w2)+[s,t]′d1(u)vd2(w2)=0. By MA-semirings identities, we further obtain [d1(u),[s,t]]Gd2(w2)={0}. By Lemma 2, we obtain either [d1(u),[s,t]]=0 or d2(w2)=0. If d2(w2)=0, then by Lemma 3, d2=0. On the other hand, if
[d1(u),[s,t]]=0 | (2.28) |
In (2.28) replacing t by st, we get [d1(u),s[s,t]]=0 and using (2.23) again [d1(u),s][s,t]=0 and therefore [d1(u),s]S[s,t]={0} and by the primeness of S, we get [S,S]={0} and hence S is commutative or [d1(u),s]=0. In view of Theorem 2.2 of [22] from [d1(u),s]=0 we have [S,S]={0} which further implies S is commutative. Hence (2.19)becomes d1(u)d2(v)=0. As above we have either d1=0 or d2=0.
Theorem 5. Let S be a 2-torsion free prime MA-semiring and G a nonzero Jordan ideal of S. If d1, d2 and d3 be nonzero. derivations such that for all u,v∈G either
1). d3(v)d1(u)+d2(u′)d3(v)=0 or
2). d3(v)d1(u)+d2(u′)d3(v)+[u,v]′=0.
Then S is commutative and d1=d2.
Proof. 1). By the hypothesis for the first part, we have
d3(v)d1(u)+d2(u′)d3(v)=0 | (2.29) |
which further implies
d3(v)d1(u)=d2(u)d3(v) | (2.30) |
In (2.29) replacing u by 4uw2, we obtain
4d3(v)d1(u)w2+4d3(v)ud1(w2)+4d2(u′)w2d3(v)+4u′d2(w2)d3(v)=0 |
and therefore by the 2-torsion freeness of S, we have
d3(v)d1(u)w2+d3(v)ud1(w2)+d2(u′)w2d3(v)+u′d2(w2)d3(v)=0 | (2.31) |
Using (2.30) into (2.31), we obtain
d2(u)[d3(v),w2]+[d3(v),u]d1(w2)=0 | (2.32) |
In (2.32) replacing u by 2u[r,jk],r∈S,j,k∈G, and using (2.32) again, we get
d2(u)[r,jk][d3(v),w2]+[d3(v),u][r,jk]d1(w2)=0 | (2.33) |
In (2.33) replacing u by 4tu2,t∈S and using 2-torsion freeness and (2.33) again, we get
d2(t)u2[r,jk][d3(v),w2]+[d3(v),t]u2[r,jk]d1(w2)=0 | (2.34) |
Taking t=d3(v)t in (2.34) and using (2.34) again we obtain
d2d3(v)tu2[r,jk][d3(v),w2]=0 | (2.35) |
We see that equation (2.35) is similar as (2.23) of the previous theorem, therefore repeating the same process we obtain the required result.
2). By the hypothesis, we have
d3(v)d1(u)+d2(u′)d3(v)+[u,v]′=0 | (2.36) |
For d3=0, we obtain [G,G]={0} and by Theorem 2.3 of [11] this proves that S is commutative. Assume that d3≠0. From (2.36), using MA-semiring identities, we can write
d3(v)d1(u)=d2(u)d3(v)+[u,v] | (2.37) |
and
d3(v)d1(u)+[u,v]′=d2(u)d3(v) | (2.38) |
In (2.36), replacing u by 4uz2, we obtain
4(d3(v)ud1(z2)+d3(v)d1(u)z2+d2(u′)z2d3(v)+u′d2(z2)d3(v)+u[z2,v]′)+[u,v]′z2)=0 |
and using (2.37) and (2.38) and then 2-torsion freeness of S, we obtain
[d3(v),u]d1(z2)+d2(u)[d3(v),z2]=0 | (2.39) |
We see that (2.39) is same as (2.32) of the previous part of this result. This proves that [S,S]={0} and hence S is commutative. Also then by the hypothesis, since d3≠0, d1=d2.
Theorem 6. Let G be nonzero Jordan ideal of a 2-torsion free prime MA-semiring S and d1 and d2 be nonzero derivations of S such that for all u,v∈G
[d2(v),d1(u)]+d1[v,u]′=0 | (2.40) |
Then S is commutative.
In (2.40) replacing u by 4uw2,w∈G and using 2-torsion freeness and again using(2.40), we obtain
[d2(v)+v′,u]d1(w2)+d1(u)[d2(v)+v′,w2]=0 | (2.41) |
In (2.41) replacing u by 2u[r,jk],j,k∈G,r∈S, we obtain
u[d2(v)+v′,2[r,jk]]d1(w2)+2[d2(v)+v′,u][r,jk]d1(w2)
+ud1(2[r,jk])[d2(v)+v′,w2]+2d1(u)[r,jk][d2(v)+v′,w2]=0 |
Using 2-torsion freeness and (2.41) again, we get
[d2(v)+v′,u][r,jk]d1(w2)+d1(u)[r,jk][d2(v)+v′,w2]=0 | (2.42) |
In(2.42) replacing u by 4tu2,t∈Sand using (2.42) again, we get
[d2(v)+v′,t]u2[r,jk]d1(w2)+d1(t)u2[r,jk][d2(v)+v′,w2]=0 | (2.43) |
In (2.43) taking t=(d2(v)+v′)t and using MA-semirings identities, we obtain
(d2(v)+v′)[d2(v)+v′,t]u2[r,jk]d1(w2)+d1(d2(v)+v′)tu2[r,jk][d2(v)+v′,w2]
+(d2(v)+v′)d1(t)u2[r,jk][d2(v)+v′,w2]=0 |
and using (2.43) again, we obtain
d1(d2(v)+v′)tu2[r,jk][d2(v)+v′,w2]=0 | (2.44) |
By the primeness we obtain either d1(d2(v)+v′)=0 or u2[r,jk][d2(v)+v′,w2]=0. If d1(d2(v)+v′)=0, then by Theorem 2 we have d1=0, which contradicts the hypothesis. Therefore we must suppose u2[r,jk][d2(v)+v′,w2]=0. By Lemma 5, we have either u2=0 or [d2(v)+v′,w2]=0. But u2=0 implies G={0} which is not possible. On the other hand applying Lemma 5, we have [d2(v)+v′,r]=0,∀r∈S and therefore taking r=v,v∈G and [d2(v),v]+[v′,v]=0 and using MA-semiring identities, we get
[d2(v),v]+[v,v]′=0 | (2.45) |
As [v,v]′=[v,v], from (2.45), we obtain [d2(v),v]+[v,v]=0 and therefore
[d2(v),v]=[v,v]′ | (2.46) |
Using (2.46) into (2.45), we get 2[d2(v),v]=0 and by the 2-torsion freeness of S, we get [d2(v),v]=0. By Theorem 2.2 of [22], we conclude that S is commutative.
Corollary 1. Let G be nonzero Jordan ideal of a 2-torsion free prime MA-semiring S and d be a nonzero derivation of S such that for all u,v∈G d[v, u] = 0. Then S is commutative
Proof. In theorem (6) taking d2=0 and d1=d, we get the required result.
Theorem 7. Let G be a nonzero Jordan ideal of a 2-torsion free prime MA-semiring and d2 be derivation of S. Then there exists no nonzero derivation d1 such that for all u,v∈G
d2(v)∘d1(u)+d1(v′∘u)=0 | (2.47) |
Proof. Suppose on the contrary that there is a nonzero derivation d1 satisfying (2.47). In (2.47) replacing u by 4uw2,w∈G and using (2.47) again, we obtain
d1(u)[w2,d2(v)+v]+[u,d2(v)]′d1(w2)+ud1(v∘w2)+(u∘v)d1(w2)′+ud1[v,w2]′=0 | (2.48) |
In (2.48), replacing u by u[r,jk],r∈S,j,k∈G and using (2.48) again, we get
d1(u)[r,jk][w2,d2(v)+v]+[u,d2(v)+v]′[r,jk]d1(w2)=0 | (2.49) |
In (2.49) replacing u by 4tu2,t∈S and using (2.49) again, we obtain
d1(t)u2[r,jk][w2,d2(v)+v]+td1(u2)[r,jk][w2,d2(v)+v]
+t[u2,d2(v)+v]′[r,jk]d1(w2)+[t,d2(v)+v]′u2[r,jk]d1(w2)=0 |
and using2-torsion freeness and (2.49) again, we obtain
d1(t)u2[r,jk][w2,d2(v)+v]+[t,d2(v)+v]′u2[r,jk]d1(w2)=0 | (2.50) |
In (2.50) taking t=(d2(v)+v)t and using MA-semirings identities, we get d1(d2(v)+v)tu2[r,jk][w2,d2(v)+v]+(d2(v)+v)d1(t)u2[r,jk][w2,d2(v)+v]
+(d2(v)+v)[t,d2(v)+v]′u2[r,jk]d1(w2)=0 |
Using (2.50) again, we obtain
d1(d2(v)+v)tu2[r,jk][w2,d2(v)+v]=0 | (2.51) |
that is d1(d2(v)+v)Su2[r,jk][w2,d2(v)+v]=0. Therefore by the primeness following the same process as above, we have either d1(d2(v)+v)=0 or u2[r,jk][w2,d2(v)+v]=0 for all u,v,j,k,w∈G,r∈S. If d1(d2(v)+v)=0. As d1≠0, therefore d2(v)+v=0. Secondly suppose that u2[r,jk][w2,d2(v)+v]=0. By Lemma 5, we have either u2=0 or [w2,d2(v)+v]=0. But u2=0 implies that G={0}, a contradiction. Therefore we consider the case when [w2,d2(v)+v]=0, which implies, by Lemma 4, that [d2(v)+v,r]=0,∀r∈S and taking in particular t=v∈G, we have
[d2(v),v]+[v,v]=0 | (2.52) |
Also by definition of MA-semirings, we have [v,v]=[v,v]′. Therefore [d2(v),v]+[v,v]′=0 and therefore
[d2(v),v]=[v,v] | (2.53) |
Using (2.53) into (2.52) and then using 2-torsion freeness of S, we obtain [d(v),v]=0. By Theorem 2.2 of [22], we conclude that S is commutative. Therefore (2.47) will be rewritten as 2d1(u)d2(v)+2(d1(v′)u+v′d1(u))=0 and hence by the 2-torsion freeness of S, we obtain
d1(u)d2(v)+d1(v′)u+v′d1(u)=0 | (2.54) |
In (2.54) replacing u by 2uw and using 2-torsion freeness of S, we get
d1(u)wd2(v)+ud1(w)d2(v)+d1(v′)uw+v′d1(u)w+v′ud1(w)=0 |
and therefore
w(d1(u)d2(v)+d1(v′)u+v′d1(u))+ud1(w)d2(v)+v′ud1(w)=0 |
Using (2.54) again, we obtain
ud1(w)d2(v)+v′ud1(w)=0 | (2.55) |
In (2.55) replacing v by 2vz, we get
ud1(w)d2(v)z+ud1(w)vd2(z)+v′zud1(w)=0 |
and therefore
z(ud1(w)d2(v)+v′ud1(w))+ud1(w)vd2(z)=0 |
and using (2.55) again, we get d1(w)uGd2(z)={0}. By the above Lemma 2, we have either d1(w)u=0 or d2(z)=0 and therefore by Remark 1, we have either d1(w)=0 or d2(z)=0. As d1≠0, therefore d2=0. Therefore our hypothesis becomes d1(u∘v)=0 and therefore d1(u2)=0, ∀u∈G. By Lemma 3, d1=0 a contraction to the assumption. Hence d1 is zero.
We have proved the results of this paper for prime semirings and it would be interesting to generalize them for semiprime semirings, we leave it as an open problem.
Taif University Researchers Supporting Project number (TURSP-2020/154), Taif University Taif, Saudi Arabia.
The authors declare that they have no conflict of interest.
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1. | Tariq Mahmood, Liaqat Ali, Muhammad Aslam, Ghulam Farid, On commutativity of quotient semirings through generalized derivations, 2023, 8, 2473-6988, 25729, 10.3934/math.20231312 |