Study of Some Degree Five Identities of Type (4,1)

Indian Journal of Advanced Mathematics (IJAM) ISSN: 2582-8932 (Online), Volume-5 Issue-2, October 2025 18 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. Abstract: In non-associative algebra, irreducible identities of degree five are the least studied. Following Osborn's studies, the only identity of type (5) has generated very little literature, as seen in ([1]) and ([2]). Hence, our interest in identities of the following type. The purpose of this study is to enable us to consider a baric case study at a later stage, such as in ([3]) and ([4]). This paper is devoted to the study of three of type (4,1), taken from the families of irreducible degree five identities of Osborn. We conduct this study in the presence of an idempotent, through a Peirce decomposition, depending on whether the Peirce polynomial is reducible or not over the base field of the algebra. In each studied case, we find two orthogonal subalgebras. Therefore, in the first two studied identities, we manage to show that there is a homomorphism over one of the subspaces of A, whose kernel is an ideal. Keywords: Idempotent, Linearization, Peirce Decomposition. I. INTRODUCTION J. M. Osborn has determined the whole family of irreducible degree five identities, not implied by commutativity: an identity is said to be irreducible if, despite the presence of a unity element, it is not a consequence of an identity of a lower degree. He finds, over a field F of characteristic not 2, 3 or 5, five parameterized identities, of which: (1) In fact, identity (1) is different from the one studied by J. M. Manuscript received on 30 July 2025 | First Revised Manuscript received on 18 August 2025 | Second Revised Manuscript received on 19 September 2025 | Manuscript Accepted on 15 October 2025 | Manuscript published on 30 October 2025. *Correspondence Author(s) Dr. Abdoulaye DEMBEGA*, Researcher, Department of Mathématiques, Université Norbert ZONGO, Koudougou, Burkina Faso. Email ID: [email protected], ORCID ID: 0009-0004-6075-555X © The Authors. Published by Lattice Science Publication (LSP). This is an open-access article under the CC-BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/) Osborn because of the factor of , but it is the same, concerning the values of the parameters he has found. In our study, we are interested in the following identities. (2) (3) and (4) which corresponds respectively to We assume that the studied algebras contain a nonzero idempotent e, and F is a commutative field of characteristic not 2, 3, 5 or 7. This will enable us to conduct our study as in ([2]). II. IDENTITY (2) In this section, A is an algebra defined by identity (2). Setting x=e in (2), we have (1.1) with and being the multiplication by e. Let's consider which factorization is . We examine two cases: A. Polynomial is Irréductible over F In this case, the Peirce decomposition of A is with ( =0, 1) and The relations between the Peirce components are stated in the following theorem. Theorem Let A be an algebra defined by (2). Assume that is irreducible over F[t]. Then the Peirce decomposition of A is , with and being orthogonal subalgebras, and Moreover, we have: i. Study of Some Degree Five Identities of Type (4,1) Abdoulaye DEMBEGA Study of Some Degree Five Identities of Type (4,1) 19 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. ii. iii. iv. v. Proof Linearizing identity (2), we have (1.2) Let's consider x=e and simplify, we have (1.3) Let's take and in equation (1.3). Since , we have 3e(ey)=3ey-y and 3e(e(e(y))) = 2ey-y. Then (1.3) becomes (1.4) In the following, for every , we shall note with (i=0,1) and , then (1.4) becomes (1.5) If λ=0 in (1.5), then we have that is and . So we have and which means . And we finally have If in (1.5), then we have , which means and . This leads to and , which means . And we have . Now let's consider in (1.3). It gives, by simplifications, (1.6) switching y and z in (1.6) we have (1.7) Difference between (1.6) and (1.7) gives which means . Let's consider Since gcd(p(t),q(t))=1 then z(ey)-y(ez)=0, that means z(ey)=y(ez). Using this last equation in (1.6), it gives which means That leads to and , so, it finally gives , (1.8) , (1.9) (1.10) Replacing y by ey in (1.8), we have that means (1.11). Difference between (1.8) and (1.11) gives Since the characteristic of F is not 2, 3 or 7, we have so, Replacing y by ey in (1.9), we have - which means (1.12) Subtracting (1.9) and (1.12), we have that means and Since then we have that means And (1.10) can be simply written (1.13) Replacing y by ey in (1.13) and simplifying, we have Indian Journal of Advanced Mathematics (IJAM) ISSN: 2582-8932 (Online), Volume-5 Issue-2, October 2025 20 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. (1.14) Subtracting (1.13) and (1.14) gives yz = 0. Finally, we have Now, we study the products of three factors. Partially linearizing (1.2) gives (1.15) Let's take x=e, and in (1.15). We have (1.16) with =1 if λ=µ and 0 if not. For λ=µ in (1.16), we have (1.17) In (1.17), let's switch y and z. It gives (1.18) Let's subtract (1.17) and (1.18). We have that means (1.19) Let's take and in (1.19), we have (i) Using (1.19) in (1.17), it follows that (1.20) If λ=µ=0 in (1.20), we have Let's multiply this last equation bye, since by simplifying, we can write By comparing the two last equations, it follows that. which means Setting and , we have result (iv). If we take λ=µ=1 in (1.20), we have Multiplying this equation by e, since , after simplification, we have Comparing these two last equations, we have , which means Using notations and , we have result (v). Let us now consider λ = 0 and µ = 1 in (1.16). We have Since the only eigen values are 0 and 1, the operator is injective. So, we can write Taking and we have In the same way, if =1 and µ=0 in (1.16), we have That means because the operator is injective. Using the following notations and , we have Example Let's consider the five-dimensional commutative ℝ-algebra A, in which the non-zero products in the basis {e, e_0, e_1, e_2, e_3} are given by: A satisfies (2), with and B= . In fact, to show that this algebra satisfies (2), we just need to show that it satisfies the total linearization of (2). In this total linearization, if four or five variables are replaced by e, then equation (2) is satisfied throughout (1.1). All other combinations show that each term cancels. B. Polynomial admits two roots and in F Let's consider in this case, with The following theorem gives relations between the Peirce components. Theorem Let A be an algebra defined by (2). Assume that in F[t]. Then the Peirce decomposition of A is given by with and orthogonal subalgebras a zero algebra, (i=1,2) and Study of Some Degree Five Identities of Type (4,1) 21 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. Moreover, we have: i. ii. iii. with the conjugate number of λ, and are respectively elements of Proof. Then taking x=e, and in (1.2) we have (1.20) Let's consider Taking , we have and Then, it gives If , then and , and that leads to . For and , we have . Since and 0, 1 are not roots of , then , and finally . Now, since is the root of , then we have , and We shall use these equations to calculate when or equals (i=1,2). Then, for and , we have which roots are different from 0, 1, and On the other hand, we have ,, which leads to , and Now let's consider and , we have which roots are different from 0, 1, and . However, we have, , that means , and . Setting we can write . The roots of this last polynomial are different from 0, 1, and . Then, we have and then Let's consider and . We can write which roots are different from 0, 1, and . Then we have , that means . Partially linearizing (1.2) gives (1.15). Taking x=e, , and in equation (1.15), we have On the other hand, if we take and , we have Since is a root of then we have , , and the last equation gives . Let's take and , it gives Considering and we have Let's take and in each case, we have the results of the theorem. □ Example Let A be the five-dimensional algebra, whose non zero products in the basics are given by This algebra satisfies (2). In fact, and Example Let A be the five-dimensional algebra, whose non zero products in the basics are given by , , To show that A satisfies (2), we need to show that, regardless of how we replace the variables by elements of the basis, the complete linearization of (2) is satisfied. If four or five variables are replaced by e, equation (2) is confident throughout (1.1). If three variables are replaced by e, and the two others chosen between and , equation (2) is satisfied throughout (1.21). If two variables are replaced by e, each term cancels. Regardless of the combination of elements in the basis that is used in the linearization of (2), each term cancels, and (2) is trivially satisfied. Finally, A satisfies identity (2). Note: In the previous theorem, the subspaces and are ideals of A. In fact, let's consider and we can wri , we have Indian Journal of Advanced Mathematics (IJAM) ISSN: 2582-8932 (Online), Volume-5 Issue-2, October 2025 22 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. Then is an ideal. Similarly, we demonstrate that it is an ideal of A. Moreover, for all , we have In fact, let's consider the F-linear map (i=1,2), with Since that means the map satisfies being the product in Then is a morphism of algebras. Its kernel is an ideal of . We can easily see that it is an ideal as an intersection of ideals. Since the kernel of the algebra morphism of into the special Jordan algebra is an element of (i=1,2), that means z is also an element of III. IDENTITY (3) Now, we study the algebra defined by identity (3). Let's take x=e in (3), we have Let's consider Then, the Peirce decomposition of A is , with Define the following subspaces: and We have Theorem Let A be an algebra satisfying (3). Then we have , with et being orthogonal subspaces, , , , , ⊆ , Proof. By partially linearizing we have (2.1) Taking x=e in equation (2.1), we have that gives (2.2) Now take and in (2.2), we can write Let's set For , we have That gives , otherwise For and , we have and , that means, If , we have and this leads to Let's consider , with Then we have , Taking , equation (2.2) becomes If and , that means and then the last equation can be written Setting , we have which means then If and which means and then we have . Since and the last equation becomes Then we have If and , that means, and we can write Let's set we have Then gives If λ=1 and that means and then Study of Some Degree Five Identities of Type (4,1) 23 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. we have . Since and this last equation gives which means and finally Now let's take and set and . We can write In the same way, we have , . Using this in (2.2), we have which means (2.3) In the following part of the proof, we shall note the component of an element in A with regard to subspaces the component in . If that means , then (2.3) becomes Then we have , which means If and , that means and then (2.3) gives . Let's set Since taking the last equation gives , because Then we have Because of, we can write , that is, From we can write , which means, On the other hand, taking and in (2.3), we have Since we can write Then, we have If and (2.3) can be written Otherwise and Equation leads to that means But gives and finally, we have . □ Products of three factors are given in the following proposition. Proposition Let A be an algebra defined by identity (3). Then we have i. ii. iii. iv. v. vi. vii. viii. ix. Proof. Partially linearizing (2.1), it gives Indian Journal of Advanced Mathematics (IJAM) ISSN: 2582-8932 (Online), Volume-5 Issue-2, October 2025 24 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. (3.4) Let's consider x=e, and (3.4), we have )- 9 that gives after simplifying (3.5) in which equals if and if not. Taking , in (3.5), we have . In the same way, let's consider in (3.5), we have On the other hand, considering et , it gives . In each case, let's note and $ the result 1) follows. Now let's consider x=e, , and . Write , . Using these equations in (3.4), we have that gives after simplifying Take that means in (2.6), we have , and If , which means in (2.6), we can write We also have . (2.7) Since switching and z in (2.7), we have The sum of (2.7) and (2.8) gives, after simplifying. (2.9) Using (2.9) in (2.7) gives . Then, (2.9) can be written. Study of Some Degree Five Identities of Type (4,1) 25 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. Let's consider and in each case, that gives result 2). Example Let's consider the five-dimensional commutative ℝ-algebra A in which the non-zero products in the basics { are given by Then A satisfies (3), but is not a power associative algebra. In fact, we have and In the complete linearization, each term cancels, then A satisfies (3). Example Let's consider the five-dimensional commutative ℝ-algebra A in which the non-zero products in the basis is given by , , If 4 or 5 variables are replaced by e, equation (3) is satisfied. In all other cases, each term cancels. Then, A satisfies (3). Remark An algebra defined by (3) does not necessarily contain any idempotents. It is well known that the two-dimensional algebra of J. M. Osborn has non-zero products in the basis is given by does not contain any idempotents. However, it satisfies (3). Note Following A. A. Albert's study, we can introduce it for all. , the F linear map , Since which means, , the following map satisfies , in which defines the product in That is, is a morphism of algebras, whose kernel is an ideal of Since is the kernel of a morphism of algebras, of into the special Jordan algebra then with being the associator. IV. IDENTITY (4) In this section, A is an algebra defined by (4). Taking x=e in (4), we have with . Let's consider the polynomial The following lemma is well-known Lemma Let A be an algebra over a field F such that every subfield of the centralizer of A is contained in F and let's consider r an element of F which is not a square in F. Then, if A is simple, so is it for , in which L=F(k), with k such that We shall use the extension of the field F if necessary, and assume that ; Then the Peirce decomposition of A is Following the study of subsection (1.2), we easily establish this theorem Theorem Let A be an algebra defined by (4) over a suitable extension of F. Then the Peirce decomposition of A, relative to idempotent e is , with and being orthogonal subalgebras, and Moreover, we have i. ii. iii. iv. with v. vi. with . V. CONCLUSION The study of these degree 5 identities, conducted in the presence of an idempotent, allowed us to determine the relationships between the Peirce components. In each case, we were able to find two orthogonal subalgebras. We also managed to define, in each case, a homomorphism on one of the subspaces of the algebra A, whose kernel is an ideal of A. These results allow us to consider further study of these identities when dealing with nil-algebras or baric algebras, as in ([3]) and ([4]). ACKNOWLEDGMENT We thank the referee for his suggestions, which have helped to improve this article. DECLARATION STATEMENT I must verify the accuracy of the following information as the article's author. ▪ Conflicts of Interest/ Competing Interests: Based on my understanding, this article has no conflicts of interest. ▪ Funding Support: This article has not been funded by any organizations or agencies. This independence ensures that the research is conducted Indian Journal of Advanced Mathematics (IJAM) ISSN: 2582-8932 (Online), Volume-5 Issue-2, October 2025 26 Retrieval Number:100.1/ijam.B121205021025 DOI: 10.54105/ijam.B1212.05021025 Journal Website: www.ijam.latticescipub.com Published By: Lattice Science Publication (LSP) © Copyright: All rights reserved. with objectivity and without any external influence. ▪ Ethical Approval and Consent to Participate: The content of this article does not necessitate ethical approval or consent to participate with supporting documentation. ▪ Data Access Statement and Material Availability: The adequate resources of this article are publicly accessible. ▪ Author’s Contributions: The authorship of this article is contributed solely. REFERENCES 1. P. BEREMWIDOUGOU and A. CONSEIBO (2022). Classification and derivations of four-dimensional almost Bernstein algebras. Far East Journal of Mathematical Sciences (FJMS), Volume 56, Number 2, P. 1-25 DOI: https://doi.org/10.17654/0972555522022 2. A. GUIRO, A. DEMBEGA and A. CONSEIBO (2023), On a Class of Algebras Satisfying an Identity of Degree Five. J. P. Journal of Algebra, Number Theory and Applications, Volume 62, Number 2, pp. 87-107 DOI: https://doi.org/10.17654/0972555523023 3. D. KABRE, A. DEMBEGA and A. CONSEIBO (2024), Classification of four-dimensional baric algebras satisfying the polynomial identity of degree six. Korean J. Math. Volume 32, Number 1, P. 163-171 DOI: https://dx.doi.org/10.11568/kjm.2024.32.1.163 4. H. OUEDRAOGO, D. KABRE and A. DEMBEGA (2025), Structure of algebras satisfying an -polynomial identity of degree six. Contemporary Mathematics, Volume 6, Number 2, P. 1914-1925.DOI: https://doi.org/10.37256/cm.6220256453 AUTHOR’S PROFILE Dr. Abdoulaye Dembeba is a teacher and researcher at the Department of Mathematics of the University Norbert Zongo in Burkina Faso. BP 376 Koudougou, Burkina Faso. He holds a PhD in Mathematics, with a specialisation in non-associative algebras. His research focuses on non-associative algebras defined by polynomial identities. Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of the Lattice Science Publication (LSP)/ journal and/ or the editor(s). The Lattice Science Publication (LSP)/ journal and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions, or products referred to in the content.