Is it possible to find for any \(n,m\in\mathbb N\) a Jordan algebra of nilpotency type \((n,1,m)\)? (Q330655)

Let us give the main definition of the paper: for a nilpotent algebra \(J\) of nilpotency index \(m\), we define the nilpotency type of the algebra \(J\) as the sequence \((n_1, n_2, n_3, \ldots, n_m)\), where \(n_i = \dim J^i /J^{i+1}\). The paper is dedicated to study of Jordan central extensions of commutative \(1\)-dimensional central extensions of an algebra with zero multiplication. Namely, they studying Jordan central extensions of commutative nilpotent algebras of type \((n,1)\). As was noted in the present paper, for every \(n\)-dimensional algebra with zero multiplication there is only one \(1\)-dimensional commutative central extension without annihilator components. It is the algebra with the following multiplication table: \(e_1^2=e_2^2= \ldots=e_n^2=e_{n+1}\). Every Jordan algebra of nilpotency type \((n,1)\) can be constructed from some algebra of this type. For the main results of the paper, they proved two very useful Lemmas: 1. Let \(J\) be an associative Jordan algebra of nilpotency type \((n, 1, m)\) then \(m = 1\). [Lemma 3.8] 2. Let \(J\) be a Jordan algebra of nilpotency type \((n, 1, m)\) then \(m\leq n\). Moreover if \(m = n\) then \(J\) has no annihilator components. [Lemma 3.10] After that, in the 4th section, using Lemma 3.10 they classified all Jordan algebras of nilpotency type \((2,1,n)\). As follows, there are only two opportunities: \(n=1\) and \(n=2\). The case \(n=1\) was given in the paper of \textit{M. E. Martin} [Int. J. Math. Game Theory Algebra 20, No. 4, 303--321 (2013; Zbl 1288.17019)] and after the study of the second case \(n=2\), they completed the proof of Theorem 4.3. \par The last (5th) section is dedicated to the classification of commutative associative algebras of nilpotency type \((n, 1, m)\). Using Lemma 3.8, they can say that this classification is reduced to the case \(m=1\) and completed the classification of all these algebras in Theorem 5.2.