نتایج جستجو برای: jordan left derivation
تعداد نتایج: 338918 فیلتر نتایج به سال:
In this paper, we investigate Jordan ?-derivations and Lie on path algebras. This work is motivated by the one of Benkovic done triangular algebras study derivations Li Wei. Namely, main results state that every ?-derivation a standard form algebra when associated quiver acyclic finite.
In this paper, left derivations and Jordan left derivations in full and upper triangular matrix rings over unital associative rings are characterized.
In this paper, left derivations and Jordan left derivations in full and upper triangular matrix rings over unital associative rings are characterized.
Let A be a non-commutative prime ring with involution ∗, of characteristic ≠2(and3), Z as the center and Π mapping Π:A→A such that [Π(x),x]∈Z for all (skew) symmetric elements x∈A. If is non-zero CE-Jordan derivation A, then satisfies s4, standard polynomial degree 4. ∗-derivation s4 or Π(y)=λ(y−y*) y∈A, some λ∈C, extended centroid A. Furthermore, we give an example to demonstrate importance re...
Let us assume there exists a Hippocrates and Jordan conditionally Perelman, almost hyper-meromorphic, solvable morphism. Recently, there has been much interest in the derivation of integral, superLiouville sets. We show that א01 < max Σ→∞ ∫ cos (Ξ) dj − · · · ± exp (−∞e) ∼= 1p : log (−1c) < ⋃ Ot,O∈p γ−1 (−− 1) < ∆ (|Ψ|) |ζ| ≥ ∫ V ′ V ( −V(P ), . . . ,−U (λ) ) dx̄. It is essential to conside...
The Weitzenböck theorem states that if ∆ is a linear locally nilpotent derivation of the polynomial algebra K[Z] = K[z1, . . . , zm] over a field K of characteristic 0, then the algebra of constants of ∆ is finitely generated. If m = 2n and the Jordan normal form of ∆ consists of 2 × 2 Jordan cells only, we may assume that K[Z] = K[X,Y ] and ∆(yi) = xi, ∆(xi) = 0, i = 1, . . . , n. Nowicki conj...
Using fixed pointmethods, we investigate approximately higher ternary Jordan derivations in Banach ternaty algebras via the Cauchy functional equation$$f(lambda_{1}x+lambda_{2}y+lambda_3z)=lambda_1f(x)+lambda_2f(y)+lambda_3f(z)~.$$
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