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"A Treatise on govt" via "The Categories"
The Politics of Aristotle is the second one a part of a treatise of which the Ethics is the 1st half. It appears again to the Ethics because the Ethics seems to be ahead to thee Politics. For Aristotle didn't separate, as we're susceptible to do, the spheres of the statesman and the moralist. within the Ethics he has defined the nature invaluable for the nice lifestyles, yet that existence is for him primarily to be lived in society, and whilst within the final chapters of the Ethics he involves the sensible program of his inquiries, that reveals expression no longer in ethical exhortations addressed to the person yet in an outline of the legislative possibilities of the statesman.
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Quaternion algebras based codes In this section, we present two code constructions, one based on Hamilton quaternions and the other on generalized quaternion algebras. 3 for the coding motivation and related deﬁnitions. M. Alamouti  published a simple code construction for a wireless system with 2 transmit antennas, given as follows. Let x0 , x1 be two complex numbers that represent the information symbols to be sent, and let the codebook C be given by the following set of matrices C= x0 x1 −x1 x0 | x0 , x1 ∈ C .
We now prove the converse, which is known as Wedderburn’s theorem. 1 (Wedderburn’s Theorem). Any simple k-algebra A is isomorphic to Mr (D), for some integer r ≥ 1 and some division k-algebra D whose center is isomorphic to the center of A. In particular, a central simple algebra is isomorphic to a matrix algebra over a central division k-algebra. Moreover, the integer r and the isomorphism class of D only depend on the isomorphism class of A. More precisely, if I is a minimal right ideal of A, we have D∼ =k EndA (I) and A ∼ =A I r .
Let A and B be two k-algebras, and let L/k be a ﬁeld extension. Then the following properties hold: (1) A ⊗k B is central over k if and only if A and B are central over k; (2) A ⊗k L is central over L if and only if A is central over k. Proof. 4 shows that Z(A ⊗k B) = Z(A) ⊗k Z(B). Therefore, dimk (Z(A ⊗k B)) = dimk (Z(A)) dimk (Z(B)). Now A ⊗k B is central over k if and only if dimk (Z(A ⊗k B)) = 1. The previous equality shows that it is equivalent to dimk (Z(A)) = dimk (Z(B)) = 1, that is A and B are central over k.