By Aristotle

"A Treatise on executive" by means of "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 to be like again to the Ethics because the Ethics appears to be like ahead to thee Politics. For Aristotle didn't separate, as we're vulnerable to do, the spheres of the statesman and the moralist. within the Ethics he has defined the nature useful for the nice existence, yet that existence is for him primarily to be lived in society, and while within the final chapters of the Ethics he involves the sensible program of his inquiries, that unearths expression now not in ethical exhortations addressed to the person yet in an outline of the legislative possibilities of the statesman.

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PD[[X]] In fact, and r is a nonnegative integer. 7. THE QUOTIENT FIELD OF D[[X]] 7. " The reason for that will become clear in this section for in this section we shall be interested in the quotient fields of power series rings over integral domains. THEOREM 27. The trivial case offers little resistance. Let K be a field. Then CO K((X)) = { a^X^ I a^ ^ K, m an integer} i=m = (K[[X]]) [1/X] is the quotient field of K[[X]]. Proof. (X). By Theorem 15, K[[X]] is a DVR with unique maximal ideal In (K[[X]])[1/X] we have inverted X and so (0) is the only sur viving prime ideal.

That 0 £ 6 < t^. J=O 'j ^t. = t. ^ then + Y* Ч. = a^ . + a + y m,i Ч. We already have that t^ £ yk^, To see that (iii) holds, suppose By assumption, we have that b^ ^ I^ Also nr> 2 Yr = 2 • n a . c. S I I since j < ô < t. < к implies that 6 1=0 v,i 6-1 S^-I I - - 28 2. -Jv,j I Consequently, PATHOLOGY IN R[[X]] =b. -l I justified. We come at last to the conceptual part of the proof. 0 = 0. Thus, let g ^ IR[[X]] П . Ь X) > Then there is a finitely generated ideal C C i of R such that the content of g, c(g), is contained in C.

H. be an Since t^Dp C qDp for I ^ i ^ n, it follows that there exists f^^^ ^ D[[X]] and an s^ ^ D\P such that f = (qf^^^)/s^. procedure. If f^^^ ^ PD[[X]], then we may repeat the Moreover, if this process failed to terminate, then for each positive integer n, we would obtain an element f^^^ G D[[X]] and an element s f e ^ D \ P such that f = (q^f^^^)/s . a^(Dp[[X]])PJ^ Therefore, n (the quotient field of D[[X]]) = (0). Thus, if f Ф 0, then there exists a positive integer n such that f = (q\)/s, where h ^ D[[X]] \ PD[[X]] and s ^ D \ P.