Calculus and linear algebra by Wilfred Kaplan, Donald J. Lewis

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By Wilfred Kaplan, Donald J. Lewis

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Thus, if ˛ ¤ 0, we have jan ˛j < j˛j for n large enough. For such n we can then write jan j D j˛j, by applying the scholium F3 on the strong triangle y j D jK j. inequality to a D an , b D an ˛. This shows that jK Completion of absolute values 51 y be the valuation rings of K and K y with respect to j j, and let p Let R and R y gives rise to a and y p be the corresponding valuation ideals. The inclusion R Â R canonical homomorphism y y R=p ! R= p y f0g there exists, of residue fields, which we must show is surjective.

Let the assumptions and the notation be as in Theorem 2. a/ > 0 for all a 2 V; there are sums of squares s and t in the ring KŒV  such that f can be represented as f D 1Cs : t Proof. Suppose not. V / is real, we can apply Lemma 4 to A D KŒV  and f 2 A. V /, and an order Ä of the fraction field F of KŒX1 ; : : : ; Xn =P such that g Ä 0, where g denotes the image of f KŒX1 ; : : : ; Xn =P. a/ Ä 0. Contradiction! Theorem 6 (Dubois Nullstellensatz). Let K be a real field admitting a unique order, and let R be a real closure of K.

K2 such that j xj2 D jxj1 for all x 2 K1 . Proof. (a) Consider the quotient ring y WD C=N K of the ring C of all j j-Cauchy sequences in K modulo the ideal N of all j j-null sequences in K. This quotient is a commutative ring with unity, and is clearly not y is a field. the zero ring. an /n 2 C be a representative of ˛. an /n converging to zero, which is impossible by statement (iv) after Definition 6. Now (20) implies, in particular, that an ¤ 0 for every n > N . bn /n is a Cauchy sequence. For m; n > N , we have ˇ ˇ ˇ an am ˇ jan am j jan am j ˇD Ä jbm bn j D ˇˇ ; an am ˇ jan jjam j "2 y Then ˛ˇ D 1, where the last inequality uses (20).

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