By Lucia M., Magtone R., Zhou H.-S.

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We are now entitled to assume that j j 1 is nontrivial and just have to show that (iii) implies (ii). So assume (iii). Then (7) jbj1 > 1 jbj2 > 1 for every b 2 K: just apply (i) to a D b 1 . Next, since j j 1 is assumed nontrivial, there exists c 2 K such that jcj1 > 1. Because of (7), we have jcj2 > 1, so jcj2 D jcj1 for some > 0 in ޒ. To show that the same relation holds for all elements of K, we take an arbitrary a 2 K and relate its absolute values to those of c. First we have (8) jaj1 D jcj˛1 for some ˛ 2 ޒ: ; that is, Suppose m 2 ޚand n 2 ގsatisfy m=n < ˛.

L/ ! ޚis an isomorphism. K/ ޚ is commutative; hence the equivalence of (i) and (ii). An element of a commutative ring with unity is nilpotent if and only if it belongs to every prime ideal; together with Theorem 2, this implies the equivalence of (ii) and (iii). Implications (v) ) (iv) and (iv) ) (ii) are trivial. Thus what is left is to show that (iii) implies (v). K/ such that (29) 2i f ¤ 0 for i D 0; 1; 2; 3; : : : In an algebraic closure of K there is, by Zorn’s Lemma, a maximal extension E over which (29) still holds (with rE=K f instead of f ).

Therefore q contains the subspace Œa; b ' Œa; ad D Œa ˝ Œ1; d: The assertion now follows easily by induction on dim q. 36 22 Orders and Quadratic Forms Theorem 5. f / D 0 for every real closure L of K. f / D 0 for every order P of K. (iii) f is nilpotent. K/. (v) There exists n 2 ގsuch that 2n f D 0. P ✲ n sg (28) ✛ sg nL Proof. L/ ! ޚis an isomorphism. K/ ޚ is commutative; hence the equivalence of (i) and (ii). An element of a commutative ring with unity is nilpotent if and only if it belongs to every prime ideal; together with Theorem 2, this implies the equivalence of (ii) and (iii).