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Additional info for Algebra I [Lecture notes]
The map x → a · x is a group automorphism on R+, hence of the type x → ra x; let r−a := −ra , then a → ra is a group morphism R+ → R+, so ra = sa, with s > 0, so x · y = rx y = sxy, rx·y = s(x · y) = sxsy = rx ry , hence x → rx is an orderring embedding. ) 32 Joseph Muscat 2015 Hence, the only order-complete linearly ordered rings are 0, Z and R; and the Dedekind-completion of any Archimedean linearly ordered field is R. Recall that these are also Cauchy-complete. 2 Surreal Numbers Every linearly ordered field is embedded in the surreal numbers.
5. A linearly ordered field is Archimedean ⇔ N is unbounded ⇔ Q is dense. ) Proof: ∀x, x ≺ y ⇒ Ny is unbounded. If 0 x < y then (y − x)−1 < n m m 1 1 N is unbounded; pick smallest 2n > x. So x < 2n x + 2n < and 2n y−x x + 2 < y. 6. The extension field F (a) ∼ = F [x]/ p (p irreducible) can be linearly ordered, if p changes sign. , F [x]/ p is not formally real, so n p2n = 0 = pq (mod p) with pn = 0; q has degree at most 2(m − 1) − m = m − 2. Since p(x)q(x) = n pn (x)2 0 yet p(x1 )p(x2 ) < 0, then q(x1 )q(x2 ) < 0; decompose q = q1 · · · qr into irreducibles, then q1 (x1 )q1 (x2 ) < 0 say, and n p2n = pq = 0 (mod q1 ), still not formally real.
So x < 2n x + 2n < and 2n y−x x + 2 < y. 6. The extension field F (a) ∼ = F [x]/ p (p irreducible) can be linearly ordered, if p changes sign. , F [x]/ p is not formally real, so n p2n = 0 = pq (mod p) with pn = 0; q has degree at most 2(m − 1) − m = m − 2. Since p(x)q(x) = n pn (x)2 0 yet p(x1 )p(x2 ) < 0, then q(x1 )q(x2 ) < 0; decompose q = q1 · · · qr into irreducibles, then q1 (x1 )q1 (x2 ) < 0 say, and n p2n = pq = 0 (mod q1 ), still not formally real. If a2 > 0 then x2 − a2 is irreducible in F and changes sign from 0 to a2 + 1.