ordered-field - There is one order-complete ordered-field...

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Unformatted text preview: There is one order-complete ordered-field Jonathan L.F. King University of Florida, Gainesville FL 32611-2082, USA squash@math.ufl.edu Webpage http://www.math.ufl.edu/ squash/ 15 September, 2010 (at 10:57 ) Contents Rings . . . . . . . . . . . . . . . . . . . . . . . 1 Totally-Ordered Sets . . . . . . . . . . . . . . . 1 Least upper-bound property [ LUBP ] . . . 1 LUBP theorem . . . . . . . . . . . . . . . . . . 1 Making a Real Assumption . . . . . . . . . 2 Making a Real Assumption . . . . . . . 2 Ordered-fields . . . . . . . . . . . . . . . . . . . . . . . 2 Ordered-field lemma . . . . . . . . . . . . . . . . 2 Archimedean fields . . . . . . . . . . . . . . . . . . . . 3 Archy lemma . . . . . . . . . . . . . . . . . . . 3 OC Archimedean theorem . . . . . . . . . . . 3 Order-dense lemma . . . . . . . . . . . . . . . . 3 Complete ordered-field(s) . . . . . . . . . . . . . . . . 4 Rings. In a ring ( , + , , , 1 ), if there is a posint n so that 1 + 1 + n ... + 1 equals , then the smallest such n is called the characteristic of the ring, and I write Char() = n . If no such posint exists, then I will write Char() = ; however , the standard term is Char() = 0, and you will see this in algebra texts and in some of my notes. A ring is commutative ( abbrev., comm-ring ) if its multiplication is commutative. In a comm-ring , a zero-divisor admits a non-zero elt ( this need not be unique ) so that = . Use ZD for zero-divisor . [ Letting denote 12 , in the Z 12 ring, 9 is a ZD, since 9 8 0, yet 8 6 0. OTOHand, even though 5 24 0, this doesnt show that 5 is a Z 12 ZD, since 24 0. ] An integral domain is a commutative ring with no ZDs except for , the trivial ZD . If the charac- teristic of an integral domain is finite, then Char() is a prime number. In particular, this holds if is a field. 1 : Fact. If is a field of finite order ( finite cardinality ) then | | = p k for some prime p and posint k . Con- versely, for each such prime p and k Z + , there exists a field of order p k , and this field is unique upto field- isomorphism. Partial proof. For the prime p := Char(), there is a copy of Z p inside , making a Z p-vectorspace. Letting k be the dimension of this vectorspace, then, we obtain | | = p k . The remaining Fact s take a fair amount of work to prove. Totally-Ordered Sets. A TOS ( , ) has an antireflexive, transitive relation so that for each 6 = in , either or ....
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ordered-field - There is one order-complete ordered-field...

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