This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: There is one ordercomplete orderedfield Jonathan L.F. King University of Florida, Gainesville FL 326112082, USA squash@math.ufl.edu Webpage http://www.math.ufl.edu/ squash/ 15 September, 2010 (at 10:57 ) Contents Rings . . . . . . . . . . . . . . . . . . . . . . . 1 TotallyOrdered Sets . . . . . . . . . . . . . . . 1 Least upperbound property [ LUBP ] . . . 1 LUBP theorem . . . . . . . . . . . . . . . . . . 1 Making a Real Assumption . . . . . . . . . 2 Making a Real Assumption . . . . . . . 2 Orderedfields . . . . . . . . . . . . . . . . . . . . . . . 2 Orderedfield lemma . . . . . . . . . . . . . . . . 2 Archimedean fields . . . . . . . . . . . . . . . . . . . . 3 Archy lemma . . . . . . . . . . . . . . . . . . . 3 OC Archimedean theorem . . . . . . . . . . . 3 Orderdense lemma . . . . . . . . . . . . . . . . 3 Complete orderedfield(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., commring ) if its multiplication is commutative. In a commring , a zerodivisor admits a nonzero elt ( this need not be unique ) so that = . Use ZD for zerodivisor . [ 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 pvectorspace. 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. TotallyOrdered Sets. A TOS ( , ) has an antireflexive, transitive relation so that for each 6 = in , either or ....
View Full
Document
 Fall '07
 JURY
 Math, Calculus

Click to edit the document details