Real Fields
Let K be a field. An ordering of K is a subset P of K having the following properties: ORD 1 Given x ∊ K, we have either x ∊ P, or x = 0, or −x ∊ P, and these three possibilities are mutually exclusive. In other words, K is the disjoint union of P, {0}, a
- PDF / 1,485,007 Bytes
- 15 Pages / 439 x 666 pts Page_size
- 25 Downloads / 179 Views
XI
Real Fields
§1 .
ORDERED FIELDS
Let K be a field. An ordering of K is a subset P of K having the following propert ies:
ORO 1. G iven x E K, we have eit her x E P, or x = 0, or - x E P, and these th ree possibilities are mutually exclusive. In other words, K is the disjoint union of P, {O}, and - P. ORO 2.
If x, Y E P, then x
+ y and
x y E P.
We shall also say that K is ordered by P , and we call P the set of positive elements . Let us assu me that K is ordered by P. Since 1 #- 0 and 1 = 12 = ( _ 1)2 we see that 1 E P. By ORO 2, it follows tha t 1 + ... + 1 E P, whence K has characteristic O. If x E P, a nd x #- 0, the n xx " 1 = 1 E P imp lies that x " 1 E P. Let x, y E K . We define x < y (or y > x) to mean that y - x E P. If x < 0 we say that x is negative. Th is means th at - x is positive. One verifies trivially the usu al relati on s for inequ a lities, for insta nce : x 0 and hence a is negative.
Proposition 2.4. Let R be afield such that R # W but Ra = R(J=T). Then R is real and hence real closed . Let P be the set of elements of R which are squares and # O. We contend that P is an ordering of R. Let a E R , a # O. Suppose that a is not a square in R. Let ex be a root of X 2 - a = O. Then R(ex) = R(.j=l), and hence there exist c, d E R such that ex = c + dJ=1. Then Proof
ex 2
= c 2 + 2cdJ=1
- d
2.
REAL FIELDS
XI, §2
453
Since 1, j=1 are linearly independent over R, it follows that c = 0 (because a ~ R 2 ) , and hence - a is a square. We shall now prove that a sum of squares is a square. For simplicity, write i = j=1. Since R(i) is algebraically closed, given a, b E R we can find c, d E R such that (c + di? = a + bi. Then a = c 2 - d2 and b = 2cd. Hence a
2
+ b2 =
(c
2
+ d2 ) 2 ,
as was to be shown. If a E R, a #- 0, then not both a and - a can be squares in R. Hence P is an ordering and our proposition is proved.
Theorem 2.5. Let R be a real closedfield, and f(X) a polynomial in R[X]. Let a, bE R and assume that f(a) < 0 and feb) > O. Then there exists c between a and b such that fCc) = o. . Proof Since R(J=1) is algebraically closed, it follows that f splits into a product of irreducible factor s of degree 1 or 2. If X 2 + «X + Pis irreducible («, PER) then it is a sum of squares, namely
and we must have 4P > rx 2 since our factor is assumed irreducible. Hence the change of sign of f must be due to the change of sign of a linear factor, which is trivially verified to be a root lying between a and b.
Lemma 2.6. Let K be a subfieldof an ordered field E. Let rx E E be algebraic over K, and a root of the polynomial f(X) =
xn +
with coefficients in K . Then Iell
an_IX n- 1 + . . . + aD
~ 1
+ lan -II + . . . + lao I.
Proof If [«] ~ 1, the assertion is obvious. If Irxl > 1, we express [«]" in terms of the terms of lower degree, divide by I«]"- I, and get a proof for our lemma . Note that the lemma implies that an element which is algebraic over an ordered field cannot be infinitely large with respect to that field. Let f(X) be a polynomial with coefficients in a real closed fi
Data Loading...
