COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

A real root calculus
HANS ZASSENHAUS
HOW can we construct a really closed algebraic extension over an algebraically
ordered field F?
We assume that F is constructively algebraically ordered (see [I]).
A real root calculus over F consists in solving the following two tasks :
(I) To assign to each polynomial P of F[X] a non-negative integer
NR(P). This number will turn out to be the maximal number of
distinct roots of P in any algebraically ordered extension of F.
(II) To assign to each polynomial U of F[X] and to each index I satisfying
the condition 0~1~ NR(P) uniquely a number SIGN( U, 1, P)
which is one of the three numbers 1, 0, - 1. It should turn out to be
the sign of the value of U for the Ith root of P in any algebraically
ordered extension of F containing NR(P) distinct roots.
This task was first solved by Vandiver [4] in case the algebraic ordering of
F was archimedean. Use had to be made of factorizations of polynomials of
F[X] into irreducible factors. The task was solved again by A. Hollkott [2]
in his. 19.41 thesis without taking recourse to Vandiver’s additional two
assumptions. Tarski [3] solved the task independently.
The real root calculus which is expounded here is based on A. Hollkott’s
thesis. For the benefit of English readers streamlined proofs of the necessary
theorems are given.
1. Here are the theorems to be proven later:
THEOREM 1. (Between value theorem.) If P c F[X], A -= B, P(A)P(B) .c 0,
then there can be constructed an algebraically ordered extension of F containing
a root R or P satisfying the inequalities
A < R < B.t (1)
t As it stands, A, B denote elements of the algebraically ordered field F. We agree,
however, that A, B also are permitted to be one of the symbols 43, - 03 which are subject
to the rules: --oo 0 and that there are elements
A, B of Fsuch that sign P( - CO) = sign P(Y), if - w=sY=sA, and sign P(m) = sign
P(Y), if B=SYGCO for Yin any algebraically ordered extension of F.
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