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Imitation of an Iteration

Imitation of an Iteration fields F , ••• , F,, _ for some n ~ 2, such that each field is a sub field of the algebraic closure of F 1 1 and F c F c · · · c F,,_ • Then let a, be an element of the algebraic closure of F which is a 1 2 1 1 root of an irreducible polynomial in F,,_ d x] and define F,, = F,,_ d a,]. Then F c F c F c · · · 1 2 1 and F = U, F,, is a countable field. Assume that K is a countable sub field of F. For each a in K, there is a smallest n(a) such that a is in F,,(a)· Since a is not in F,,<,, , a is a generator of the cyclic group of nonzero elements in F,(a)· Hence we have F,,<a) c K. Since there are countably many elements inK, sup{n(a): aEK}=oo. The nesting of the F,,'s now shows that F= UaF,r(a) = K. References [ 1] S. Lang. Algebra, Addison-Wesley. Reading. Mass .. 1965. pp. 253-255. DANIEL A. RAWSTHORNE Wheaton, MD :!09(>:! A most fascinating and frustrating problem is the Collatz 3x + http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Mathematics Magazine Taylor & Francis

Imitation of an Iteration

Mathematics Magazine , Volume 58 (3): 5 – May 1, 1985

Imitation of an Iteration

Mathematics Magazine , Volume 58 (3): 5 – May 1, 1985

Abstract

fields F , ••• , F,, _ for some n ~ 2, such that each field is a sub field of the algebraic closure of F 1 1 and F c F c · · · c F,,_ • Then let a, be an element of the algebraic closure of F which is a 1 2 1 1 root of an irreducible polynomial in F,,_ d x] and define F,, = F,,_ d a,]. Then F c F c F c · · · 1 2 1 and F = U, F,, is a countable field. Assume that K is a countable sub field of F. For each a in K, there is a smallest n(a) such that a is in F,,(a)· Since a is not in F,,<,, , a is a generator of the cyclic group of nonzero elements in F,(a)· Hence we have F,,<a) c K. Since there are countably many elements inK, sup{n(a): aEK}=oo. The nesting of the F,,'s now shows that F= UaF,r(a) = K. References [ 1] S. Lang. Algebra, Addison-Wesley. Reading. Mass .. 1965. pp. 253-255. DANIEL A. RAWSTHORNE Wheaton, MD :!09(>:! A most fascinating and frustrating problem is the Collatz 3x +

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References (8)

Publisher
Taylor & Francis
Copyright
Copyright Taylor & Francis
ISSN
1930-0980
eISSN
0025-570X
DOI
10.1080/0025570X.1985.11977179
Publisher site
See Article on Publisher Site

Abstract

fields F , ••• , F,, _ for some n ~ 2, such that each field is a sub field of the algebraic closure of F 1 1 and F c F c · · · c F,,_ • Then let a, be an element of the algebraic closure of F which is a 1 2 1 1 root of an irreducible polynomial in F,,_ d x] and define F,, = F,,_ d a,]. Then F c F c F c · · · 1 2 1 and F = U, F,, is a countable field. Assume that K is a countable sub field of F. For each a in K, there is a smallest n(a) such that a is in F,,(a)· Since a is not in F,,<,, , a is a generator of the cyclic group of nonzero elements in F,(a)· Hence we have F,,<a) c K. Since there are countably many elements inK, sup{n(a): aEK}=oo. The nesting of the F,,'s now shows that F= UaF,r(a) = K. References [ 1] S. Lang. Algebra, Addison-Wesley. Reading. Mass .. 1965. pp. 253-255. DANIEL A. RAWSTHORNE Wheaton, MD :!09(>:! A most fascinating and frustrating problem is the Collatz 3x +

Journal

Mathematics MagazineTaylor & Francis

Published: May 1, 1985

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