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Additional resources for 4-Manifolds which embed in R5 R6, and Seifert manifolds for fibered knots

Sample text

Proof The first statement is obvious. Suppose that f is a non-zero element of F[x]. We clear denominators: there exists () in R such that bf E R[x]. Let y be the content of bf Then bf=yg, where g is primitive in R[x], and so f=(b- 1 y)g=f3g. Suppose that f = {3'g' is another such expression. We again clear denominators: there exists a in R such that af3 and cx/3' are in R. Then af = (af3)g = (af3')g'. ), and so af3 and af3' are 32 Rings associates in R. f3g', so that g=sg' and g and g' are associates in R[x].

1 Constructible points There are many constructions that one can carry out with ruler (straight-edge) and compasses alone. Many children, on first being given a pair of compasses, find out for themselves how to construct a regular hexagon (and so construct the angle n/3). I hope that you remember enough school geometry to know how to bisect an angle, to drop a perpendicular from a point to a line, to draw a line through a point parallel to a given line, and so divide an interval into a given rational ratio, using ruler and compasses alone.

As F[x] is a unique factorization domain, m = k and there exists a permutation p of { 1, ... , k} and non-zero elements t: 1 , ... , t:k ofF such that fi = t:if~ for 1 ~ i:::;; k. 11. Thus R[x] is a unique factorization domain. Corollary 1 Suppose that f is a primitive element of R[x], that g is a nonzero element of R[x] and that f divides gin F[x]. Then f divides gin R[x]. Proof. jg1 ... i are irreducible in Rand the gi are irreducible elements of R[x] of positive degree. By Gauss' lemma, each gi is primitive and irreducible in F[x].

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4-Manifolds which embed in R5 R6, and Seifert manifolds for fibered knots by Cochran T.


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