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From: Kent Holing on 30 Apr 2010 03:47
Let f(x) be a monic irreducible polynomial with rational coefficients and with a root r. If g(r) is also a root of the given equation in Q[r] (different from r), why is g(g(r)) also a root?

Kent Holing,
NORWAY
From: Pubkeybreaker on 30 Apr 2010 08:23
On Apr 30, 7:47 am, Kent Holing <K...(a)statoil.com> wrote:
> Let f(x) be a monic irreducible  polynomial with rational coefficients and with a root r. If g(r) is also a root of the given equation in Q[r] (different from r), why is g(g(r)) also a root?

Huh? We are given that r and r1 = g(r) are both roots. But then
g(g(r) = g(r1) is also trivially a root by the assumptions.
This is just a change of variable --> first year algebra.
From: Kent Holing on 30 Apr 2010 05:11
I did not assume that g(.) for any root was a root, just that g(r) is a root for a given root r.
From: David Hartley on 30 Apr 2010 11:10
In message
<9ad2f2d8-cda1-4335-aa25-f8d58b23e4cd(a)g21g2000yqk.googlegroups.com>,
Pubkeybreaker <pubkeybreaker(a)aol.com> writes
>On Apr 30, 7:47�am, Kent Holing <K...(a)statoil.com> wrote:
>> Let f(x) be a monic irreducible �polynomial with rational
>>coefficients and with a root r. If g(r) is also a root of the given
>>equation in Q[r] (different from r), why is g(g(r)) also a root?
>
>Huh? We are given that r and r1 = g(r) are both roots. But then
>g(g(r) = g(r1) is also trivially a root by the assumptions.
>This is just a change of variable --> first year algebra.

It's not true if f is not irreducible, so it's not just a matter of
changing variables.

If f is irreducible, then it divides any polynomial of which r is a
root, in particular f(g(x)) - f(x). But then any root of f must also be
a root of this poly, including r1. Hence f(g(r1) = f(r1) = 0; g(r1) is
also a root of f.
--
David Hartley
From: Rob Johnson on 30 Apr 2010 12:01
In article <847439184.49060.1272628101670.JavaMail.root(a)gallium.mathforum.org>,
Kent Holing <KHO(a)statoil.com> wrote:
>Let f(x) be a monic irreducible polynomial with rational coefficients
>and with a root r. If g(r) is also a root of the given equation in
>Q[r] (different from r), why is g(g(r)) also a root?

If f has rational coefficients, being monic tells us nothing in this
context.

The conclusion about g(g(r)) is not true, at least not without some
constraints on g. Take for example f(x) = x^2 + 1, r = i, and
g(x) = x - 2i. We have that g(r) = -i is a root, but g(g(r)) = -3i
is not a root.

Rob Johnson <rob(a)trash.whim.org>
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