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From: Robert Israel on 30 Apr 2010 13:51

> 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.

.... it divides any polynomial with rational coefficients of which r is a root.
So you must assume g is a polynomial with rational coefficients (which
rules out Rob Johnson's example).
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: David Hartley on 30 Apr 2010 13:51
In message <20100430.083131(a)whim.org>, Rob Johnson <rob(a)trash.whim.org>
writes
>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.

I assume the OP wanted g to be a polynomial with rational coefficients.
--
David Hartley
From: Pubkeybreaker on 30 Apr 2010 18:14
On Apr 30, 12:01�pm, r...(a)trash.whim.org (Rob Johnson) wrote:
> In article <847439184.49060.1272628101670.JavaMail.r...(a)gallium.mathforum..org>,
>
> 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?
>
> 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.


For THIS particular g, yes. But:

The original problem said nothing about g. As such, all we assume is
the
existence of such a function.


And it did not say for "some root r, g(r) is also a root". When it
said, for 'a root' r,
I assumed that it meant "for 'any root r, g(r) is also a root".

Perhaps the original problem needs to be clearer about quantifiers?

for f(x) = x^2 + 1, the following function g works:

g(r) = r - 2i for r = i
g(r) = r + 2i for r = -i.

There were no limitations on g in the original problem. I only
assumed
that existence of a function g such that if r is any root, then
g(r) is
also a root.

Nowhere did it state any requirements on g.
From: Pubkeybreaker on 30 Apr 2010 18:15
On Apr 30, 1:51�pm, David Hartley <m...(a)privacy.net> wrote:
> In message <20100430.083...(a)whim.org>, Rob Johnson <r...(a)trash.whim.org>
> writes
>
> >In article
> ><847439184.49060.1272628101670.JavaMail.r...(a)gallium.mathforum.org>,
> >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?
>
> >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.
>
> I assume the OP wanted g to be a polynomial with rational coefficients.

Why?? I saw no such restriction. I assume that problems are fully
and
completely stated.
From: David Hartley on 30 Apr 2010 19:32
In message
<10098b8f-48a6-43d4-a0f3-dfaa2703676c(a)u21g2000vbr.googlegroups.com>,
Pubkeybreaker <pubkeybreaker(a)aol.com> writes
>There were no limitations on g in the original problem. I only assumed
>that existence of a function g such that if r is any root, then
>g(r) is
>also a root.
>
>Nowhere did it state any requirements on g.


For such a g, the original problem is trivial, even if f is not
irreducible, indeed even if f is not a polynomial. I agree the OP was
unclear but think it very likely that he intended that g be a polynomial
with rational coefficients and that only one root r is required to have
the property that g(r) is also a root of f.
--
David Hartley
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