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From: Rob Johnson on 30 Apr 2010 19:46
In article <0Tp5rmP1h22LFwgA(a)212648.invalid>,
David Hartley <me9(a)privacy.net> wrote:
>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.

Like David, I agree that the question posed to the OP was stated with
the intent that g would be a polynomial with rational coefficients.
However, since the OP did not specify this, I wanted to get them to
think about this and see why it was important. I was not sure if
this was homework, so I didn't want to say too much.

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
From: Kent Holing on 30 Apr 2010 16:04
To be precise:
f(x) = 0 is irreducible with rational coefficents with roots r and g(r) for g(r) in Q[r] for a FIXED root r. (This means that g is a polynomial with RATIONAL coefficients).
We can prove more than asked for in the first place. In fact, if f(r) = f(g(r)) = 0 for the fixed root r, then explain why g(t) is a root of f(x) = 0 for ANY root t of
f(x) = 0.
Kent Holing
From: Pubkeybreaker on 30 Apr 2010 21:09
On Apr 30, 7:32�pm, David Hartley <m...(a)privacy.net> wrote:
> In message
> <10098b8f-48a6-43d4-a0f3-dfaa27036...(a)u21g2000vbr.googlegroups.com>,
> Pubkeybreaker <pubkeybrea...(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

The difficulty is that within this forum we do not know who people are
or what their background is. Perhaps the problem was presented by a
high school freshman, or even someone in junior high.
From: Kent Holing on 1 May 2010 08:12
It is important that f(x) is irreducible. Since fg and f
in Q[x] have the common root r and f(x) is irreducible over Q then f(x) must divide fg(x). So, fg(x) = f(x) H(x)
for all x.

For t any root of f(x) = 0, we have f(g(t)) = fg(t) =
f(t) H(t) = 0 since f(t) = 0. So, g(t) is a root of f(x) = 0 for any root t of f(x) = 0, as claimed.

This is related to the Galois group of f(x) = 0.

Take the irreducible quartic as an example. Then |G| = 4 if and only all roots are in Q[r] for r a root.

If the quartic has both real and complex roots (i.e. negative discriminant) then G = D4 if all real roots are in Q[r] (r real); otherwise G = S4.

In the case above, if the two real roots are r and g(r) then for the complex (conjugate) roots of the quartic r3 and r4, we also have r4 = g(r3).
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