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From: Kusanagi on 14 Nov 2009 00:43
Let f(x)=ax^5 + bx^4 + cx^3 + dx^2 +e^x +1 \in Q[x].

If f(x) has three distinct nonzero real roots and two distinct complex roots (not reals), is it true that a Galois group of f(x) is S_5 that is not a solvable group?
From: TPiezas on 14 Nov 2009 11:16
On Nov 14, 9:43 am, Kusanagi <Kusan...(a)hotmail.com> wrote:
> Let f(x)=ax^5 + bx^4 + cx^3 + dx^2 +e^x +1 \in Q[x].
>
> If f(x) has three distinct nonzero real roots and two distinct complex roots (not reals), is it true that a Galois group of f(x) is S_5 that is not a solvable group?

Yes, it is S_5. See here: http://www.math.uni-duesseldorf.de/~klueners/minimum/node14.html
with a table showing that if an irreducible eqn has 3 real roots, then
it must have group S_5.

The general result, in case you are interested, is: If an irreducible
equation f(x) of PRIME degree p has a solvable group, then f(x) only
has either 1 or p real roots (never the values in between.)

- Titus
From: TPiezas on 14 Nov 2009 11:21
On Nov 14, 10:16 am, TPiezas <tpie...(a)gmail.com> wrote:
> On Nov 14, 9:43 am, Kusanagi <Kusan...(a)hotmail.com> wrote:
>
> > Let f(x)=ax^5 + bx^4 + cx^3 + dx^2 +e^x +1 \in Q[x].
>
> > If f(x) has three distinct nonzero real roots and two distinct complex roots (not reals), is it true that a Galois group of f(x) is S_5 that is not a solvable group?
>
> Yes, it is S_5.  See here:http://www.math.uni-duesseldorf.de/~klueners/minimum/node14.html
> with a table showing that if an irreducible eqn has 3 real roots, then
> it must have group S_5.
>
> The general result, in case you are interested, is:  If an irreducible
> equation f(x) of PRIME degree p has a solvable group, then f(x) only
> has either 1 or p real roots (never the values in between.)
>
> - Titus

.... but, if I may add, the theorem, of course, does NOT imply if an
eqn has 1 or p real roots, then it solvable. (After all, a quintic
with group A_5 has 1 or 5 real roots, but is not a solvable group.)

- Titus
From: AP on 15 Nov 2009 04:00
On Sat, 14 Nov 2009 10:43:58 EST, Kusanagi <Kusanagi(a)hotmail.com>
wrote:

>Let f(x)=ax^5 + bx^4 + cx^3 + dx^2 +e^x +1 \in Q[x].
>
>If f(x) has three distinct nonzero real roots and two distinct complex roots (not reals), is it true that a Galois group of f(x) is S_5 that is not a solvable group?

? f(x)=(x-1)(x-2)(x-3)(x^2+1) is solvable

but it is true if also f is irreducible
the proof is exactly the same for the example X^5-10X+5 ( in every
book on Galois )

and

this proof is valid for prove
if P irred , d�P=p prime >=5 , P has exactly two not reals roots
then Gal(P) is S_p

ex X^7-10X^5+10X+5


From: AP on 12 Dec 2009 01:34
On Sat, 14 Nov 2009 08:16:19 -0800 (PST), TPiezas <tpiezas(a)gmail.com>
wrote:

>On Nov 14, 9:43�am, Kusanagi <Kusan...(a)hotmail.com> wrote:
>> Let f(x)=ax^5 + bx^4 + cx^3 + dx^2 +e^x +1 \in Q[x].
>>
>> If f(x) has three distinct nonzero real roots and two distinct complex roots (not reals), is it true that a Galois group of f(x) is S_5 that is not a solvable group?
>
>Yes, it is S_5. See here: http://www.math.uni-duesseldorf.de/~klueners/minimum/node14.html
>with a table showing that if an irreducible eqn has 3 real roots, then
>it must have group S_5.
>
>The general result, in case you are interested, is: If an irreducible
>equation f(x) of PRIME degree p has a solvable group, then f(x) only
>has either 1 or p real roots (never the values in between.)
if f(x)=x^2+1 , it is false?
>- Titus

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