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From: achille on 10 Aug 2010 03:57
On Aug 10, 3:57 pm, quasi <qu...(a)null.set> wrote:
> On Mon, 9 Aug 2010 23:38:39 -0700 (PDT), achille
>
>
>
> <achille_...(a)yahoo.com.hk> wrote:
> >On Aug 10, 2:41 pm, quasi <qu...(a)null.set> wrote:
> >> On Mon, 09 Aug 2010 13:05:17 -0500, quasi <qu...(a)null.set> wrote:
> >> >On Mon, 09 Aug 2010 02:23:50 -0500, quasi <qu...(a)null.set> wrote:
>
> >> >>Prove or disprove:
>
> >> >>If f in K(x,y,z) is a symmetric rational function such that
>
> >> >>   f = g^2 + h^2
>
> >> >>for some g,h in K(x,y,z), then g^2 and h^2 are symmetric.
>
> >> >I'll conjecture that the answer is "yes".
>
> >> But Tim Murphy pointed the way to a counterexample, and achille
> >> followed up and actually produced one, so the conjecture, as stated
> >> above, is dead.
>
> >> But what if we switch from rational functions to polynomials and then
> >> also require that f is irreducible?
>
> >> So here's a revised version ...
>
> >> Prove or disprove:
>
> >> Let K be a field. If f in K[x,y,z] is an irreducible symmetric
> >> polynomial such that
>
> >>    f = g^2 + h^2
>
> >> for some g,h in K[x,y,z], then g^2 and h^2 are symmetric.
>
> >> Remark:
>
> >> I think my previously posted "fuzzy plan of attack" might be work for
> >> this version.
>
> >> quasi
>
> >My mistake, I thought you mean g, h are totally symmetric
> >in x,y,z. In any event, let
>
> >   g(x,y,z) = y*z^2+x*y*z+x^2*z+x*y^2
> >   h(x,y,z) = x*z^2+y^2*z+x*y*z+x^2*y
> >   f(x,y,z) = g(x,y,z)^2 + h(x,y,z)^2
>
> >then g and h are neither symmetric nor antisymmetric but
> >f is irreducible and symmetric in x,y,z.
>
> Yes, very nice.
>
> I think I understand where my intuition went wrong.
>
> But maybe 3 variables is the border case?
>
> What if the problem required 4 variables, w,x,y,z? Could a
> counterexample be created analagous to the one you gave above?
>
> quasi

I'm not 100% sure the following works or not:

p(w,x,y,z) = (w+x)*(w+y)*(w+z)*(x+y)*(x+z)*(y+z);
q(w,x,y,z) = (w-x)*(w-y)*(w-z)*(x-y)*(x-z)*(y-z);

g(w,x,y,z) = (p(w,x,y,z)+q(w,x,y,z))/2;
h(w,x,y,z) = (p(w,x,y,z)-q(w,x,y,z))/2;

f(w,x,y,z) = g(w,x,y,z)^2 + h(w,x,y,z)%2.

again g, h is symmetric under all even permutations
and g maps to h, h maps to g under all odd permutations.

My computer fails to factorize polynomial like

f(w,w+1,1,2) =
18*w^10+306*w^9+2273*w^8+9680*w^7+26158*w^6+46544*w^5
+55213*w^4+42794*w^3+20702*w^2+5616*w+648

but I'm not sure whether it is the limitation of my computer
or f(w,w+1,1,2) and hence f(w,x,y,z) is irreducible...
From: quasi on 10 Aug 2010 13:24
Based on achille's succession of counterexamples, it suggests to me
that I need to block the potential of the alternating group. So maybe
what I need is not more variables, but instead, odd exponents, or at
least exponents that are not powers of 2.

To keep it simple, let's use 3 variables, exponents equal to 3, and
take the field Q instead of an arbitrary field K. The conclusion gets
a slight boost.

Here's the latest version ...

Prove or disprove:

If f in Q[x,y,z] is an irreducible symmetric polynomial such that

f = g^3 + h^3

for some g,h in Q[x,y,z], then g and h are symmetric.

quasi
From: quasi on 10 Aug 2010 13:30
On Tue, 10 Aug 2010 12:24:02 -0500, quasi <quasi(a)null.set> wrote:

>Based on achille's succession of counterexamples, it suggests to me
>that I need to block the potential of the alternating group. So maybe
>what I need is not more variables, but instead, odd exponents, or at
>least exponents that are not powers of 2.
>
>To keep it simple, let's use 3 variables, exponents equal to 3, and
>take the field Q instead of an arbitrary field K. The conclusion gets
>a slight boost.
>
>Here's the latest version ...
>
>Prove or disprove:
>
>If f in Q[x,y,z] is an irreducible symmetric polynomial such that
>
> f = g^3 + h^3
>
>for some g,h in Q[x,y,z], then g and h are symmetric.

Oops.

I better not require that f is irreducible!

Ok, quick fix:

Prove or disprove:

If f in Q[x,y,z] is a symmetric polynomial such that

f = g^3 + h^3

for some g,h in Q[x,y,z], then g and h are symmetric.

quasi
From: quasi on 10 Aug 2010 16:57
On Tue, 10 Aug 2010 12:30:11 -0500, quasi <quasi(a)null.set> wrote:

>On Tue, 10 Aug 2010 12:24:02 -0500, quasi <quasi(a)null.set> wrote:
>
>>Based on achille's succession of counterexamples, it suggests to me
>>that I need to block the potential of the alternating group. So maybe
>>what I need is not more variables, but instead, odd exponents, or at
>>least exponents that are not powers of 2.
>>
>>To keep it simple, let's use 3 variables, exponents equal to 3, and
>>take the field Q instead of an arbitrary field K. The conclusion gets
>>a slight boost.
>>
>>Here's the latest version ...
>>
>>Prove or disprove:
>>
>>If f in Q[x,y,z] is an irreducible symmetric polynomial such that
>>
>> f = g^3 + h^3
>>
>>for some g,h in Q[x,y,z], then g and h are symmetric.
>
>Oops.
>
>I better not require that f is irreducible!
>
>Ok, quick fix:
>
>Prove or disprove:
>
>If f in Q[x,y,z] is a symmetric polynomial such that
>
> f = g^3 + h^3
>
>for some g,h in Q[x,y,z], then g and h are symmetric.

Since it's now cubes, not squares, I guess I should rename the thread.

Ok, so then for any followups on the cubic version, see the new thread
(to be posted shortly).

Thanks.

quasi
From: Robert Israel on 10 Aug 2010 17:05
achille <achille_hui(a)yahoo.com.hk> writes:

> On Aug 10, 3:57=A0pm, quasi <qu...(a)null.set> wrote:
> > On Mon, 9 Aug 2010 23:38:39 -0700 (PDT), achille
> >
> >
> >
> > <achille_...(a)yahoo.com.hk> wrote:
> > >On Aug 10, 2:41=A0pm, quasi <qu...(a)null.set> wrote:
> > >> On Mon, 09 Aug 2010 13:05:17 -0500, quasi <qu...(a)null.set> wrote:
> > >> >On Mon, 09 Aug 2010 02:23:50 -0500, quasi <qu...(a)null.set> wrote:
> >
> > >> >>Prove or disprove:
> >
> > >> >>If f in K(x,y,z) is a symmetric rational function such that
> >
> > >> >> =A0 f =3D g^2 + h^2
> >
> > >> >>for some g,h in K(x,y,z), then g^2 and h^2 are symmetric.
> >
> > >> >I'll conjecture that the answer is "yes".
> >
> > >> But Tim Murphy pointed the way to a counterexample, and achille
> > >> followed up and actually produced one, so the conjecture, as stated
> > >> above, is dead.
> >
> > >> But what if we switch from rational functions to polynomials and then
> > >> also require that f is irreducible?
> >
> > >> So here's a revised version ...
> >
> > >> Prove or disprove:
> >
> > >> Let K be a field. If f in K[x,y,z] is an irreducible symmetric
> > >> polynomial such that
> >
> > >> =A0 =A0f =3D g^2 + h^2
> >
> > >> for some g,h in K[x,y,z], then g^2 and h^2 are symmetric.
> >
> > >> Remark:
> >
> > >> I think my previously posted "fuzzy plan of attack" might be work for
> > >> this version.
> >
> > >> quasi
> >
> > >My mistake, I thought you mean g, h are totally symmetric
> > >in x,y,z. In any event, let
> >
> > > =A0 g(x,y,z) =3D y*z^2+x*y*z+x^2*z+x*y^2
> > > =A0 h(x,y,z) =3D x*z^2+y^2*z+x*y*z+x^2*y
> > > =A0 f(x,y,z) =3D g(x,y,z)^2 + h(x,y,z)^2
> >
> > >then g and h are neither symmetric nor antisymmetric but
> > >f is irreducible and symmetric in x,y,z.
> >
> > Yes, very nice.
> >
> > I think I understand where my intuition went wrong.
> >
> > But maybe 3 variables is the border case?
> >
> > What if the problem required 4 variables, w,x,y,z? Could a
> > counterexample be created analagous to the one you gave above?
> >
> > quasi
>
> I'm not 100% sure the following works or not:
>
> p(w,x,y,z) =3D (w+x)*(w+y)*(w+z)*(x+y)*(x+z)*(y+z);
> q(w,x,y,z) =3D (w-x)*(w-y)*(w-z)*(x-y)*(x-z)*(y-z);
>
> g(w,x,y,z) =3D (p(w,x,y,z)+q(w,x,y,z))/2;
> h(w,x,y,z) =3D (p(w,x,y,z)-q(w,x,y,z))/2;
>
> f(w,x,y,z) =3D g(w,x,y,z)^2 + h(w,x,y,z)%2.
>
> again g, h is symmetric under all even permutations
> and g maps to h, h maps to g under all odd permutations.
>
> My computer fails to factorize polynomial like
>
> f(w,w+1,1,2) =3D
> 18*w^10+306*w^9+2273*w^8+9680*w^7+26158*w^6+46544*w^5
> +55213*w^4+42794*w^3+20702*w^2+5616*w+648
>
> but I'm not sure whether it is the limitation of my computer
> or f(w,w+1,1,2) and hence f(w,x,y,z) is irreducible...

Maple says this is irreducible.

More generally, for n variables x_i and coefficients c_i, consider
g(x_1,...,x_n) = product_{pi in A(n)} sum_{i=1}^n c_i x_{pi(i)}
h(x_1,...,x_n) = product_{pi in S(n) \ A(n)} sum_{i=1}^n c_i x_{pi(i)}

where S(n) is the symmetric group and A(n) the alternating group on {1,...,n}.
Then f(x_1,...,x_n) = g(x_1,...,x_n)^2 + h(x_1,...,x_n)^2 is a symmetric
function, but g(x_1,...,x_n)^2 and h(x_1,...,x_n)^2 are not, being
interchanged by any odd permutation. I'm not sure, but I strongly suspect
that (for at least some choice of coefficients) f will be irreducible.
You could also replace the square by some other power (of course if
p is odd, g^p + h^p is not irreducible, being divisible by g + h).
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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