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From: achille on 10 Aug 2010 02:38
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. [Note 1]

In general, I find that for any f, g, h \in Z[x,y,z] such that
f irreducible and f = g^2 + h^2, then there are 2 possible
cases:

1) g and h can be symmetric or antisymmetric independently.
OR 2) by a suitable choice of sign of h,

g(x,y,z) = g(y,z,x) = g(z,x,y)
= h(y,x,z) = h(x,z,y) = h(z,y,x)

ie g, h are symmetric under cyclic permutations and
g maps to h, h maps to g under any odd permutations.

This is equivalent to 2f can be rewritten as

2f = (g+h)^2 + (g-h)^2

where g+h are symmetric and g-h antisymmetric in x,y,z.

---Notes---
[1] If f is reducible over Z[x,y,z], polynomials like f(x,1,2)
will be reducible over Z[x]. However,

f(x,1,2) = 5*x^4+24*x^3+65*x^2+48*x+20

is irreducible over Z[x], so does f over Z[x,y,z].
From: quasi on 10 Aug 2010 03:57
On Mon, 9 Aug 2010 23:38:39 -0700 (PDT), achille
<achille_hui(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
From: Robert Israel on 10 Aug 2010 02:58
quasi <quasi(a)null.set> writes:


> 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.

Try g = (2*x-y)*(2*y-z)*(2*z-x), h = (x-2*y)*(y-2*x)*(z-2*x)
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: quasi on 10 Aug 2010 04:01
On Tue, 10 Aug 2010 01:58:26 -0500, Robert Israel
<israel(a)math.MyUniversitysInitials.ca> wrote:

>quasi <quasi(a)null.set> writes:
>
>
>> 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.
>
>Try g = (2*x-y)*(2*y-z)*(2*z-x), h = (x-2*y)*(y-2*x)*(z-2*x)

Another nice counterexample.

Thanks.

quasi
From: master1729 on 9 Aug 2010 23:28
> On Mon, 09 Aug 2010 13:05:17 -0500, quasi
> <quasi(a)null.set> wrote:
>
> >On Mon, 09 Aug 2010 02:23:50 -0500, quasi
> <quasi(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?

thats close to my idea ...

>
> 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
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