Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Polynomials' Zeros
Replies: 7   Last Post: Apr 26, 2013 1:07 PM

 Messages: [ Previous | Next ]
 Maury Barbato Posts: 792 From: University Federico II of Naples Registered: 3/15/05
Re: Polynomials' Zeros
Posted: Apr 25, 2013 8:52 AM

Butch Malahide:

> On Apr 23, 1:42 pm, Butch Malahide
> <fred.gal...@gmail.com> wrote:

> > On Apr 23, 12:25 pm, dullr...@sprynet.com wrote:
> >
> >
> >
> >
> >

> > > On Tue, 23 Apr 2013 09:52:05 -0700 (PDT),
> sputaro...@alice.it wrote:
> > > >Hello,
> > > >I have the following question.
> > > >Let P(X_1,...,X_n) be a non zero polynomial in

> X_1,...,X_n with coefficients in an infinite field K.
> > > >Does there exist n infinite subsets J_1,...,J_n
> of K such that
> > > >P(x_1,...,x_n) =/= 0
> > > >for every (x_1,...,x_n) in J_1 x J_2 x...x J_n?
> > > >The answer is urely yes if K is a subfield of C:

> just consider a neighborhood
> > > >of a point in which the poynomial has a non zero
> value.
> > > >But I don't know the answer for the general
> case.
> > > >What do you think about?
> >
> > > Induction on n.
> >
> > > Consider the case n =  2 to make things easier to
> type.
> > > Say p(x,y) = 0 for all (x,y) in IxJ, where I and
> J are infinite.
> > > [. . .] So p = 0.
> >
> > I think it's a little bit more complicated. How do

> you get from the
> > negation of the statement
> >
> > "p(x,y) = 0 for all (x,y) in IxJ, where I and J are

> infinite"
> >
> > to the OP's statement
> >
> > "p(x,y) =/= 0 for all (x,y) in IxJ, where I and J

> are infinite"?
> >
> > I think something like the following will work. I

> believe the argument
> > you gave actually proves the stronger result:
> >
> > [(for infinitely many x) (for infinitely many y)

> p(x,y) = 0] implies p
> > = 0.
> >
> > That is, if there exist an infinite set I and

> infinite sets J_x (x in
> > I) such that p(x,y) = 0 whenever x in I and y in
> J_x, then p = 0.
> >
> > In other words, if p =/= 0, then (writing "a.e."

> for "all but finitely
> > many"):
> >
> > (1) (for a.e. x) (for a.e. y) p(x,y) =/= 0.
> >
> > Also, interchanging the roles of x and y,
> >
> > (2) (for a.e. y) (for a.e. x) p(x,y) =/= 0.
> >
> > From (1) and (2) we can construct infinite sets I

> and J such that
> > p(x,y) =/= 0 whenever x in I and y in J.
>
> Moreover, if |K| = aleph_{nu}, we can make |I| = |J|
> = aleph_{nu}.

Dear Butch, as you said, prof. Ullrich's argument
actually proves that (1) and (2) are true, but
I don't see how to use these two statements in
order to prove that there exist two sets I and J
with the same cardinality of K, such that
p(x,y) =/= = for every x in I and y in J.
Could you give some hint, please?

Thank you very very much for your invaluable help.
I couldn't have really realized alone how to use
prof. Ullirch's argument to give an answer to
my question.
My Best Regards,
Maurizio Barbato

A person who never made a mistake never tried anything new.
A. Einstein

Date Subject Author
4/23/13 Maury Barbato
4/23/13 David C. Ullrich
4/23/13 Maury Barbato
4/23/13 Butch Malahide
4/23/13 Butch Malahide
4/25/13 Maury Barbato
4/26/13 Maury Barbato
4/24/13 David C. Ullrich