Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

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


Math Forum » Discussions » sci.math.* » sci.math

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Maury Barbato

Posts: 791
From: University Federico II of Naples
Registered: 3/15/05
Re: Polynomials' Zeros
Posted: Apr 26, 2013 1:07 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

I wrote:

> 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


I have found a way to construct countable I and J.
Let {x_1,x_2,?} be a countable subset of K such that for
every x_i we have p(x_i,y)=0 for finitely many y in K
(this subset exists for property (1)). Analogously, let
{y_1,y_2,?} be a countable subset of K such that for every
y_i p(x,y_i)=0 for finitely many x in K.

Let Z_{+}={1,2,3,?} be the set of positive integers.
For every positive integer n, define the (finite) set

S_n={ m in Z_{+} | p(x_n,y_m)=0}.

Analogously, let T_n be the set

T_n={m in Z_{+} | p(x_m,y_n)=0}.

We construct I={x_{i_1}, x_{i_2},?} and J={y_{j_1},
y_{j_2},?} by induction in the following way.
Define i_1=1. Suppose that we have defined i_1,?,i_k,
and j_1,?,j_{k-1}. Then we define

j_k = min Z_{+} \ ({j_1,?,j_{k-1}} \/ T_{i_1} \/ T_{i_2} \/ ?\/ T_{i_k}),

and

i_{k+1}= min Z_{+} \({i_1,?,i_k} \/ S_{j_1} \/ S_{j_2} \/ ? \/ S_{j_k}).

It is immediate from construction that I and J have the desired property.
I think the same construction can be used to show that
we can find I and J with |I|=|J|=|K| by using
transfinite induction insted of ordinary induction.
But I don't know the theory of ordinal numbers so well,
so I am not sure. What do you think?
Thank you very very much for your attention.
My Best Regards,
Maurizio Barbato



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.