The Math Forum

Search All of the Math Forum:

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

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

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

Advanced Search

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

Posts: 792
From: University Federico II of Naples
Registered: 3/15/05
Re: Polynomials' Zeros
Posted: Apr 23, 2013 1:58 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

David C. Ullrich wrote:

> On Tue, 23 Apr 2013 09:52:05 -0700 (PDT),
> 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.
> For each fixed y, consider the polynomial p_y(x),
> defined by
> p_y(x) = p(x,y).
> For each y in J we have p_y = 0, since p_y has
> infinitely
> many zeroes.
> But there are polynomials q_j so that
> p(x,y) = p_y(x) = sum_j q_j(y) x^j.
> If y is in J then all the coefficients of p_y vanish.
> This says that q_j(y) = 0 for all y in J, and hence
> q_j = 0 since J is infinite. So p = 0.

> >My Best Regards,
> >Maurizio Barbato


Dear Prof. Ullrich, first of all thank you very much
for answering me. Your argument is correct, but
it does not answer my question.
What you prove is that if p(x,y)=0 for every
(x,y) in IxJ, with I,J infinite, then p(X,Y) is
the zero polynomial.
My original question is different. I ask if
there exists some infinite sets I,J, such that
p(x,y) has always a non zero value in IxJ.
As I said in my original post, if K is a subfield
of C, then the answer is yes. We know that
p(x,y) is non zero for some (x,y) in K^2 (this
immediately follows from the statement you proved).
Then consider a neighborhood of (x,y).
By continuity, p(x,y) is non zero in this
neighborhood. So for some real intervals
I, J, we get p(x,y) =/= 0 for every (x,y) in IxJ.
Since K /\ I and K /\ J are infinite (they contain
all the rationals in I and J respectively)
we are done.
What about a generic infinite field K?
I don't know ...
Thank you very much again 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]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.