9
$\begingroup$

For which fields $k$ does the following hold for all $n \geq 1$? Let $(a_1,\ldots,a_n) \in k^n$. Then there exists a polynomial $f(x_1,\ldots,x_n) \in k[x_1,\ldots,x_n]$ such that $f(a_1,\ldots,a_n) = 0$ but $f(b_1,\ldots,b_n) \neq 0$ for all $(b_1,\ldots,b_n) \neq (a_1,\ldots,a_n)$?

This is clearly impossible for $k$ algebraically closed.

It does hold for $k$ a subfield of $\mathbb{R}$; indeed, in that case we can take

$$f(x_1,\ldots,x_n) = (x_1-a_1)^2 + (x_2-a_2)^2 + \cdots + (x_n-a_n)^2.$$

$\endgroup$
19
$\begingroup$

If $k$ is not algebraically closed, such a polynomial always exists (the opposite is also true and is mentioned in the post).

We may assume that $a_i=0$ for all $i$. Take an irreducible polynomial $g(x)$ of degree $d>1$, then for the homogeneous form $G(x,y)=y^dg(x/y)$ we have $G(x,y)=0$ if only if $x=y=0$. This solves the case $n=2$, for $n=3$ consider the polynomial $G(G(x,y),z)$, it takes zero value only when $x=y=z=0$, and so on.

$\endgroup$
  • $\begingroup$ alternatively take $G(G(x,y),z^d)$ (and so on) to produce homegeneous polynomials. $\endgroup$ – YCor Nov 15 '16 at 7:05
2
$\begingroup$

I guess finite fields $k = \mathbb{F}_{q}$ satisfy this property, namely we can take $$ f = ((x_{1}-a_{1})^{q-1}-1) \dotsb ((x_{n}-a_{n})^{q-1}-1) - (-1)^{n} $$ for any $n$.

In fact we have a Lagrange interpolation formula for finite fields, see this answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.