9
$\begingroup$

This is a refinement of my question asked earlier, which is answered beautifully in the negative by Thomas Browning. The example he gave was geometrically reducible. Now I want to ask the same question, but with the extra assumptions that $f(0,0) = 0$ and $f$ is geometrically irreducible for all $c \in \mathbb{R}$.

For convenience, the question is: does there exist a polynomial $f \in \mathbb{Z}[x,y]$ such that $f(0,0) = 0$ and $f$ is geometrically irreducible, and such that

$$\displaystyle f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$

and

$$\displaystyle \liminf_{\mathbb{R}^2} f(x,y) = -\infty?$$

$\endgroup$
2
  • 3
    $\begingroup$ How about taking that example and adding a random positive number? $\endgroup$
    – pinaki
    Nov 24 at 22:39
  • $\begingroup$ Yes, I want to avoid examples like $f(x,y) = x^2 - y^2 - 2$ and such, and hoping that the assumption $f(0,0) = 0$ would eliminate such trivialities. If this is not enough then we may further assume that $f(x,y) + c$ is geometrically irreducible for all $c \in \mathbb{R}$. $\endgroup$ Nov 24 at 22:44

1 Answer 1

20
$\begingroup$

Just a slight modification of the previous example: $$f(x,y):=\left(5 y^2+5 y+1\right) \left(x^2+y^2\right)+\left(10 y^2+10 y+1\right) y^2.$$


Your conditions $f(0,0) = 0$ and $$f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$ contradict each other. So, I was assuming that you meant $f(0,0) = 0$ and $$f(a,b) > 0 \text{ for all } (a,b) \in \mathbb{Z}^2\setminus\{(0,0)\}.$$

$\endgroup$
1
  • 2
    $\begingroup$ I think this example captures what I was looking for. Thanks! $\endgroup$ Nov 24 at 22:46

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.