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

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


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

  • 3
    $\begingroup$ How about taking that example and adding a random positive number? $\endgroup$
    – pinaki
    Nov 24, 2022 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, 2022 at 22:44

1 Answer 1


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 $$f(a,b) > 0 \text{ for all } (a,b) \in \mathbb{Z}^2\setminus\{(0,0)\}.$$

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    $\begingroup$ I think this example captures what I was looking for. Thanks! $\endgroup$ Nov 24, 2022 at 22:46

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