Let $f(x,y)$ be a polynomial with integer coefficients, and let $\alpha=(\alpha\_1,\alpha\_2)\in \mathbb{C}^2$ be a complex point. I want to show that $f$ cannot vanish at $\alpha$ to high order unless $\alpha$ is ``simple''. More precisely, suppose that


* f(x,y) vanishes to order at least $0.99\cdot \deg f$ at $(\alpha\_1,\alpha\_2)$, and
* f(1,1)=1



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> Does it imply existence of an integer polynomial $g(x,y)$ satisfying $g(\alpha\_1,\alpha\_2)=0$ and $g(1,1)=1$ that is *linear*, i.e., of degree $1$?
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``Vanishing to order $m$'' at $\alpha$ means that all possible partial derivatives of order at most $m-1$ vanish at point $\alpha$. 


Note that the single-variable version is a simple consequence of [Gauss's lemma](https://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29).


I previously asked an [analogous question over $\mathbb{C}$](https://mathoverflow.net/q/209720/806), but the answer turned out to be contrary to what I expected. This is the original question that I am interested in.


I do not know how to answer this question even if one replaces $\mathbb{Z}$ by $\mathbb{Q}$ throughout.



Order of vanishing of an integer polynomial at a point
The solution to this question appeared in Section 4 of [my paper on the ranks of matrices with few distinct entries](http://arxiv.org/abs/1508.00145). The solution borrows an idea from the Dracula's answer. Please upvote his answer. Below is a sketch of the solution.


**Step 1)**
Let $V$ be the set of points where $f$ vanishes to order greater than $\tfrac{2}{3}\deg f$. If $l$ is any secant of $V$, then $f$ vanishes to the order exceeding $\tfrac{1}{3}\deg f$ on each point of $l$. Hence, $f$ vanishes to the order $\mathord{\geq} \tfrac{1}{3}\deg f$ on the secant variety $\operatorname{Sec}\_1(V)$. Similarly $f$ vanishes on any line passing through a point of $V$ and a point of $\operatorname{Sec}\_1(V)$. So, $f$ vanishes on the $2$'nd secant variety $\operatorname{Sec}\_2(V)$. It is clear that $\operatorname{Sec}\_2(V)$ is an affine subspace defined over $\mathbb{Q}$.


**Step 2)** If there is no linear polynomial $g$ that vanishes on $\operatorname{Sec}\_2(V)$ and satisfies $g(1,1)=1$, then in fact there is a rational point $v\in \operatorname{Sec}\_2(V)$ for which no such polynomial vanishing at $v$ exists. This is proved by using a theorem of van der Waerden characterizing when a system of linear equations in integers admits no solution.


**Step 3)** One then shows that there is no integer polynomial $g$ satisfying $g(1,1)=1$ vanishing at $v\in \mathbb{Q}^2$ if and only if $v\equiv (1,1)\pmod p$ for some prime $p$. In that case, however, $f(v)\equiv f(1,1)\equiv 1 \pmod p$, contradicting the fact that $f$ vanishes on $\operatorname{Sec}\_2(V)$.


**Furthermore** The bound $\tfrac{2}{3}\deg f$ is sharp. Given a cubic Galois extension $F/\mathbb{Q}$, one can find three points $p\_1,p\_2,p\_3\in F^2$ and three linear polynomials $f\_1,f\_2,f\_3$ with the following properties:


* The points $p\_1,p\_2,p\_3$ are Galois conjugates of one another,
* The linear polynomials $f\_1,f\_2,f\_3$ are Galois conjugates of one another,
* The linear polynomial $f\_1$ vanishes on the line $p\_2p\_3$, the $f\_2$ vanishes on $p\_1p\_3$, and $f\_3$ --- on $p\_1p\_2$,
* The coefficients of $f\_1,f\_2,f\_3$ are algebraic integers, and $f\_1(1,1)=f\_2(1,1)=f\_3(1,1)=1$.


Then the polynomial $f=f\_1f\_2f\_3$ has integer coefficients, and vanishes to order $2$ at each of $p\_1,p\_2,p\_3$. It also satisfies $f(1,1)=1$. However, no integer linear polynomial $g$ vanishes at $p\_1$, say.


A concrete example that one obtains using the splitting field of $x^3-3x+1$ consists of the polynomial $f(x,y)=x^3+3 x^2 y-6 x^2-3 x y+3 x-y^3+3 y+1$ and points $p\_i=(\rho\_i^2-2,3-\rho\_i^2-\rho\_i)$ where $\rho\_1,\rho\_2,\rho\_3$ are the roots of $x^3-3x+1$. 


