Order of vanishing of an integer polynomial at a point 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



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 $1$?

``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.
I previously asked an analogous question over $\mathbb{C}$, 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.
 A: Over $\mathbf{Q}$ you can argue as follows.
Case I. $a$ is the unique point where $f$ vanishes to high order. Then $a$ is invariant under $\text{Aut}(\mathbf{C})$ and hence defined over $\mathbf{Q}$ and hence there is a line defined over $\mathbf{Q}$ passing through $a$.
Case II. There are exactly $2$ points $a, b$ where $f$ vanishes to high order. Then the line $ab$ is defined over $\mathbf{Q}$.
Case III. There are finitely many $> 2$ points where $f$ vanishes to high degree. This is impossible because then every conic passing through these points must be contained in $f = 0$. (Use Bezout.)
Case IV. There is an infinite nr of points where $f$ vanishes to high degree. The locus of these points has to be an algebraic curve $g = 0$ and $g$ has to have degee $1$ because a high power of $g$ divides $f$.
Since $\mathbf{Z} \subset \mathbf{Q}$ this also proves something over $\mathbf{Z}$ and with a variant of Gauss's lemma you can get a suitable statement, but I haven't thought it through. Maybe Case I is the most interesting one from this point of view.
A: The solution to this question appeared in Section 4 of my paper on the ranks of matrices with few distinct entries. 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 $x^3-3x+1$. 
