[Math] Order of vanishing of an integer polynomial at a point

ac.commutative-algebraag.algebraic-geometryarithmetic-geometrynt.number-theory

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.

Best Answer

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.

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