A NULLSTELLENSATZ FOR AMOEBAS

412 KEVIN PURBHOO
For n large, the terms { ···} are small in comparison with the other terms, and so
this is approximately
g n (z) =±(z d ) n ∓ (α 1 z d−1 ) n ± (α 1 α 2 z d−2 ) n ∓···−(α 1 ···α d−1 z) n + (α 1 ···α d ) n .
Suppose that |α k+1 | < |z| < |α k |. Consider g n (z)/(α 1 ···α d−k z d−k ) n .Asn →∞,
every term tends to zero except for the constant term, which is ±1. Thus, for n large,
there is a single term in g n (z), and likewise in ˜f n , which is much bigger in absolute
value than all the others.
2.2. The one-variable lemmas
We now formalise this heuristic argument in a way that is useful in proving Theorem 1.
At the crux of the heuristic argument are the following three key facts about ˜f n .
(1) It has no roots inside a certain annulus. (In the heuristic argument, the annulus
is {z ∈ C ||α k+1 | < |z| < |α k |}.)
(2) The only nonzero terms that appear are of the form cz nk .
(3) The number of terms is not too large. (This approach fails if instead of ˜f n (z),
we try to use ˜f n (z) D with D ≫ n.)
To get a result that we can apply to the multivariable case, we need to be able to
make a uniform statement about polynomials with these properties. This is precisely
captured by the next two lemmas. By applying Lemma 2.2 directly to the family of
functions ˜f n (z), one immediately obtains a complete proof of Theorem 1 in the r = 1
case.
LEMMA 2.1
Let A ={z | β 0 < |z|