A NULLSTELLENSATZ FOR AMOEBAS

A NULLSTELLENSATZ FOR AMOEBAS 417
γ → 1. The best general answer for the question is of the same form; that is,
n log γ −1 ≤ (D 0 + D 1 ) log n + log (c 0 c 1 ).
One can see this by performing the requisite analysis on the polynomials h n (z) =
(z n + 1) c 0n D 0
. As a general heuristic, the more closely the roots of f n are packed, the
larger n has to be; thus the family of polynomials h n (z), where every root has as high
a multiplicity as possible, is where we expect our worst-case behaviour to occur.
Suppose we want n large enough to guarantee that h n {log |z 0 |} is (c 1 n D 1 )-superlopsided
for |z 0 | < γ < 1. Write h n (z) = 1 + c 0 n D 0 zn + ···. We know that
1 is the dominant term as n gets large (since 1 = lim n→∞ h n (z 0 )); thus we need
(c 1 n D 1 )(c 0n D 0 zn 0 ) < 1 or, equivalently,
n log γ −1 ≥ (D 0 + D 1 ) log n + log(c 0 c 1 ).
If we only want to guarantee that f n {a} is lopsided for a ∈ K, we need n large
enough so that
∑
|m l (z 0 )|≤(1 + γ n ) d − 1 < 2 − (1 + γ n ) d ≤|m k (z 0 )|
l≠k
or, equivalently, (1 + γ n ) c 0n D 0
≤ 3/2. This holds if we have
(
n log γ −1 c
)
0
≥ D 0 log n + log .
log 3/2
So n needs to be only about half as big to guarantee that f n {a} is lopsided as it does
to guarantee that f n {a} is superlopsided. Again, we can see that this is fairly close to
the best answer by considering (z n + 1) c 0n D 0
.
3. The hypersurface case
3.1. Preliminaries
In this section, we prove our main theorem characterising the amoeba of a hypersurface.
If f (z) ∈ C[z 1 ,z −1
1 ,...,z r,z −1 ], we consider the Laurent polynomials
˜f n (z) =
∏n−1
k 1 =0
r
···
∏n−1
k r =0
f (e 2πi k 1/n z 1 ,...,e 2πi k r /n z r ).
Theorem 1 states that for ε>0 and for a point a ∈ R r in the complement of the
amoeba A f whose distance from A f is at least ε, we can choose n large enough so
that ˜f n {a} is superlopsided. Moreover, the theorem gives an upper bound on how large
n needs to be, based only on ε and the Newton polytope of f .