A NULLSTELLENSATZ FOR AMOEBAS

A NULLSTELLENSATZ FOR AMOEBAS 419
Clearly, we have ( ˜f n ) = n r (f ). This gives us an upper bound on the number
of terms that ˜f n can have.
PROPOSITION 3.3
Let d = d((f )). Then ˜f n has at most dn r2 −r terms.
Proof
By Proposition 3.2, the number of terms in ˜f n is at most the number of integral points
in (1/n)( ˜f n ) = n r−1 (f ). This is less than or equal to dn r2−r .
Finally, we need to know something about maxdeg( ˜f i,ζ
n
c i (f ):= max x i
(
(f )
)
− min xi
(
(f )
)
,
where x i denotes the ith coordinate function on R r .
) − mindeg( ˜f
n
i,ζ).Let
PROPOSITION 3.4
We have maxdeg( ˜f i,ζ
n
Proof
We have
maxdeg( ˜f i,ζ
n
) − mindeg( ˜f i,ζ
n ) = c i(f )n r .
) − mindeg( ˜f i,ζ
n
) = max x (
i ( ˜f n ) ) (
− min x i ( ˜f n ) )
(
= n r max x i ( ˜f n ) ) (
− n r min x i ( ˜f n ) )
= d i n r .
3.2. Proof of Theorem 1
Armed with these facts and Lemmas 2.1 and 2.2, we are now in a position to precisely
state and prove our main result for amoebas of hypersurfaces.
THEOREM 1
Let ε>0. Suppose that a = (a 1 ,...,a r ) ∈ R r \ A f is a point in the amoeba
complement whose distance from A f is at least ε. Letd = d((f )), and let c =
max{c i (f ) | 1 ≤ i ≤ r}.
(1) If n is large enough so that
then ˜f n {a} is lopsided.
nε ≥ (r − 1) log n + log ( (r + 3)2 r+1 c ) , (3.1)