A NULLSTELLENSATZ FOR AMOEBAS

432 KEVIN PURBHOO
Clearly, f a is a Laurent polynomial and is in I. Moreover, if we restrict z to Log −1 (a),
then z i ¯z i = e 2a i
,so
f a (z) =
=
=
k∑
f i (z 1 ,...,z r ) ¯f i (¯z 1 ,...,¯z r )
i=1
k∑
f i (z 1 ,...,z r )f i (z 1 ,...,z r )
i=1
k∑
|f i (z)| 2 .
i=1
Thus f a (z) = 0 if and only if f i (z) = 0 for all i.
This result is also true for ideals in C[z 1 ,...,z r ]: one can find a suitable monomial
m(z) such that
m(z 1 ,...,z r ) ¯f i (e 2a 1
z −1
1 ,...,e2a r
z −1
r
)
is a polynomial for all i, and a similar argument holds if
f a (z) =
k∑
i=1
f i (z 1 ,...,z r ) ( m(z 1 ,...,z r ) ¯f i (e 2a 1
z −1
1 ,...,e2a r
z −1
r
) ) .
As an immediate consequence of Proposition 5.1, we have the following.
COROLLARY 5.2
For any ideal I ⊂ C[z 1 ,z −1
1 ,...,z r,z −1
r
],
A I = ⋂ f ∈I
A f .
It is now a simple task to prove our second main result.
THEOREM 2
Let I ⊂ C[z 1 ,z −1
1 ,...,z r,zr
−1 ] be an ideal. A point a ∈ R r is in the amoeba A I if
and only if g{a} is not (super)lopsided for every g ∈ I.
Proof
If a ∈ A I ,thenf {a} cannot be lopsided for any f ∈ I since a ∈ A f for every
f ∈ I. On the other hand, suppose that a /∈ A I . Then by Proposition 5.1, ifwetake