1824 CHAPTER 58. ELLIPTIC FUNCTIONS
and (w1w2
)=
(e fg h
)(w′1w′2
).
Therefore, (w′1 w′1w′2 w′2
)=
(a bc d
)(e fg h
)(w′1 w′1w′2 w′2
).
However, since w′1/w′2 is not real, it is routine to verify that
det(
w′1 w′1w′2 w′2
)̸= 0.
Therefore, (1 00 1
)=
(a bc d
)(e fg h
)and so det
(a bc d
)det(
e fg h
)= 1. But the two matrices have all integer entries and
so both determinants must equal either 1 or −1.Next suppose (
w′1w′2
)=
(a bc d
)(w1w2
)(58.1.5)
where(
a bc d
)is unimodular. I need to verify that {w′1,w′2} is a basis. If w ∈M, there
exist integers, m,n such that
w = mw1 +nw2 =(
m n)( w1
w2
)From 58.1.5
±(
d −b−c a
)(w′1w′2
)=
(w1w2
)and so
w =±(
m n)( d −b−c a
)(w′1w′2
)which is an integer linear combination of {w′1,w′2} . It only remains to verify that w′1/w′2 isnot real.
Claim: Let w1 and w2 be nonzero complex numbers. Then w2/w1 is not real if andonly if
w1w2−w1w2 = det(
w1 w1w2 w2
)̸= 0
Proof of the claim: Let λ = w2/w1. Then
w1w2−w1w2 = λw1w1−w1λw1 =(
λ −λ
)|w1|2