 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2010/b/6/a/solution

D = \sum_{p} n_{p} [p]

with

n_{p} \in Z

. Its degree is

de g D = \sum_{p} n_{p}

(f) = \sum_{p} ord_{p} (f) [p]

for a nonzero meromorphic function, with positive order at a zero and negative order at a pole. Two divisors are linearly equivalent if their difference is principal. Principal divisors form a subgroup because

(f g) = (f) + (g)

and

(1/ f) = - (f)

For the doubly periodic function, put

h = f^{'} / f

. It is periodic with the same period lattice, and its residue at

p_{i}

equals

n_{i}

. Opposite sides of the fundamental parallelogram are translates by a period with opposite directions, so their integrals of

h

cancel. The residue theorem gives

0 = \oint_{Γ_{a}} h (w) d w = 2 πi \sum_{i} n_{i} .

Thus

i \sum n_{i} = 0.

In particular an elliptic function has equally many zeros and poles counted with multiplicity in a period cell.

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