By Froberg R.

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**Sample text**

Ak ], −[a1 ], −[a2 ], . . , −[ak ] is the set of zeroes of h counted with multiplicity. We may also choose a sequence [b1 ], [b2 ], . . , [bk ] which does the same thing for the poles (and these sequences have the same length as the number of poles and zeroes are the same). Now put g(z) := ℘ (z) − ℘ (ai ) . i ℘ (z) − ℘ (bi ) i This elliptic function has the same pole and zero sets (counted, of course, with multiplicities) as h, as the poles at [0] from the different factors cancel, which means that h(z)/g(z) has neither zeroes nor poles and is hence a constant which finishes the proof.

Exercise 24. Work through all the special cases of the proof. When it comes to describing all elliptic functions we shall now see that they are all rational functions in ℘ (z) and ℘ (z). 10. All elliptic functions of periods ω1 and ω2 are of the form f (℘ (z))+ g(℘ (z))℘ (z), where f (x) and g(x) are rational functions. Proof. Let h be such an elliptic function. We start by writing it as a sum of an even and an odd function: h(z) = h(z) + h(−z) h(z) − h(−z) + . 2 2 Both parts are still elliptic with the same periods so we are reduced to the case when h is odd or even.

The lines in question are those for which x is constant. This fits very well with what we were doing when we were discussing the addition theorem where we postulated that lines parallel with the y-axis intersected the curve in the point at infinity (apart from the intersection with X). Furthermore, we also introduced a fictitious line that only intersected X in the point at infinity. From the point of view of the projective plane these postulations become facts. Furthermore, we can make more aspects of that discussion precise.