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The doubly periodic functions for Λ form a field, which the next two propositions determine. 8. There is the following relation between ℘ and ℘ : ℘ (z)2 = 4℘(z)3 − g2 ℘(z) − g3 where g2 = 60G2 (Λ) and g3 = 140G3 (Λ). Proof. We compute the Laurent expansion of ℘(z) near 0. Recall (from Math 115) that for |t| < 1, 1 = 1 + t + t2 + · · · . 1−t On differentiating this, we find that 1 = ntn−1 = (n + 1)tn . 2 (1 − t) n≥1 n≥0 Hence, for |z| < |ω|, 1 1 1 1 − 2 = 2 2 (z − ω) ω ω 1 − ωz 2 − 1 = (n + 1) n≥1 zn .

The map z → (℘(z) : ℘ (z) : 1) : C/Λ → E(Λ) 0 → (0 : 1 : 0) is an isomorphism of Riemann surfaces. Proof. It is certainly a well-defined map. The function ℘(z) is 2 : 1 in a period parallelogram 2 containing 0, except at the points ω21 , ω22 , ω1 +ω , where it is one-to-one. Since the function 2 (x : y : 1) → x : E(Λ) \ {O} → C has the same property, and both maps have image the whole of C, this shows that the map in z → (℘(z) : ℘ (z) : 1) is one-to-one. Finally, one can verify that it induces isomorphisms on the tangent spaces.

Note that Gk (cΛ) = c Gk (Λ) for c ∈ C× . S. MILNE The field of doubly periodic functions. Let Λ be a lattice in C. The doubly periodic functions for Λ form a field, which the next two propositions determine. 8. There is the following relation between ℘ and ℘ : ℘ (z)2 = 4℘(z)3 − g2 ℘(z) − g3 where g2 = 60G2 (Λ) and g3 = 140G3 (Λ). Proof. We compute the Laurent expansion of ℘(z) near 0. Recall (from Math 115) that for |t| < 1, 1 = 1 + t + t2 + · · · . 1−t On differentiating this, we find that 1 = ntn−1 = (n + 1)tn .