By Haskell Curry
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Extra info for A theory of formal deducibility
What about σ? From the preceding, we see that log σ(z + ω) = log σ(z) + η(ω)z + c(ω) for some function c on the lattice. It is convenient to write this as σ(z + ω) = ψ(ω)eη(ω)(z+ω/2) σ(z) thereby deﬁning ψ(ω). Suppose ﬁrst that ω/2 ∈ / L. Setting z = −ω/2 above and using the fact that σ is odd, we see at once that ψ(ω) = −1. On the other hand, σ(z + 2ω) σ(z + ω) σ(z + 2ω) = σ(z) σ(z + ω) σ(z) Elliptic Functions 45 and so by applying the functional equation twice and using the fact that η(2ω) = 2η(ω), we get ψ(2ω) = ψ(ω)2 .
We will write P ≺ Q if Q dominates P . It is easily veriﬁed that if P1 ≺ Q1 and P2 ≺ Q2 , then P1 + P2 ≺ Q1 + Q2 and P1 P2 ≺ Q1 Q2 . Moreover, if Di is the derivative operator with respect to the i-th variable and P ≺ Q, then Di P ≺ Di Q. If the total degree of a polynomial P in n variables is r, then P ≺ size(P )(1 + x1 + · · · + xn )r . We also need some facts about derivations. Recall that a derivation D of a ring R is a map D : R → R such that D(x + y) = D(x) + D(y) and which satisﬁes D(xy) = D(x)y + xD(y).
We end this chapter by noting the following result for future reference. 4 The numbers ℘(ω1 /2), ℘(ω2 /2) and ℘((ω1 + ω2 )/2) are distinct. Proof. Suppose not. Let L be the lattice spanned by ω1 , ω2 which are linearly independent over R. Let us consider the function f1 (z) = ℘(z) − ℘(ω1 /2). This has a double order zero at z = ω1 /2 since ℘ (ω1 /2) = 0. 4. It follows that any zero must be congruent to ω1 /2 modulo L. If ℘(ω2 /2) = ℘(ω1 /2), then we would have ω1 ≡ ω2 modulo L, contrary to their linear independence over R.