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Unformatted text preview: 56 LECTURE 6. MODULAR FORMS enters only in the first degree. Thus we can express y in terms of x,t and obtain that E is isomorphic to P 1 ( C ). If d = 0 we obtain that f could be chosen of degree 2. Again this is impossible. Note that we also have in (6.13) 4 A 3 + 27 B 2 . 6 = 0 (6.16) This is the condition that the polynomial x 3 + Ax 2 + B does not have a multiple root. If it has, (6.13) does not define a Riemann surface. A cubic equation of the form (6.15) with the condition (6.16) is called a Weierstrass cubic equation . We know from Lecture 3 that T = 1 2 1 2 ( z ; ) has simple zeroes at the points z = . Since X does not vanish at these points (it is a linear combinations of T 2 and 00 ( z ; ) 2 ), ( z ) has poles of order 2 at z . Differentiating (6.14), we obtain 2 1 ( z ) 1 ( z ) = (3 ( z ) 2 + A ) ( z ) . Let 1 ( z 1 ) = 0. If 3 ( z 1 ) 2 + A = 0, the polynomial x 3 + Ax + B is reducible since ( z 1 ) must be its double root. So,) must be its double root....
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.
- Fall '09