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288    14. Elliptic Curves over Local Fields

           The two curves C(ev) and C(od) are treated as disjoint spaces with the com-
        pactifying points also distinct. Returning to the diagram in (4.2), we describe J :

        C(ev) ∪ C(od) = E   (R) → P 1 (R).

        (4.9) Theorem. The function J : E   (R) → P 1 (R) is 2:1 and the separate restric-
        tions of J to C(ev) and C(od) have the form:
           (1) J ev = J|C(ev) : C(ev) → [1, ∞] ⊂ P 1 (R) and corresponds to  (τ) > 0.
        Over (∞, 1) the function J ev is a smooth 2: 1 mapping with exactly on one point in
        each of the inverse images of 1 and ∞.
           (2) J od = J|C(od) : C(od) → [∞, 1] ⊂ P 1 (R) and corresponds to  (τ) < 0.
        Over (∞, 1) the function J od is a smooth 2: 1 mapping except over the point 0 ∈
        [∞, 1]. There is exactly one point in each of the inverse images of ∞ and 1.

                                                                        2
        Proof. The 2:1 character of the restrictions of J follow from (4.7) applied to y =
         3
                                                                     4
        x − 3E 4 (τ)x + 2E 6 (τ). In the case (1) we have the relations E 4 (i/t) = t E 4 (it)
                       6
        and E 6 (i/t) =−t E 6 (it) for τ = it which up to rescaling give the two signs for
        E 6 , and except for the two special points give a 2:1 mapping.
           In the case (2) we have the two signs at ζ 6 = T (ζ 6 ) and T (ζ 3 ) for τ = (1/2)+ti.

        By continuity this sign difference extends over pairs of points τ = T (β ) and τ =



        T (β ) where β and β are two boundary points on the fundamental domain giving

        the same elliptic curve over C. since they are nonisomorphic at {β ,β }={ζ 3 ,ζ 6 },

        they are nonisomorphic in general. This proves the theorem.
        (4.10) Remark. At the two points over 0 where J od is not smooth, we have vertical
        tangents.
                                                       2
                                                          3
                                                      y  = x  + x 2
                                                          3
                                                       2
                                                      y  = x  − x 2
                       c (od)                                 c (od)
                                               c (ev)
                                                              ∆ < 0
                                ∆ < 0
                J
                                               ∆ > 0
                                                                  (   )
                                                                A
                              0          1              ∞
                                        2
                            y  = x  + 1  y  = x  + x  ∆ > 0
                                           3
                               3
                            2
                            y  = x  − 1  y  = x  − x  ∆ < 0
                                        2
                                           3
                               3
                            2
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