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280    14. Elliptic Curves over Local Fields

                      2
                                            3
                                                    2
         (a)         y + a 1,0 x 1 y 1 + a 3,1 y 1 = πx + a 2,0 x + a 4,1 x 1 + a 6,2 .
                      1                     1       1
        The singular point on the fibre, whose local ring is A = R[x, y] m , blows up into the
        conic
                                                2
                          2
         (b)             y + a 1,0 xy + a 3,1 y = a 2,0 x + a 4,1 x + a 6,2 .
        whose discriminant is b 8,2 =−π −2 b 8 for (A2). Further, in (A3) the conic becomes
                                   2
         (c)                       t + a 3,1 t + a 6,2 = 0,
                                           2
                                      2
                                                       2
        where t = y −ax is given by (y −ax) = y +a 1 xy −a 2 x . The discriminant of the
        quadratic (c) is b 6,2 , and thus under the assumptions of (A3) the conic degenerates
                                        3
        into two distinct lines. Dividing (a) by x we obtain an equation F(u,v) = π which
                                        1
        modulo π factors into L 1 L 2 L 3 . Since the local ring generated by any two factors L 1 ,
        L 2 ,or L 3 is regular, equation (a) gives a regular scheme over R with fibre the three
             ¯
        lines L i = 0. Finally, c is the number lines rational over k. One line is rational over
                                 2
        k and the others are given by t + a 3,1 t + a 6,2 = 0. Hence the value c = 1or3,but
        it is 3 over a large enough k.
           Step III. We assume that v(a 6 ) ≥ 2, v(b 8 ) ≥ 3, and v(b 6 ) ≥ 3. With the sin-
                         ¯
        gularity at (0, 0) on E this leads to the relations v(a 1 ) ≥ 1, v(a 2 ) ≥ 1, v(a 3 ) ≥ 2,
                                                         3
                                                               2
        v(a 4 ) ≥ 2, and v(a 6 ) ≥ 3. For the cubic polynomial P(t) = t +a 2,1 t +a 4,2 t +a 6,3 ,
        the equation of the curve becomes
                             2
                            y + a 1,1 x 1 y 1 + a 3,2 y 2 = P(x 1 ).
                             2
        There are three cases depending on the multiplicity of the roots of P(t). If there is
        a double root, there are two subcases depending on the multiplicity of the roots of
                             2
        a quadratic polynomials y + a 3,3 y − a 6,4 . If there is a triple root, there are three
                                                       2
        subcases depending on the multiplicity of the roots of y + a 3,2 y − a 6,4 and the
        divisibility of a 4 .
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                                                               ∗
                                                     ∗
                                              ∗
                                                 ∗
           For details and a description of how Cases I ,I ,IV , III , and II come up, see
                                              0  m
        Tate [1975, LN 476, pp. 50–52] and for a related analysis Ogg [1967]. We conclude
        with Table 1.
                                                                 ˇ
        §3. Galois Criterion of Good Reduction of N´ eron–Ogg–Safareviˇ c
        Let K be a local field with valuation ring R and residue class field R → R/Rπ = k.
        There is a criterion in terms of the action of Gal(K s /K) on E(K s ) and the subgroups
        N E(K s ) for an elliptic curve E over K to have good reduction at k. The idea behind
                                                          2
        this condition is that the N E(K s ) is isomorphic to (Z/ZN) for N prime to the
        characteristic of K, and the following are equivalent:
                                        2
        (a) N E(k s ) ns is isomorphic to (Z/ZN) for all N prime to char(k).
              ¯
        (b) E(k s ) ns is complete, i.e., E = E ns .
                                     ¯
                                ¯
            ¯
        (c) E has good reduction.
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