5   Isogenies  Let E 1   E 2 be elliptic curves   Definition   Isogeny   An isogeny      E 1   E 2 is a nonconstant morphism with     0 E 1     0 E 2  by Theorem 2 8  a morphism is nonconstant if and only if surjective on K   points    We say E 1 and E 2 are isogenous in this case   Definition  Hom   E 1   E 2       isogenies E 1   E 2       0     This is an abelian group under                  P         P         P         If E 1     E 2     E 3 are isogenies then       is an isogeny   Tower Law implies deg           deg       deg         Proposition 5 1  Assuming that   0   n      Then    n     E   E is an isogeny   Proof    n   is a morphism by Theorem 4 4  We must show   n       0     Assume char K   2   Case n   2  Lemma 4 5 implies  E   2     E implies   2       0     Case n odd  Lemma 4 5 implies    0   T   E   2    Then n T   T   0 which gives   n       0     Now use   m n       m       n     If char K   2  then could rpelace Lemma 4 5 with an explicit lemma about 3 torsion points     Corollary 5 2  Hom   E 1   E 2   is a torsion free   module   Theorem 5 3  Assuming that       E 1   E 2 an isogeny  Then      P   Q         P         Q   for all P   Q   E 1   Proof  sketch      incudes           Div 0   E 1     Div 0   E 2     P   E 1 n P P     P   E 1 n P     P       Recall       K   E 2     K   E 1     Fact   If f   K     E 1     then     div   N K   E 1     K   E 2   f           div f         So     sends principal divisors to principal divisors   Since     0 E 1     0 E 2  the following diagram commutes        a group homomorphism implies   is a group homomorphism     Lemma 5 4  Assuming that       E 1   E 2 an isogeny  Then  there exists a morphism   making the following diagram commute     x i   x coordinate on a Weierstrass equation for E i   Moreover if     t     r   t   s   t    r   s   K   t    coprime  then      deg     deg     max   deg   r     deg   s           Proof  For i   1   2  K   E i     K   x i   is a degree 2 Galois extension  with Galois group generated by     1       Theorem 5 3 implies that         1         1         So if f   K   x 2   then          1         f           1     f       f       Therefore     f   K   x 1     In particular            x 1   for some rational function     Tower Law implies deg     deg     Now K   x 2     K   x 1    x 2       x 1     r   x 1   s   x 1    r   s   K   t   coprime   We claim the minimal polynomial of x 1 over K   x 2   is     F   t     r   t     s   t   x 2   K   x 2     t         Check    F   x 1     0  true by the definition of our embedding   F is irreducible in K   x 2   t    since r   s coprime   hence irreducible in K   x 2     t   by Gauss s lemma   Therefore      deg       K   x 1     K   x 2       deg F   max   deg r   deg s           Lemma 5 5  deg   2     4   Proof  Assume char K   2   3  the result is true even in the case of char K     2   3    but we will only prove the simpler case    E  y 2   x 3   a x   b   f   x     If P     x   y     E  then     x   2 P       3 x 2   a 2 y   2   2 x     3 x 2   a   2   8 x f   x   4 f   x     x 4     f   x       The numerator and denominator are coprime  Indeed  otherwise there exists     K   with f         f           0 and hence f has a multiple root  contradiction    Now Lemma 5 4 implies deg   2     max   4   3     4     Definition   Quadratic form in an abelian group   Let A be an abelian group  We say q   A     is a quadratic form  if   i  q   n x     n 2 q   x   for all n      x   A    ii    x   y     q   x   y     q   x     q   y   is   bilinear   Lemma 5 6  q   A     is a quadratic form  if and only if it satisfies the parallelogram law       q   x   y     q   x   y     2 q   x     2 q   y     x   y   A       Proof      Let   x   y     q   x   y     q   x     q   y    Then        x   x     q   2 x     2 q   x     2 q   x        the equality in the last step is using property  i  with n   2    But by  ii          x   y   x   y       x   y   x   y     2   x   x     2   y   y         Hence  upon dividing by 2       q   x   y     q   x   y     2 q   x     2 q   y             See Example Sheet 2     Theorem 5 7  deg   Hom   E 1   E 2       is a quadratic form    Note   We define deg   0     0   For the proof we assume char K   2   3  Write E 2  y 2   x 3   a x   b   Let P   Q   E 2 with P   Q   P   Q   P   Q   0  Let x 1       x 4 be the x coordinates of the 4 points   Lemma 5 8  There exists W 0   W 1   W 2       a   b     x 1   x 2   of degree in x 1 and of degree   2 in x 2 such that        1   x 3   x 4   x 3 x 4       W 0   W 1   W 2         Proof  Two methods    1    Direct calculation      W 0     x 1   x 2   2 W 1     W 2         see formula sheet    2    Let y     x     be the line through P and Q      x 3   a x   b       x       2     x   x 1     x   x 2     x   x 3     x 3   s 1 x 2   x 2 x   s 3     where s i is the i th elementary symmetric polynomial in x 1   x 2   x 3   Comparing coefficients        2   s 1   2       s 2   a   2   s 3   b     Eliminating   and   gives        s 2   a   2   4 s 1   s 3   b     0     F   x 1   x 2   x 3       where F has degree   2 in x i   x 3 is a root of the quadratic      W   t     F   x 1   x 2   t         Repeating for the line through P and   Q shows that x 4 is the other root  Therefore      W 0   t   x 3     t   x 4     W 0 t 2   W 1 t   W 2       hence        1   x 3   x 4   x 3 x 4       W 0   W 1   W 2           We show that if         Hom   E 1   E 2    then      deg             deg             2 deg         2 deg             We may assume                         0  otherwise trivial  or use deg     1     1  deg   2     4             x   y         1   x               x   y         2   x                   x   y         3   x                   x   y         4   x             Lemma 5 8 implies        1     3     4     3   4           1     2   2             Put   i   r i s i  r i   s i   K   t   coprime         s 3 s 4   r 3 s 4   r 4 s 4   r 3 r 4     coprime       r 1 s 2   r 2 s 1   2                 Therefore     deg             deg             max   deg   r 3     deg   s 3       max   deg   r 4     deg   s 4       max   deg   s 3 s 4     deg   r 3 s 4   r 3 s 3     deg   r 3 r 4       2 max   deg   r 1     deg   s 1       2 max   deg   r 2     deg   s 2       2 deg         2 deg        1      Now replace        by               to get     deg   2       deg   2       2 deg             2 deg           4 deg         4 deg         2 deg             2 deg           2 deg         2 deg         deg             deg            2       1   and  2   give that deg satisfies the parallelogram law  hence deg is a quadratic form   This proves Theorem 5 7   Corollary 5 9  deg   n       n 2 deg       for all n          Hom   E 1   E 2    In particular  deg   n     n 2   Example 5 10  Let E   K be an elliptic curve  Suppose char K   2  Let 0   T   E   K     2     Without loss of generality E  y 2   x   x 2   a x   b    a   b   K  b   a 2   4 b     0  and T     0   0     If P     x   y   and P     P   T     x     y     then     x       y x   2   a   x   x 2   a x   b x   a   x   b x y       y x   x       b y x     Let         x   x     a     y x   2     y   y     y x   x   b x     2     y x   2     x   b x   2   4 b               a   2   4 b           2   2 a     a 2   4 b       Let E    y 2   x   x 2   a   x   b      where a       2 a  b     a 2   4 b  There is an isogeny         E   E     x   y         y x   2   y   x 2   b   x 2   1     2   3 0 ord 0 E       1 0 3   3 0 E     0   1   0     0 E       To compute the degree    y x   2   x 2   a x   b x  coprime  numerator and denominator as b   0   Lemma 5 4 gives deg     2   We say   is a 2 isogeny