3   Weierstrass Equations  In this section  we drop the assumption that K   K    but we instead assume that K is a perfect field   Definition   Elliptic curve   An elliptic curve  E   K is a smooth projective curve of genus 1  defined over K with a specified K rational point 0 E   E   K     Example    X 3   p Y 3   p 2 Z 3   0       2 is not  an elliptic curve over    since it has no   rational points    Theorem 3 1  Assuming that   E is an elliptic curve  Then  E is isomorphic over K to a curve in Weierstrass form via an isomorphism taking 0 E to   0   1   0     Remark   Proposition 2 7 treated the special case E is a smooth plane cubic and 0 E is a point of inflection    Fact   If D   Div   E   is defined over K  i e  fixed by Gal   K     K    then L   D   has a basis in K   E    not just K     E      Proof of Theorem  3 1     L   2   0 E     L   3   0 E    Pick basis 1   x for L   2   0 E   and 1   x   y for L   3   0 E     Note  ord 0 E   x       2  ord 0 E   y       3   The 7 elements 1   x   y   x 2   x y   x 3   y 2 in the 6 dimensional space L   6   0 E   must satisfy a dependence relation   Leaving out x 3 or y 2 gives a basis for L   6   0 E   since each term has a different order pole at 0 E   Therefore the coefficients of x 3 and y 2 are non zero  Rescaling x and y  if necessary  we get      y 2   a 1 x y   a 3 y   x 3   a 2 x 2   a 4 x   a 6     for some a i   K   Let E   be the curve defined by this equation  or rather its projective closure   There is a morphism         E   E       2 P     x   P     y   P     1       x y   P     1   1 y   P     0 E     0   1   0       Then       K   E     K   x       deg   E   x   1     ord 0 E   1 x     2   K   E     K   y       deg   E   y   1     ord 0 E   1 y     3     This gives us a diagram of field extensions    By the Tower Law  since 2   3 are coprime   we get that   K   E     K   x   y       1  Hence K   E     K   x   y         K   E      so deg     1  so   is birational   If E   is singular then E and E   are rational  contradiction   So E   is smooth  Then Remark 2 9 implies that     1 is a morphism  So   is an isomorphism     Proposition 3 2  Assuming that   K is perfect  E  E   are elliptic curves over K  E   E   are in Weierstrass form  Then  E   E   over K  if and only if the equations are related by a change of variables     x   u 2 x     r y   u 3 y     u 2 s x     t     where u   r   s   t   K  with u   0   Proof      Obvious         1   x     L   2   0 E       1   x     hence x     x     r for some     r   K with     0    1   x   y     L   3   0 E       1   x     y     implies y     y       x     t for some         t   K with     0   Looking at coefficients of x 3 and y 2  we get   3     2  So     u 2      u 3 for some 0   u   K   Put s     u 2     A Weierstrass form equation defines an elliptic curve if and only if it defines a smooth curve  which happens if and only if     a 1       a 6     0  where         a 1       a 6   is a certain polynomial   If char K   2   3  we may reduce  to the case E   y 2   x 3   a x   b  with discriminant       1 6   4 a 3   2 7 b 2     Corollary 3 3  Assuming that   char K   2   3 and K is perfect  elliptic curves E   y 2   x 3   a x   b  E     y 2   x 3   a   x   b    Then  E and E   are isomorphic over K  if and only if a     u 4 a  b     u 6 b for some u   K     Proof  E and E   are related by a substitution as in Proposition 3 2 with r   s   t   0     Definition   j invariant   The j invariant  is      j   E     1 7 2 8   4 a 3   4 a 3   2 7 b 2       Corollary 3 4  Assuming that   E   E    Then   j   E     j   E      Moreover  the converse holds if K   K     Proof     E   E       a     u 4 a b     u 6 b for some u   K       a 3   b 2         a     3     b     2     j   E     j   E         and the converse holds if K   K    to go backwards on the   step  we only  need to take some kind of n th root