Show that every maximal ideal is one of the following:
* for some ;
* for some with ;
* for some with .
this is a nice problem! i don't know what you know and what you don't, so i'll assume that you know that if is a field, then any maximal ideal of can be generated by polynomials
such that is a polynomial in so if is a maximal ideal of then now we may assume that is irreducible, because if
then since is prime, either or if then this proves our claim that we may assume that is
irreducible. thus either or is an even number. [note that cannot be a constant because then wouldn't be maximal] now if is even, then is a product of some
irreducible quadratic polynomials in thus in this case, we may assume that is an irreducible quadratic polynomial in so you have two cases:
case 1: we may assume that now write where then dividing each by gives us for some
hence again, as we already showed, we may assume that either or is an irreducible quadratic
polynomial in if then we'll get and if is an irreducible quadratic polynomial, we'll get where of course
because otherwise would have a real root and hence it wouldn't be irreducible anymore.
case 2: is an irreducible quadratic polynomial. this case is left for you to try!