The cover is easy to understand because it is defined by the single equation . To give a maximal ideal of such that is the same as giving a homomorphism (up to automorphisms of the image), which is in turn the same as giving a root of in (up to automorphism), which is the same as giving an irreducible factor of the reduction of modulo .

As suggested in the proof of the lemma, we find all homomorphisms by finding all homomorphism . In terms of ideals, if is a maximal ideal of , then the ideal of is also maximal, since

We formalize the above discussion in the following theorem:

We return to the example from above, in which , where is a root of . According to , the maximal order has discriminant :

> Discriminant(MaximalOrder(K)); 2945785The order has the same discriminant as , so and we can apply the above theorem.

> Discriminant(x^5 + 7*x^4 + 3*x^2 - x + 1); 2945785We have

If we replace by , then the index of in will be a power of , which is coprime to , so the above method will still work.

> f:=MinimalPolynomial(7*a); > f; x^5 + 49*x^4 + 1029*x^2 - 2401*x + 16807 > Discriminant(f); 235050861175510968365785 > Discriminant(f)/Discriminant(MaximalOrder(K)); 79792266297612001 // coprime to 5 > S<t> := PolynomialRing(GF(5)); > Factorization(S!f); [ <t + 1, 2>, <t + 4, 1>, <t^2 + 3*t + 3, 1> ]Thus factors in as

> f:=MinimalPolynomial(5*a); > f; x^5 + 35*x^4 + 375*x^2 - 625*x + 3125 > Discriminant(f) / Discriminant(MaximalOrder(K)); 95367431640625 // divisible by 5 > Factorization(S!f); [ <t, 5> ]

William Stein 2004-05-06