Let be a number field and be an integral ideal in , then can we find the structure of in terms of cyclic groups?

By the chinese remainder theorem, if we have , we can decompose the quotient as

Which reduces the problem of finding the structure of the unit group into when .

A useful way to understand what is to study the completion of at , given by .

Suppose the residue field of has order and let be the ramification index of . Let be the th unit groups and let be the ring of integers and let be a uniformizing element.

As the map has kernel , we have the isomorphism

By Hensel's lemma, contains the roots of unity, hence it contains . Since the map from has kernel , we have the isomorphism

Since , we only need to figure out the structure of .

To work out the structure of this group, we first determine the structure of .

When , we have the isomorphisms between the multiplicative group and the additive group (Neukirch Proposition II.5.5)

Note that we have and we have, as additive groups, . The only other elements that are missing comes from the th roots of unity, thus giving us

In the ramified case, we have . For now, lets assume that . The maximum order of each generator is . I claim that there must exist a generator with order by counting arguments:

The local ring has elements in it, and elements in the maximal ideal , hence it has a total of elements in its unit group.

Let . Suppose that every generator has order , which will produce the largest unit group, then this largest group has order

Since , we have is negative, so we have less elements in this group than in the unit group of , which cannot be possible, hence we need at least of the generators to have maximal order . Hence we can deduce that

where is a group with generators of order a power of but at most . Let be the generators, by the same counting argument, we also need

Notice that when , we have , rederiving the unramified case.

If , we could possibly modify the result slightly by having and as long as , the result holds except we have

where has generators satisfying

Example

Specializing in the case of the Gaussian integers, we have

• , we have and the unit group is isomorphic to

• , we have and the unit group is isomorphic to

• , since , could be non-zero and we are not able to apply the results directly.. (turns out ?)

Some code

K.<a> = NumberField(x^4-x^3-x^2-2*x+4)
R = K.ring_of_integers()
p = 7
n = 3
for P,e in K.factor(p):
    f = P.residue_class_degree()
    d = e*f
    a = e//(p-1)
    r = (1-n)%e
    if r*f-d+a>=0:
        print("a could be non-zero, skipping this case")
        continue
    print(f"Checking the prime {P} with ramification index {e} and inertia degree {f}")
    RqI = R.quo(P^n,'b')
    od = p^(1+(n-2)//e)*(p^f-1)  # order
    for p_fac in [i for i,j in list(factor(p^f-1))]+[p]:
        not_od = ZZ(od/p_fac)
        print(f"Checkng {p_fac}")
        while True:
            k = R.random_element()
            if K.ideal(k) + P == K.ideal(1): # coprime
                assert RqI(k)^od == RqI(1):
                if RqI(k)^not_od != RqI(1):
                    print(f"{k} does not have order {not_od}")
                    break
    print(f"{od} is minimal order for R/IR")