Unit group of quotients [WIP]

Let \(K\) be a number field and \(I\) be an integral ideal in \(K\), then can we find the structure of \(\left(\frac{\mc O_K}{I\mc O_K}\right)^*\) in terms of cyclic groups?

By the chinese remainder theorem, if we have \(I=\prod_i\mf P_i^{n_i}\), we can decompose the quotient as \[\frac{\mc O_K}{I\mc O_K}\cong\bigoplus_i\frac{\mc O_K}{\mf P_i^{n_i}}\] Which reduces the problem of finding the structure of the unit group into when \(I=\mf P^n\).

A useful way to understand what \(\frac{\mc O_K}{\mf P^n}\) is to study the completion of \(K\) at \(\mf P_i\), given by \(L=\hat K_{\mf P}\).

Suppose the residue field \(k\) of \(L\) has order \(q=p^f\) and let \(e\) be the ramification index of \(L/\mb Q_p\). Let \(U^{(n)}=1+\mf P^n\) be the \(n\)th unit groups and let \(\mc O=\mc O_L\) be the ring of integers and let \(\pi\) be a uniformizing element.

As the map \(\mc O^*\to\left(\frac{\mc O}{\mf P^n}\right)^*\) has kernel \(U^{(n)}\), we have the isomorphism \[\frac{\mc O^*}{U^{(n)}}\cong\left(\frac{\mc O}{\mf P^n}\right)^*\]

By Hensel’s lemma, \(\mc O^*\) contains the \(q-1\) roots of unity, hence it contains \(\mu_{q-1}\). Since the map from \(\frac{\mc O^*}{U^{(n)}}\to k^*\cong\mu_{q-1}\) has kernel \(\frac{U^{(1)}}{U^{(n)}}\), we have the isomorphism

\[\left(\frac{\mc O}{\mf P^n}\right)^*\cong\frac{\mc O^*}{U^{(n)}}\cong\mu_{q-1}\oplus\frac{U^{(1)}}{U^{(n)}}\]

Since \(\mu_{q-1}\cong C_{q-1}\), we only need to figure out the structure of \(\frac{U^{(1)}}{U^{(n)}}\).

To work out the structure of this group, we first determine the structure of \(U^{(1)}\).

When \(n>\frac e{p-1}\), we have the isomorphisms between the multiplicative group \(U^{(n)}\) and the additive group \(\mf P^n\)(Neukirch Proposition II.5.5) \[\log:U^{(n)}\to\mf P^n\] \[\exp:\mf P^n\to U^{(n)}\]

Note that we have \(\mf P^n=\pi^n\mc O\) and we have, as additive groups, \(\mc O\cong\mb Z_p^d\). The only other elements that are missing comes from the \(p^a\)th roots of unity, thus giving us \[U^{(1)}=\mu_{p^a}\oplus\mb Z_p^d\] for some \(a\leq\frac e{p-1}\). Finally, this shows that we have \(d+1\) generators for \(U^{(1)}\), with one of them having order \(p^a\) for \(n\geq a\) in \(\frac{U^{(1)}}{U^{(n)}}\).

First we study the non-ramified case, we have \(a=0\) and \((p)=(\pi)\), hence the generators form cyclic groups of order \(p^{n-1}\) and the structure of the unit group is given by

\[\left(\frac{\mc O}{\mf P^n}\right)^*\cong\frac{\mc O^*}{U^{(n)}}\cong C_{q-1}\oplus C_{p^{n-1}}^d\]

In the ramified case, we have \((p)=(\pi^e)\). For now, lets assume that \(a=0\). The maximum order of each generator is \(p^{\left\lceil\frac{n-1}e\right\rceil}\). I claim that there must exist a generator with order \(p^{\left\lceil\frac{n-1}e\right\rceil}\) by counting arguments:

The local ring \(\frac{\mc O}{\mf P^n}\) has \(p^{fn}\) elements in it, and \(p^{f(n-1)}\) elements in the maximal ideal \(\frac{\mf P\mc O}{\mf P^n}\), hence it has a total of \(p^{f(n-1)}\left(p^f-1\right)\) elements in its unit group.

Let \(r=e\left\lceil\frac{n-1}e\right\rceil-(n-1)<e\). Suppose that every generator has order \(p^{\left\lceil\frac{n-1}e\right\rceil-1}\), which will produce the largest unit group, then this largest group has order \[\left(p^f-1\right)\left(p^{\left\lceil\frac{n-1}e\right\rceil-1}\right)^d=\left(p^f-1\right)p^{f(n-1+r)-d}=p^{f(n-1)}\left(p^f-1\right)p^{rf-d}\]

Since \(r<e\), we have \(rf-d\) is negative, so we have less elements in this group than in the unit group of \(\frac{\mc O}{\mf P^n}\), which cannot be possible, hence we need at least \(d-rf\) of the generators to have maximal order \(p^{\left\lceil\frac{n-1}e\right\rceil}\). Hence we can deduce that

\[\left(\frac{\mc O}{\mf P^n}\right)^*\cong\frac{\mc O^*}{U^{(n)}}\cong C_{q-1}\oplus C_{p^{\left\lceil\frac{n-1}e\right\rceil}}^{d-rf}\oplus G\]

where \(G\) is a group with \(rf\) generators of order a power of \(p\) but at most \(p^{n-1}\). Let \(g_i\) be the generators, by the same counting argument, we also need \[\prod\text{ord}\left(g_i\right)=p^{\left(\left\lceil\frac{n-1}e-1\right\rceil\right)rf}\]

Notice that when \(e=1\), we have \(r=0\), rederiving the unramified case.

If \(a\neq0\), we could possibly modify the result slightly by having \(p^{rf-d+a}\) and as long as \(rf-d+a<0\), the result holds except we have \[\left(\frac{\mc O}{\mf P^n}\right)^*\cong\frac{\mc O^*}{U^{(n)}}\cong C_{q-1}\oplus C_{p^{\left\lceil\frac{n-1}e\right\rceil}}^{d-rf+a}\oplus G\] where \(G\) has \(rf-a\) generators satisfying \[\prod\text{ord}\left(g_i\right)=p^{\left(\left\lceil\frac{n-1}e-1\right\rceil\right)(rf-a)}\]


Specializing in the case of the Gaussian integers, we have

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")
    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}")
    print(f"{od} is minimal order for R/IR")