# 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 $\prod\text{ord}\left(g_i\right)=p^{\left(\left\lceil\frac{n-1}e-1\right\rceil\right)(rf-a)}$

## Example

Specializing in the case of the Gaussian integers, we have

• $$p\equiv1\pmod 4$$, we have $$e=f=1$$ and the unit group is isomorphic to $$C_{p-1}\oplus C_{p^{n-1}}=C_{p^n-p^{n-1}}$$

• $$p\equiv3\pmod 4$$, we have $$e=1,f=2$$ and the unit group is isomorphic to $$C_{p^2-1}\oplus C_{p^{n-1}}\oplus C_{p^{n-1}}$$

• $$p=2,e=2,f=1$$, since $$\frac e{p-1}=2\geq1$$, $$a$$ could be non-zero and we are not able to apply the results directly.. (turns out $$a=2$$?)

## 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")