extension of Euler's totient function to number fields

Question

It is well known that the Euler totient function $\varphi$ satisfies the formula $n = \sum_{d | n}\varphi(d)$. This follows for example from the fact that $\mathbb Z / n \mathbb Z$ can be written (as monoid) as the disjoint union $\coprod_{d|n} (\mathbb Z / d \mathbb Z)^\times$.

My question is now if this can be generalized to an arbitrary number field $K$? More precisely, if $\mathcal O _K$ denotes the ring of integers of $K$ and $\mathfrak f$ a non-zero integral ideal of $\mathcal O _K$, do we have $\mathcal O _K / \mathfrak f \cong \coprod_{\mathfrak d | \ \mathfrak f} (\mathcal O _K / \mathfrak d)^\times$?

If I am not mistakes then my question would have an affirmative answer, if the number of units of $\mathcal O _K / \mathfrak p ^k$, for $\mathfrak p$ a prime ideal and $k \geq 1$ a natural number, is equal to $\mathcal N (\mathfrak p)^{k-1}(\mathcal N (\mathfrak p) - 1)$.

Edit: Reading a little bit in Zariski/Samuel I, the rings $\mathcal O _K / \mathfrak p ^k$ should be local principle ideal rings, the maximal ideal given by $\mathfrak p / \mathfrak p ^k$, i.e. in order to verify my question one only has to show that the number $\# (\mathfrak p / \mathfrak p ^k)$ equals $N(\mathfrak p ^{k-1})$.

Edit2: If I am not mistaken, then the last assertion follows directly from Noether's third Isomorphism Theorem.

Thank you very much in advance.

Answer 1

First of all, as you guessed, it is true that $(\mathcal O_K/\mathfrak p^k)^\times$ has order $\mathcal N(\mathfrak p)^{k-1}(\mathcal N(\mathfrak p)-1)$. Indeed, there is a short exact sequence $$1\to 1+\mathfrak p/(1+\mathfrak p^{k})\to (\mathcal O_K/\mathfrak p^k)^\times\to (\mathcal O_K/\mathfrak p)^\times\to 1.$$Since $\mathcal O/\mathfrak p$ is a field, $(\mathcal O/\mathfrak p)^\times$ has size $\mathcal N(\mathfrak p)-1$. Thus, we are left to check $1+\mathfrak p/(1+\mathfrak p^{k})$ has size $\mathcal N(p)^{k-1}$. But this is clear by induction, since the following is exact: $$0\to\mathfrak p^{k-1}/\mathfrak p^k\to (1+\mathfrak p)/(1+\mathfrak p^k)\to (1+\mathfrak p)/(1+\mathfrak p^{k-1})\to 1,$$ with the first map being $x\mapsto 1+x$.

As you noted, this implies $\mathcal O_K/\mathfrak f$ has the same number of elements as $\bigsqcup_{\mathfrak d|\mathfrak f}(\mathcal O_K/\mathfrak d)^\times$.