# extension of Euler's totient function to number fields

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.

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The last isomorphism identity looks indeed quite familiar to me; so I am looking forward to its developments... In any case, I think positively of this question. – awllower Feb 10 '12 at 13:44