Generalized Euler phi function

Let $n$ be an integer, there is a well-known formula for $\varphi(n)$ where $\varphi$ is the Euler phi function. Essentially, $\varphi(n)$ gives the number of invertible elements in $\mathbb{Z}/n\mathbb{Z}$. My questions are:

1) Since Dedekind domains have the same factorization theorem for ideals analogous to that of the integers, can one define a generalized Euler phi function type for an ideal of a Dedekind domain, i.e, $\varphi(I)$ shall give the number of invertible elements in $R/I$, and is there a nice formula for it? It makes sense to me that perhaps the formula should resemble that of the integer, using the factorization of $I$ into prime ideals. But I do not have a concrete idea of what it should be.

2) What about domains that are not Dedekind, more specifically, what are the minimum hypotheses that one can impose on a domain so that one can have perhaps a formula for Euler phi function type on the ideals? I am not sure if this even makes sense at this point.

• In a Dedekind doman, $I$ factors uniquely into the product $P_1^{r_1} \cdots P_k^{r_k}$. By the Chinese remainder theorem, $R/I$ is ring-isomorphic to the product of the $k$ rings $R/P_k^{r_k}$. Therefore your $\phi(I)$ is multiplicative in this sense. I believe that the number of invertible elements in these more basic rings is related to the norm of the prime ideal $P_k$, but I don't know the exact value offhand. – Greg Martin Jan 21 '11 at 7:05
• Well, in a general Dedekind domain, the problem is that the Euler phi function may be infinite-valued: think e.g. about $\mathbb{C}[t]$. So what you want is...ahem...a Dedekind abstract number ring as in www.math.uga.edu/~pete/aant.pdf. In the familiar cases of $S$-integer rings in global fields, I do believe that the analogue has been well pursued, including analytic questions about distribution of values. I seem to recall discussing this with Carl Pomerance at one point and that he mentioned that a student of his was working on this sort of thing. – Pete L. Clark Jan 21 '11 at 8:41
• @Pete : Interesting! – François Brunault Jan 21 '11 at 13:40
• @Pete : Can you do an analogous theory for $k$-algebras which satisfy the following finiteness condition : every nonzero ideal has finite codimension ? – François Brunault Jan 21 '11 at 14:27
• @François: at the moment, I am having plenty of trouble with the case of finite residue fields. (I have made some recent "progress" in the sense of becoming more familiar with the literature, and I will upload a longer version of aant.pdf relatively soon.) So...I can't, no. – Pete L. Clark Jan 21 '11 at 14:32

Yes, there is a formula for $\varphi(I)$ in the case of number fields. Let $R$ be the ring of integers of a number field. As mentioned in Greg's comment, it suffices to consider the case $I=\mathfrak{p}^n$ where $\mathfrak{p}$ is a maximal ideal of $R$. Then we have a surjective ring morphism

$$\frac{R}{\mathfrak{p}^n} \to \frac{R}{\mathfrak{p}}$$ such that the preimage of $(R/\mathfrak{p})^{\times}$ is exactly $(R/\mathfrak{p}^n)^{\times}$ (this is because $R/\mathfrak{p}^n$ is local). Thus $\varphi(\mathfrak{p}^n) = q^{n-1}(q-1)$ where $q=\operatorname{Card} (R/\mathfrak{p})$.

Note that there are Dedekind domains $R$ such that $R/I$ is never finite for $I \neq R$ : for example take $R=\mathbf{C}[T]$.

To define a function $\varphi$ for general rings, one would obviously need the hypothesis that $(R/I)^{\times}$ is finite, but then it is only clear that $\varphi$ is (weakly) multiplicative in the sense that $\varphi(I\cdot J) = \varphi(I) \varphi(J)$ if $I+J=R$.

• @François: (Sorry, I made my comment above before I read your answer.) Note that even in the case of $R = \mathbb{Z}$, $\varphi$ is only "weakly multiplicative" in your sense! – Pete L. Clark Jan 21 '11 at 8:43
• @Pete: You're right, in the case $R=\mathbf{Z}$ one definitely cannot ask more than this multiplicativity. I wasn't sure of the english terminology, it seems that "complete multiplicativity" means that we can remove the "coprime" condition. – François Brunault Jan 21 '11 at 13:40

Let $R$ be a Dedekind domain and $\zeta_R$ it's zeta function. Using Mobius inversion on $|R/I|=\sum _{J|I}|(R/J)^*|$, and the fact that $\zeta_R(s)^{-1}=\sum \mu(I)/N(I)^s$ you get the identity $$\frac{\zeta_R(s-1)}{\zeta_R(s)}=\sum \frac{\phi(I)}{N(I)^s}$$ Euler's formula for the generalized totient function follows from the Euler product and you get $$\phi(I)=N(I)\prod_{P|I}(1-N(P)^{-1})$$ where the product ranges over all prime ideals dividing $I$. If you're not in a Dedekind domain, the only part which doesn't generalize is the Euler product for the zeta function, or equivalently unique factorization into prime ideals, without which there is not much hope for a formula for this totient function.

• @Gjergji: This is a nice point of view. Do you need to assume something like $R$ is finitely generated over $\mathbf{Z}$ to define the zeta function of $R$, or am I missing something? – François Brunault Jan 21 '11 at 13:48
• Gjergi is also using the finiteness of the quotient rings $R/I$ ("abstract number rings" again). Of course the zeta function can be defined on the much larger class of finitely generated $\mathbb{Z}$-algebras since it only looks at residue fields, but I don't see a $\varphi$ function in Krull dimension greater than one. – Pete L. Clark Jan 21 '11 at 14:38
• By the way, there has also been some work on computing the zeta function of a non-Dedekind (once again, sorry for the apparently pretentious terminology, which is actually meant to be self-effacing) abstract number ring -- or better put, of a non-normal integral $\mathbb{Z}$-scheme of dimension one. You'll still get a factorization into local factors, finitely many of which will be "bad" and thus more complicated. (If anyone is interested, I can dig up references...) So I don't think that a $\varphi$ function here is totally out of reach. – Pete L. Clark Jan 21 '11 at 14:43
• @Pete, that sounds interesting, and is relevant to the question, so if you find time to dig those references, please share them here. – Gjergji Zaimi Jan 21 '11 at 17:01
• @Pete, I'd be also interested. And, does the zeta function you refer to coincide with the zeta function of the scheme $\operatorname{Spec} R$ (which naturally has an Euler product) ? – François Brunault Jan 21 '11 at 18:34