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Here's a question that comes up every now and then. Of course, the quotient of a number ring (ring of integers of a number field) by an ideal $I$ is a finite (Artin) ring. If we take $I$ to be the power of a prime, we obtain a finite local (Artinian) ring. Is there a characterization of finite local rings which arise in this way?

In particular, can we obtain the quotient rings $k[x]/x^n$, where $k$ is a finite field?

Much thanks!

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This question really comes up now and then? – KConrad Aug 6 '11 at 10:15
The structure of $\mathcal{O}_K / \mathfrak{P}^n$ for a number field $K$ and a prime ideal $\mathfrak{P}$ in the maximal order $\mathcal{O}_K$ was studied in M. Elia, J. Carmelo Interlando, R. Rosenbaum: On the structure of residue rings of prime ideals in algebraic number fields - part I and II. You can find the papers freely available online, just ask your favourite search engine. Part I says that with an unramified prime, you cannot obtain $k[x]/(x^n)$ (for $n > 1$). So you might want to look at Part II directly. – felix Aug 6 '11 at 10:42
It is easy to see you can't use unramified primes for $n > 1$ (see my answer below) and the authors of those papers are aiming to avoid completions. Personally I think completions make the situation a lot clearer in this question. – KConrad Aug 6 '11 at 11:37
Is there any reason to think you can't get all finite rings in this way? – KConrad Aug 6 '11 at 11:46
@KConrad: Since the ring of integers of a number field is a regular one dimensional ring, the Zariski tangent space at any maximal ideal is one dimensional, so this is an upper bound for the Zariski tangent space for any quotient ring. Thus, most finite rings do not occur as quotients but perhaps the necessary condition on the Zariski tangent space is also sufficient. – ulrich Aug 6 '11 at 12:19
up vote 18 down vote accepted

If ${\mathcal O}_K/{\mathfrak p}^r$ has characteristic $p$ then ${\mathfrak p}^r|p{\mathcal O}_K$, so $r \leq e({\mathfrak p}|p)$. In particular, if ${\mathfrak p}$ is unramified then you can't use it to produce the examples you seek with $n > 1$.

Let $K/{\mathbf Q}$ have degree $n$ and be totally ramified at a prime $p$. (Example: $K = {\mathbf Q}(\sqrt[n]{p})$, or more generally $K = {\mathbf Q}(\alpha)$ where $\alpha$ is the root of any monic in ${\mathbf Z}[x]$ which is Eisenstein at $p$.) Then $p{\mathcal O}_K = {\mathfrak p}^n$. We'll show ${\mathcal O}_K/(p) \cong {\mathbf F}_p[x]/(x^n)$. The quotient ring ${\mathcal O}_K/{\mathfrak p}^r$ for any $r$ is unchanged if we pass to the completion: ${\mathcal O}_K/{\mathfrak p}^r \cong {\widehat{\mathcal O}}/(\pi)^r$, where the $\widehat{\mathcal O}$ denotes the ${\mathfrak p}$-adic completion of ${\mathcal O}_K$ and $\pi$ generates the maximal ideal of $\widehat{\mathcal O}$.

The ${\mathfrak p}$-adic completion $K_{\mathfrak p}$ is a totally ramified extension of ${\mathbf Q}_p$ with degree $n$, so its ring of integers, which is $\widehat{\mathcal O}$, has the form ${\mathbf Z}_p[\pi]$. (The chapter on completions in Lang's Algebraic Number Theory has a theorem that the integers in any finite extension $F$ of ${\mathbf Q}_p$ has the form ${\mathbf Z}_p[\alpha]$ and if you look carefully at the proof then you see that in the totally ramified case you can take as $\alpha$ any generator of the maximal ideal of the integers of $F$.) Therefore ${\mathcal O}_K/{\mathfrak p}^r \cong {\mathbf Z}_p[\pi]/(\pi^r)$ for any $r$.

When $r \leq n$, the ring ${\mathbf Z}_p[\pi]/(\pi^r)$ has characteristic $p$, so there is a surjective ring homomorphism ${\mathbf F}_p[x] \rightarrow {\mathbf Z}_p[\pi]/(\pi^r)$ by sending $x$ to $\pi \bmod \pi^r$. The map kills $x^r$, so we get a surjective ring homomorphism ${\mathbf F}_p[x]/(x^r) \rightarrow {\mathbf Z}_p[\pi]/(\pi^r)$. It is left to the reader to check ${\mathbf Z}_p[\pi]/(\pi^r)$ has size $p^r$, so we have an isomorphism.

To realize $k[x]/(x^n)$ for finite fields $k$ of non-prime size, pick your favorite number field $F$ whose ring of integers has a prime ideal at which the residue field is isomorphic to $k$ (example: given any prime power $q = p^f$, the number field ${\mathbf Q}(\zeta_{q-1})$ at any prime lying over $p$ has a residue field of order $q$). Then let $E$ be any totally ramified extension of $F$ with degree $n$, e.g., the extension obtained by adjoining to $F$ the root of a monic of degree $n$ in ${\mathcal O}_F[X]$ which is Eisenstein at the prime whose residue field is isomorphic to $k$. Letting ${\mathfrak p}$ denote the prime in $E$ that lies over your chosen prime in $F$, an argument like the one above that works with $F$ and its completions as base field instead of ${\mathbf Q}$ and ${\mathbf Q}_p$ implies ${\mathcal O}_E/{\mathfrak p}^n \cong k[x]/(x^n)$.

When $q$ is a power of $p$, the prime $p$ is unramified in ${\mathbf Q}(\zeta_{q-1})$, so the polynomial $X^n - p$ is Eisenstein at any prime lying over $p$ in ${\mathbf Q}(\zeta_{q-1})$. Therefore we can use the number field $E := {\mathbf Q}(\zeta_{q-1},\sqrt[n]{p})$ to generate the desired example: if ${\mathfrak p}$ is any prime which lies over $p$ in ${\mathcal O}_E$ then ${\mathcal O}_E/{\mathfrak p}^n \cong {\mathbf F}_q[x]/(x^n)$.

The ring of integers of $E$ may be more than ${\mathbf Z}[\zeta_{q-1},\sqrt[n]{p}]$, so this is a situation where working in a ${\mathfrak p}$-adic completion (which has isomorphic quotient rings as the ring of integers for ideals that are powers of that prime) makes life easier.

[Edit: As someone pointed out to me by email, there is an obvious obstruction to realizing general finite commutative rings as quotient rings of integers of number fields: the local factors of the quotient of a ring of integers of a number field by a nonzero ideal are quotients of a DVR, hence the local factors have principal maximal ideals (and finite residue field). So one never gets, for example, the finite ring ${\mathbf F}_p[x,y]/(x,y)^2$ as the quotient ring of a ring of integers.

Moreover, that obstruction is the only one. That is, a finite commutative ring is isomorphic to the quotient ring of the integers of some number field iff its local factors have principal maximal ideals. I showed above how to realize any finite local ring of the form $k[x]/(x^n)$ using suitable $p$-adic fields and one can show more generally that any finite local commutative ring is a quotient ring of the integers of some $p$-adic field. Then it remains to put together a product of such finite rings. If $K_1,...,K_n$ are non-archimedean local fields then by weak approximation (treating together the places over a common rational prime, etc.) there exists a number field $F$ with pairwise distinct places $v_1,...,v_n$ such that $K_i = F_{v_i}$, so every finite quotient of ${\mathcal O}_{K_i}$ at powers of primes corresponding to $v_i$ is a finite quotient of ${\mathcal O}_F$ and any finite product of such quotients is a finite quotient of ${\mathcal O}_F$ by the Chinese Remainder Theorem.]

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Incidentally, if you are interested in the structure of the multiplicative group of $({\mathfrak o}/{\mathfrak p}^n)^\times$, where $\mathfrak o$ is the ring of integers in a number field and ${\mathfrak p}\subset{\mathfrak o}$ is a prime ideal, see Nakagoshi (1979). Here it is even more useful to think locally. – Chandan Singh Dalawat Aug 7 '11 at 3:27
Thank you; this is very helpful. To answer your comment above, as a student I was curious about the classification of finite rings, and this method of production suggested itself. More recently, I am hoping to learn about methods to transfer characteristic zero results to characteristic p ones, and such isomorphisms are evidently used. Of course, one may also pose the analogous question for integers of division algebras... – Steven Spallone Aug 7 '11 at 7:29

This answer consists of a few remarks to put the "necessary condition" in context. Recall:

Theorem: If $R$ is a Dedekind domain and $I$ is a nonzero ideal of $R$, then $R/I$ is a principal Artinian ring (i.e., an Artinian ring in which every ideal is principal).

Using the factorization of ideals into primes, the Chinese Remainder Theorem, and the easy fact that a finite product of rings is principal Artinian iff each factor is principal Artinian, one reduces to the case of $I = \mathfrak{p}^a$ a prime power. The ideals of $R/I$ correspond to the ideals of $R$ containing $\mathfrak{p}^a$, i.e., $R, \mathfrak{p},\ldots,\mathfrak{p}^a$. So $R/I$ is certainly Artinian. Moreover, $R/I = R_{\mathfrak{p}}/I_{\mathfrak{p}}$ is also a quotient of a DVR, hence of a PID, and any quotient of a principal ring is principal.

In my experience it is traditional to emphasize not the previous result per se but the following consequence.

Theorem (C.-H. Sah) For an integral domain $R$, the following are equivalent:
(i) $R$ is a Dedekind domain.
(ii) Every ideal $I$ of $R$ can be generated by "$1+\epsilon$" elements: for all $0 \neq a \in I$, there exists $b \in I$ such that $I = \langle a,b \rangle$.

The implication (i) $\implies$ (ii) is immediate from the above. The converse implication is also short but somewhat tricky, so I refer to $\S 20.5$ of my commutative algebra notes for the proof.

So it should be "well known" that the answer to the OP's question is not all finite rings. Note also that the example $\mathbb{F}_q[x,y]/\langle x,y \rangle^2$ of a finite non-principal ring appears as an Exercise in $\S 16.3$ of my notes. I am very sorry to say that my 264 pages of notes have no discussion of Zariski tangent spaces whatsoever: caveat emptor, to be sure.

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Why do you write $1 + \varepsilon$ generators and not two generators in your statement of the theorem? – KConrad Aug 7 '11 at 15:42
@KConrad: what comes before the colon is meant to be a paraphrase of what comes after. "Every ideal can be generated by two elements" is much weaker than what comes after so would not be a good paraphrase. The idea is that this is as close as possible to every ideal being principal: once you throw in any old "small" nonzero element of the ideal, it becomes principal. Two final comments: (i) This is certainly an informal use of mathematical language, and it is not included in the statement of the theorem in my notes. (ii) This is not due to me, though I can't remember where I got it from. – Pete L. Clark Aug 7 '11 at 21:26
Pete: I see what the intention was. Thanks. – KConrad Aug 11 '11 at 16:22

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