Let $\mathcal{O}_K$ be the ring of algebraic integers of a number field defined by a cubic polynomial $P$, and let $\mathfrak{p}=(p)$ be a principal ideal generated by a rational prime number p. A well known theorem by Dedekind states that if $P \mod p$ factors in irreducible polynomials $P \equiv P_1^{e_1} \ldots P_k^{e_k} \mod p$ then the ideal $\mathfrak{p}$ factors accordingly in prime ideals: $\mathfrak{p}=\mathfrak{p}_1^{e_1} \ldots \mathfrak{p}_k^{e_k}$. This under certain conditions that have to be met by $p$ (see Dedekindf.pdf).

Thus in this case the factorization is predictible and so the quotient $\mathcal{O}_K/\mathfrak{p}$ is the direct sum of finite fields. But in the opposite case the factorization is not predictible, and rings other than fields can appear. As an example consider the case $P=x^3-5$. We have that $\mathcal{O}_K=\mathbb{Z}[\theta]$, with $\theta$ a root of $P$ The discriminant of $\mathcal{O}_K$ is $-625$, so for $p=5$ the factorization is inpredictable. The only idempotent elements of $Q=\mathcal{O}_K/(5)$ are $0$ and $1$, so $Q$ is not a direct sum, but it is not a field since there are non trivial nilpotent elements (e.g. $\theta$). If the Dedekind theorem would apply then $(5)$ would be the sum of three fields $GF(5)$. I claim that in all similar cases following theorem applies (is it correct? is the proof correct?):

THEOREM: If $Q$ is the quotient of the ring of integers of a cubic extention of degree 3 by a principal ideal (p), where p is a rational prime, and if $Q$ is not a field, nor a direct sum of non trivial ideals then $Q$ is isomorphic to the ring $\mathbb{Z}[X]/(p,X^3)$.

Proof: The proof is based on the following two articles Rings of order $p^3$ and Rings of order $p^2$. Since the underlying additive group is of type (1,1,1) and the ring $Q$ is commutative , is indecomposable and has a unit, it falls under the classification II.C.1.a for the first paper, which gives three possibilities. The first of them is a field, so it can be excluded. The last of them is $\mathbb{Z}[X,Y]/(p, X^2, XY, Y^2)$. Now this last ring has as a maximal (unique) ideal that corresponds to the classification J of the second paper. But this ring is the direct sum of two trivial identical rings of order $p$, say $I$, and we have $I.I=0$. Taking the inverse image of the quotient map $\pi:\mathcal{O}_K \rightarrow \mathcal{O}_K/(p)$ this would mean that $\pi^{-1}(I.I)=\pi^{-1}(0) \Leftrightarrow \pi^{-1}(I).\pi^{-1}(I) = (p)$. In other words $(p)$ is the product of two ideals of norm $p$. But $Q$ has norm $p^3$ so this possibility is excluded. The remaining possibility is the ring $\mathbb{Z}[X]/(p, X^3)$.$\square$

Remark that in our example the unique maximal ideal is the kernel of the homomorphism $q \mapsto q^p$. (For p>=3 this reduces to the set ${q: q^3=0})$. And the ring $\mathbb{Z}[X]/(p, X^3)$ also has a maximal ideal that corresponds to the classification I of the second paper.

Marc Bogaerts

  • $\begingroup$ Dear Nimda: The ring $\mathcal O_K/\mathfrak p$ is not necessarily a direct product of finite fields. If there is ramification, then $\mathcal O_K/\mathfrak p$ has nilpotent elements. $\endgroup$ – Bruno Joyal Aug 25 '14 at 17:19
  • $\begingroup$ Indeed, I should have added non-ramification to the condition of predictability. But this does not alter the statement in the theorem. $\endgroup$ – Marc Bogaerts Aug 26 '14 at 8:57

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