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Let $f(x)$ be a monic, irreducible polynomial in $\mathbb{Z}[x]$ with $\theta\in\mathbb{C}$ as root. Does every non-zero prime ideal of $\mathbb{Z}[\theta]$ factorize uniquely, up to order, in prime ideals of $\mathbb{Z}[\theta]$?

I know that $\mathcal{O}_{\mathbb{Q}(\theta)}$ is a Dedekind domain and in $\mathcal{O}_{\mathbb{Q}(\theta)}$ the factorization of ideals in prime ideal is unique. Furthermore $\mathbb{Z}[\theta]$ is a subring of $\mathcal{O}_{\mathbb{Q}(\theta)}$, but how can I prove or disprove that the factorization in prime ideals in $\mathbb{Z}[\theta]$ is unique?

EDIT: what I really care is that principle ideals of $\mathbb{Z}[\theta]$ factorizes uniquely in prime ideals of $\mathbb{Z}[\theta]$. What I can say is that the principal ideals of $\mathbb{Z}[\theta]$ factorizes uniquely in prime ideals of $\mathcal{O}_{\mathbb{Q}(\theta)}$ simply because $\mathbb{Z}[\theta]$ is a subring of $\mathcal{O}_{\mathbb{Q}(\theta)}$.

Can I conclude that if $\alpha \in \mathbb{Z}[\theta]$ then $\alpha \in \mathcal{O}_{\mathbb{Q}(\theta)}$ and $\langle \alpha \rangle = \mathcal{P}_1\cdot\ldots\cdot\mathcal{P}_k$ where $\mathcal{P}_i$ are prime ideals of $\mathcal{O}_{\mathbb{Q}(\theta)}$. Hence $\langle \alpha \rangle$ factorize uniquely in $\mathbb{Z}[\theta]$ as $ \mathcal{Q}_1\cdot\ldots\cdot\mathcal{Q}_k$ where $\mathcal{Q}_i = \mathcal{P}_i \cap \mathbb{Z}[\theta]$ ?

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    $\begingroup$ I think you have unique factorization for ideals prime to the index $[\mathcal{O}_{\mathbb{Q}(\theta)}:\mathbb{Z}[\theta]]$ For the other case consider something like 4 in the ring $\Bbb{Z}[\sqrt{5}]$ Maybe search for factorization of ideals in orders of a number ring. $\endgroup$ – sharding4 Jun 22 '17 at 19:42
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A subring of a Dedekind domain is not necessarily a Dedekind domain. A theory of "orders" has even been developped to study this deficiency. For simplification, let us stick to number theory : $K$ is a number field of degree $n$, $O=O_K$ its ring of integers; a subring $R \subset O$ is called an order of $K$ if $(R, +)$ is a free $\mathbf Z$-module of rank $n$. It is not difficult to show that $O$ is the maximal order of $K$. A strict order $R \neq O$ cannot be Dedekind because it is not integrally closed in $K$. Example : for a quadratic field $K=\mathbf Q (\sqrt d)$ s.t. $d \equiv 1$ mod $4$, $\mathbf Z [\sqrt d]$ is a strict order.

To answer more precisely to your edit, let us now stick to the particular case of a quadratic field $K=\mathbf Q (\sqrt d)$ and an order $R= \mathbf Z + fO$, where $f := [O : R]$ is the conductor of $R$. Another important invariant is the discriminant of $R$, defined by $D = f^2 . disc(K)$. It can be shown that unique factorization holds for ideals of $R$ which are prime to $f$ . To remedy to the general lack of unique factorization of ideals in $R$ , let us introduce the notion of a proper fractional ideal $J$ of $R$, which is such that $R$ = {$x \in K ; xJ \subset J$} (= the stabilizer of $J$; in general, there is only a strict inclusion). The key result : " A fractional ideal $J$ of $R$ is proper iff it is invertible " allows to define the ideal class group $C(R)$ in the usual manner.

The study of orders in quadratic fields is "justified" by its astonishing connection with Gauss' genus theory of quadratic forms. Consider an order $R$ of discriminant $D$ in an imaginary quadratic field $K$. If $f(x,y) = ax^2 + bxy + cy^2$ is a primitive definite quadratic form of discriminant $D$, then the ideal $J$ with $\mathbf Z$-basis {$a, (-b + \sqrt D)/2$} is a proper ideal of $R$, and the map sending $f(x,y)$ to $J$ induces an isomorphism between the class group $C(D)$ of quadratic forms of discriminant $D$ and the ideal class group $C(R)$.

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The answer is no in general: this would imply $\mathbf Z[\theta]$ is itself a Dedekind domain, hence would be $\mathcal O_{\mathbf Q(\theta)}$, which is not necessarily the case.

For instance the ring of integers of the quadratic field $\mathbf Q(\sqrt d)$ is $\mathbf Z\biggl[\dfrac{1+\sqrt d}2\biggr]$.

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  • $\begingroup$ Thank you. Can you have also a look to my edit,please? $\endgroup$ – ilmarchese Jun 22 '17 at 19:46
  • $\begingroup$ I'll have to think about it, but a priori, I don't think so for one reason: in general you don't have $\;a\mathcal{O}_{\mathbf{Q}(\theta)}\cap\mathbf Z[\theta]=a\mathbf{Z}[\theta]$ (it's true is the morphism $\;\mathbf Z[\theta]\hookrightarrow\mathcal{O}_{\mathbf{Q}(\theta)}$ is flat). $\endgroup$ – Bernard Jun 22 '17 at 19:56

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