I'm reading through Complex Multiplication by Reinhard Schertz, and I'm stuck at Theorem 3.1.8. 


Let $\mathfrak{O}\_t$ be the order of conductor $t$ in an imaginary quadratic field $K$.


He defines the groups:
$\mathfrak{I}\_t$ is the group generated by invrtible ideals of $\mathfrak{O}\_t$ (i.e., the group of fractional ideals). Let $\mathfrak{f}$ be an ideal of $\mathfrak{O}\_t$:
$$\mathfrak{I}\_{t,\mathfrak{f}} = \{\mathfrak{a}\in \mathfrak{I}\_t \;|\; \mathfrak{a} = \frac{\mathfrak{a\_1}}{\mathfrak{a\_2}}, \mathfrak{a}\_i \in \mathfrak{I},\mathfrak{a}\_i + \mathfrak{f} = \mathfrak{O}\_t \} $$
$$\mathfrak{S}\_{t,\mathfrak{f}} = \{\mathfrak{a} = \frac{\alpha\_1}{\alpha\_2}\mathfrak{O}\_t \;|\; \alpha\_i\mathfrak{O}\_t + \mathfrak{f} = \mathfrak{O}\_t, \alpha\_1\equiv \alpha\_2 \pmod{\mathfrak{f}}\} $$.


Theorem 3.1.8 claims that the quotient $\mathfrak{I}\_{t,\mathfrak{f}}/\mathfrak{S}\_{t,\mathfrak{f}}$ is isomorphic to a quotient $\mathfrak{A}^{t\mathfrak{f}}/\mathfrak{A}\_{t,\mathfrak{f}}$ where $\mathfrak{A^{t\mathfrak{f}}}$ is the group of fractional ideals of the maximal order prime to $t\mathfrak{f}$.


To do this, he shows that there is a regular ideal in every ideal class of $\mathfrak{I}\_{t,\mathfrak{f}}$ modulo $\mathfrak{S}\_{t,\mathfrak{f}}$.


In the proof he makes the following claim:



> 
> we observe that $\mathfrak{O}\_t$ is a noetherian ring, in which every ideal $\neq (0)$ is maximal. This follows from the facts that every ideal $\neq (0)$ is a free rank two module over $\mathbb{Z}$, which implies that $\mathfrak{O}\_t/\mathfrak{a}$ is finite. Hence, the two ideals $\mathfrak{f}$ and $\mathfrak{ft}$ have a decomposition as a product of primary ideals $$\mathfrak{f} = \mathfrak{q}\_1\cdots \mathfrak{q}\_m \quad \mbox{resp.} \quad \mathfrak{ft} = \mathfrak{q}\_1'\cdots \mathfrak{q}\_n'$$
> 
> 
> 


I don't believe that every ideal not equal to zero is maximal. This just seems wrong.


Even assuming the statement, I don't know which theorem gives the product of primary ideals. 



Proof in Schertz's Complex Multiplication
The statement that seems odd to you is indeed false, what (I believe) he wants to say is that $\mathfrak{D}\_t$ is a Noetherian ring in which every *prime* ideal $\mathfrak{p}\neq 0$ is maximal.


I believe that all you need is in Atiyah-Macdonald's *Introduction to Commutative Algebra*, to which I will refer as [AM69].


1. Since $\mathfrak{D}\_t$ is Noetherian, every ideal $\mathfrak{a}$ has a finite primary decomposition, by [AM69], Theorem 7.13, so
$$
\mathfrak{a}=\cap\_{i=1}^n\mathfrak{q}\_i
$$
where the $\mathfrak{q}\_i$ are primary.
2. The radical of a primary ideal is prime by [AM69], Proposition 4.1. Hence, $r(\mathfrak{q}\_i)=\mathfrak{m}\_i$ are maximal, by the remark that all non-zero primes in $\mathfrak{D}\_t$ are maximal.
3. Since the decomposition is minimal, the $\mathfrak{m}\_i$ are distinct (by definition of "having a primary decomposition", paragraph before [AM69], Theorem 4.5).
4. From the fact that $\mathfrak{m}\_i\neq\mathfrak{m}\_j$ for $i\neq j$, which means that they are coprime, we obtain that $\mathfrak{q}\_i,\mathfrak{q}\_j$ are also coprime for $i\neq j$ by [AM69], Proposition 1.16.
5. From this, we obtain that the intersection is in fact a product by [AM69], Proposition 1.10.


