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.