# Proof in Schertz's Complex Multiplication

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.

## 2 Answers

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.

Presumably, the author aimed only at showing that the order $$\mathfrak{O}_t$$ is a one-dimensional Noetherian domain, i.e., it is a Noetherian domain and every of its non-zero prime ideals is maximal. He succeeded in doing so.

The subsequent claim about the primary decomposition of a non-zero ideal in a one-dimensional Noetherian integral domain holds true and seems to be a well-known fact of commutative algebra:

Claim. Let $$R$$ be a Noetherian integral domain of Krull dimension $$1$$. Let $$\mathfrak{a}$$ be a non-zero ideal of $$R$$. Then $$\mathfrak{a}$$ is the product of finitely many primary ideals of $$R$$.

Proof. By [Theorem 3.3, 1], that is E. Lasker - E. Noether's Primary Decomposition Theorem, the ideal $$\mathfrak{a}$$ is the intersection of finitely many primary ideals of $$R$$ with pairwise distinct radicals (consider a primary decomposition of shortest length). These primary ideals are pairwise co-maximal because their radicals are pairwise co-maximal (this is where the dimension one comes into play). Indeed, if any pair of these primary ideals is contained in some maximal ideal, then this ideal contains the of sum of the two radicals, a contradiction. As a result, the intersection is a product.

In the above proof, we agreed on the following definition. Two ideals $$I$$ and $$J$$ of $$R$$ are co-maximal if $$I + J = R$$.

I don't know of any textbook where this claim is both stated and proved. My closest match is [Proposition I.12.3: "Strong version of Chinese Remainder Theorem", 2], but in my opinion the proof cannot be understood without appealing to Lasker-Noether's theorem, and this theorem is not cited there.

[1] D. Eisenbud, "Commutative Algebra with a View to Algebraic Geometry", 1995.
[2] J. Neukirch, "Algebraic Number Theory", 1999.

• Well, see $\S$ 10.8 of math.uga.edu/~pete/integral2015.pdf. – Pete L. Clark Nov 21 '19 at 3:39
• @PeteL.Clark Thanks, I appreciate precise written references to facts that can be annoying to re-prove without full knowledge/memory of their context. – Luc Guyot Nov 21 '19 at 11:31