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I am trying to understand a proof of the following statement:

Let $R$ be a Dedekind ring and let $I$ be an ideal of $R$. $I$ is contained in only a finite number of prime ideals $\mathfrak{p}_1,...,\mathfrak{p}_n$ of $R$ and $I=\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}$ for some $a_1,\cdots,a_n \in\mathbb{N}^+$.

The proof, which was taken from Janusz's Algebraic Number Fields and modified by me, goes as follows:

Consider the ring $S=R/I$. Note that since $R$ is of Krull dimension 1, $S$ is a commutative ring in which every prime ideal is maximal. Therefore every ideal of $S$ contains a product of prime ideals. The prime ideals of $S$ correspond bijectively with the prime ideals of $R$ which contain $I$. Note that there can be only finitely many prime ideals of $S$ and hence only finitely many prime ideals $\mathfrak{p}_1,\cdots,\mathfrak{p}_n$ of $R$ which contain $I$. In particular, there exists $a_1,\cdots, a_n \in \mathbb{N}$ such that $({\mathfrak{p}_1/I})^{a_1}\cdots ({\mathfrak{p}_n/I})^{a_n}=0$. This implies $\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}\subseteq I$.

Let $\widetilde{R}=R/\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}$. Since the ideals $\mathfrak{p}_i$ and $\mathfrak{p}_j$ are maximal for all $i$ and $j$, they are comaximal for all $i\neq j$. Thus by the Chinese remainder theorem we have $$ \widetilde{R}\cong R/{\mathfrak{p}_1}^{a_1}\times \cdots \times R/{\mathfrak{p}_n}^{a_n}. $$ The ideals of $\widetilde{R}$ are of the form $I_1\times \cdots \times I_n$, where $I_k$ is an ideal of $R/{\mathfrak{p}_k}^{a_k}$. Since the ideals of $R/\mathfrak{p}_k^{a_k}$ are all powers of $\mathfrak{p}_k/\mathfrak{p}_k^{a_k}$, the ideals of $\widetilde{R}$ are the image of some product $\mathfrak{p}_1^{b_1}\cdots \mathfrak{p}_n^{b_n}$ for some $b_i \leq a_i$. In particular, this means $I$ has the same image as some $\mathfrak{p}_1^{c_1}\cdots \mathfrak{p}_n^{c_n}$, where $c_i \leq a_i$ for all $i$. Since both $I$ and $\mathfrak{p}_1^{c_1}\cdots \mathfrak{p}_n^{c_n}$ contain $\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}$ and map to the same element ideal of $\widetilde{R}$, we must have $I=\mathfrak{p}_1^{c_1}\cdots \mathfrak{p}_n^{c_n}$. $\square$

The part which I do not understand is the final step:

Since both $I$ and $\mathfrak{p}_1^{c_1}\cdots \mathfrak{p}_n^{c_n}$ contain $\mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}$ and map to the same element ideal of $\widetilde{R}$, we must have $I=\mathfrak{p}_1^{c_1}\cdots \mathfrak{p}_n^{c_n}$.

I'm sure the reason is staring me right in the face, but I cannot see it.


What is the reasoning behind the final step?

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I have to say that the proof given in Janusz (at least from what I see above in your post) is shorter than either what I've seen in Neukirch of Marcus' Number Fields. – user38268 Jan 2 '13 at 12:37
@benjalim well, he builds it up with a series of lemmas which are used implicitly in the proof. So his exposition is a bit longer than this. – Holdsworth88 Jan 2 '13 at 12:44
Sorry, but who are referring to as "he" in your comment above? Are you referring to Janusz, Marcus or Neukirch? – user38268 Jan 2 '13 at 12:49
up vote 3 down vote accepted

I believe this comes from the correspondence theorem: If you have two ideals $P$ and $Q$ of $R$ that contain $J$ such that under the canonical homomorphism $\pi : R \to R/J$ we have $\pi(P) = \pi(Q)$ then in fact $P = Q$.

In your case above we have $P = I$, $Q = \mathfrak{p}_1^{c_1}\ldots\mathfrak{p}_n^{c_n}$ and $J = \mathfrak{p}_1^{a_1}\ldots \mathfrak{p}_n^{a_n}$.

A lecturer I had once phrased this to me as follows: "The two ideals are equal upstairs iff they are equal downstairs."

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