Let $R$ be a Dedekind domain, let $\mathfrak{i}$ be a non-zero ideal of $R$. By factorization theorem we can write $$\mathfrak{i}=\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}$$ for distinct non-zero primes $\mathfrak{p_i}$ and positive integers $a_i$.

Now consider the localization $R_{\mathfrak{p_i}}$ of $R$ at the prime $\mathfrak{p_i}$. Consider the extended ideal $\mathfrak{i}R_{\mathfrak{p}_i}$.

I want to prove that $$\mathfrak{i}R_{\mathfrak{p}_i}=\mathfrak{p}_i^{a_i}R_{\mathfrak{p}_i}.$$

I'm convinced I actually have a proof of this: every non-zero prime ideal in a Dedekind is maximal, hence two distinct non-zero primes are coprime, and the same holds true for powers of distinct non-zero-prime. From this I can prove that $\mathfrak{p_j^{a_j}}$ is not contained in $\mathfrak{p}_i$ for every $j\neq i$, otherwise $\mathfrak{p_j^{a_j}}+\mathfrak{p}_i=\mathfrak{p}_i\neq R$, contradicting coprimality, hence $\mathfrak{p}_j^{a_j}$ contains invertible element of the localization $R_{\mathfrak{p}_i}$ hence the result.

The fact is that I'm looking for a different proof. I want to use the fact that in a Dedekind domain every localization outside a prime is a DVR and that every non-zero ideal of a DVR is a power of the unique maximal ideal, so in this case I have $$\mathfrak{i}R_{\mathfrak{p}_i}=\mathfrak{p_i}^kR_{\mathfrak{p}_i}.$$

Why $k$ should actually be equal to $a_i$?


2 Answers 2


Let $R$ be a commutative ring and $\mathfrak p_1,\dots,\mathfrak p_n$ prime ideals such that $\mathfrak p_j\not\subseteq\mathfrak p_i$ for all $i\neq j$. Then $$\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}R_{\mathfrak{p}_i}=\mathfrak{p}_i^{a_i}R_{\mathfrak{p}_i}.$$

$\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}R_{\mathfrak p_i}=(\mathfrak{p}_i^{a_i}R_{\mathfrak p_i})(\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_{{i-1}}^{a_{i-1}}\mathfrak{p}_{i+1}^{a_{i+1}}\cdots\mathfrak{p}_n^{a_n}R_{\mathfrak p_i})$. One knows that $\mathfrak aR_{\mathfrak p}\neq R_{\mathfrak p}$ iff $\mathfrak a\subseteq\mathfrak p.$ If $\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_{{i-1}}^{a_{i-1}}\mathfrak{p}_{i+1}^{a_{i+1}}\cdots\mathfrak{p}_n^{a_n}\subseteq \mathfrak p_i$, then there exists $j\neq i$ such that $\mathfrak p_j\subseteq\mathfrak p_i$, a contradiction. This shows that $\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_{{i-1}}^{a_{i-1}}\mathfrak{p}_{i+1}^{a_{i+1}}\cdots\mathfrak{p}_n^{a_n}R_{\mathfrak p_i}=R_{\mathfrak p_i}$ and we are done.

  • 2
    $\begingroup$ why do $\mathfrak{p}_j\subseteq\mathfrak{p}_i$ is a contradiction? Is it because they are supposed to be distinct? I can't understand this, i mean, if $\mathfrak{p}_i,\mathfrak{p}_j$ were maximal,(and in the Dedekind case they are) certainly inclusion implies equality, but you said this holds in every commutative ring. $\endgroup$ Apr 10, 2013 at 17:50
  • $\begingroup$ i can't still see why $\mathfrak{p}_j\subseteq\mathfrak{p}_i$ for some $j\neq i$. $\endgroup$ Apr 10, 2013 at 21:59
  • 1
    $\begingroup$ @FedericaMaggioni Use the definition of a prime ideal: if a prime ideal contains a product of ideals, then it contains (at least) one of them. $\endgroup$
    – user26857
    Apr 10, 2013 at 23:10
  • $\begingroup$ @BenjaLim No need to use Nakayama Lemma. Use the following fact: $S^{-1}I=S^{-1}R$ iff $I\cap S\neq\emptyset$, where $S$ is a multiplicative system in $R$. $\endgroup$
    – user26857
    Apr 10, 2013 at 23:44

Firstly it is clear that $\mathfrak{i}R_{\mathfrak{p}_i} \subseteq \mathfrak{p_i}^{a_i}R_{\mathfrak{p}_i}$. To show that this is actually an equality it will suffice to show that the quotient

$$\mathfrak{p}_i^{a_i}R_{\mathfrak{p}_i}/\mathfrak{i}R_{\mathfrak{p}_i} \cong (\mathfrak{p}_i^{a_i}/\mathfrak{i})_{\mathfrak{p}_i}$$

is actually zero. To see this apply the Chinese Remainder Theorem for modules to the R.H.S. to get

$$\mathfrak{p}_i^{a_i}/\mathfrak{i} \cong \mathfrak{p}_i^{a_i}/\mathfrak{p}_1^{a_1} \times \ldots \times \mathfrak{p}_i^{a_i}/\mathfrak{p}_n^{a_n}.$$

Now when you localize the right hand side at $\overline{\mathfrak{p}_i} = \mathfrak{p}_i/\mathfrak{i}$ we claim that everything gets killed. Firstly we know $\mathfrak{p}_i^{a_i}/\mathfrak{p}_i^{a_i} = 0$ gets killed. Now when $j \neq i$ we know that there is an element $x \in \mathfrak{p}_j^{a_j}$ that is not in $\mathfrak{p}_i$ for $\mathfrak{p}_j^{a_j} \subsetneqq \mathfrak{p}_i$. This means to say that the complement

$$R/\mathfrak{i} - \overline{\mathfrak{p}_i}$$

contains zero so that the localization $(\mathfrak{p}_i^{a_i}/\mathfrak{p}_j^{a_j})_{\overline{\mathfrak{p}_i}}$ is actually zero for all $i \neq j$. Thus we have shown that

$$(\mathfrak{p}_i^{a_i}/\mathfrak{i} )_{\overline{\mathfrak{p}_i}} = 0$$ and consequently your result follows.


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