Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a Dedeking 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 $a_i$?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
2  
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. –  Federica Maggioni Apr 10 '13 at 17:50
    
i can't still see why $\mathfrak{p}_j\subseteq\mathfrak{p}_i$ for some $j\neq i$. –  Federica Maggioni Apr 10 '13 at 21:59
1  
@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. –  user26857 Apr 10 '13 at 23:10
    
@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 $R$. –  user26857 Apr 10 '13 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.