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 $D$ be a Dedeking domain, $\mathfrak{i}$ a nonzero ideal of $D$ and let $B=D/\mathfrak{i}$ be the quotient ring. Then $B$ is a noetherian ring, and every prime ideal of $B$ is maximal. I have proved the following facts:

1) In $B$ every ideal contains a product of prime ideals;

2) Applying 1) to the zero ideal, we get that there exist distinct prime ideals $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ of $B$ such that $$\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}=(0)$$ for suitable $a_j\geq 1$;

3) Applying Chinese remainder theorem for rings, we get an isomorphism of rings $$B\cong \frac{B}{\mathfrak{p}_1^{a_n}}\oplus\ldots\oplus\frac{B}{\mathfrak{p}_n^{a_n}}$$

What I can't prove is the following:

4) ideals $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$ in 2) are all the primes in $B$.

What I did is this: let $\mathfrak{q}$ be a prime ideal of $B$. I want to show that $\mathfrak{q}$ is actually one of the $\mathfrak{p}_i$'s. If we are in the case: $\mathfrak{q}\supseteq\mathfrak{p}_i^{a_i}$ for some $i$, then i've proved that $\mathfrak{q}=\mathfrak{p}_i$, for some $i$. What I can't prove is the other case: suppose that $\mathfrak{q}$ does not contain any of the $\mathfrak{p}_i$'s. Then $\mathfrak{q}\neq \mathfrak{p}_i$ for every $i$. Then $\mathfrak{q}$ and $\mathfrak{p}_i$ are distinct maximal, hence coprime. Then by CRT $$\frac{B}{\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}\cdot\mathfrak{q}}\cong \frac{B}{\mathfrak{p}_1^{a_n}}\oplus\ldots\oplus\frac{B}{\mathfrak{p}_n^{a_n}}\oplus\frac{B}{\mathfrak{q}}$$ But $\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}=(0)$, thus $\mathfrak{p}_1^{a_1}\cdots\mathfrak{p}_n^{a_n}\cdot\mathfrak{q}=(0)$ so that $$B\cong\frac{B}{(0)}\cong \frac{B}{\mathfrak{p}_1^{a_n}}\oplus\ldots\oplus\frac{B}{\mathfrak{p}_n^{a_n}}\oplus\frac{B}{\mathfrak{q}}$$ Putting together i find $$\frac{B}{\mathfrak{p}_1^{a_n}}\oplus\ldots\oplus\frac{B}{\mathfrak{p}_n^{a_n}}\cong \frac{B}{\mathfrak{p}_1^{a_n}}\oplus\ldots\oplus\frac{B}{\mathfrak{p}_n^{a_n}}\oplus\frac{B}{\mathfrak{q}}$$

I want to deduce from this last equation that $\frac{B}{\mathfrak{q}}=(0)$ so that $B=\mathfrak{q}$, a contraddiction, but i don't know how to deduce this. Could someone help me please?

share|improve this question
add comment

1 Answer

Well ${p_1}^{a_1}\cdot ... {p_n}^{a_n}=(0)$ and so is contained in $\mathfrak{q}$. $\mathfrak{q}$ is prime, and so doesn't it immediately follow that for some $i$, $p_i \subset \mathfrak{q}$?

share|improve this answer
add comment

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.