# Proving that a ring has only finitely many prime ideals

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?

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## 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 $i$, $p_i \subset \mathfrak{q}$?

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