The intersection of an infinite number of prime ideals in a ring of integers Let $\mathcal{O}$ be the ring of integers of a number field, $\{\mathfrak{p}_i,\,i \in \mathbb{N}\}$ a sequence of two-by-two pairwise distinct prime ideals. Does it follow that$$\bigcap_i \mathfrak{p}_i = \{0\}?$$
 A: This is true. If $I$ denotes your ideal then $I \subseteq \mathfrak p_i$ for each $i$. Suppose that $I$ is not the zero ideal. If $I= \mathfrak q_1 \dots \mathfrak q_j$ is the decomposition of $I$ as a product of prime ideals, then $ \mathfrak q_1 \dots \mathfrak q_j \subseteq \mathfrak p_i$ implies that $\mathfrak q_{r_i} \subseteq \mathfrak p_i$ for some $1 \leq r_i \leq j$, because $\mathfrak p_i$ is a prime ideal. Since $\mathcal O$ has dimension $1$, we actually have $\mathfrak q_{r_i} = \mathfrak p_i$. But since there are infinitely many $i$'s and finitely many possible choices for $r_i$, there are two distinct $i, i'$ with $r_i = r_i'$; but then $\mathfrak p_i = \mathfrak p_{i'}$ which contradicts the assumption made on the $\mathfrak p$'s. 
A: Let $A$ be any ideal in $\mathcal{O}$. For $\alpha$ in $A$, $\alpha$ satisfies a monic polynomial of smallest degree, say $\alpha^n+b_1\alpha^{n-1}+\cdots+b_n=0$ with $b_n\neq 0$. Then $b_n=\alpha(-\alpha^{n-1}-b_1\alpha^{n-2}-\cdots-b_{n-1})$, which implies that the ideal containing $\alpha$ contains some non-zero rational integer $b_n$. 
In particular, each prime ideal $\mathfrak{p}_i$ contains a prime integer in $\mathbb{Z}$. 
If $\cap_i\mathfrak{p}_i$ is non-zero ideal, then it will contain an integer $n$ in $\mathbb{Z}$, and hence $n$ will be divisible by infinitely many prime integers  (chosen from $\mathfrak{p}_i\cap\mathbb{Z}$), 
 in $\mathbb{Z}$, a contradiction.

(Modified answer after the comment from Jyrki.)
A: The intersection of ideals is an ideal. Since yours is divisible by infinitely many primes, it must be that it has no presentation as a product of primes. But in a Dedekind domain, every non-zero ideal has a unique expression as the finite product of prime ideals (to some power). Hence it must be that your ideal is the only ideal not covered by this theorem:  the zero ideal.
