# Does every ideal of the ring of all algebraic integers contain a finite product of prime ideals?

I am studying the proof of unique factorisation of ideals into prime ideals in the ring of integers $\mathcal{O}$ of a number field $K$. The first step is to show that given any proper ideal of $I$ of $\mathcal{O}$ there exist non-zero prime ideals $P_1,\dots,P_r$ such that $I \supseteq P_1\cdots P_r$. The proof uses the fact that $\mathcal{O}$ is noetherian.

The set of algebraic integers $\mathbb{B}$ is not noetherian, so the same proof does not fall through for $\mathbb{B}$.

However, can we still show that every proper ideal of $\mathbb{B}$ contains a product of finitely many non-zero prime ideals? If not, can we construct a specific example of a proper ideal $I \subset \mathbb{B}$ such that it does not contain $P_1 \cdots P_r$ for any $r$ non-zero prime ideals?

• I think all prime ideals are maximal, corresponding to homomorphisms $\mathbb{B} \to \overline{\mathbb{F}}_p$? I imagine no finitely generated (and thus principal) ideal can contain a product of finitely many primes.
– user14972
Aug 9, 2016 at 12:44
• I'm not sure how the homomorphisms $\mathbb{B} \to \overline{\mathbb{F}}_p$ come into the picture, but this question - math.stackexchange.com/questions/1451676/… - says that all prime ideals are maximal in $\mathbb{B}$.
– user279515
Aug 9, 2016 at 14:33
• @Hurkyl What is the intuition for guessing that a finitely generated ideal does not contain a product of finitely many primes? Also, thank you for your answer, it was very clear.
– user279515
Aug 9, 2016 at 14:35
• If $I$ is a maximal ideal, then it contains some integer prime $p$, and $\mathbb{B} / I \cong \overline{\mathbb{F}}_p$. My intuition was that any non-unit algebraic integer has to vanish in infinitely many of these, (because the behavior of completely unrelated algebraic integers would ensure there are many different prime ideals). Like how a prime ideal in one field can split into many prime ideals in its extension fields.
– user14972
Aug 9, 2016 at 14:43

Consider any finitely generated (and thus principal) ideal $(\alpha)$.

Suppose we had $P_1 P_2 \cdots P_r \subseteq (\alpha)$. Pick any nonzero elements $\beta_i \in P_i$, and let $\beta = \beta_1 \beta_2 \cdots \beta_r$.

Since $P_i$ is prime, $\beta_i^{1/n} \in P_i$, and thus $\beta^{1/n} \in (\alpha)$.

Now, consider the fields $K_n = \mathbb{Q}(\alpha, \beta^{1/n})$. We have

$$N_{K_n / \mathbb{Q}}(\alpha) = N_{K_1 / \mathbb{Q}}(\alpha)^{[K_n : K_1]}$$ $$N_{K_n / \mathbb{Q}}(\beta) = N_{K_1 / \mathbb{Q}}(\beta)^{[K_n : K_1]}$$ $$N_{K_n / \mathbb{Q}}(\beta^{1/n}) = N_{K_1 / \mathbb{Q}}(\beta)^{[K_n : K_1] / n}$$

And, in particular,

$$N_{K_n / \mathbb{Q}}\left( \frac{\beta^{1/n}}{\alpha} \right) = \left( \frac{N_{K_1 / \mathbb{Q}}(\beta)^{1/n}}{N_{K_1 / \mathbb{Q}}(\alpha)} \right)^{[K_n : K_1]}$$

Thus, for some sufficiently large $n$, the norm of $\beta^{1/n} / \alpha$ has magnitude less than one, and so $\beta^{1/n} / \alpha$ cannot be an algebraic integer, contradicting the fact that $\beta^{1/n} \in (\alpha)$.