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Given an algebraic number field $K/\mathbb Q$, let $R = \mathbb Z_K$ be the ring the algebraic integers of that fields. Is it possible to say how many prime ideals there are in $R$? I suspect we always have infinitely many, as this is certainly the case for $K=\mathbb Q$ (because then $R = \mathbb Z$).

I was already able to show that an Ideal $I\subseteq \mathbb R$ ($I\neq0$) is prime if and only if it is maximal. One idea would have been showing that you can generate infinitely many fields via $R/I$ but I was never able to show what $R/I$ would be isomorphic to, as I do not really know what the prime ideals $I$ look like.

So have no idea how to go from there, and I'd be happy for every hint!

flawr
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  • Try to show that for any distinct $p,q\in \mathbb{N}$ prime numbers, $p$ and $q$ cannot belong to the same prime ideal of $\mathbb{Z}_K$. – Captain Lama Mar 09 '16 at 18:13
  • I just had this very idea, thank you very much for that hint. If $p,q$ are both in that prime ideal. Because they are coprime, we know that also $1$ will be in that ideal.and therefore so is $\mathbb Z$, but now I am just stuck at finding a contradiction, any further hint? Am I on the right track? – flawr Mar 09 '16 at 19:59
  • Actually in any ring, an ideal that contains $1$ must be the whole ring. – Captain Lama Mar 09 '16 at 22:50
  • Do'h, I am not sure why I didn't see this. It was probably way too late=) Now I could complete the proof, thank you very much! – flawr Mar 10 '16 at 19:18

2 Answers2

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There are infinitely many primes in $\mathbb{Z}_K$. You can prove it by showing that for every prime $p$ of $\mathbb{Z}$, there exists a prime ideal $P$ of $\mathbb{Z}_K$ such that $P \cap \mathbb{Z} = p\mathbb{Z}$. This is called the lying over theorem. There are infinitely many prime ideals of $\mathbb{Z}$, hence infinitely many prime ideals of $\mathbb{Z}_K$.

D_S
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In the specific case of rings of integers, you can mimic Euclid's proof over $\mathbb Q$. The key point is that every ideal $\mathfrak a \subset \mathcal O_K$ factors (uniquely) as a product of prime ideals.

Suppose that we have a finite set of non-zero prime ideals $\mathfrak p_1,\ldots,\mathfrak p_n\subset\mathcal O_K$. Then $\mathfrak p_1\cdots\mathfrak p_n\cap\mathbb Z$ is a non-trivial ideal of $\mathbb Z$, so is equal to $(n)$ for some $n\in\mathbb Z\setminus\{0,1\}$. In particular, $n\in \mathfrak p_i$ for all $i$

Now consider the ideal $(n+1)\mathcal O_K$. If $\mathfrak p_i\mid (n+1)\mathcal O_K$ for any $i$, then $n, n+1\in \mathfrak p_i$ and hence $1\in \mathfrak p_i$, a contradiction.

In particular, $(n+1)\mathcal O_K$ is divisible by a prime ideal not in $\{\mathfrak p_1,\ldots,\mathfrak p_n\}$ and the result follows.

For a different proof mimicking Euler's proof for $\mathbb Q$, see this question.

Mathmo123
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