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I am working on a problem that has a completely different point and I didn't work with algebraic number fields much before, so I was wondering if someone could point me in the right direction for this:

How can prime ideals of the ring of integers of an algebraic number field be characterised? Since the ring is Dedekind they are all maximal, but what are their generators?

user26857
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baltazar
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  • What are you thinking of? For example, it is known that every ideal in a Dedekind ring is generated by two elements, see page 8 here. – Fredrik Meyer Nov 22 '14 at 17:44
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    Can you clarify the question? Are you asking for a criterion to distinguish prime ideals from ideals that are not prime by looking at their generators? – Ben Blum-Smith Nov 22 '14 at 17:48
  • @BenBlum-Smith I am trying to study the spectrum of $\mathcal{O}_k[x]$, for an arbitrary alg. number field $k$. It would be helpful to know something about the primes of the ring of integers, as in $k[x]$ for k alg. closed you know that the non-zero primes are generated by irreducible polynomials etc. – baltazar Nov 23 '14 at 20:38
  • Okay I think I understand what you are asking for, see if the below answer helps. – Ben Blum-Smith Nov 25 '14 at 00:27

1 Answers1

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Let $K$ be an algebraic number field, and let $\mathcal{O}_K$ be its ring of integers. $K$ contains $\mathbb{Q}$ and $\mathcal{O}_K$ contains $\mathbb{Z}$; for any prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$, $\mathfrak{p}\cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$, which therefore has the form $(p)$ for a prime integer $p$. The prime $\mathfrak{p}$ of $\mathcal{O}_K$ is said to "lie over" $(p)$.

So, all the prime ideals of $\mathcal{O}_K$ are lying over the primes of $\mathbb{Z}$. The question is then how to find the primes that are lying over a given $(p)$ explicitly. For all but a finite number of primes $p$, this can be done as follows:

Find an $\alpha\in\mathcal{O}_K$ that generates $K$ over $\mathbb{Q}$ as a field. Then find a minimal polynomial $f$ for $\alpha$ over $\mathbb{Q}$. $f$ is a monic integer polynomial because of how $\alpha$ was chosen. Now consider $f$'s image mod $p$, and factor it into irreducible factors. If $g$ is any irreducible factor of $f$ mod $p$, then the ideal generated by $p$ and $g(\alpha)$ is a prime ideal of $\mathcal{O}_K$ lying over $p$. Thus the primes over $p$ correspond in a lovely way with irreducible factors of $f$ mod $p$.

This is guaranteed to work unless $p$ is one of the primes dividing the index of the ring $\mathbb{Z}[\alpha]$ as an additive subgroup of $\mathcal{O}_K$, of which there are only finitely many and in favorable cases there may be none (i.e. when you can find $\alpha$ such that $\mathcal{O}_K = \mathbb{Z}[\alpha]$).

Even in the cases where this algorithm doesn't work, if you have an explicit description of $\mathcal{O}_K$, you may still be able to find the primes lying over $p$ by finding the prime ideals in $\mathcal{O}_K/p\mathcal{O}_K$, which is a finite ring, and pulling them back to $\mathcal{O}_K$.

Ben Blum-Smith
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    Excellent. It is exactly the kind of thing I was asking for! I am just wondering, are all the primes of $\mathcal{O}_K$ given in this way, I mean by irreducible factors of $f$ mod $p$, for some $p$ excluding the ones that divide the index of $\mathbb{Z}[\alpha]$? – baltazar Nov 25 '14 at 22:35
  • I am asking because in my problem, I am supposed to describe the fibers of the map from the spectrum of $\mathcal{O}_K[x]$ to the spectrum of $\mathcal{O}_K$, for any alg. number field (so no explicit description). If there isn't a way to nicely describe all the primes of $\mathcal{O}_K$ maybe the problem is not set up correctly. – baltazar Nov 25 '14 at 22:42
  • @baltazar - yes they can all be described that way except for the primes dividing the index of $\mathbb{Z}[\alpha]$. But it seems to me that for your problem, you actually don't need much explicitness in your description of the primes of $\mathcal{O}_K$. I bet it is sufficient to know that $\mathcal{O}_K$ is Dedekind and its quotients by nonzero primes are finite fields. – Ben Blum-Smith Nov 26 '14 at 16:44
  • @baltazar - back to the theorem I was describing, though: in fact, it's even better than "all the primes over $p$ are obtainable this way": if $p\nmid[\mathcal{O}_K:\mathbb{Z}[\alpha]]$, then the factorization of $f$ into irreducible factors mod $p$ exactly gives you the factorization of the ideal $(p)\triangleleft\mathcal{O}_K$ into primes of $\mathcal{O}_K$. – Ben Blum-Smith Nov 26 '14 at 16:47
  • @baltazar - proofs for everything I've said would be in any standard book on algebraic number theory, but I have in mind the proof in Marcus' Number Fields, which I believe is in chapter 3. – Ben Blum-Smith Nov 26 '14 at 16:49
  • Excellent. Thank you very much. This is much more than I hoped to get from my vaguely posed question. Actually the second part of my problem is to order the primes of $\mathcal{O}_K[x]$ by inclusion, this is why I am looking for descriptions. – baltazar Nov 26 '14 at 18:39
  • I was wondering if you could take a look at the original problem I was trying to solve, since I haven't been able to complete it, except maybe in the case when the ring of integers is generated by a single element? I posted it here http://math.stackexchange.com/questions/1065676/describing-spec-mathcalo-kx. Any suggestions would be much appreciated. – baltazar Dec 13 '14 at 16:00