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I apologize if my abstract algebra is so shaky that I might have glossed over an answer to this problem.

Yesterday I suddenly got the motivation to do some number theory again - as a challenge, I wanted to see if I could figure out when and how rational primes could be factored in the ring of integers $\mathcal{O}$ of an arbitrary number field, inspired for example by $$17 = (1-\alpha)(1+\alpha)$$ where $\alpha = 4i$. To figure this out on my own, I selected $\mathcal{O}_{\mathbb{Q}(\alpha)}$ with a number satisfying $\alpha^3-\alpha-2=0$ as my ring of choice. After some searching around I found the neat Dedekind-Kummer theorem, which settles the factorization problem of ideals into prime ideals:

$$\langle p\rangle=\bigl\langle p,g_1(\alpha)\bigr\rangle ^{e_1}\,\bigl\langle p,g_2(\alpha)\bigl\rangle^{e_2}\cdots\bigl\langle p,g_k(\alpha)\bigl\rangle^{e_k}.$$

where the polynomial factors and exponents are corresponding to the factorization of a given polynomial $\bmod p$. At this point I thought I was done and tried to factor the prime $p = 11$ in $\mathcal{O}_{\mathbb{Q}(\alpha)}$ and got

$$\langle 11 \rangle = \langle 11,\alpha+8 \rangle \langle 11,\alpha^2+3\alpha+8 \rangle$$

but after reading up on some basic ring theory again, I realised that this isn't quite what I wanted to achieve and I started questioning myself if a concrete answer to this is even feasible. Since "primes" is often taken synonymously with "prime ideals" in this context, it is really hard to filter the search results for what I want to read up on here. My one/two questions are as follows:

Is "When is a given rational prime decomposable in a fixed arbitrary number field?" answerable? If this is possible, how can one algorithmically find the splitting of such a prime? How about my example with $\alpha^3-\alpha-2=0$?

Thank you.

TheOutZ
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    See https://math.stackexchange.com/questions/989494/how-to-factorise-a-number-in-mathbb-z-sqrt-5/993868#993868 for a worked example. – Ricardo Buring Aug 30 '24 at 17:08
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    Well number rings in general are not UFDs... – user1090793 Aug 30 '24 at 17:08
  • @user1090793 Yes, I am aware of this, the question is whether there are any factorizations at all, being a UFD seems much more restrictive than this. – TheOutZ Aug 30 '24 at 17:11
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    In any Noetherian integral domain, it is possible to factor elements as a products of irreducible elements. Is this what you're looking for, or something else? In any case, this question of uniqueness of factorization is moot in your example, since the ring of integers of $\mathbb{Q}(\alpha)$ is a UFD: https://www.lmfdb.org/NumberField/3.1.104.1 . And in fact we have $11 = (\alpha^2 + \alpha - 1)(\alpha^2 + 3 \alpha - 3)$. – Viktor Vaughn Aug 30 '24 at 17:17
  • @ViktorVaughn If every ring of integers is a noetherian integral domain (as it seems to be), then this question can also be rephrased as "When is a rational prime in a given ring of integers (ir)reducible?". – TheOutZ Aug 30 '24 at 17:33

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The answer to your question is indeed not easy and is subject to still ongoing research, compare for example http://faculty.fairfield.edu/pbaginski/Papers/FactoringAlgIntegersMonthly.pdf The reason is, that primes don‘t stay prime in the Ring of integers in many different forms. A rational prime may split even into products of a different number of different irreducible elements. The cited article presents examples for this on the very first pages.