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Kummer-Dedekind says the following:

Let $K = \mathbb{Q}(\theta)$ for some $\theta \in \mathcal{O}_K$ with minimal polynomial $f \in \mathbb{Z}[T]$. Let $p$ be a prime with $p \not \mid [\mathcal{O}_K : \mathbb{Z}[\theta]]$. If $f$ reduced mod $p$ factors as $\bar{f} = \prod_{i = 1}^k \bar{g_i}^{e_i}, e_i \geq 1$ where each $\bar{g}_i \in \mathbb{F}_p[T]$ are distinct, monic and irreducible, then $(p) = \prod_{i = 0}^k P_i^{e_i}$ where $P_i = (p, g_i(\theta)) \leq \mathcal{O}_K$ are distinct primes in $\mathcal{O}_K$, where $g_i \in \mathbb{Z}[T]$ are monic with reduction $\bar{g}_i$ mod $p$.

What happens when I need to factorise $(2)$? For example, I'm trying to compute the cardinality of the ideal class group of $\mathbb{Q}(\sqrt{-23})$ and it would be good to be able to factorise $(2)$, but $-23 \equiv 1 \mod 4$ so $[\mathcal{O}_K : \mathbb{Z}[\sqrt{-23}]] = 2$, so can't use Kummer-Dedekind.

Is there a fix to this?

Tom Misch
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  • For how to compute the class number of $\Bbb Q(\sqrt{-23})$ , see here. This post also has a factorization of $(2)$. – Dietrich Burde Apr 01 '22 at 14:13
  • Perhaps my algebra is really shaky, but it seems to me that Minkowski bound gives you what prime ideals you need to check by bounding their norm, so you still need to identify (in this case) which prime ideals have norm 2. In the thread you linked, the OP factorised $(2)$ into prime ideals as $(2, \omega)(2, \omega + 1)$ where $\omega = (1 + \sqrt{-23})/2$. How did they do this? – Tom Misch Apr 01 '22 at 14:18
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    They worked with the correct ring of integers. The field is generated by a root of $x^2 + x + 6 = 0$. –  Apr 02 '22 at 06:59
  • @franzlemmermeyer ah I think I see it! $\mathbb{Q}(\sqrt{d}) = \mathbb{Q}(\omega)$ so apply the theorem as stated but with $\theta = \omega$ instead! – Tom Misch Apr 02 '22 at 15:09
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    Keith Conrad has some good notes on applying Dedekind-Kummer here: https://kconrad.math.uconn.edu/blurbs/gradnumthy/dedekindf.pdf In particular, he discusses issues that arise when the element adjoined doesn't generate the full ring of integers; cf., Examples 2 and 5. – Viktor Vaughn Apr 11 '22 at 20:27

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