Let $K$ be a number field and let $\mathcal{O}_K$ be its ring of integers. Since $\mathcal{O}_K$ is a Dedekind domain, every ideal has a unique factorisation into a product of prime ideals. Let $(p)$ be a prime ideal in $\Bbb Z$ and let $\mathfrak{P}_1, \dots, \mathfrak{P}_g$ be the primes lying above $(p)$ in $\mathcal{O}_K$ so that

$$(p) = \mathfrak{P}_1^{e_1}\dots\mathfrak{P}_g^{e_g}.$$

Is it possible for $e_i$ to be greater than $1$ for more than one of the $e_i$ (say $(p) = \mathfrak{P}_1^2 \mathfrak{P}_2^3$ or something) and if so, how does one describe the decomposition of that ideal in $\mathcal{O}_K$? Would one say that it is both ramified and split? If so, what is the ramification index in this case? Is it $2$ or $3$?

  • 3
    $\begingroup$ Yes. The simplest case happens when $K = \mathbb Q[\sqrt a, \sqrt b]$ is a biquadratic field. If $(p)$ ramifies in $\mathbb Q[\sqrt a]$ and splits in $\mathbb Q[\sqrt b]$, then $(p) = \mathfrak P_1^2 \mathfrak P_2^2$ in $K$. At least to my knowledge there is no specific terminology for those case between totally ramified and totally split. We say the ramification index for both $\mathfrak P_1$ and $\mathfrak P_2$ are 2. $\endgroup$
    – Hw Chu
    Jan 4, 2018 at 14:41
  • $\begingroup$ @HwChu Oh I see, the ramification index refers to the primes in the extension, thanks $\endgroup$ Jan 4, 2018 at 14:42

1 Answer 1


Yes, it is possible that $e_i>1$ for more than one index $i$. However, the ramification indices $e_i$ and the residue class degrees $f_i$ have to satisfy the degree relation $$ \sum_{i=1}^g e_if_i=[K:\mathbb{Q}]=n. $$ For $n=4$ there are examples, mentioned in the comment.

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    $\begingroup$ And the interesting cases arise when the extension is not normal, so that behavior like $(p)=\mathfrak p_1\mathfrak p_2^2$ is perfectly possible. $\endgroup$
    – Lubin
    Jan 4, 2018 at 14:48

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