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 $3$?

  • 2
    $\begingroup$ 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 '18 at 14:41
  • $\begingroup$ @HwChu Oh I see, the ramification index refers to the primes in the extension, thanks $\endgroup$ – Edward Evans Jan 4 '18 at 14:42

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.

  • 3
    $\begingroup$ 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 '18 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.