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Let $K$ be a number field, let $\mathscr{O}_K$ denote the ring of algebraic integers of $K$ and let $d_K$ be the discriminant of $\mathscr{O}_K$, i.e. the discriminant of a $\mathbb{Z}$-basis for $\mathscr{O}_K$. I know that if a rational prime $p$ ramifies in $\mathscr{O}_K$, then $p$ divides $d_K$. Is it true also the converse? If a prime $p$ divides $d_K$, then $p$ ramifies in $\mathscr{O}_K$?

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Yes, this is true.

More generally, if you define the discriminant ideal $\Delta_{L/K}$ of an extension of number fields $L/K$ to be the ideal of $\mathcal{O}_K$ generated by elements of the form $\text{disc}(\omega_1,\ldots,\omega_n)$ where $\{\omega_1,\ldots,\omega_n\}$ is a basis of $L/K$ contained in $\mathcal{O}_L$, then a prime $\mathfrak{p}$ of $\mathcal{O}_K$ ramifies in $L$ if and only if $\mathfrak{p}\mid \Delta_{L/K}$.

In the special case when $K=\mathbb{Q}$, you have that $\Delta_{L/\mathbb{Q}}=d_L\mathbb{Z}$, and so your result follows.

You can look up a proof of the above, but I think the most informative case is when $L/\mathbb{Q}$ is monogeneic. In other words, $\mathcal{O}_L=\mathbb{Z}[\alpha]$. In this case, the Dedekind-Kummer theorem tells you that a prime $p$ ramifying in $L$ is the same thing as $\overline{m_{\alpha,\mathbb{Q}}(T)}$ having multiple roots in $\mathbb{F}_p[T]$ (where $m_{\alpha,\mathbb{Q}}\in\mathbb{Z}[T]$ is the minimal polynomial for $\alpha$ and the bar denotes passing to $\mathbb{F}_p[T]$). But, $\overline{m_{\alpha,\mathbb{Q}}(T)}$ having multiple roots means that the polynomial discriminant $\text{disc}(\overline{m_{\alpha,\mathbb{Q}}(T)})\in\mathbb{F}_p$ is zero. But, it's not hard to check that $\text{disc}(\overline{m_{\alpha,\mathbb{Q}}(T)}=\overline{\text{disc}(m_{\alpha,\mathbb{Q}}(T)})$. Thus, $p$ should ramify if and only if $\text{disc}(m_{\alpha,\mathbb{Q}}(T))$ is in $(p)$ ,or that $(p)\mid (\text{disc}(m_{\alpha,\mathbb{Q}}(T)))$. But, a quick check shows that $d_L=\pm \text{disc}(m_{\alpha,\mathbb{Q}}(T))$, and so we see that $p$ ramifies if and only if $(p)\mid (d_L)$.

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  • $\begingroup$ I just realized that my calling of the theorem about factoring in monogeneic extensions might the "Dedekind-Kummer theorem" may be non-standard. It seems most places just call it Dedekind's theorem. I'm not sure why I've alway called it Dedekind-Kummer. I guess Kummer secretly did most things--secretly. I hope you understood though! $\endgroup$ – Alex Youcis Sep 18 '13 at 7:27
  • $\begingroup$ My teacher calls it Kummer thm, without any mention to Dedekind. $\endgroup$ – bateman Sep 18 '13 at 7:31
  • $\begingroup$ @bateman Well, then, perhaps Dedekind-Kummer is a happy medium :) $\endgroup$ – Alex Youcis Sep 18 '13 at 7:32
  • $\begingroup$ @AlexYoucis This also helps me a lot! But can you tell me how to prove $\text{disc}(\overline{m_{\alpha,\mathbb{Q}}(T)}=\overline{\text{disc}(m_{\alpha,\mathbb{Q}}(T)})$? I have no idea on that. $\endgroup$ – user112564 Jun 5 '14 at 8:23

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