I'm stuck on the proof of Theorem 4.14 in Washington - Introduction to Cyclotomic Fields.

We take a prime power $n = p^m$ and define $\pi = \zeta_n - 1$, $\zeta_n$ a primitive $n$-th root of unity. Let $\mathbb{Q}(\zeta_n)^+ = \mathbb{Q}(\zeta_n + \zeta_n^{-1})$ be the maximal real subfield. The proof claims that the $\pi$-adic valuation of $\mathbb{Q}(\zeta_n)$ only takes on even values on $\mathbb{Q}(\zeta_n)^+$. I'm not sure how to show this.

Thanks for any help.


The key is to notice $\mathcal{O}_{\mathbb{Q}(\zeta_n)}$ is totally ramified at $p$ and there is a unique prime of $\mathcal{O}_{\mathbb{Q}(\zeta_n)}$ which lies above $p,$ namely $\pi.$ It follows that the same holds for any intermediate extension $L$ satisfying $\mathbb{Q} \subset L \subset \mathbb{Q}(\zeta_n).$ So if $\mathfrak{p} _L$ is the unique prime of $\mathcal{O}_L$ above $p,$

$$e(\mathfrak{p}_L | p) = [L:\mathbb{Q}].$$

And thus,

$$e((\pi) | \mathfrak{p}_L) = [\mathbb{Q}(\zeta_n):L].$$

It follows

$$v_{\pi}(\mathbb{Q}(\zeta_n)^+) = v_{\pi}((\pi) \cap \mathcal{O}_{\mathbb{Q}(\zeta_n)^+})\mathbb{Z} = v_{\pi}(\mathfrak{p}_{\mathbb{Q}(\zeta_n)^+})\mathbb{Z} = e(\mathfrak{p}_{\mathbb{Q}(\zeta_n)} | \mathfrak{p}_{\mathbb{Q}(\zeta_n)^+}) \mathbb{Z} = 2\mathbb{Z}.$$


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.