Given a prime $p$, the $p$-adic valuation on the field $\mathbb{Q}$ is the map $\nu:\mathbb{Q}^*\to\mathbb{Z}$ given by $\nu(p^ka/b)=k$, where $a,b$ are prime to $p$.
I want to consider extensions of this valuation to number fields; in particular I want to extend $\nu$ to a valuation on $\mathbb{Q}(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.
Let $\mathcal{O}$ denote the ring $\mathbb{Z}[\zeta_p]$ of integers in $\mathbb{Q}(\zeta_p)$.
- Is it true that every prime ideal $\mathfrak{p}$ dividing $p\mathcal{O}$ determines an extension of the valuation?
- If so, is this a special case of a general principle?
- How is this new extended valuation defined?
- Why is it a valuation?
- Is there only such prime ideal $\mathfrak{p}$, so we have a unique extended valuation to choose?
Thanks in advance!
Best Answer
$\newcommand{\p}{\mathfrak{p}} \newcommand{\O}{\mathcal{O}}$ Let $K$ be any number field. Any prime ideal $\p$ of $\O_K$ determines a discrete valuation $v_\p$ on $K$: for $x\in \O_K$, define $v_\p(x) = n$, where $n$ is the highest integer such that $x\in \p^n$, where $\p^0=\O_K$. For arbitrary $\alpha\in K$, write $\alpha=\frac xy$ for $x,y\in \O_K$, define $v_\p(\alpha)=v_\p(x)-v_\p(y)$. You can normalise this differently: pick any $k\in \mathbb{N}$, define $v_\p(x) = n/k$, where $n$ is as above.
If you set $k=1$, then the valuation is normalised in such a way as to have image $\mathbb{Z}$. But in general, its restriction to $\mathbb{Q}$ will not have image $\mathbb{Z}$, so such a normalisation will not extend a normalised valuation from $\mathbb{Q}$. If you want $v_\p$ to extend $v_p$, where $p$ is the rational prime lying below $\p$, then you need to take $k=e_\p$, the ramification index of $\p/(p)$. In this case, of course, the image of $v_\p$ is not $\mathbb{Z}$, but $\frac 1e\mathbb{Z}$.
In the example you give, $p$ is totally ramified, so there is exactly one prime in $\O_K$ lying above $p$, generated by $\zeta_p-1$ (to convince yourself of that, compute the norm of this elements as the constant coefficient in its minimal polynomial), and the extension of $v_p$ to $K$ is unique. In general, there are as many non-equivalent extensions of $v_p$ to $K$, as there are prime ideals of $\O_K$ lying above $p$.