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!

  • 1
    $\begingroup$ If $\mathfrak{p}$ is a prime ideal in (the ring of integers of) a number field, then the $\mathfrak{p}$-adic valuation of a non-zero element $x$ is simply the exponent on $\mathfrak{p}$ in the prime factorization of the ideal $x\mathcal{O}$. (and, of course, you can get equivalent valuations by multiplying by a constant) Can you work out everything you need from there? $\endgroup$ – user14972 Nov 22 '11 at 0:01
  • $\begingroup$ Not quite, sorry. Could you be a little more explicit? $\endgroup$ – Clinton Boys Nov 22 '11 at 0:05
  • $\begingroup$ Take $x \in \mathbb{Q}(\zeta_p)$. Since $\mathcal{O}$ is a Dedekind domain, the fractional ideal $x\mathcal{O}$ factors uniquely as a product of prime ideals (some might appear with negative powers). If $\mathfrak{p}$ is such a prime ideal, we denote $v_{\mathfrak{p}}(x)$ the exponent of $\mathfrak{p}$ in the decomposition. From the definition, it's easy to see it's a valuation. Moreover, if $\mathfrak{p}$ divides $p$, then $(v_{\mathfrak{p}})_{| \mathbb{Q}}$ is just a multiple of $v_p$ (the coefficient is $v_{\mathfrak{p}}(p)$), so we can rescale the valuation so that it extends $v_p$. $\endgroup$ – Joel Cohen Nov 22 '11 at 0:13
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    $\begingroup$ Try to understand first the case of valuations on the quadratic field ${\mathbf Q}(i)$ and how they are related to valuations on ${\mathbf Q}$ before looking at valuations on ${\mathbf Q}(\zeta_p)$. That is, learn well a simple case before trying to figure out a more complicated case. $\endgroup$ – KCd Nov 22 '11 at 1:12

$\newcommand{\p}{\mathfrak{p}} \newcommand{\O}{\mathcal{O}}$ 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 $p$.

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  • $\begingroup$ You say ''any prime ideal of $\mathcal{O}_K$ determines a discrete valuation on $K$'', but then you say we need to choose a prime ideal in $\mathcal{O}_K$ lying above $p$ to get a valuation? $\endgroup$ – Clinton Boys Nov 22 '11 at 0:33
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    $\begingroup$ @ClintonBoys : Any prime ideal determines a valuation, but if you want your valuation to extend the $p$-adic valuation (for $p$ fixed), you have to chose a prime ideal above $p$. $\endgroup$ – Joel Cohen Nov 22 '11 at 1:03
  • $\begingroup$ Just to clarify, is this what you are saying? 1. Any prime ideal of $\mathcal{O}_K$ determines a valuation on $K$. 2. If the chosen prime ideal lies above $p\mathcal{O}_K$, it extends the $p$-adic valuation. 3. In the particular $K$ I have chosen (i.e. the cyclotomic number field $\mathbb{Q}(\zeta_p)$, there is a unique such ideal (what is it?) and so a unique such extension. $\endgroup$ – Clinton Boys Nov 22 '11 at 3:07
  • $\begingroup$ 1. Any prime ideal of $\mathcal{O}_K$ determines a valuation on $K$, up to a scalar. 2. If the chosen prime ideal lies above $(p)$, it extends the $p$-adic valuation, if you choose the right scalar in 1. (More details is in Alex B.'s answer) 3. In the particular $K$ you have chosen, $p$ is totally ramified, i.e. only one prime ideal lying above $p$. (Do you know what the factorization of $p$ in $\mathbb{Z}[\zeta_p]$ is?) So there is a unique such ideal, and unique such extension. $\endgroup$ – Soarer Nov 22 '11 at 4:11
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    $\begingroup$ Hint: $x^{p-1} + ... + x + 1 \equiv (x - \zeta_p)...(x - \zeta_p^{p-1})$ now let x=1 and take ideals. It can be shown that all of the ideals on the right have to be the same as $\langle 1-\zeta_p\rangle$ and that this ideal is prime. $\endgroup$ – fretty Mar 24 '12 at 13:36

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