In cyclotomic fields, how does one define the generalised notion of absolute value?

For context, in $\mathbb{C}$ the relationship between the field norm and the absolute value is simply $|z| = \sqrt{\prod_{i=1}^2\sigma_i(z)}=\sqrt{z \sigma(z)} = \sqrt{z\bar{z}}$ where $\sigma$ is the only non-trivial Galois map of the extension $\mathbb{C}/\mathbb{R}$ corresponding to complex conjugation.

From this, I would hazard a guess that in the $n^{th}$-cyclotomic field (as there maybe more that one Galois map corresponding to a complex conjugation), we could define the absolute value to be $|\alpha| = \sqrt[k]{\prod_{i=1}^k \sigma_i (\alpha)}$ where $k = |\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})|$. Could someone clarify this, or offer an accepted definition.

  • $\begingroup$ An absolute value extending the usual one on $\mathbb{Q}$ will be $|\sigma(\alpha)|$ where $\sigma$ is an embedding $\mathbb{Q}(\zeta) \to \mathbb{C}$ (ie. since it is a Galois extension $\sigma \in Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$) and $|.|$ the absolute value on $\mathbb{C}$. Now what you wrote is related to the field norm $N_{\mathbb{Q}(\zeta)/\mathbb{Q}}(\alpha)=\prod_{\sigma \in Gal(\mathbb{Q}(\zeta)/\mathbb{Q})} \sigma(\alpha)$ an homomorphism $\mathbb{Q}(\zeta)^* \to \mathbb{Q}^*$ $\endgroup$ – reuns Jul 30 '17 at 23:14

TL;DR: As such there is not just one single absolute value on a number field $K$; there are many of them. In the case of $K=\mathbb Q(\zeta_n)$, it turns out that there are $\frac{\varphi(n)}{2}$ good candidates, where $\varphi$ is Euler's totient function. Read ahead for details.

Let $K$ be a number field of degree $n$. We denote by $r_1$ the number of real embeddings of $K$, i.e. non-zero homomorphisms $\sigma_i:K\to \mathbb R$ and by $2r_2$, the number of complex embeddings, $\tau_j:K\to \mathbb C$ such that the image is not entirely real. There are always an even number of complex embeddings because we can compose such a map with complex conjugation to get a different one.

Then $n=r_1+2r_2$.

So with each embedding, you can use the usual notion of absolute value you are used to (in either $\mathbb R$ or $\mathbb C$ accordingly) to get $r_1+r_2$ different absolute values on $K$.

Why only $r_1+r_2$ absolute values even though there are $r_1+2r_2$ embeddings? Because the pairs of complex places that only differ by complex-conjugation give the same absolute value because any complex number $a+bi$ and $a-bi$ have the same absolute value $a^2+b^2$.

So to answer the question in your special case, notice that to embed our abstract $K=\mathbb Q(\zeta_n)$ into $\mathbb C$, there are $\phi(n)$ primitive $n$-th roots of unity in $\mathbb C$ that our abstract $\zeta_n$ can map to and these all define and determine an embedding. So there are $\frac{varphi(n)}{2}$ such choices.

These $r_1+r_2$ absolute values are sometimes called the infinite places of $K$ or the infinite primes of $K$.

This terminology is used because, for every prime $\mathfrak p$ of $O_K$ the ring of integers of $K$, you can define an absolute value first for $\alpha \in O_K$. This is how it goes. First define the $\mathfrak p$-adic valuation of $\alpha$ by $$v_{\mathfrak p}(\alpha )= \sup_{n}\{n|\alpha\in \mathfrak p^n\}$$ and use this to define an absolute value on $O_K$ by $$|\alpha|_{\mathfrak p}=2^{-v_{\mathfrak p}(\alpha)} .$$

Then you can extend this to $\frac{\alpha}{\beta}\in K$ with $\alpha, \beta \in O_K$ by $v_{\mathfrak p}(\frac{\alpha}{\beta})=v_{\mathfrak p}(\alpha)-v_{\mathfrak p}(\beta)$. The $2$ used in the definition doesn't really matter. Any real number $a>1$ can be used in it's stead will yield equivalent absolute values (in the sense that they yield the same topology on $K$).

But now I'm getting ahead of myself and if I don't stop this will turn into a course in Algebraic Number Theory.

  • $\begingroup$ Okay so just to clarify by way of example: for $\alpha \in \mathbb{Q}(\zeta_n)$, such that $\alpha = a_0 + a_1 \zeta_n +..+a_k \zeta^k$, where $k = \varphi(n)$, (I take you to mean) we can define $|\sigma_i (\alpha)| = \sqrt{\sigma_i (\alpha) \sigma_j (\bar \alpha)|} = \sqrt{\sigma_i(a_0 + a_1 \zeta_n +..+a_k \zeta^k)\sigma_i(a_0 - a_1 \zeta_n -..-a_k \zeta^k)}$ for each $\sigma_i \in \text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$? $\endgroup$ – user3343452 Jul 31 '17 at 10:22

I would like to state at first that I am not major in number theory. If there is any mistake in this answer, please straightly edit this or devote me if this answer is incurable.

This answer is just a mixture by two materials by some people, where the first part is an addendum for Mr. Ravi's answer, the second part is an example, and we will only talk about $\mathbb{Q}(\zeta_{m})$ rather than replacing $\mathbb{Q}$ by other fields.

Admit the fact that $\mathbb{Q}(\zeta_{p^{r}q^{s}})=\mathbb{Q}(\zeta_{p^{r}})\mathbb{Q}(\zeta_{q^{s}})$; $\mathcal{O}_{\mathbb{Q}(\zeta)}=\mathbb{Z}(\zeta);$ $[\mathbb{Q}(\zeta):\mathbb{Q}]=\phi(p^{r}).$

(1) $\bf{Claim}$ For $\mathbb{Q}(\zeta)/\mathbb{Q}$, $(1-\zeta)\mathbb{Z}(\zeta)$ is the unique prime ideal over $p.$

Proof $\text{ }$ By $\mathbb{Z}(\zeta)/(1-\zeta)=\mathbb{Z},$ $(1-\zeta)\mathbb{Z}(\zeta)$ is a prime ideal. It is totally ramified over $p$. To see this, we prove $$(1-\zeta)^{\phi(p^{r})}\mathbb{Z}(\zeta)=p\mathbb{Z}(\zeta).$$

The LHS equals to $\prod_{i=1}^{\phi(p^{r})}(1-\zeta)(\sum_{j=1}^{s_{i}-1}\zeta^{j})\mathbb{Z}(\zeta)=\prod_{i=1}^{\phi(p^{r})}(1-\zeta_{i})\mathbb{Z}(\zeta)$ with $\zeta_{i}$ being all $p^{r}$-th primitive roots of unit and $s_{i}$ s.t. $\zeta_{i}=\zeta^{s_{i}},$ with $\frac{1-\zeta_{i}}{1-\zeta}=\sum_{j=1}^{s_{i}-1}\zeta^{j}\in \mathbb{Z}(\zeta).$

In fact $\prod_{i=1}^{\phi(p^{r})}(1-\zeta_{i})=\Phi_{p^{r}}(1)$ with $\Phi_{p^{r}}$ the cyclotomic polynomial

(see https://en.wikipedia.org/wiki/Cyclotomic_polynomial)

and thus this equals to RHS. This proof is finished because we already know $[\mathbb{Q}(\zeta):\mathbb{Q}]=\phi(p^{r}),$ which means the ramification index is the degree of extension. Q.E.D.

(2)Applying the material written by Ravi, someone could see the calculations by Mr.carmichael561 in Cyclotomic fields and valuation. are reasonable.


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