Absolute Value in Cyclotomic Fields

535 Views Asked by At

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.

1

There are 1 best solutions below

1
On

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.