# A non-trivial absolute value on an algebraic number field $K$ restricts to $Q$ non-trivially.

Let $|\cdot|$ be a non-trivial absolute value over an algebraic number field over $K$. Then restriction of $|\cdot|$ to $Q$ is non-trivial.(i.e. $i:Z\to K$ embedding allows $Q\to K$ embedding. $|\cdot|$ is restricted to the image of $Q\to K$ embedding.)

$\textbf{Q:}$ Since non-trivial absolute values over $Q$ are either equivalent to the absolute value defined by $p-$valuation($p$ is a prime in $Z$) or equivalent absolute value in $R$, can I say this restricted absolute value must be the absolute value coming from $R$(real number)?(In particular, do I know this absolute value ultrametric or not?) I do not see any reason this has to be the case.

This is related to Taylor-Frohlich Pg 70. (2.21) statement.

• The absolute values on $K$ are either Archimedean or $\mathfrak p$-adic, where $\mathfrak p$ is a prime of $\mathcal O_K$ above $p$. In the latter case, the restriction to $\mathbb Q$ will be $p$-adic. By the way, blackboard bold letters can be produced by writing $\mathbb Q$ for example. – Mathmo123 Apr 14 '18 at 23:24
• What is your intuition for saying $| \cdot |$ would induce the archimedean absolute value on $\mathbb Q$? Why do you think it could not induce a $p$-adic absolute value? – D_S Apr 16 '18 at 1:29
• @D_S I do not have any goo reason to say that. However, on $R$, I have a very good "absolute value" which is standard absolute value. There is no reason to guess that I will get $p-$adic absolute value here. So I cannot see any good reason to induce $p-$adic value from $R$. Of course, irrational may have $p-$adic expansion but there is no guarantee that they will converge in $p-$adic sense. At this time point, hensel lifting has not been covered by the book. I do not know why it should be obvious. – user45765 Apr 16 '18 at 13:56

Here is a counterexample where an absolute value on $K$ restricts to a nonarchimedean absolute. Up to equivalence of absolute value, it is the only counterexample.

Let $p$ be a prime number, and let $\mathfrak p$ be a prime of $A = \mathcal O_K$ lying over $p\mathbb Z$. Since $A$ is a Dedekind domain, the localization $A_{\mathfrak p}$ is a discrete valuation ring with unique maximal ideal $\mathfrak p A_{\mathfrak p}$. Let $\varpi$ be a generator of the unique maximal ideal of $A_{\mathfrak p}$. Then

$$pA_{\mathfrak p} = \mathfrak p^e A_{\mathfrak p} = \varpi^e A_{\mathfrak p}$$

where $e \geq 1$ is the ramification index of $\mathfrak p$ over $p$. Every nonzero element of $K$ can be written uniquely as $u \omega^n$, for $u \in A_{\mathfrak p}^{\ast}$ and $n \in \mathbb{Z}$. Define an absolute value on $K$ by

$$|u \omega^n| = \rho^{n}$$

where $\rho = p^{\frac{-1}{e}} \in (0,1)$. This absolute value is nonarchimedean, since it is coming from a DVR. We can write $p = \varpi^e u_1$, for $u_1 \in A_{\mathfrak p}^{\ast}$.

If we restrict $| \cdot |$ to $\mathbb{Q}$, we get the $p$-adic absolute value: every nonzero rational number can be uniquely written as $ap^n$, for $a$ a rational number relatively prime to $p$. Then

$$|ap^n| = |a \varpi^{en} u_1^n| = \rho^{en} = p^{-n}$$

which is the $p$-adic absolute value.

Using Ostrowski's theorem, the absolute values on $K$ can be completely classified by how they restrict to $\mathbb Q$; if the restriction is archimedean, the absolute value comes from an embedding of $K$ into $\mathbb C$. If it is archimedean, the absolute value comes from a unique prime $\mathfrak p$ of $\mathcal O_K$ as above.