Let $K$ be a number field. The place $\nu$ is an equivalence class of non-trivial absolute values. Moreover, the absolute value $| \quad | : K \to \mathbb{R}$ is called non-archimedean if $$|x+ y| \leq \max(|x|, |y|), \forall \ x, y \in K.$$ Then, one has
"there is only one archimedean place $\infty$ on $\mathbb{Q}$ given by the ordinary absolute value $| \quad |$."
I am reading this proposition, but I can't prove its. Can you help me to send a proof for that? Thank you so much.
$\newcommand{\abs}[1]{\lvert #1 \rvert}$ Obviously, the absolute value is an archimedean place.
Furthermore, you should try to prove that any absolute value $\abs{~}\colon K → [0..∞)$ on any field $K$ is non-archimedean if and only if it is bounded on $ℕ$ (and then it’s bounded from below by $1$).
One direction is pretty clear by induction, for the other you should use the binomial theorem: Prove for $x, y ∈ K$ with $x ≠ 0$, $y ≠ 0$ and $\abs{x} ≤ \abs{y}$, $$\abs{x+y} = \sqrt[n]{\abs{x+y}^n} \quad\text{and}\quad\abs{x+y}^n ≤ (n+1)\abs{y}^n.$$ You will need $\abs{~}$ being bounded on $ℕ$ for the last inequality. Then take the limit $n → ∞$.
Now, being multiplicative, an archimedean absolute value on $ℚ$ restricts to a multiplicative monoid homomorphism $ℕ → [0..∞)$. Since it’s unbounded (as we now know), there must be a prime $p ∈ ℕ$ with $\abs{p} > 1$. Without loss of generality (by replacing $\abs{~}$ by $\abs{~}^{\log_{\abs p}~p}$) we may assume $\abs{p} = p$, so $\abs{p^k} = p^k$ for all $k ∈ ℕ$.
Also $\abs{1} = 1$. By induction, $\abs{n} ≤ n$. Now if there was any $m ∈ ℕ$ with $\abs{m} < m$, then, by induction $\abs{n} < n$, for all $n ≥ m$. But $\abs{p^k} = p^k$ for all $k ∈ ℕ$, so we have $\abs{n} = n$ for all $n ∈ ℕ$.
Since $\pm ℕ$ generates $ℚ^×$ as group multiplicatively, this determines the group homomorphism $\abs{~} \colon ℚ^× → (0..∞)$ completely. And $\abs{0} = 0$.