Inequality between archimedean and non-archimedean height function on number fields

Question

Let $K$ be a number field, $V=V(K)$ be the set of valuations of $K$, $V_0$ be the set of non-archimedean valuations of $K$, $V_1$ be the set of archimedean valuations of $K$. For any $x\in K^\times$, we have the product formula $$\prod_{v\in V}|x|_v = 1 .$$

Define the height functions $h_0(x)=\sum_{v\in V_0} \log(\max(|x|_v,|x^{-1}|_v))$ and $h_1(x)=\sum_{v\in V_1} \log(\max(|x|_v,|x^{-1}|_v))$.

Then for $K=\mathbb{Q}$ and rational number $x=\pm\frac{a}{b}\neq 0$ (where $a,b$ are coprime positive integers), we have $$0 \le h_1(x) = |\log(a)-\log(b)|\le h_0(x)=\log(a)+\log(b).$$ That is to say, the archimedean height funcetion $h_1(x)$ is "controlled" by the non-archimedean height funcetion $h_0(x)$.

I‘m wondering for general number fields $K$, does there exists real number $C=C(K)>0$, such that for any $x\in K^\times$, we have $0\le h_1(x) \le C\cdot h_0(x)$? If exists, how to find this $C=C(K)$ from $K$? Thank you for reading this post.

Answer 1

Such a bound cannot exist if $K$ has infinitely many units, because for a unit $h_0=0$ but $h_1$ can be large. Hence assume $K=\mathbb Q(\sqrt{-d})\subset\mathbb C$ is imaginary quadratic. Then $h_1^K(x)=h_1^{\mathbb Q}(N(x))\le h_0^{\mathbb Q}(N(x))\le h_0^{K}(x)$.