fractional ideals are discrete in $K_\infty$

Question

Let $K$ be a numbe field, $S_\infty$ the set of infinite places of $K$ and $S$ a finite set of places of $K$ containing $S_\infty$. Define a fractional ideal $I$ over the ring of integer $\mathfrak{o}_K$ of $K$ by $$I=\prod_{v\in S-S_\infty} P_v^{k_v}$$ where $k_v$ are some integers and $P_v$ the prime of $K_v$. How to show that $I$ is a discrete in $K_\infty:=\prod_{v\in S_\infty}K_v$? I want to use the fact that $K$ is discrete in the adele $A_K$ of $K$, so $I$ is discrete in $A_K$, but what I want is regard $I\subseteq K$ as a subset in $K_\infty$, not in $A_K$.

Answer 1

Let $n=[K:\Bbb{Q}]$ and $\sigma_1,\ldots,\sigma_n$ be the embeddings $K\to \Bbb{C}$.

$O_K$ is discrete in $\Bbb{C}^n$ because for $a,b\in O_K$, if $\sup_j |\sigma_j(a-b)|< 1/[K:\Bbb{Q}]! $ then the coefficients of the polynomial $\prod_{j=1}^n (x-\sigma_j(a-b))\in \Bbb{Z}[x]$ all are $< 1$ so $\prod_{j=1}^n (x-\sigma_j(a-b))=x^n$ and hence $a=b$.

This implies that any fractional ideal is discrete in $K_\infty\subset \Bbb{C}^n$ as well.