Is the quotient group of $2$-th root of unity compact?

Question

Let $F$ be a number field and $\mathbb{A}$ its adele ring.

Let $\mu_2$ be the group scheme of 2-th root of unity defined over $F$.

I guess that $\mu_2(F) \backslash \mu_2(\mathbb{A})$ is compact because the special maximal compact subgroup of $\mu_2(F_v)$ is itself for all finite places of $F$.

Is this true?

Answer 1

Since $\mathbb A=\prod'_v(F_v,\mathcal O_v)$ is the restricted product, $\mu_2(\mathbb A)=\prod_{v}\mu_2(F_v)$. By Tychonoff's theorem, this is compact. Taking quotient by $\mu_2(F)$ does not change this.