= Solution
For each <place of a number field> $v$, let $K_v$ be the corresponding completion, and for finite $v$ let $\mathcal O_v$ be its <valuation ring>. The <adele ring> is the <restricted product>
$$
\mathbb A_K=\prod_v'K_v
=\left\{(x_v)_v:x_v\in\mathcal O_v\text{ for all but finitely many finite }v\right\}.
$$
Its <restricted product topology> has basic open sets $\prod_vU_v$, where every $U_v\subseteq K_v$ is open and $U_v=\mathcal O_v$ at all but finitely many finite places.
First take $K=\mathbb Q$. The neighborhood
$$
(-1/2,1/2)\times\prod_p\mathbb Z_p
$$
of zero meets the diagonal copy of $\mathbb Q$ only in zero: a rational number lying in every $\mathbb Z_p$ is an <integer>, and the only integer in the indicated real interval is zero. Thus $\mathbb Q$ is discrete in $\mathbb A_{\mathbb Q}$.
Every rational adele is congruent modulo $\mathbb Q$ to an element of
$$
[0,1]\times\prod_p\mathbb Z_p.
$$
Indeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the <Chinese remainder theorem>; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every $\mathbb Z_p$, and the <Tychonoff theorem>. Its image covers the quotient, so $\mathbb A_{\mathbb Q}/\mathbb Q$ is compact.
Now choose a $\mathbb Q$-basis of the <number field> $K$. The given topological isomorphism
$$
\mathbb A_{\mathbb Q}\otimes_{\mathbb Q}K\simeq\mathbb A_K
$$
identifies the additive pair $(\mathbb A_K,K)$ with $(\mathbb A_{\mathbb Q}^n,\mathbb Q^n)$. A finite product of discrete subgroups is discrete, and
$$
\mathbb A_K/K\simeq(\mathbb A_{\mathbb Q}/\mathbb Q)^n
$$
is compact.
The <idele group> is
$$
J_K=\prod_v'K_v^\times,
$$
where the distinguished subgroup at a finite place is $\mathcal O_v^\times$. It carries the corresponding <restricted product topology on the idele group>. The inclusion $j:J_K\to\mathbb A_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^\times$, and outside finitely many places every idele component already belongs to $\mathcal O_v^\times\subseteq\mathcal O_v$.
It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in\mathbb Z_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhood
$$
\prod_{v\mid\infty}K_v^\times\times\prod_{v\nmid\infty}\mathcal O_v^\times
$$
contains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.
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