# Adelic topology on the group of ideles

The topology on $\mathbb{A}^\times$ is the subspace topology with respect to $\prod_v \mathbb{Q}_v^\times$ and a basis is given by the sets $$\prod_v\Omega_v$$ with $\Omega_v\subset\mathbb{Q}_v^\times$ open, equal to $\mathbb{Z}_v^\times$ for almost every valuation $v$.

I've read everywhere that this topology is stronger than the subspace topology with respect to $\mathbb{A}^\times\subset\mathbb{A}$. However, I don't understand why, because: a basis element of this latter topology is of the form $$B=\mathbb{A}^\times\cap\prod_v \Omega_v$$ for $\Omega_v\subset\mathbb{Q}_v$ open, equal to $\mathbb{Z}_v$ for almost every valuation $v$. If $x\in B$, then $$x\in (\Omega_\infty-0)\times\dots\times(\Omega_{p_n}-0)\times(\mathbb{Z}_{p_{n+1}}-0)\times\dots\times(\mathbb{Z}_{p_{n+r}}-0)\times\mathbb{Z}_{p_{n+r+1}}^\times\times\mathbb{Z_{p_{n+r+2}}}^\times\times\dots\subset B.$$ Thus, another basis for the second topology is given by the open sets $$\prod_v\Omega_v$$ with $\Omega_v\subset\mathbb{Q}_v^\times$ open, equal to $\mathbb{Z}_v^\times$ for almost every valuation $v$. This is the same basis as before.

I had the same problem as you, and I'm writing this answer primarily for my own understanding. We work over the adele ring and idele group for a number field $K$, rather than just over $\mathbb Q$, but the idea is the same.

We aim to study the adelic topology on the ideles and understand why it is not generated by the same basis of open sets.

Recall that, as abstract groups, $\mathbb I_K\cong\mathbb A_K^\times$. Now

$$\mathbb A_K^\times=\{\left(x_v\right)_{v\in M_K}\colon x_v\ne0\forall v,\left|x_v\right|_v=1\textrm{ for almost all }v\}$$

A basic open set in the topology in the adeles is given by $$\prod_v U_v$$ where $U_v$ is open in $K_v$ for each $v$ and almost all $U_v$ are equal to $\mathcal{O}_{K_v}$.

Intersecting this with $\mathbb A_K^\times$ we get the set $$\left\{\left(x_v\right)_{v\in M_K}\colon x_v\ne0\textrm{ for all }v,\;\left|x_v\right|_v=1\textrm{ for almost all }v,\;x_v\in U_v\textrm{ for all }v\right\}$$ Now any such set is the union of sets of the form $$\prod_v W_v$$ where $W_v\subset K_v^\times$ is open and $W_v=\mathcal{O}_{K_v}^\times$ for almost all $v$. However, these sets aren't open in the subspace topology! The reason is actually fairly subtle: although any element $(x_v)$ of our open set above satisfies $\left|x_v\right|_v=1$ for almost all $v$, these elements of absolute value $1$ can occur anywhere in the sequence $(x_v)$ - i.e., the set of places $v$ such that $\left|x_v\right|_v\ne1$ can depend on the element $\left(x_v\right)_v$. The sets of the form $\prod_v W_v$ with $W_v=\mathcal{O}_{K_v}^\times$ for almost all $v$ are only allowed to have elements $x_v$ with norm not equal to $1$ in some fixed finite set of places $v$ independent of the individual elements.

• In other words, any open subset of $A_K^{\times}$ with respect to the subspace topology (induced by the adelic topology) is the union of basic open sets of the idelic topology (which are not individually open with respect to the subspace topology). In particular, the idelic topology is finer than the subspace topology. – Watson Jan 4 '18 at 15:08

Asking questions helps to clarify things... If I'm right, the mistake is that $$(\Omega_\infty-0)\times\dots\times(\Omega_{p_n}-0)\times(\mathbb{Z}_{p_{n+1}}-0)\times\dots\times(\mathbb{Z}_{p_{n+r}}-0)\times\mathbb{Z}_{p_{n+r+1}}^\times\times\mathbb{Z_{p_{n+r+2}}}^\times\times\dots=\mathbb{A}^\times\cap(\Omega_\infty\times\dots\times\Omega_{p_n}\times\mathbb{Z}_{p_{n+1}}\times\dots\times\mathbb{Z}_{p_{n+r}}\times\mathbb{Z}_{p_{n+r+1}}^\times\times\dots),$$ is not open in the second topology since we end up with infinitely many $\mathbb{Z}_{...}^\times$