Adeles under base change Let $ K \subset L $ be a  finite  separable extension of  global fields with set of places $ M_{K} $, $ M_{L} $ and adele rings $ \mathbb{ A}_{K}$, $ \mathbb{A}_{L} $ respectively. 
I have trouble understanding the proof of Proposition 22.10 in these notes. The first line says that $$ \mathbb{A}_{K} \otimes L    $$ is equal to the restricted direct product of $ K_{v} \otimes_{K} L  $ with respect to $ \mathcal{O}_{v} \otimes _ { \mathcal{O}_{K}}   \mathcal{O } _ { L }  $. 
First, why does $ a_{v} \otimes x $ lies in $  \mathcal{O}_{v} \otimes_{\mathcal{O}_{K}}  \mathcal{O}_{L} $ for all but finitely many places of $ v \in M_{K} $? 
Second, how can I rigorously convince myself this is a ring isomorphism (kernel is zero) and a homeomorphism of topological spaces?      
 A: Your notation $O_v$ means $v$-adic completion $O_{K,v}$ of the integer ring $O_K$, right?
There are isomorphisms $\sigma_v:K_v\otimes_K L \simeq \prod_{w|v} L_w$ and $\sigma_{v,res}:O_{K,v}\otimes_{O_K}O_L \simeq \prod_{w|v} O_w$ for each $v \in M_K$(Surely, $w \in M_L$).
I don't know whether the isomorphisms are canonical or not. But It doesn't matter.
Note that $K_v\otimes_K L\simeq K_v \otimes _{O_K} O_L$ and $\mathbb{A}_K\otimes_{K}L\simeq\mathbb{A}_K\otimes_{O_K}O_L$.
Also we have $\mathbb{A}_K= \widehat{O_K}\oplus K_{\infty}+K$ (See Lang's 'Algebraic number theory Ch.7, section2, Theorem2) where $\widehat{O_K}=\prod_v O_{K,v}$ and $K_{\infty}$ is the Archimedean part. Here $K$ is a subgroup of $\mathbb{A}_K$ via the diagonal embedding.
Consdier the exact sequence $0\longrightarrow O_K\longrightarrow K\times (\widehat{O_K}\oplus  K_{\infty}) \longrightarrow \mathbb{A}_K\longrightarrow0 $ and tensoring with $O_L$ induces
$0\longrightarrow O_K \otimes_{O_K}O_L\longrightarrow(K\otimes _{O_K}O_L) \times \big((\widehat{O_K}\oplus K_{\infty})\otimes_{O_K}O_L \big)\longrightarrow \mathbb{A}_K\otimes_{O_K}O_L \longrightarrow 0$.
Surely $K\otimes_{O_K}O_L \simeq L$ and $K_{\infty} \otimes _{O_K} O_L \simeq L_{\infty}$, so it suffices to check $\widehat{O_K}\otimes_{O_K}O_L\simeq \widehat{O_L}$.
Since $\widehat{\mathbb{Z}}\otimes O_K=  (\varprojlim \mathbb{Z}/m\mathbb{Z}) \otimes O_K \simeq \varprojlim (O_K/ mO_K)=\widehat{O_K} $ (Inverse limit usually does not commute with tensor products, but in this case, $O_K$ is a free $\mathbb{Z}$-module, so commutes.), $\widehat{O_K}\otimes_{O_K}O_L\simeq (\widehat{\mathbb{Z}}\otimes O_K)\otimes_{O_K}O_L\simeq \widehat{O_L}$.
All of morphisms are topological homeomorphisms, because $\widehat{O_K} \otimes O_L$ and $\widehat{O_L}$ have both profinite topologies.
