Maybe this is rather trivial, but I could not solve this (actually, I think this is not true, however I'm not sure). Is $\widehat{\mathbb{Z}} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{A}_{\mathbb{Q}}^f$ as topological rings?
Here I'm assuming that $\mathbb{Q}$ have the discrete topology and $\widehat{\mathbb{Z}}$ have the profinite topology. The tensor product topology is defined in page 244 of Bourbaki's "Algèbre commutative" (see for instance, https://books.google.com.br/books?id=Bb30CjGW7EAC&pg=PA244&lpg=PA244&dq=tensor+product+topological+modules&source=bl&ots=BtP4jdGaeO&sig=mp9FWlAmYIPgNyoF6EHre-03XvQ&hl=pt-BR&sa=X&ved=0CGUQ6AEwCGoVChMIgsiOtrGSxwIVBkuQCh1zYgPJ#v=onepage&q=tensor%20product%20topological%20modules&f=false).
Thanks in advance.
Good question! My feeling is that these topologies ought to be the same, but I also get something different. Perhaps we are making the same error?
Following the linked Bourbaki definition, a basis of open neighborhoods of $1$ on the tensor product is given by $$ E \otimes \mathbb{Q} + \widehat{\mathbb{Z}} \otimes F $$ as $E$ and $F$ vary over neighborhood bases of 0 in $\widehat{\mathbb{Z}}$ respectively $\mathbb{Q}$. It suffices to consider $F = \{0\}$ as we are giving $\mathbb{Q}$ the discrete topology; meanwhile, $E$ will range over subgroups of $\widehat{\mathbb{Z}}$.
So we are getting as our open neighborhoods of $0$ anything of the form $n\widehat{\mathbb{Z}} \otimes \mathbb{Q}$. But for $n \neq 0$, the natural map $n\widehat{\mathbb{Z}} \otimes \mathbb{\mathbb{Q}} \to \widehat{\mathbb{Z}} \otimes \mathbb{Q}$ is surjective and so these open neighborhoods are just the whole space.