Is always true that $\hat{A} \otimes_A A_f/A \cong \hat{A}_f/ \hat{A}$? Let $A$ be a commutative ring. Let $f \in A$. Let $A_f= A\left [ \frac{1}{f}\right ]$.
Let $\hat{A}$ the $f$-adic completion of $A$. Is it always (even when $A$ is not noetherian) true that
$$\hat{A} \otimes_A \left (A_f/A \right ) \cong \hat{A}_f/ \hat{A}$$
 A: Here's (after some thinking) how I would approach my own question. Still, there are some details that I am not sure about. So I'd be happy about answers/comments!
Let $f \in A$ a non zerodivisor (else is trivial).
$\hat{A}$ has no $f$-torsion (this should follow directly from the definition of completion, but I would like to have a feedback that I'm not mistaken).
Hence $\operatorname{Tor}_1^A(A/f^nA, \hat{A})=0$ for all $n \geq 0$ and $\varinjlim A/f^nA \cong A_f/A$. (Here again, I am not sure how to see this nicely, but the idea is taking the map arising from $A/f^nA \rightarrow A_f/A$, $u \mapsto u/f^n$. The induced map should be well defined and an isomorphism)
We consider the exact sequence
$$0 \rightarrow A \rightarrow A_f \rightarrow A_f/A \rightarrow 0.$$
As $\operatorname{Tor}$ commutes with colimits, we have just seen that tensoring with $\hat{A}$ is now exact, giving us
$$0 \rightarrow \hat{A} \rightarrow \hat{A}_f \rightarrow \left( A_f/A \right) \otimes \hat{A} \rightarrow 0$$
which is what we wanted to show.
