Let $A$ be a commutative ring with 1, $I$ an ideal in $A$, $B$ an $A$-algebra. I am trying to prove the following isomorphism of $A$-algebras:
$$ \big( A^\* \otimes \_A B \big) ^\* \cong B^\* $$
"$^\*$" denotes the $I$-adic completion: every $A$-algebra $X$ may be endowed with the $I$-adic topology, defined by the ideal $IX$ in $X$, and the $I$-adic completion of the algebra is the completion with respect to this (uniform) topology.


I have so far been able to prove that the image of $B$ in $T :=A^\*\otimes\_A B$ under the map $1 \otimes id \colon B \to T$ is dense in $A$. At this stage I considered the map $(1 \otimes id)^\* \colon B^\* \to T^\*$, and tried to show that it is an isomorphism, but I'm having troubles both with the injectivity and the surjectivity of this mapping. 


Are these $A$-algebras always isomorphic? If so, how can this be proven? If not, how can a counterexample be constructed, and what do I have to require (Noetherity? Flatness? Finiteness?) for them to be isomorphic?



Completion and Tensor Product of Algebras
It's true if $A$ is Noetherian.


For any $A$-algebra $C$ and any ideal $J$ in $A$, note that $C/JC$ is isomorphic to the tensor product algebra $C \otimes\_A A/J$.


Now for any $n \geq 0$, $A/I^n$ is a module over $A^\*$, and the multiplication map $A/I^n \otimes\_A A^\* \to A/I^n$ is an isomorphism, since $A$ is Noetherian --- see for example Proposition 10.13 of Atiyah and Macdonald's "Introduction to Commutative Algebra".


Let $T = A^\* \otimes\_A B$. Then we get isomorphisms


$T/I^n T \cong A/I^n \otimes\_A (A^\* \otimes\_A B) \cong (A/I^n \otimes\_A A^\*) \otimes\_A B \cong (A/I^n) \otimes\_A B \cong B/I^nB$.


Now pass to the inverse limit to obtain an isomorphism $T^\* \stackrel{\cong}{\longrightarrow} B^\*$.


