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. Indeed, 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 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?) for them to be isomorphic?

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    $\begingroup$ Try \otimes instead of @ :) Detexify might also help in the future. $\endgroup$ – t.b. Jul 18 '12 at 7:57
  • $\begingroup$ Well, \tensor always worked for me in the past. Thanks for the tip! :) $\endgroup$ – Quercus Jul 18 '12 at 8:02
  • $\begingroup$ $A^*\otimes_A B\cong B^*$, so your left hand side becomes $(B^*)^*\cong B^*$. $\endgroup$ – messi Jul 18 '12 at 8:36
  • $\begingroup$ I doubt that this has a chance to be true without any finiteness conditions. $\endgroup$ – Martin Brandenburg Jul 18 '12 at 8:43
  • $\begingroup$ @messi If you assume $A$ Noeth and $B$ finite over $A$, then sure. $\endgroup$ – Dylan Moreland Jul 18 '12 at 9:07

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