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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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    Try `\otimes` instead of `@` :) [Detexify](http://detexify.kirelabs.org/classify.html) might also help in the future.2012-07-18
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    Well, `\tensor` always worked for me in the past. Thanks for the tip! :)2012-07-18
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    $A^*\otimes_A B\cong B^*$, so your left hand side becomes $(B^*)^*\cong B^*$.2012-07-18
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    I doubt that this has a chance to be true without any finiteness conditions.2012-07-18
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    @messi If you assume $A$ Noeth and $B$ finite over $A$, then sure.2012-07-18
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    @messi - the isomorphism you wrote is not true, for at least two reasons: 1. You need to takt the image of $B$ in the tensor product on the right hand side. 2. The left hand side is not complete.2012-07-18
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    @DylanMoreland - how would you go about proving that?2012-07-18
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    @Dylan you are right, i was assuming that, sorry.2012-07-18

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