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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: 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?

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2 Answers 2

up vote 3 down vote accepted

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 $T^* \stackrel{\cong}{\longrightarrow} B^*$.

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Thank you for your answer, Konstantin!

In fact, since asking the question, my advisor, Prof. Dan Haran, has found a proof which does not use any Noetherity conditions:

For any $A$-algebra $C$, we denote by $\varphi_C$ the canonical mapping from $C$ to its $I$-adic completion $C^* $, and for an $A$-homomorphism $\psi \colon B \to C$ we denote by $\psi^* $ the induced $A$-homomorphism from $B^* $ to $C^* $.

Using the fact that the image of $A$ under $\varphi_A$ is dense in $ A^* $, one can directly see that the image of $B$ under $1 \otimes id$ is dense in $T : = A^* \otimes_A B$. Therefore the image of $ B^* $ under the induced mapping $ (1 \otimes id)^* $ must also be dense in $ T^* $.

The algebra mapping $\beta \colon A \to B$ induces the mapping $\beta^* \colon A^* \to B^* $, and using the universal property of the tensor product (as a co-product in the category of commutative $A$-algebras), we obtain a unique $A$-homomorphism $\psi \colon T \to B^* $ such that $\psi \circ (1 \otimes id) = \varphi_B $ and $\psi \circ (id \otimes 1) = \beta^* $.

Now $ \varphi_B^* = \varphi_{B^* } \colon B^* \to (B^* )^* $ is an isomorphism, and $ \psi^* \circ (id \otimes 1 )^* = \varphi_B^* $ shows that $ (id \otimes 1 )^* $ is an injection. Therefore we can view $ B^* $ as a complete dense $A$-subalgebra of $ T^* $, which means that: $$ T^* = \overline{B^* } \cong (B^* )^* \cong B^* $$ (where $\overline{B^* }$ is the closure of $B^* $ in $T^* $ ).

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