Let $A$ be a commutative ring (not necessarily noetherian).

Let $I\subseteq J\subseteq A\,$ be two finitely generated ideals.

Let us denote the completion functor by $\Lambda_K (M) = \varprojlim_n M/K^nM$.

I would like to compare the two rings:

$\Lambda_J(A)$ and $\Lambda_{J'} (\Lambda_I(A))$ where $J' = J\Lambda_I(A)$.

Are they the same? are they isomorphic?

I thought about showing that in the right hand side, $A$ is dense in the $J$-adic topology, and that it is $J$-adically complete. Is this true? and are those two facts enough to show that it is isomorphic to the left hand side?

Edit: I should mention that in my application, the two rings $\Lambda_I(A)$ and $\Lambda_J(A)$ are noetherian.


Let $A$ be a commutative ring, $I\subset A$ be an ideal. Denote by $\hat{A}$ the completion with respect to the $I$-adic filtration.

One can consider the kernel $\hat{I}_n$ of the natural projection $\hat{A}\to A/I^n$. The obvious observation is: $\hat{A}/\hat{I}_n=A/I^n$. Thus, $\hat{A}=\varprojlim_n\hat{A}/\hat{I}_n$.

Now, it' clear that $I^n\hat{A}\subseteq\hat{I}_n\subseteq(\hat{I}_1)^n$. In Noetherian case they are always equal. This gives an affirmative answer to your question. In general they can differ. However, I can not come up with an example immediately.

  • 1
    $\begingroup$ Just to be sure, are you claiming that my question has a negative answer in general? $\endgroup$ – DARK Oct 10 '12 at 9:49
  • $\begingroup$ If $I$ is of finite type then $\widehat{I} = I \widehat{A}$ (more generally if $M$ is a f.g. $A$-module then $M \otimes \widehat{A} \to \widehat{M}$ is surjective. This is indeed false in general, for example you can take $A=k[X_1,X_2,\ldots]$ and $I=\langle X_1,X_2,\ldots \rangle$. Then $F = X_1+X_2^2+\cdots \in \widehat{I}$ doesn't belong to the image of $I$. $\endgroup$ – François Brunault Oct 10 '12 at 11:09

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