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Let $R$ be a commutative Noetherian ring, $I$ is an ideal of $R$. It is well known that the completion of $R$ is complete and Hausdorff.

In most books of commutative algebra, the proof is as follows:

1.$\hat R \otimes I\cong \hat I$;2. $\hat {I^n}\cong \hat I^n$；3.$\hat{R/{I^n}}\cong R/{I^n}$.

So we know $\hat{\hat R}\cong lim_{\leftarrow}\hat R/\hat I^n\cong \hat R$.

My question is how to prove the natural morphism $\hat R\rightarrow \hat{\hat R}$ is isomorphism?

I can not write down the above composition clearly.

Thank you in advance.

Jian
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1 Answers1

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Show that for all $n$, there is a commutative diagram: $$\require{AMScd} \begin{CD} \vdots @. \vdots\\ @VVV @VVV\\ R/I^{n+1} @>{\cong}>> \hat{R}/\hat{I}^{n+1}\\ @VVV @VVV\\ R/I^n @>{\cong}>> \hat{R}/\hat{I}^n\\ @VVV @VVV\\ \vdots @. \vdots \end{CD} $$ where the isomorphisms $\varphi_n:R/I^n\cong \hat{R}/\hat{I}^n$ is induced by the natural homomorphism $\varphi:R\to \hat{R}$. These isomorphisms form an isomorphism of inverse systems (two columns), which results in an isomorphism of inverse limits. This is nothing other than the natural morphism $\hat{\varphi}:\hat{R}\to \hat{\hat{R}}$.

withoutfeather
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