1

How one can do the problem 1.3.8 from Qing Liu's Algebraic Geometry and Arithmetical Curves. Namely,

Let $A$ be a Noetherian ring, and $I,J$ ideals of $A$. Let $\widehat{A}$ be the $I$-adic completion of $A$ and $(A/J)^\widehat{\hskip1ex}$ the completion of $A/J$ for the $(I+J)/J$-adic topology. Why do there is a canonical isomorphism $\hat A/J\hat A\simeq (A/J)^\widehat{\hskip1ex}$?

user 1
  • 7,447
Jaakko Seppälä
  • 2,691
  • A very useful result says the following: if $A$ is noetherian, and $I=(a_1,\dots,a_n)$, then $\hat A\simeq A[[X_1,\dots,X_n]]/(X_1-a_1,\dots,X_n-a_n)$. By using this the isomorphism in your question is pretty clear. – user26857 Feb 15 '15 at 22:02

2 Answers2

1

Since the given hint is too general, let me try an answer.

We have an exact sequence of $A$-modules $$0\to J\to A\to A/J\to 0,$$ and consider their $I$-adic completions. Since $A$ is noetherian $0\to\hat J\to\hat A\to\widehat{A/J}\to 0$ is an exact sequence of $A$-modules. Thus we get $\hat A/\hat J\simeq\widehat{A/J}$. By using again that $A$ is noetherian we get $\hat J=J\hat A$, so $\hat A/J\hat A\simeq\widehat{A/J}$. It's not hard to check that this is also a ring isomorphism. Moreover, $\widehat{A/J}$ is the same as the completion of $A/J$ in the $(I+J)/J$-adic topology since $\dfrac{A/J}{I^n(A/J)}=\dfrac{A}{I^n+J}=\dfrac{A/J}{((I+J)/J)^n}$.

user26857
  • 52,094
0

Hint. in general there is a theorem:

Let $R$ be a noetherian ring, $I$ an ideal of $R$, and $0 \to M' \to M \to M'' \to 0$ be an exact sequence of finitely generated $R$-modules. Then the sequence of $I$-adic completions $0 \to \widehat{M'}\to \widehat{M} \to \widehat{M''} \to 0 $ is also exact.

note that since $R$ is noetherian, an ideal $J$ of it will be a finitely generated $R$-module.

user26857
  • 52,094
user 1
  • 7,447