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Here is the problem:

Let $A$ be a Noetherian ring and $a$, $b$ be ideals in $A$. If $M$ is any $A$-module, let $M^a$, $M^b$ denote its $a$-adic and $b$-adic completions respectively. If $M$ is finitely generated, prove that $(M^a)^b\cong M^{a+b}$.

Since $A$ is Noetherian, by proposition 10.13, we only need to prove $(A^a)^b\cong A^{a+b}$. Now I don't know how to prove one of the hints.

How to prove: $$\varprojlim_m ( \varprojlim_n A/(a^nA+b^mA))= \varprojlim_n A/(a^nA+b^nA)$$

So this is my question.

Update: My guess is that: $$\varprojlim_m ( \varprojlim_n A/(a^nA+b^mA))= \varprojlim_{n,m} A/(a^nA+b^mA) =\varprojlim_n A/(a^nA+b^nA)$$

WWK
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1 Answers1

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Sketch.

Facts. For a flat $A$-algebra $B$, let $I$ be an ideal of $A$, $M$ an $A$-module. Then $B\otimes_AIM\cong I(B\otimes_AM)$ (considering them as submodules of $B\otimes_AM$)

So when $B=\widehat{A}$ and $A$ is Noetherian, $M$ is finite, $\mathfrak{b}$ is ideal of $A$, we obtain $(\mathfrak{b}^mM)^{\mathfrak{a}}=\mathfrak{b}^mM^{\mathfrak{a}}$.

Take $\mathfrak{a}$-adic completions of the following exact sequence: $$ 0\to \mathfrak{b}^mM\to M\to M/\mathfrak{b}^mM\to 0, $$ we get a canonical isomorphism $$ (M/\mathfrak{b}^mM)^{\mathfrak{a}}\cong M^{\mathfrak{a}}/(\mathfrak{b}^mM)^{\mathfrak{a}}\cong M^{\mathfrak{a}}/\mathfrak{b}^mM^{\mathfrak{a}} $$

Thus we get the canonical isomorphism $$ (M^{\mathfrak{a}})^{\mathfrak{b}}\cong \varprojlim_m(M/\mathfrak{b}^mM)^{\mathfrak{a}}=\varprojlim_m\varprojlim_n M/(\mathfrak{b}^m+\mathfrak{a}^n)M=\varprojlim M/(\mathfrak{a}+\mathfrak{b})^kM. $$

Unfortunately, the map is NOT an isomorphism of topological groups in general.

Example. Consider the polynomial ring $k[x,y]$ in two variables, the ideal $(x,y)=(x)+(y)$, so the two rings $k[[x]][[y]]$ and $k[[x,y]]$ are isomorphic as rings, but one can check the isomorphism is not an isomorphism as topological rings.

user119882
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  • I have question about last step, why $$\varprojlim_m ( \varprojlim_n M/(a^nM+b^mM))= \varprojlim_n M/(a^nM+b^nM)$$ – WWK Apr 15 '14 at 02:01
  • The last step is in general true, one can just write down the canonical map and verify the map is an isomorphism. In general, we have the following lemma. Let $\mathcal{I}$, $\mathcal{J}$ be index categories. Let $M : \mathcal{I} \times \mathcal{J} \to \mathcal{C}$ be a functor. We have $$ \colim_i \colim_j M_{i, j} = \colim_{i, j} M_{i, j} = \colim_j \colim_i M_{i, j} $$ provided all the indicated colimits exist. Similar for limits. – user119882 Apr 23 '14 at 16:39
  • See http://stacks.math.columbia.edu/tag/002M . The limits (here the inverse limit) or colimits in the category of sets are easy to describe, you can see the section "4.15. Limits and colimits in the category of sets" in The Stacks Project. (stacks.math.columbia.edu/) – user119882 Apr 23 '14 at 16:47