10 Completions

Highlight: topological group/ring/module; $A$ Noetherian→ $A[[x]]$ Noetherian.

• Completion is exact, see ^z8znz5.
• if $A$ Noetherian, then $\hat{A}$ is also Noetherian, see ^b6b2a0.

Example. Product topology on ${\prod }_{i\in I}{X}_i$: its topological basis is generated by ${\prod }_{i\in I}{U}_i$ with $U_i\subseteq X_i$ for finitely many $U_i$ and $U_i=X_i$ for almost every $i$.

In our course, we only use “open neighborhood”.

Topologies and Completions

Definition

A topological abelian group is a abelian group $G$ equipped with topology such that $G\times G\to G,(a,b)\mapsto ab$ is continuous, and $G\to G,g\mapsto g^{-1}$ is continuous.

Examples. They are topological groups:

• $(\mathbb{R},+)$ with usual topology
• any group equipped with discrete topology
• $({\mathrm{GL}}_n(\mathbb{R}),\times )$

Remark. The category of topological abelian groups is NOT an abelian category. Consider

$$(\mathbb{R},+,\text{discrete topology})\to (\mathbb{R},+,\text{usual topology}),a\mapsto a$$

which is a homomorphism of topological groups with trivial kernel and cokernel, but they are not isomorphism. Clausen-Scloze: {condensed abelian groups} $\supset$ {topological abelian groups}.

In the following, only consider topological abelian groups. Notation: for $x\in X$, let

$$N(x)=\{U\subseteq X:U\text{ open, }x\in U\}.$$

Lemma

Let $G$ be a topological abelian group, and let $a\in G$. For any $V\in N(a)$, $V=a+U$ with $U\in N(0)$.

\begin{proof} Consider the composite map

$$T_a:G\to G\times G\to G,g\mapsto (g,a)\mapsto g+a,$$

which is continuous. Its inverse map is $T_{-a}:g\mapsto g-a$, which is also continuous. Hence $T_a,T_{-a}$ are homeomorphism. Hence $T_a$ maps $N(0)$ bijectively to $N(a)$. \end{proof}

Lemma

Let $U\in N(0)$, then there exists $V\in N(0)$ such that $V+V\subseteq U$ and $V-V\subseteq U$.

\begin{proof} Consider the continuous map

$$G\times G\to G\to G,(a,b)\mapsto (a,-b)\mapsto a-b$$

which deduces that $f^{-1}(U)$ is open. Since $(0,0)\in f^{-1}(U)$, we have $f^{-1}(U)\in N((0,0))$. Then there exist $U_1,U_2$ containing $0$ such that $f^{-1}(U)\supseteq U_1\times U_2$. Let $V=U_1\cap U_2\in N(0)$. One can check $f(V\times V)\subseteq U$ and so $V-V\in U$.

Similarly we can get $V^{\prime }$ such that $V^{\prime }+V^{\prime }\subseteq U$. Then $V\cap V^{\prime }$ is what we desired, \end{proof}

Lemma

Let $G$ be a topological group. Let $H={\cap }_{U\in N(0)}U$. Then

• $H$ is a subgroup;
• $H=\overline{\{0\}}$, the closure of $\{0\}$;
• $G/H$ is Hausdorff.
• $G$ is Hausdorff iff $H=\{0\}$.

\begin{proof} i) Suppose $h_1,h_2\in H$, we aim to show $h_1\pm h_2\in H$. For any given $U\in N(0)$, it suffices to show $h_1\pm h_2\in U$. Note that $-U\in N(0)$ and $h_1,h_2\in \pm U$. Then $h_1+U\in N(0)$ and so $h_2\in h_1+U$, which yields $h_2-h_1\in U$. Similarly, we can prove that $h_1\pm h_2\in U$.

ii) Note that $x\notin \overline{\{0\}}$ iff there exists $V\in N(x)$ such that $0\notin V$. But $V=x+V_0$ for some $V_0\in N(0)$. So $0\notin x+V_0$ yields $-x\notin V_0$. Therefore, $x\notin \overline{\{0\}}$ iff $-x\notin H$ iff $x\notin H$ by i) iff $\overline{\{0\}}=H$.

iii) By ii), $H$ is closed. Recall quotient topology on $G/H$ is defined by $G\stackrel{\pi }{\to }G/H$, where $V\subseteq G/H$ open iff ${\pi }^{-1}(V)$ open. Since ${\pi }^{-1}(0+H)=H$ is closed, we know $\overline{0}\in G/H$ is a closed point. Let $x,y\in G/H$ be distinct points. Then $x-y\ne \overline{0}$, and there exists $U\in N(x-y)$ such that $0\notin U$. Also $U=(x-y)+U_0$ for $U_0\in N(0)$. Take $V_0\in N(0)$ such that $V_0-V_0\subseteq U_0$ by ^36d8qm. Since $0\notin U$, we have $0\notin (x-y)+(V_0-V_0)$. It deduces that $(x+V_0)\cap (y+V_0)=\varnothing$. Therefore, $G/H$ is Hausdorff.

iv) "→" If $G$ is Hausdorff, then $0$ is closed and so $H=\overline{\{0\}}=\{0\}$. "←" by iii). \end{proof}

Remark. Let $G$ be a group, and let $H<G$ be a subgroup. Then $G/H$ Hausdorff iff $H$ is closed. The proof is similar as above.

Completion of Topological Abelian Group

Definition

Say a sequence $\{x_n{\}}_{n⩾1}\subseteq G$ is a Cauchy sequence, if for any $U\in N(0)$, there exists $n_U>0$ such that $x_m-x_{m^{\prime }}\in U$ for any $m,m^{\prime }>n_U$.

Say two Cauchy sequences $\{x_n\},\{y_n\}$ are equivalent, if $\{x_n-y_n\}\to 0$.

Let $\hat{G}$ be the set of equivalent classes of Cauchy sequence, which is called it the completion of $G$.

Facts.

• $\hat{G}$ is an abelian group, by defining $\{x_n\}+\{y_n\}=\{x_n+y_n\}$. It suffices to show $\{x_n+y_n\}$ is a Cauchy sequence.
  • Proof is easy
• $\hat{G}$ is still a topological abelian group

  • Sketch of proof. To define topology on $\hat{G}$, it suffices to define $N(\hat{0})$. For $U\in N(0)\subseteq G$, define

    $$\hat{U}=\{\hat{x}\in \hat{G}:\text{for any representative }\hat{x}=\{x_n\},\exists N>0\text{ such that }{x}_m\in U,\forall m⩾N\}$$

    and then define $N(\hat{0})=\{\hat{U}:U\in N(0)\}$. One can check $\{\hat{x}+N(\hat{0}):\hat{x}\in \hat{G}\}$ is a topology basis.

• The map $\varphi :G\to \hat{G},g\mapsto (g,g,\cdots )$ is continuous, whose kernel is $\overline{\{0\}}$. Thus $\varphi$ injective iff $G$ Hausdorff.

Remark. Similar to Completion.

Definition

We say topological group $G$ is complete, if $\varphi$ is an isomorphism.

Examples.

• $G=(\mathbb{Q},+)$, then $\hat{G}=(\mathbb{R},+)$.
• $G=(\mathbb{Z},+)$ with discrete topology, then $\hat{G}=G$.
• $G=(\mathbb{Z},+)$ with $p$-adic topology
  • $N(0)=\{B(0,r):\text{open balls}\}$ by ^34rum4 where $B(0,r)=\{n:v_p(n)⩾-\log r/\log p\}$. Let $s=-\log r/\log p$. Then $B(0,r)=\{n:v_p(n)⩾[s]+1\}=p^{[s]+1}\mathbb{Z}$.
  • $N(0)=\{\mathbb{Z},p\mathbb{Z},p^2\mathbb{Z},\cdots \}$.
  • Let ${\mathbb{Z}}_p$ be the completion of $(\mathbb{Z},+)$ under $p$-adic topology. For $\hat{x}\in {\mathbb{Z}}_p$ with $\hat{x}=\{x_n\}$, it means for any $k>0$ and $n\gg 0$ such that $p^k\mid x_m-x_{m^{\prime }}$.

Use a sequence of subgroups to define a topology

Let $G$ be an abelian group (not topology yet), and let $G=G_0\supseteq G_1\supseteq \cdots \supseteq G_n\supseteq \cdots$ be a sequence of subgroups.

Fact. The subsets $\{g+G_n{\}}_{g\in G,n⩾0}$ forms a basis of topology.

\begin{proof} If $x\in (a+G_m)\cap (b+G_n)$, then we aim to find $x\in c+G_k$ such that $c+G_k\subseteq (a+G_m)\cap (b+G_m)$. WLOG suppose $m⩽n$, then $G_m\supseteq G_n$, which yields $x\in (a+G_m)\cap (b+G_m)$ and so $a+G_m=b+G_m$. Then $x\in (b+G_m)\cap (b+G_n)=b+G_n$. So we can let $c+G_k=b+G_n$. \end{proof}

Remarks.

• So one can use above to define a topology on $G$. It is relatively easy to check this makes $G$ a topology group.
• Fact. For $G=(\mathbb{Z},+)$, use $\mathbb{Z}\supseteq p\mathbb{Z}\supseteq p^2\mathbb{Z}\supseteq \cdots$ and the induces topology is precisely the $p$-adic topology.

Fact. For above topology on $G$, $G_n$ is both open and closed.

\begin{proof} By definition, $G_n$ is open. But $G={\bigsqcup }_{\overline{a}\in G/G_a}(G_n+a)$ yields $G_n$ is closed. \end{proof}

Fact. These $G_n$ forms a fundamental system of open neighborhood of $0$. That is, for any $U\in N(0)$, there exists $G_n$ such that $G_n\subseteq U$.

\begin{proof} Since $U$ is open, $U={\cup }_{i\in I}(g_i+G_{n_i})$. Thus $0\in g_i+G_{n_i}$ for some $i$ and so $-g_i\in G_{n_i}$. Thus $g_i+G_{n_i}=G_{n_i}\subseteq U$. Now we finish the proof. \end{proof}

In above set-up, we can use “topological inver limit” to define $\hat{G}$

Definition

An inverse system of groups is a sequence $\{A_n{\}}_{n⩾1}$ of groups together with group homomorphisms $A_{n+1}\stackrel{{\theta }_{n+1}}{\to }{A}_n$. If ${\theta }_n$ is always surjective, call it surjective system.

Define ${\lim }_{\leftarrow }{A}_n=\{(a_n{)}_{n⩾1}:{\theta }_{n+1}(a_{n+1})=a_n,\forall n\}$, call it the inverse limit.

If the above $A_n$ are topological groups, then one can equip ${\lim }_{\leftarrow }{A}_n\subseteq {\prod }_{i=1}^n{A}_n$ with induced topology.

Now let $G=G_0\supseteq \cdots \supseteq G_n\supseteq \cdots$, equip $G/G_n$ with discrete topology, and note $\{G/G_{n+1}\stackrel{{\theta }_{n+1}}{\twoheadrightarrow }G/G_n{\}}_n$ form an inverse system. We can define a topological group ${\lim }_{\leftarrow }G/G_n$.

Theorem

There is a homeomorphism of topology groups ${\lim }_{\leftarrow }G/G_n\simeq \hat{G}$.

\begin{proof} Step 1. Define $\iota :\hat{G}\to {\lim }_{\leftarrow }G/G_n$.

Let $\hat{x}=\{x_m{\}}_{m⩾1}\in \hat{G}$. For each $n$, there exists $N_n\gg 0$ such that $x_m-x_{m^{\prime }}\in G_n$ if $m,m^{\prime }>N_n$. This implies $x_m+G_n=x_{m^{\prime }}+G_n$. Thus $x_m+G_n$ with $m\gg 0$ define a unique element in $G/G_n$. We let ${\xi }_n=x_m+G_n$ for $m\gg 0$. Now verify ${\theta }_{n+1}({\xi }_{n+1})={\xi }_n$. This is easy, because $x_m+G_{n+1}={\xi }_{n+1}$ and $x_m+G_n={\xi }_n$ for $m\gg 0$. Now we have defined $\iota$.

It is easy to check $\iota$ is a group homomorphism. If $({\xi }_n)=(0)$, then for any $n$, there exists $x_m\in G_n$ for $m\gg 0$. Thus $\{x_m\}=\{0\}$ in $\hat{G}$ and so $\iota$ is injective.

Step 2. Check $\iota$ is surjective.

Let $({\alpha }_n{)}_{n⩾1}\in {\lim }_{\leftarrow }G/G_n$, so ${\alpha }_n=y_n+G_n$ for some non-unique $y_n$. One can easily check $\{y_n\}$ is a Cauchy sequence. So it defines a element in $\hat{G}$, which is independent of choices of $y_n$.

Step 3. Check $\iota$ and ${\iota }^{-1}$ are both continuous.

(omit) \end{proof}

Example. ${\mathbb{Z}}_p\simeq li{m}_{\leftarrow }\mathbb{Z}/p^n\mathbb{Z}$. It has been proved by

• Fact. For $G=(\mathbb{Z},+)$, use $\mathbb{Z}\supseteq p\mathbb{Z}\supseteq p^2\mathbb{Z}\supseteq \cdots$ and the induces topology is precisely the $p$-adic topology.

Link to original

Definition

Suppose $\{A_n\}$, $\{B_n\}$ and $\{C_n\}$ are three inverse systems of abelian groups. If for each $n$ have a commutative diagram

where each row is short exact, then say they form an exact sequence of inverse systems. This actually induces a sequence

$$0\to \underleftarrow{\lim }{A}_n\to \underleftarrow{\lim }{B}_n\to \underleftarrow{\lim }{C}_n.\text{\hspace{1em}(*)}$$

In fact, it is always left exact; but not right exact in general. The following proposition shows that if the $\{A_n\}$ is a surjective system, then $(\ast )$ is short exact.

Proposition

If $0\to \left\{A_n\right\}\to \left\{B_n\right\}\to \left\{C_n\right\}\to 0$ is an exact sequence of inverse systems then

$$0\to \underleftarrow{\lim }{A}_n\to \underleftarrow{\lim }{B}_n\to \underleftarrow{\lim }{C}_n$$

is always exact. If, moreover, $\left\{A_n\right\}$ is a surjective system then

$$0\to \underleftarrow{\lim }{A}_n\to \underleftarrow{\lim }{B}_n\to \underleftarrow{\lim }{C}_n\to 0$$

is exact.

\begin{proof} See homework. \end{proof}

Corollary

Let $0\to G^{\prime }\to G\stackrel{p}{\to }{G}^{\prime \prime }\to 0$ be an exact sequence of abelian groups. Let $\{G_n\}$ be a decreasing sequence of subgroups of $G$, which induces a topology on $G$. So $\{G_n\cap G^{\prime }\}$ induces a topology on $G^{\prime }$, and $\{p(G_n)\}$ induces a topology on $G^{\prime \prime }$. Then $0\to \widehat{G^{\prime }}\to \hat{G}\to \widehat{G^{\prime \prime }}\to 0$ is short exact.

\begin{proof} Apply proposition 10.2 to the following sequence

$$0\to G^{\prime }/(G_n\cap G^{\prime })\to G/G_n\to G^{\prime \prime }/p(G_n)\to 0,$$

which is exact for each $n$. Also note $G^{\prime }/(G_{n+1}\cap G^{\prime })\to G^{\prime }/(G_n\cap G^{\prime })$ is surjective for all $n$. \end{proof}

Proposition

Let $G=G_0\supseteq G_1\supseteq \cdots \supseteq G_n\supseteq \cdots$, which gives a topology on $G$. Then $\hat{\hat{G}}=\hat{G}$.

\begin{proof} In ^5f86e5, consider $0\to G_n\to G\to G/G_n\to 0$, and we have $0\to {\hat{G}}_n\to \hat{G}\to \widehat{G/G_n}$, where

$$\widehat{G/G_n}=\underleftarrow{\lim }(G/G_n)/(G_m/G_n)=(0,G/G_1,\cdots ,G/G_n,G/G_n,G/G_n,\cdots )=G/G_n.$$

So $\hat{G}/{\hat{G}}_n\simeq G/G_n$ for all $n$. Take inverse system, then $\hat{\hat{G}}\simeq \hat{G}$. \end{proof}

We now define a special type topology on a ring

Let $\alpha \subseteq A$ be an ideal. Then the sequence $A\supseteq \alpha \supseteq {\alpha }^2\supseteq \cdots$ is a sequence of subgroups. This makes $A$ an topology abelian group.

Fact. This actually makes $A$ a topology ring, that is, $A\times A\stackrel{f}{\to }A,(a,b)\mapsto ab$ is continuous w.r.t. this topology.

\begin{proof} It suffices to show $f^{-1}(a+{\alpha }^n)$ is open. Suppose $(x,y)\in f^{-1}(a+{\alpha }^n)$, then it is easy to see $(x+{\alpha }^n,y+{\alpha }^n)\in f^{-1}(a+{\alpha }^n)$. \end{proof}

Fact. $(A,\alpha )⟶\hat{A}={\lim }_{\leftarrow }A/{\alpha }^n$ is still a topology ring.

Definition

We say $A$ is $\alpha$-adic compete if the natural map $A\to \hat{A}$ is an isomorphism.

Example. Using $(\mathbb{Z},(p))$, its completion ${\mathbb{Z}}_p$ is also a topology ring. So ${\mathbb{Z}}_p$ is $p$-adic complete.

Remark. If $\alpha =(x)$ is principal, then also write $x$-adic complete.

Example. If $A=k[x]$ with $\alpha =(x)$, then $\hat{A}={\lim }_{\leftarrow }k[x]/x^n\simeq k[[x]]$.

Definition

Let $M$ be an $A$-module. Let $\alpha \subseteq A$ be an ideal. So we have a sequence $M\supseteq \alpha M\supseteq \cdots$ giving a topology on abelian group $M$.

Fact. With this topology, $M$ becomes a topology module over the topology ring $A$, that is, $A\times M\to M,(a,m)\mapsto am$ is continuous. Furthermore, $\hat{M}$ is also a topology module over $\hat{A}$.

Fact. (fix $\alpha$-adic topology as above)

• Let $f:M\to N$ be a homomorphism of $A$-modules, then $f$ is automatically continuous w.r.t. $\alpha$-adic topology.
  • Proof. For any $y\in f^{-1}(x+{\alpha }^{k}N)$, then it is easy to see $y+{\alpha }^{k}M\subseteq f^{-1}(x+{\alpha }^{k}M)$.
• Above $f$ induces a continuous map $\hat{f}:\hat{M}\to \hat{N}$.
  • Proof. $\hat{M}={\lim }_{\leftarrow }M/{\alpha }^{n}M$.

Definition

Let $M$ be an $A$-module, and let $\alpha \subseteq A$ be an ideal. A sequence $M=M_0\supseteq M_1\supseteq \cdots \supseteq M_n\supseteq \cdots$ of submodule is called a filtration of $M$. Call it an $\alpha$-filtration if $\alpha M_n\subseteq M_{n+1}$ for all $n$; call it a stable $\alpha$-filtration if it is an $\alpha$-filtration and $\alpha M_n=M_{n+1}$ for $n\gg 0$.

Example. $M\supseteq \alpha M\supseteq \cdots$ is a stable $\alpha$-filtration.

very useful

It $(M_n),(M_n^{\prime })$ are stable $\alpha$-filtrations of $M$, then they have bounded differences, that is, there exists $n_0$ such that $M_{n+n_0}\subseteq M_n^{\prime }\subseteq M_{n-n_0}$ for all $n\gg 0$. So they induced the same topology on $M$, namely, the $\alpha$-adic topology.

\begin{proof} It suffices to consider the case $M_n^{\prime }={\alpha }^{n}M$.

Since $M_n$ is $\alpha$-filtration, there is $M_n\supseteq {\alpha }^{n}M$ and so $M_n\supseteq M_n^{\prime }$. Note that $M_n$ is stable, then there exists $n_0$ such that $M_{n+n_0}={\alpha }^n{M}_{n_0}$ for all $n⩾0$, and so $M_{n+n_0}={\alpha }^n{M}_{n_0}\subseteq {\alpha }^{n}M=M_n^{\prime }$. In summary, $M_{n+2n_0}^{\prime }\subseteq M_{n+n_0}\subseteq M_n^{\prime }$. \end{proof}

Graded Rings and Modules

Definition

A graded ring is a ring $A$ together with $(A_n{)}_{n⩾0}$, where $A_n<A$ are subgroups, such that as additive groups, $A$ is a direct sum of $A_n$, that is,

$$A={\oplus }_{n=0}^{\infty }{A}_n$$

and $A_m{A}_n\subseteq A_{m+n}$. Here ${\oplus }_{n=0}^{\infty }{A}_n=\{(a_0,a_1,\cdots ,a_k,0,\cdots ,0,\cdots )\}$.

Examples.

• $k[x,y]={\oplus }_{n=0}^{\infty }\{\text{homogeneous polynomials of degree }m\}$.
• $k[[x]]$ does not have natural grading.

Fact. (non-trivial) $A_0$ is in fact a subring (recall in our class subring contains $1$). The proof is as follows.

• Since $A_0{A}_0\subseteq A_0$ and $A_0$ is an additive subgroup, it is enough to show $1\in A_0$.

• Suppose $1=x_0+x_1+\cdots +x_n$ for some $x_i\in A_i$. Note that this composition is unique, and we aim to show $x_0=1$. Note that

  $$x_i=x_0{x}_i+\cdots +x_n{x}_i\text{ with }{x}_i{x}_j\in A_{i+j}$$

  and $x_i=x_i$, then $x_i{x}_0=x_i$ and $x_i{x}_1=\cdots =x_i{x}_n=0$.

• Thus $x_0=x_0\cdot 1=x_0+\cdots +x_n$, and so $x_0=1$.

Fact. Each $A_n$ is in fact an $A_0$-module.

Notation. Let $A_{+}={\oplus }_{n>0}{A}_n$. It is clear $A_{+}\subseteq A$ is an ideal, and $A/A_{+}\simeq A_0$.

Definition

Let $A$ be a graded ring. A graded $A$-module is an $A$-module $M$, together with a grading $M={\oplus }_{n=0}^{\infty }{M}_n$, where $M_n\subseteq M$ is a subgroup (not necessary a $A$-submodule), such that $A_m{M}_n\subseteq M_{m+n}$ for any $m,n⩾0$.

Remark.

• Each $M_n$ is an $A_0$-submodule.
• Say $x\in M$ is homogenous of degree $n$ if $x\in M_n$.
• Each $y\in M$ has a unique decomposition $y={\sum }_{i=0}^{\infty }{y}_i$, say $y_i$ is the $i$-th homogeneous component of $y$.

Definition

A homomorphism $f:M\to N$ of $A$-module is called a homomorphism of graded $A$-module, if $f(M_n)\subseteq N_n$ for all $n$.

Proposition

Let $A$ be a graded ring. TFAE:

• $A$ is Noetherian;
• $A_0$ is Noetherian, and $A$ is finitely generated as $A_0$-algebra.

\begin{proof} ii)→i). There is a surjective map $A_0[x_1,\cdots ,x_n]\twoheadrightarrow A$. By ^75xtmd, $A$ is Noetherian.

i)→ii) Recall $A/A_{+}\simeq A_0$, so $A_0$ is Noetherian. Also $A_{+}$ is an ideal of $A$, so $A_{+}$ is finitely generated. Assume that $A_{+}=(x_1,\cdots ,x_s)$ with $\mathrm{deg}{x}_i=k_i$.

Note each $x_n$ can be written as some homogeneous components, so WLOG one can assume each $x_i$ is homogeneous. Define $A^{\prime }=A_0[x_1,\cdots ,x_s]\subseteq A$, and we aim to show $A^{\prime }=A$.

It suffices to prove for any $n$, $A_n\subseteq A^{\prime }$. Note that $n=0$ is trivial, and suppose it holds for all $A_k$ with $k<n$.

Let $y\in A_n\subseteq A_{+}$, then $y={\sum }_{i=1}^s{a}_i{x}_i$ with $a_i\in A$. Break $a_i=b_0+b_1+\cdots +b_{n-k_i}+\cdots$, then

$$a_i{x}_i=b_0{x}_i+\cdots +b_{n-k_i}{x}_i+\cdots =(\mathrm{deg}<n)+b_{n-k_i}{x}_i+(\mathrm{deg}>n).$$

Sum up, and we get $y={\sum }_{i=1}^s{b}_{n-k_i}{x}_i$, as $y$ is homogeneous of degree $n$.

So WLOG, one can assume $a_i\in A_{n-k_i}$ already. By induction hypothesis, $A_{n-k_i}\subseteq A^{\prime }$, thus $y\in A^{\prime }$. Now we finish the proof. \end{proof}

$A^{\ast }$ and $M^{\ast }$

Definition

Let $\alpha \subseteq A$ be an ideal. Define a graded ring $A^{\ast }={\oplus }_{n=0}^{\infty }{\alpha }^n$, so $A\hookrightarrow A^{\ast }$. Suppose $M$ is an $A$-module, with an $\alpha$-filtration $\{M_n{\}}_{n=0}^{\infty }$, then one can define $M^{\ast }={\oplus }_{n=0}^{\infty }{M}_n$, then it is a graded $A^{\ast }$-module.

Fact. If $A$ is Noetherian, then $\alpha$ is finitely generated. Assume that $\alpha =(a_1,\cdots ,a_s)$. Then there is a surjective map $A[x_1,\cdots ,x_s]\twoheadrightarrow A^{\ast },x_i\mapsto a_i$.

Lemma

For a Noetherian ring $A$, a finitely generated $A$-module $M$, and an $\alpha$-filtration $\{M_n{\}}_{n=0}^{\infty }$, TFAE:

• $M^{\ast }$ is finitely generated $A^{\ast }$-module;
• $\{M_n\}$ is $\alpha$-stable.

\begin{proof} For each $n$, define $Q_n={\oplus }_{i=0}^n{M}_i\subseteq M^{\ast }$, which is a subgroup but not a submodule. The $A^{\ast }$-submodule generated by $Q_n$ is

$$M_n^{\ast }=M_0\oplus \cdots \oplus M_n\oplus \alpha M_n\oplus {\alpha }^2{M}_n\oplus \cdots$$

Since $Q_n$ is a finitely generated $A$-module, $M_n^{\ast }$ is also a finitely generated $A^{\ast }$-module. Note that $M_n^{\ast }$ is increasing and ${\cup }_{n=0}^{\infty }{M}_n^{\ast }=M^{\ast }$.

So $M^{\ast }$ is a finitely generated $A^{\ast }$-module iff $M^{\ast }$ is a Noetherian $A^{\ast }$-module iff the increasing sequence $\{M_n^{\ast }\}$ is stable iff when $n_0\gg 0$, $M_{n_0}^{\ast }=M^{\ast }$ iff for $n_0\gg 0$, ${\alpha }^k{M}_{n_0}=M_{n_0+k}$ for any $k>0$ iff $\{M_n\}$ is $\alpha$-stable. \end{proof}

Generalized Artin-Rees Lemma

For a Noetherian ring $A$, a finitely generated $A$-module $M$, and a stable $\alpha$-filtration $\{M_n{\}}_{n=0}^{\infty }$, if $M^{\prime }\subseteq M$ is an submodule, then $\{M_n\cap M^{\prime }{\}}_{n=0}^{\infty }$ is also a stable $\alpha$-filtration.

\begin{proof} First note $\alpha (M_n\cap M^{\prime })\subseteq M_{n+1}\cap M^{\prime }$, so it is an $\alpha$-filtration. Denote $M^{\ast }={\oplus }_{n=0}^{\infty }(M_n\cap M^{\prime })\subseteq M^{\ast }$. By ^m9pvrs, since $M^{\ast }$ is finitely generated and $A^{\ast }$ is Noetherian, $M^{\ast }$ is finitely generated and so $\{M_n\cap M^{\prime }\}$ is $\alpha$-stable. \end{proof}

Artin-Rees Lemma

Let $A,\alpha ,M,M^{\prime }$ as above, then there exists $k\gg 0$ such that ${\alpha }^{n}M\cap M^{\prime }={\alpha }^{n-k}({\alpha }^{k}M\cap M^{\prime })$ for any $n⩾k$.

\begin{proof} Apply $M_n={\alpha }^{n}M$ to ^r7klzn. \end{proof}

Remark. The motivation of this lemma is, for a Noetherian ring, its induced topology and the alpha-adic topology are the same. With this lemma, the exactness is preserved under completion, and ultimately the completion is flat. See ^z8znz5.

Theorem

For a Noetherian ring $A$ and its ideal $\alpha$, define $A$-modules $M\supseteq M^{\prime }$, then the filtration $\{{\alpha }^k{M}^{\prime }{\}}_n$ and $\{{\alpha }^{n}M\cap M^{\prime }{\}}_n$ are both stable $\alpha$-filtration, so they induce the same topology on $M^{\prime }$.

\begin{proof} Apply ^2c426a. \end{proof}

Proposition

Some notation as above, write $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, take $\alpha$-adic completion, then $0\to \widehat{M^{\prime }}\to \hat{M}\to \widehat{M^{\prime \prime }}\to 0$ is a short exact sequence of $\hat{A}$-module.

\begin{proof} Apply ^4t4tgc and ^5f86e5. \end{proof}

Proposition

For a ring $A$ and a finitely generated $A$-module $M$, $\hat{A}\otimes M\to \hat{M}$ is surjective. If $A$ is Noetherian, then $\hat{A}{\otimes }_{A}M\to \hat{M}$ is an isomorphism.

\begin{proof} Since $M$ is finitely generated, there exists a short exact sequence $0\to N\to F\to M\to 0$ with $F=A^{\oplus n}$. Consider

By ^5f86e5, they are right exact sequence. It is easy to check $\alpha$ is surjective, because $\beta$ is an isomorphism and $\delta$ is surjective.

Now assume $A$ is Noetherian. By ^umxmsx, $ϵ$ is injective, and $\gamma$ is surjective by arguments above. Use diagram chasing, $\alpha$ is injective. \end{proof}

It is the main theorem of this section.

Proposition

For a Noetherian ring $A$ and an ideal $\alpha$, $\hat{A}$ ($\alpha$-adic completion) is a flat $A$-algebra.

\begin{proof} Recall ^e6ct2u, to see an $A$-module is flat, it suffices to check for each short exact sequence $(\ast )$, $(\ast )\otimes \hat{A}$ is still flat.

Now let $0\to M\to N\to P\to 0$ is a short exact sequence of finitely generated $A$-module, by ^umxmsx, $0\to \hat{M}\to \hat{N}\to \hat{P}\to 0$ is exact. By ^wc26ph, $0\to M\otimes \hat{A}\to N\otimes \hat{A}\to P\otimes \hat{A}\to 0$ is exact. Hence $\hat{A}$ is a flat $A$-module. \end{proof}

Finally, we give some properties of $\hat{A}$ and $\hat{\alpha }$.

Proposition

For a Noetherian ring $A$ and its ideal $\alpha$, define $\hat{A}$ as $\alpha$-adic completion of $A$, that is, $\hat{A}={\lim }_{\leftarrow }A/{\alpha }^n$. Then:

• $\hat{\alpha }=\hat{A}\alpha \simeq \hat{A}{\otimes }_A\alpha$
• $\widehat{({\alpha }^n)}=(\hat{\alpha }{)}^n$
• ${\alpha }^n/{\alpha }^{n+1}\simeq \widehat{({\alpha }^n)}/\widehat{({\alpha }^{n+1})}$
• $\hat{\alpha }\subseteq \mathrm{Jac}(\hat{A})$.

\begin{proof} i) We have $\hat{A}{\otimes }_A\alpha \simeq \hat{\alpha }$ by ^wc26ph, and consider the following composition

$$\hat{A}{\otimes }_A\alpha \simeq \hat{\alpha }\hookrightarrow \hat{A}$$

where $\hat{\alpha }\to \hat{A}$ injective is guaranteed by ^umxmsx. Note that the image of the composition above is $\hat{A}\alpha$, then we finish the proof.

ii) Apply i) to ${\alpha }^n$, then we get $\widehat{({\alpha }^n)}=\hat{A}{\alpha }^n$. It is easy to check $\hat{A}{\alpha }^n=(\hat{A}\alpha {)}^n$. Then by i) $(\hat{A}\alpha {)}^n=(\hat{\alpha }{)}^n$.

iii) By the following commute diagram

take kernels of these surjective maps, and we get ${\alpha }^n/{\alpha }^{n+1}\simeq \widehat{({\alpha }^n)}/\widehat{({\alpha }^{n+1})}$.

iv) By ii) and iii) we have $\hat{A}={\lim }_{\leftarrow }A/{\alpha }^n\simeq {\lim }_{\leftarrow }\hat{A}/\widehat{({\alpha }^n)}\simeq {\lim }_{\leftarrow }\hat{A}/(\hat{\alpha }{)}^n$, which means $\hat{A}$ is $\hat{\alpha }$-adic complete. Let $x\in \hat{\alpha }$ for any $y\in A$, then $(1-xy{)}^{-1}=1+xy+(xy{)}^2+\cdots$ converges to an element in $\hat{A}$ (since $\hat{A}$ is $\hat{\alpha }$-adic complete). Thus $1-xy\in A^{\times }$. By ^h2e8bp, $x\in \mathrm{Jac}(A)$. \end{proof}

Associated Graded Ring

In this section, we aim to show for a Noetherian ring $A$ and its ideal $\alpha$, $\hat{A}={\lim }_{\leftarrow }A/{\alpha }^n$ is Noetherian.

Definition

For a Noetherian ring $A$ and its ideal $\alpha$, define (the associated graded ring)

$$G(A)=G_{\alpha }(A)={\oplus }_{n=0}^{\infty }{\alpha }^n/{\alpha }^{n+1},$$

where ${\alpha }^0=A$. It is easy to check $G(A)$ is a graded ring.

For an $A$-module $M$ with an $\alpha$-filtration $(M_n)$, define $G(M)={\oplus }_{n=0}^{\infty }{M}_n/M_{n+1}$, then it is a graded $G(A)$-module.

Remark. We define $A^{\ast }={\oplus }_{i=0}^{\infty }{\alpha }^n$ and $M^{\ast }={\oplus }_{i=0}^{\infty }{M}_n$ earlier. They are different with $G(A)$ and $G(M)$!

Proposition

For a Noetherian $A$ and an ideal $\alpha \subseteq A$, we have:

• $G_{\alpha }(A)$ is Noetherian
• $G_{\alpha }(A)\simeq G_{\hat{\alpha }}(\hat{A})$
• if $M$ is a finitely generated $A$-module and $(M_n)$ is a stable $\alpha$-field, then $G(M)$ is a finitely generated $G(A)$-module.

\begin{proof} i) Since $A$ is Noetherian, $\alpha =(x_1,\cdots ,x_s)$ is finitely generated. Let ${\overline{x}}_i=x_i+{\alpha }^2$ in $\alpha /{\alpha }^2$, then one can see $G(A)=A/\alpha [{\overline{x}}_1,\cdots ,{\overline{x}}_s]$ is Noetherian.

ii) Use ^wvvi9p iii).

iii) Since $(M_n)$ is $\alpha$-stable, there exists $n_0$ such that $M_{n_0+r}={\alpha }^r{M}_{n_0}$ for any $r⩾0$. Then $G(M)$ is generated over $G(A)$ by ${\oplus }_{n=0}^{n_0}{M}_n/M_{n+1}$, because

$$M_{n_0}/M_{n_0+1}\otimes \alpha /{\alpha }^2\twoheadrightarrow M_{n_0+1}/M_{n_0+2}=\alpha M_{n_0}/\alpha M_{n_0+1}.$$

Thus $G(M)$ is finitely generated over $G(A)$ by $A$ is Noetherian. \end{proof}

Lemma

Let $A,B$ be abelian groups with filtration. For $A=A_0\supseteq A_1\supseteq \cdots$ and $B=B_0\supseteq B_1\supseteq \cdots$, define $G(A)={\oplus }_{n=0}^{\infty }{A}_n/A_{n+1}$ and $G(B)={\oplus }_{n=0}^{\infty }{B}_n/B_{n+1}$, and define $\hat{A}={\lim }_{\leftarrow }A/A_n$ and $\hat{B}={\lim }_{\leftarrow }B/B_n$. Let $\varphi :A\to B$ be a group homomorphism such that $\varphi (A_n)\subseteq \varphi (B_n)$. So it induces a group homomorphism $G(\varphi ):G(A)\to G(B)$ and $\hat{\varphi }:\hat{A}\to \hat{B}$. Then:

• $G(\varphi )$ injective yields $\hat{\varphi }$ injective;
• $G(\varphi )$ surjective yields $\hat{\varphi }$ surjective.

\begin{proof} Consider the following commutative diagram

and by ^jgbcg4 we get a exact sequence

$$0\to \ker G_n(\varphi )\to \ker {\alpha }_{n+1}\to \ker {\alpha }_n\to \mathrm{coker}{G}_n(\varphi )\to \mathrm{coker}{\alpha }_{n+1}\to \mathrm{coker}{\alpha }_n\to 0.\text{\hspace{1em}(*)}$$

i) If $G(\varphi )$ is injective, then $\ker G_n(\varphi )=0$ for all $n$. Consider $n=0$, the diagram becomes

and so $G_0(\varphi )={\alpha }_1$. Thus ${\alpha }_1$ is injective and $\ker {\alpha }_1=0$. We get an exact sequence $0\to 0\to \ker {\alpha }_2\to 0\to \cdots$ and so $\ker {\alpha }_2=0$. By induction, one can check that $\ker {\alpha }_n=0$ for all $n$, then $\hat{\varphi }$ is injective.

ii) If $G(\varphi )$ is surjective, then $G_n(\varphi )$ is surjective and ${\alpha }_1$ is surjective. Use right half of above long exact sequence, one can see ${\alpha }_n$ is surjective for all $n$. Note that it is not enough to see $\hat{\varphi }={\lim }_{\leftarrow }{\alpha }_n$ is surjective. But consider the exact sequence of inverse systems

Since the left column is a surjective system by $(\ast )$, we know

$$0\to \underleftarrow{\lim }\ker {\alpha }_n\to \hat{A}\to \hat{B}\to 0$$

is short exact by ^bxmt3l. Hence $\hat{\varphi }$ is surjective. \end{proof}

Proposition

For a ring $A$ and its ideal $\alpha$, suppose $M$ is a $A$-module and $(M_n{)}_{n⩾0}$ is an $\alpha$-filtration. Define $\hat{M}={\lim }_{\leftarrow }M/M_n$ Suppose $A=\hat{A}={\lim }_{\leftarrow }A/{\alpha }^n$ and ${\cap }_{n⩾0}{M}_n=0$, and suppose $G(M)={\oplus }_{n⩾0}{M}_n/M_{n+1}$ is a finitely generated $G(A)={\oplus }_{n⩾0}{\alpha }^n/{\alpha }^{n+1}$-module. Then $M$ is a finitely generated $A$-module.

\begin{proof} Choose a finite set of generators of $G(M)$ over $G(A)$. By breaking elements into homogeneous elements, we can assume

$$G(M)={⟨{\xi }_1,\cdots ,{\xi }_s⟩}_{G(A)}$$

with ${\xi }_i\in M_{n(i)}/M_{n(i)+1}$. Choose a lift $x_i\in M_{n(i)}$ of ${\xi }_i$.

Let $F$ be a free $A$-module of rank $s$, with basis $(f_1,\cdots ,f_s)$. Define $\varphi :F\to M,f_i\mapsto x_i$. We aim to show $\varphi$ is surjective. Since $A$ is complete, $F$ is also complete. Also $M\to \hat{M}={\lim }_{\leftarrow }M/M_n$ is injective because $\ker =\cap M_n=0$. So it suffices to show the composite $\hat{\varphi }:F\to M\hookrightarrow \hat{M}$ is surjective. Via ^xmmn5a, it suffices to construct some surjective $G(\varphi )$.

First, define a filtration on $F={\oplus }_{i=1}^{s}A{f}_i$, so that the submodule $F_k={\oplus }_{i=1}^s{\alpha }^{k-n(i)}{f}_i$ with convention ${\alpha }^{<0}=A$. This makes sure the map $\varphi :F\to M,f_i\to x_i$ is compatible with filtration, because $\varphi ({\alpha }^{k-n(i)}{f}_i)\subseteq {\alpha }^{k-n(i)}{M}_{n(i)}\subseteq M_k$. ^xmmn5a tells us to show $\varphi$ surjective, it is suffices to show $G(\varphi )$ is surjective. But

$$G(\varphi ):{⟨{\overline{f}}_1,\cdots ,{\overline{f}}_s⟩}_{G(A)}\stackrel{(\ast )}{=}{\oplus }_{n⩾0}{F}_k/F_{k+1}=G(F)\twoheadrightarrow G(M)={\oplus }_{n⩾0}{M}_k/M_{k+1}={⟨{\xi }_1,\cdots ,{\xi }_s⟩}_{G(A)},{\overline{f}}_i\mapsto {\xi }_i$$

is surjective.

To explain $(\ast )$, there is an example. Say $s=2$, $n(1)=1$ and $n(2)=3$. There is $F_1=A{f}_1\oplus A{f}_2$, $F_2=\alpha f_1\oplus A{f}_2$, $F_3={\alpha }^2{f}_1\oplus A{f}_2$, $F_4={\alpha }^3{f}_1\oplus {\alpha }^2{f}_2$. Then $G(F)=0\oplus (A/\alpha ){\overline{f}}_1\oplus (\alpha /{\alpha }^2{\overline{f}}_1)\oplus ({\alpha }^2/{\alpha }^3{\overline{f}}_1\oplus A/\alpha {\overline{f}}_2)$. Note that ${\overline{f}}_1$ appears in the first summand and ${\overline{f}}_2$ appears in the third summand, which is coincides with $n(1)=1$ and $n(2)=3$. \end{proof}

Remark. As a corollary, $M\simeq \hat{M}$ as $\hat{\varphi }$ is surjective.

Corollary

Use the same hypothesis as above. Suppose further that $G(M)$ is Noetherian $G(A)$-module, then $M$ s Noetherian $A$-module.

\begin{proof} Let $M^{\prime }\subseteq M$ be a submodule. We want to prove $M^{\prime }$ is finitely generated over $A$.

Define a filtration $M_n^{\prime }=M_n\cap M^{\prime }$, which is still an $\alpha$-filtration and $\cap M_n^{\prime }\subseteq \cap M_n=0$. Note that $M_n^{\prime }/M_{n+1}^{\prime }\to M_n/M_{n+1}$ is injective by definition. Thus $G(M^{\prime })\hookrightarrow G(M)$ is a submodule.

Since $G(M)$ is a Noetherian $G(A)$-module, $G(M^{\prime })$ is a finitely generated $G(A)$-module. By ^yad12c, $M^{\prime }$ is a finitely generated $A$-module. \end{proof}

This is the main theorem of this section.

Theorem

For a Noetherian ring $B$ and its ideal $\beta$, $\hat{B}={\lim }_{\leftarrow }B/{\beta }^n$ is a Noetherian ring.

\begin{proof} Apply ^yad12c and ^5fncr8 with $A=\hat{B}$, $M=\hat{B}$ filtered by ${\hat{\beta }}^n\hat{B}$, then it suffices to show "$G(M)$" is a Noetherian "$G(A)$"-module. But "$G(A)$"$=G_{\hat{\beta }}(\hat{B})$ is a Noetherian ring by ^7r78vs. So the proof is complete. \end{proof}

Corollary

If $A$ is Noetherian, then $A[[x]]$ is Noetherian.

\begin{proof} By ^75xtmd $A[x]$ is Noetherian. Then by ^b6b2a0, ${\lim }_{\leftarrow }A[x]/x^n$ is Noetherian. \end{proof}