Inverse Limits and Noetherian Completion

If is $R$-linear, define $f_n:={\pi }_n\circ f\text{\hspace{0.17em}⁣}:N\to M_n\mathrm{.}$ For each $x\in N$, the element $f(x)$ lies in the compatible submodule from [L1], so its coordinates satisfy ${\phi }_n(f_n(x))=f_{n-1}(x)$ for every $n\ge 2$. Thus the family $(f_n)$ is compatible.

Conversely, let $(f_n)$ be a compatible family and define $f(x):=(f_n(x){)}_{n\ge 1}\in \prod M_n\mathrm{.}$ Compatibility says ${\phi }_n(f_n(x))=f_{n-1}(x)$ for every $n\ge 2$, so $f(x)$ actually lies in . Since products and coordinate maps are $R$-linear, $f$ is $R$-linear.

The two constructions are inverse to each other: starting from $f$ and then taking coordinates recovers each $f_n$, while starting from $(f_n)$ and then forming $f$ gives the unique map whose $n$th coordinate is $f_n$.

Therefore maps are in bijection with compatible families $(f_n)$, which is exactly the terminal-cone universal property.

For each $n$, let $K_n:=\ker (f_n)$. Compatibility of the squares ${\psi }_n\circ f_n=f_{n-1}\circ {\phi }_n$ shows that ${\phi }_n(K_n)\subseteq K_{n-1}$, so the kernels form an inverse system.

An element lies in the kernel of exactly when $f_n(x_n)=0$ for every $n$. By [L2], that is equivalent to $x_n\in K_n$ for every $n$. Together with the compatibility condition from [L1], this says precisely that .

Thus the underlying subsets of coincide inside $\prod M_n$, and the module structures also agree componentwise. Hence these two modules are canonically equal, in particular canonically isomorphic.

For an exact sequence $0\to A_n\to B_n\to C_n$, exactness means $A_n=\ker (B_n\to C_n)$ for every $n$. Applying step 2.1 to the maps $B_n\to C_n$ gives which is exactly left exactness.

By [L1], the sequence of inverse limits is already exact at and at . It remains to prove surjectivity of

Let . For each $n$, set $E_n:=g_n^{-1}(c_n)\subseteq B_n\mathrm{.}$ Since $g_n$ is surjective, each $E_n$ is nonempty. Compatibility of the inverse system and of the family $(c_n)$ makes every transition map $B_{n+1}\to B_n$ restrict to a map $E_{n+1}\to E_n$.

The system $(E_n)$ is Mittag-Leffler as a system of sets. Fix $m$. By [L2] choose $c(m)\ge m$ such that $\mathrm{im}(A_n\to A_m)=\mathrm{im}(A_{c(m)}\to A_m)(n\ge c(m))\mathrm{.}$ For $n\ge c(m)$ the inclusion $\mathrm{im}(E_n\to E_m)\subseteq \mathrm{im}(E_{c(m)}\to E_m)$ is automatic. For the reverse inclusion, take $y\in \mathrm{im}(E_{c(m)}\to E_m)$ and choose $e_c\in E_{c(m)}$ mapping to $y$. Choose any $e_n^0\in E_n$. The images of $e_c$ and $e_n^0$ in $C_{c(m)}$ are both $c_{c(m)}$, so their difference lies in $A_{c(m)}$. By the stabilization choice there is $a_n\in A_n$ whose image in $A_m$ equals the image of $e_c-e_n^0$. Then $e_n:=e_n^0+a_n$ lies in $E_n$ and maps to $y$ in $E_m$. Hence the images stabilize.

For each $n$, let $E_n^{\mathrm{\prime }}:={\bigcap }_{m\ge n}\mathrm{im}(E_m\to E_n)\mathrm{.}$ Because $(E_n)$ is Mittag-Leffler and nonempty, $E_n^{\mathrm{\prime }}$ is equal to one stable image and is therefore nonempty. The restricted maps $E_{n+1}^{\mathrm{\prime }}\to E_n^{\mathrm{\prime }}$ are surjective: if $y\in E_n^{\mathrm{\prime }}$, then $y$ comes from some sufficiently high stage $E_m$ with $m\ge n+1$, and the image of that same element in $E_{n+1}$ lies in $E_{n+1}^{\mathrm{\prime }}$ and maps to $y$.

By the Axiom of Choice, choose $x_1\in E_1^{\mathrm{\prime }}$, and after $x_n$ has been chosen choose $x_{n+1}\in E_{n+1}^{\mathrm{\prime }}$ mapping to $x_n$; this is possible by surjectivity from step 2.1. Then $(x_n{)}_{n\ge 1}$ is an element of , hence of , and by construction it maps to .

Therefore is surjective. Combined with step 1.1, this proves exactness of

The completion map ${\kappa }_M$ is the unique map into the inverse limit of [L1] whose $n$th component, for $n\ge 1$, is the quotient map $q_n\text{\hspace{0.17em}⁣}:M\to M\mathrm{/}{I}^{n}M\mathrm{.}$

An element lies in $\ker ({\kappa }_M)$ exactly when every quotient map $q_n$, $n\ge 1$, kills it, equivalently when it belongs to every $I^{n}M$ for $n\ge 1$. Since $I^{0}M=M$, this gives $\ker ({\kappa }_M)={\bigcap }_{n\ge 1}{I}^{n}M={\bigcap }_{n\ge 0}{I}^{n}M\mathrm{.}$

Let $f\text{\hspace{0.17em}⁣}:M\to N$ be $R$-linear with $N$ complete. Since $f$ is $R$-linear, one has $f(I^{n}M)\subseteq I^{n}N$ for every $n\ge 1$, so $f$ induces maps $f_n\text{\hspace{0.17em}⁣}:M\mathrm{/}{I}^{n}M\to N\mathrm{/}{I}^{n}N\mathrm{.}$ These maps are compatible with the quotient transition maps.

By [L4], the compatible family $(f_n)$ yields a unique map such that the $n$th projection of $\tilde{f}{\kappa }_M$ is $f_n{q}_n$. These are also the coordinates of ${\kappa }_{N}f$, so uniqueness in [L4] gives $\tilde{f}{\kappa }_M={\kappa }_{N}f\mathrm{.}$

The image of ${\kappa }_M$ is dense for the inverse-limit topology. Indeed, for $x\in \widehat{M}$ and $n\ge 1$, choose $m\in M$ representing ${\pi }_n^M(x)\in M\mathrm{/}{I}^{n}M$; then $x-{\kappa }_M(m)\in K_n$. Now let $g:\widehat{M}\to N$ be continuous and vanish on ${\kappa }_M(M)$. If $g(x)\mathrm{\ne }0$, separatedness of $N$ from [L2] gives $n$ with $g(x)\mathrm{\notin }{I}^{n}N$. The inverse image $g^{-1}(g(x)+I^{n}N)$ is an open neighbourhood of $x$ by [L3], so density gives some ${\kappa }_M(m)$ in it. But $g{\kappa }_M(m)=0$, which would put $g(x)$ in $I^{n}N$, a contradiction. Hence $g=0$. Applying this to the difference of two continuous extensions proves uniqueness.

The last equality shows that ${\kappa }_M$ is injective exactly when ${\bigcap }_{n\ge 0}{I}^{n}M=0$, which is exactly the definition of being $I$-adically separated.

Write ${\pi }_n^M:\widehat{M}\to M\mathrm{/}{I}^{n}M$ and . Step 1.4 gives ${\pi }_n^N\tilde{f}=f_n{\pi }_n^M,$ so $\tilde{f}(K_n)\subseteq \ker ({\pi }_n^N)$ for every $n\ge 1$. Hence $\tilde{f}$ is continuous for the two inverse-limit topologies. Because $N$ is complete, [L2] identifies under this isomorphism the preimage of $\ker ({\pi }_n^N)$ is $I^{n}N$, so [L3] makes ${\kappa }_N$ a homeomorphism from the $I$-adic topology to the inverse-limit topology. Thus $\widehat{f}:={\kappa }_N^{-1}\tilde{f}:\widehat{M}\to N$ is continuous and satisfies $\widehat{f}{\kappa }_M=f$.

Parts 1 and 2 are proved by steps 1.2, 1.5, 2.1, and 2.2.

For each $n\ge 1$, the finite geometric sum $v_n:=1+a+a^2+\cdots +a^{n-1}$ satisfies $(1-a)v_n=1-a^n\mathrm{.}$ Since $a\in I$, one has $a^n\in I^n$, so the image of $v_n$ in $R\mathrm{/}{I}^n$ is an inverse to the image of $u=1-a$.

The residue classes $(v_n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>I^n{)}_n$ are compatible: the image of $v_{n+1}$ in $R\mathrm{/}{I}^n$ equals the image of $v_n$ because $v_{n+1}-v_n=a^n\in I^n$. Therefore they define an element

By [L1], there is a unique element $w\in R$ corresponding to $v$. Since each component of $v$ is an inverse to the image of $u$, the products $uw$ and $wu$ map to $1$ in every quotient $R\mathrm{/}{I}^n$. Completeness includes separatedness, so the kernel of is $0$; hence $uw=wu=1.$ Thus $u$ is a unit.

Let $x\in I$ and $r\in R$. Then $1-rx\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$, so step 3.1 shows $1-rx$ is a unit. The elementary ideal characterization of the Jacobson radical now gives $x\in J(R)$: if a maximal ideal omitted $x$, its image would generate the residue field, contradicting invertibility of every $1-rx$. Therefore $I\subseteq J(R)$.

For each $n\ge 0$, the given short exact sequence induces an exact sequence $0\to M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)\to M\mathrm{/}{I}^{n}M\to M^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{I}^n{M}^{\mathrm{\prime }\mathrm{\prime }}\to 0.$ The first map is injective because its kernel is $(M^{\mathrm{\prime }}\cap I^{n}M)\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)$, and the second map is surjective because every class in $M^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{I}^n{M}^{\mathrm{\prime }\mathrm{\prime }}$ lifts to a class in $M\mathrm{/}{I}^{n}M$.

The left inverse system $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)$ has surjective transition maps, hence is Mittag-Leffler. Indeed, if $n\ge m$ and $x\in M^{\mathrm{\prime }}$, then the class of $x$ in $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{m}M)$ is the image of the class of the same $x$ in $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)\mathrm{.}$ Therefore every transition map $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)\to M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{m}M)$ is surjective.

Taking inverse limits in the sequences of step 1.1 and using [L1] yields an exact sequence

By [L2], the middle and right inverse limits are $\widehat{M}$ and $\widehat{M^{\mathrm{\prime }\mathrm{\prime }}}$. For the left inverse limit, the induced filtration $M^{\mathrm{\prime }}\cap I^{n}M$ on $M^{\mathrm{\prime }}$ is equivalent to the intrinsic $I$-adic filtration $I^n{M}^{\mathrm{\prime }}$ by The filtration induced on a submodule is equivalent to its intrinsic ideal-adic filtration, so its completion is $\widehat{M^{\mathrm{\prime }}}$.

Substituting these identifications into step 3.1 gives $0\to \widehat{M^{\mathrm{\prime }}}\to \widehat{M}\to \widehat{M^{\mathrm{\prime }\mathrm{\prime }}}\to 0,$ which is the claimed exactness.

If $M=R$, then $R{\otimes }_R\widehat{R}\cong \widehat{R}$ by [L3], and this is exactly ${\theta }_R$. Therefore ${\theta }_R$ is an isomorphism.

Choose an exact sequence $F_1\stackrel{u}{\to }{F}_0\to M\to 0$ with $F_0$ and $F_1$ finite free. Tensoring with $\widehat{R}$ and using [L2] gives an exact row $F_1{\otimes }_R\widehat{R}\stackrel{u\otimes 1}{\to }{F}_0{\otimes }_R\widehat{R}\to M{\otimes }_R\widehat{R}\to 0.$ Completing the exact sequence $F_1\stackrel{u}{\to }{F}_0\to M\to 0$ and using [L1] gives another exact row $\widehat{F_1}\stackrel{\widehat{u}}{\to }\widehat{F_0}\to \widehat{M}\to 0.$

Since tensor products commute with finite direct sums by [L3], the same is true for finite free modules: $R^m{\otimes }_R\widehat{R}\cong {\widehat{R}}^m\cong \widehat{R^m}\mathrm{.}$ Thus ${\theta }_F$ is an isomorphism for every finite free module $F$. In particular, ${\theta }_{F_0}$ and ${\theta }_{F_1}$ are isomorphisms in step 1.2.

The maps ${\theta }_{F_1}$, ${\theta }_{F_0}$, and ${\theta }_M$ form a commutative diagram between the two exact rows of step 1.2. Since the first two vertical maps are isomorphisms by step 2.1, the two rows present $M{\otimes }_R\widehat{R}\text{and}\widehat{M}$ as cokernels of isomorphic maps. Therefore ${\theta }_M$ is an isomorphism.

So the completion of a finite module is obtained by extension of scalars from $R$ to $\widehat{R}$.

Apply [L1] to the short exact sequence $0\to N\to M\to M\mathrm{/}N\to 0.$ This gives an exact sequence $0\to \widehat{N}\to \widehat{M}\to \widehat{M\mathrm{/}N}\to 0.$ Therefore $\widehat{M}\mathrm{/}\widehat{N}\cong \widehat{M\mathrm{/}N},$ proving part 1.

By [L2], the map $\widehat{N}\to \widehat{M}$ identifies with $N{\otimes }_R\widehat{R}⟶M{\otimes }_R\widehat{R}\mathrm{.}$ Its image is, by definition, the $\widehat{R}$-submodule generated by the image of $N$, namely $N\widehat{R}$. Hence the image of $\widehat{N}$ in $\widehat{M}$ is $N\widehat{R}$. Taking $N=JM$ yields $\widehat{JM}\cong J\widehat{M}\mathrm{.}$

Apply part 1 to the submodule $I^{n}M\subseteq M$. Then $\widehat{M}\mathrm{/}\widehat{I^{n}M}\cong \widehat{M\mathrm{/}{I}^{n}M}\mathrm{.}$ By step 1.2, $\widehat{I^{n}M}$ identifies with $I^n\widehat{M}$. Also $M\mathrm{/}{I}^{n}M$ is annihilated by $I^n$, so its $I$-adic filtration reaches $0$ after stage $n$ and its completion is canonically itself. Hence $\widehat{M\mathrm{/}{I}^{n}M}\cong M\mathrm{/}{I}^{n}M,$ which proves part 3.

Parts 1, 2, and 3 are exactly the three displayed conclusions above.

Let $J\subseteq R$ be an ideal. It is a finite $R$-module by [L4]. Hence [L1] identifies $J{\otimes }_R\widehat{R}$ with the completion $\widehat{J}$, and under this identification the multiplication map $J{\otimes }_R\widehat{R}\to \widehat{R}$ becomes the natural map $\widehat{J}\to \widehat{R}\mathrm{.}$

The map $\widehat{J}\to \widehat{R}$ is injective because [L3] applies to the short exact sequence $0\to J\to R\to R\mathrm{/}J\to 0.$ Therefore $J{\otimes }_R\widehat{R}\to \widehat{R}$ is injective for every ideal $J$.

Applying [L2], we conclude that $\widehat{R}$ is flat as an $R$-module.

Let $\mathfrak{m}\subset R$ be a maximal ideal. Because $I\subseteq J(R)$, one has $I\subseteq \mathfrak{m}$, so [L2] gives $\widehat{R}\mathrm{/}\mathfrak{m}\widehat{R}\cong R\mathrm{/}\mathfrak{m}\mathrm{\ne }0.$

The ring $R\mathrm{/}\mathfrak{m}$ is a field, so step 1.1 shows directly that $\mathfrak{m}\widehat{R}$ is a maximal, hence prime, ideal of $\widehat{R}$. Its contraction to $R$ contains $\mathfrak{m}$; the contraction is proper because it is the preimage of a prime ideal, so maximality of $\mathfrak{m}$ forces the contraction to equal $\mathfrak{m}$. Thus every maximal ideal of $R$ has a prime ideal of $\widehat{R}$ above it.

The completion map $R\to \widehat{R}$ is flat by [L1]. Together with step 2.1, [L3] now shows that $R\to \widehat{R}$ is faithfully flat.

By [L2], choose generators $f_1,\dots ,f_r$ of $I$. The quotient $R\mathrm{/}I$ is Noetherian by [L1], hence the polynomial ring $(R\mathrm{/}I)[T_1,\dots ,T_r]$ is Noetherian by [L3].

The graded ring ${\mathrm{gr}}_I(R):={⨁}_{n\ge 0}{I}^n\mathrm{/}{I}^{n+1}$ is a quotient of $(R\mathrm{/}I)[T_1,\dots ,T_r]$ via $T_j\mapsto \overline{f_j}$, so ${\mathrm{gr}}_I(R)$ is Noetherian.