Inverse Limits and Noetherian Completion — Examples

By [L2], applying the adic-completion definition to the ring $\mathbb{Z}$ and the ideal $(p)$ gives exactly the inverse system displayed above. Therefore its inverse limit is canonically the $(p)$-adic completion of $\mathbb{Z}$.

Under this identification, the completion map is the familiar residue map $\mathbb{Z}\to {\mathbb{Z}}_p,m\mapsto (m<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p^n{)}_{n\ge 1}\mathrm{.}$ Its kernel is ${\bigcap }_{n\ge 0}{p}^n\mathbb{Z}$, which is $0$ because the only integer divisible by every power of $p$ is $0$.

So the compatible-residue construction of ${\mathbb{Z}}_p$ and the $(p)$-adic-completion construction agree canonically.

Applying [L1] to $M=\mathbb{Z}$ and $I=(p)$ gives the displayed formula for ${\kappa }_{\mathbb{Z}}$. Compatibility is automatic because reduction modulo $p^r$ followed by reduction modulo $p^s$ agrees with direct reduction modulo $p^s$ whenever $r\ge s$.

By [L2], $\ker ({\kappa }_{\mathbb{Z}})={\bigcap }_{r\ge 1}{p}^r\mathbb{Z}\mathrm{.}$ If $n\mathrm{\ne }0$, then for sufficiently large $r$ one has $p^r>\mathrm{\mid }n\mathrm{\mid }$, so $p^r$ cannot divide $n$. Thus the intersection is $0$, and ${\kappa }_{\mathbb{Z}}$ is injective.

The image of ${\kappa }_{\mathbb{Z}}$ is therefore exactly the copy of ordinary integers inside the completion, written componentwise as their residue systems modulo $p^r$.

For each $k\ge 1$, the inclusion $I^c\subseteq J$ gives $I^{ck}M\subseteq J^{k}M$ and hence a quotient map $M\mathrm{/}{I}^{ck}M\to M\mathrm{/}{J}^{k}M\mathrm{.}$ The powers $ck$ form a cofinal subsystem of the $I$-adic tower. The displayed maps are compatible, so [L2] induces an $R$-linear map $\mathrm{\Phi }:{\widehat{M}}^I\to {\widehat{M}}^J\mathrm{.}$

Similarly, $J^d\subseteq I$ gives compatible quotient maps $M\mathrm{/}{J}^{dk}M\to M\mathrm{/}{I}^{k}M$ on a cofinal subsystem and hence a map $\mathrm{\Psi }:{\widehat{M}}^J\to {\widehat{M}}^I\mathrm{.}$

On either side, composing the two systems of quotient maps eventually reduces modulo a larger and larger power of the same ideal. Hence the composites $\mathrm{\Psi }\mathrm{\Phi }$ and $\mathrm{\Phi }\mathrm{\Psi }$ induce the identity on every finite stage, so they are the identity on the inverse limits. Therefore $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse isomorphisms.

Thus equivalent adic filtrations have canonically isomorphic completions.

The map $f$ is injective because, if $f\text{\hspace{0.17em}⁣}\left({\sum }_{i=1}^N{a}_i{e}_i\right)={\sum }_{i=1}^N{a}_i{p}^i{e}_i=0,$ then every coefficient $a_i{p}^i$ is $0$ and hence every $a_i$ is $0$. Therefore $0\to M\stackrel{f}{\to }M\to \mathrm{coker}(f)\to 0$ is exact.

For each $N\ge 1$, set $s_N:={\sum }_{n=1}^N{p}^n{e}_n\in M\mathrm{.}$ If $N^{\mathrm{\prime }}>N$, then $s_{N^{\mathrm{\prime }}}-s_N={\sum }_{n=N+1}^{N^{\mathrm{\prime }}}{p}^n{e}_n\in p^{N+1}M\mathrm{.}$ For each $r\ge 1$, define $x_r$ to be the class of $s_N$ modulo $p^{r}M$ for any $N\ge r$. The displayed containment makes this independent of $N$, and the classes $x_r$ are compatible under reduction. Hence $x=(x_r{)}_r$ is an element of $\widehat{M}$ by [L1].

Suppose $x=\widehat{f}(y)$ for some $y\in \widehat{M}$. Let $\bar{y}$ be the image of $y$ in $M\mathrm{/}pM={⨁}_{n\ge 1}(\mathbb{Z}\mathrm{/}p\mathbb{Z})e_n\mathrm{.}$ Because this is a direct sum, $\bar{y}$ has finite support. But for each $n\ge 1$, comparing the $e_n$-component modulo $p^{n+1}$ shows that the $e_n$-coefficient of $\bar{y}$ must be $1\in \mathbb{Z}\mathrm{/}p\mathbb{Z}$, since $f$ multiplies the $e_n$-coordinate by $p^n$ and $x$ has $e_n$-coefficient $p^n$ modulo $p^{n+1}$. Thus $\bar{y}$ would have infinitely many nonzero coordinates, a contradiction. So $x\mathrm{\notin }\mathrm{im}(\widehat{f})$.

Every partial sum $s_N$ from step 1.2 lies in $\mathrm{im}(f)$, so its image in $\mathrm{coker}(f)$ is $0$. Therefore the compatible family $x\in \widehat{M}$ maps to $0$ in every quotient $\mathrm{coker}(f)\mathrm{/}{p}^r\mathrm{coker}(f)$, hence to $0$ in $\widehat{\mathrm{coker}(f)}$. Step 2.1 showed that $x\mathrm{\notin }\mathrm{im}(\widehat{f})$, so the completed sequence fails exactness at the middle term. This is exactly the failure excluded by the finite-generation hypothesis in [L2].

The polynomial ring $k[x,y]$ is Noetherian, so [L1] makes $k[x,y{]}_{(x,y)}$ and its quotient $A$ Noetherian. The quotient is local because it is a quotient of the local ring $k[x,y{]}_{(x,y)}$.

In the formal power-series ring $k[\text{\hspace{0.17em}⁣}[x]\text{\hspace{0.17em}⁣}]$, the binomial series gives an element $u=(1+x{)}^{1\mathrm{/}2}\in k[\text{\hspace{0.17em}⁣}[x]\text{\hspace{0.17em}⁣}]$ with $u^2=1+x$. Consequently, in $k[\text{\hspace{0.17em}⁣}[x,y]\text{\hspace{0.17em}⁣}]$, $y^2-x^2-x^3=(y-xu)(y+xu)\mathrm{.}$ Neither factor is a unit because each has zero constant term.

The polynomial $f(y):=y^2-x^2-x^3=y^2-x^2(1+x)$ is irreducible in $k(x)[y]$: it is quadratic in $y$, so reducibility would force $x^2(1+x)$ to be a square in $k(x)$, but $1+x$ is not a square in $k(x)$ because its divisor has the simple zero $x=-1$. Hence the ideal $(f)$ is prime in $k[x,y]$, and localizing preserves primality. Therefore $A$ is a domain.

For every $n$, localization away from $(x,y)$ does not change the quotient modulo $(x,y{)}^n$, so $A\mathrm{/}{\mathfrak{m}}^n\cong k[x,y]\mathrm{/}((x,y{)}^n,f)\mathrm{.}$ Passing to the inverse limit identifies the completion coefficientwise with $\widehat{A}\cong k[\text{\hspace{0.17em}⁣}[x,y]\text{\hspace{0.17em}⁣}]\mathrm{/}(f)\mathrm{.}$ Let $\alpha ,\beta$ be the images of $y-xu$ and $y+xu$ in $\widehat{A}$. Their product is $0$ by step 1.2. If $\alpha =0$, then in the domain $k[\text{\hspace{0.17em}⁣}[x,y]\text{\hspace{0.17em}⁣}]$ one would have $(y-xu)=(y-xu)(y+xu)h$ for some $h$, hence $1=(y+xu)h,$ which is impossible because $y+xu$ lies in the maximal ideal. So $\alpha \mathrm{\ne }0$, and similarly $\beta \mathrm{\ne }0$. Thus $\widehat{A}$ has nonzero zero divisors and is not a domain.

Therefore a Noetherian local domain can have a completion that is not a domain.

Apply [L1] with $J=I^r$. The inclusions $I^r\subseteq I\text{and}{I}^r\subseteq I^r$ give the hypotheses with $c=r$ and $d=1$, so ${\widehat{M}}^I\stackrel{\sim }{\to }{\widehat{M}}^{I^r}\mathrm{.}$

Since $I+I^r=I$, the completion for the combined ideal is exactly ${\widehat{M}}^I$. Combining this identity with step 1.1 shows that the $I$-adic, $I^r$-adic, and $(I+I^r)$-adic one-step completions are canonically the same module.

Thus passing from $I$ to the power $I^r$, or replacing the pair $(I,I^r)$ by the combined ideal $I+I^r$, does not change the one-step completion.

Distinct maximal ideals are pairwise comaximal. If $I+J=R$, expanding $(a+b{)}^{2n-1}=1$ for $a\in I$, $b\in J$, and $a+b=1$ shows that $I^n+J^n=R$; hence the powers ${\mathfrak{m}}_i^n$ are again pairwise comaximal. Applying [L2] first to the ${\mathfrak{m}}_i$ and then to their powers gives, for every $n\ge 1$, ${\mathfrak{m}}^n={\left({\prod }_{i=1}^t{\mathfrak{m}}_i\right)}^n={\prod }_{i=1}^t{\mathfrak{m}}_i^n={\bigcap }_{i=1}^t{\mathfrak{m}}_i^n$ and $R\mathrm{/}{\mathfrak{m}}^n\cong {\prod }_{i=1}^{t}R\mathrm{/}{\mathfrak{m}}_i^n\mathrm{.}$

Localizing $R\mathrm{/}{\mathfrak{m}}_i^n$ at $R\setminus {\mathfrak{m}}_i$ changes nothing. Indeed, if $s\mathrm{\notin }{\mathfrak{m}}_i$, maximality gives $a\in R$ and $u\in {\mathfrak{m}}_i$ with $as+u=1$, and $as(1+u+\cdots +u^{n-1})=1-u^n\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{\mathfrak{m}}_i^n)\mathrm{.}$ Thus every such $s$ is already a unit modulo ${\mathfrak{m}}_i^n$, and $R\mathrm{/}{\mathfrak{m}}_i^n\cong R_{{\mathfrak{m}}_i}\mathrm{/}{\mathfrak{m}}_i^n{R}_{{\mathfrak{m}}_i}\mathrm{.}$ Taking inverse limits and using [L1] yields

A compatible tuple in the inverse limit of the finite products in step 2.1 is exactly a choice, for each $i$, of a compatible tuple in the $i$th quotient tower. Therefore inverse limit commutes with this finite product, and This is the claimed decomposition.