The Stacks project

110.7 Noncomplete completion

Let $R$ be a ring and let $\mathfrak m$ be a maximal ideal. Consider the completion

\[ R^\wedge = \mathop{\mathrm{lim}}\nolimits R/\mathfrak m^ n. \]

Note that $R^\wedge $ is a local ring with maximal ideal $\mathfrak m' = \mathop{\mathrm{Ker}}(R^\wedge \to R/\mathfrak m)$. Namely, if $x = (x_ n) \in R^\wedge $ is not in $\mathfrak m'$, then $y = (x_ n^{-1}) \in R^\wedge $ satisfies $xy = 1$, whence $R^\wedge $ is local by Algebra, Lemma 10.18.3. Now it is always true that $R^\wedge $ complete in its limit topology (see the discussion in More on Algebra, Section 15.36). But beyond that, we have the following questions:

  1. Is it true that $\mathfrak m R^\wedge = \mathfrak m'$?

  2. Is $R^\wedge $ viewed as an $R^\wedge $-module $\mathfrak m'$-adically complete?

  3. Is $R^\wedge $ viewed as an $R$-module $\mathfrak m$-adically complete?

It turns out that these questions all have a negative answer. The example below was taken from an unpublished note of Bart de Smit and Hendrik Lenstra. See also [Exercise III.2.12, Bourbaki-CA] and [Example 1.8, Yekutieli]

Let $k$ be a field, $R = k[x_1, x_2, x_3, \ldots ]$, and $\mathfrak m = (x_1, x_2, x_3, \ldots )$. We will think of an element $f$ of $R^\wedge $ as a (possibly) infinite sum

\[ f = \sum a_ I x^ I \]

(using multi-index notation) such that for each $d \geq 0$ there are only finitely many nonzero $a_ I$ for $|I| = d$. The maximal ideal $\mathfrak m' \subset R^\wedge $ is the collection of $f$ with zero constant term. In particular, the element

\[ f = x_1 + x_2^2 + x_3^3 + \ldots \]

is in $\mathfrak m'$ but not in $\mathfrak m R^\wedge $ which shows that (1) is false in this example. However, if (1) is false, then (3) is necessarily false because $\mathfrak m' = \mathop{\mathrm{Ker}}(R^\wedge \to R/\mathfrak m)$ and we can apply Algebra, Lemma 10.96.5 with $n = 1$.

To finish we prove that $R^\wedge $ is not $\mathfrak m'$-adically complete. For $n \geq 1$ let $K_ n = \mathop{\mathrm{Ker}}(R^\wedge \to R/\mathfrak m^ n)$. Then we have short exact sequences

\[ 0 \to K_ n/(\mathfrak m')^ n \to R^\wedge /(\mathfrak m')^ n \to R/\mathfrak m^ n \to 0 \]

The projection map $R^\wedge \to R/\mathfrak m^{n + 1}$ sends $(\mathfrak m')^ n$ onto $\mathfrak m^ n/\mathfrak m^{n + 1}$. It follows that $K_{n + 1} \to K_ n/(\mathfrak m')^ n$ is surjective. Hence the inverse system $\left(K_ n/(\mathfrak m')^ n\right)$ has surjective transition maps and taking inverse limits we obtain an exact sequence

\[ 0 \to \mathop{\mathrm{lim}}\nolimits K_ n/(\mathfrak m')^ n \to \mathop{\mathrm{lim}}\nolimits R^\wedge /(\mathfrak m')^ n \to \mathop{\mathrm{lim}}\nolimits R/\mathfrak m^ n \to 0 \]

by Algebra, Lemma 10.87.1. Thus we see that $R^\wedge $ is complete with respect to $\mathfrak m'$ if and only if $K_ n = (\mathfrak m')^ n$ for all $n \geq 1$.

To show that $R^\wedge $ is not $\mathfrak m'$-adically complete in our example we show that $K_2 = \mathop{\mathrm{Ker}}(R^\wedge \to R/\mathfrak m^2)$ is not equal to $(\mathfrak m')^2$. Note that an element of $(\mathfrak m')^2$ can be written as a finite sum
\begin{equation} \label{examples-equation-sum} \sum \nolimits _{i = 1, \ldots , t} f_ i g_ i \end{equation}

with $f_ i, g_ i \in R^\wedge $ having vanishing constant terms. To get an example we are going to choose an $z \in K_2$ of the form

\[ z = z_1 + z_2 + z_3 + \ldots \]

with the following properties

  1. there exist sequences $1 < d_1 < d_2 < d_3 < \ldots $ and $0 < n_1 < n_2 < n_3 < \ldots $ such that $z_ i \in k[x_{n_ i}, x_{n_ i + 1}, \ldots , x_{n_{i + 1} - 1}]$ homogeneous of degree $d_ i$, and

  2. in the ring $k[[x_{n_ i}, x_{n_ i + 1}, \ldots , x_{n_{i + 1} - 1}]]$ the element $z_ i$ cannot be written as a sum ( with $t \leq i$.

Clearly this implies that $z$ is not in $(\mathfrak m')^2$ because the image of the relation ( in the ring $k[[x_{n_ i}, x_{n_ i + 1}, \ldots , x_{n_{i + 1} - 1}]]$ for $i$ large enough would produce a contradiction. Hence it suffices to prove that for all $t > 0$ there exists a $d \gg 0$ and an integer $n$ such that we can find an homogeneous element $z \in k[x_1, \ldots , x_ n]$ of degree $d$ which cannot be written as a sum ( for the given $t$ in $k[[x_1, \ldots , x_ n]]$. Take $n > 2t$ and any $d > 1$ prime to the characteristic of $k$ and set $z = \sum _{i = 1, \ldots , n} x_ i^ d$. Then the vanishing locus of the ideal

\[ (\frac{\partial z}{\partial x_1}, \ldots , \frac{\partial z}{\partial x_ n}) = (dx_1^{d - 1}, \ldots , dx_ n^{d - 1}) \]

consists of one point. On the other hand,

\[ \frac{\partial ( \sum \nolimits _{i = 1, \ldots , t} f_ i g_ i ) }{\partial x_ j} \in (f_1, \ldots , f_ t, g_1, \ldots , g_ t) \]

by the Leibniz rule and hence the vanishing locus of these derivatives contains at least

\[ V(f_1, \ldots , f_ t, g_1, \ldots , g_ t) \subset \mathop{\mathrm{Spec}}(k[[x_1, \ldots , x_ n]]). \]

Hence this is a contradiction as the dimension of $V(f_1, \ldots , f_ t, g_1, \ldots , g_ t)$ is at least $n - 2t \geq 1$.

Lemma 110.7.1. There exists a local ring $R$ and a maximal ideal $\mathfrak m$ such that the completion $R^\wedge $ of $R$ with respect to $\mathfrak m$ has the following properties

  1. $R^\wedge $ is local, but its maximal ideal is not equal to $\mathfrak m R^\wedge $,

  2. $R^\wedge $ is not a complete local ring, and

  3. $R^\wedge $ is not $\mathfrak m$-adically complete as an $R$-module.

Proof. This follows from the discussion above as (with $R = k[x_1, x_2, x_3, \ldots ]$) the completion of the localization $R_{\mathfrak m}$ is equal to the completion of $R$. $\square$

Comments (6)

Comment #195 by on

"upublished" should be "unpublished". In the sentence after that the word order is a bit strange too.

Comment #1917 by Matthieu Romagny on

Missing 'if' in the sentence : Hence, if R ∧ is not 𝔪 -adically complete as an R -module...

Comment #5136 by Christopher Chiu on

In the sentence "Take n>2t and any d>1 prime to the characteristic of p and" p should probably be k.

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