The Stacks project

110.7 Noncomplete completion

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

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 The example below was taken from an unpublished note of Bart de Smit and Hendrik Lenstra.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

(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

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

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

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

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

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 (110.7.0.1) with $t \leq i$.

Clearly this implies that $z$ is not in $(\mathfrak m')^2$ because the image of the relation (110.7.0.1) 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 (110.7.0.1) 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

consists of one point. On the other hand,

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

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