The Stacks project

109.17 Another local ring with nonreduced completion

In this section we make an example of a Noetherian local domain of dimension $2$ complete with respect to a principal ideal such that the recompletion of a localization is nonreduced.

Let $p$ be a prime number. Let $k$ be a field of characteristic $p$ such that $k$ has infinite degree over its subfield $k^ p$ of $p$th powers. For example $k = \mathbf{F}_ p(t_1, t_2, t_3, \ldots )$. Consider the ring

\[ A = \left\{ \begin{matrix} \sum a_{i, j} x^ iy^ j \in k[[x, y]] \text{ such that for all }n \geq 0 \text{ we have } \\ [k^ p(a_{n, n}, a_{n, n + 1}, a_{n + 1, n}, a_{n, n + 2}, a_{n + 2, n}, \ldots ) : k^ p] < \infty \end{matrix} \right\} \]

As a set we have

\[ k^ p[[x, y]] \subset A \subset k[[x, y]] \]

Every element $f$ of $A$ can be uniquely written as a series

\[ f = f_0 + f_1 xy + f_2 (xy)^2 + f_3 (xy)^3 + \ldots \]

with

\[ f_ n = a_{n, n} + a_{n, n + 1} y + a_{n + 1, n} x + a_{n, n + 2} y^2 + a_{n + 2, n} x^2 + \ldots \]

and the condition in the formula defining $A$ means that the coefficients of $f_ n$ generate a finite extension of $k^ p$. From this presentation it is clear that $A$ is an $k^ p[[x, y]]$-subalgebra of $k[[x, y]]$ complete with respect to the ideal $xy$. Moreover, we clearly have

\[ A/xy A = C \times _ k D \]

where $k^ p[[x]] \subset C \subset k[[x]]$ and $k^ p[[y]] \subset D \subset k[[y]]$ are the subrings of power series from Algebra, Example 10.119.5. Hence $C$ and $D$ are dvrs and we see that $A/ xy A$ is Noetherian. By Algebra, Lemma 10.97.5 we conclude that $A$ is Noetherian. Since $\dim (k[[x, y]]) = 2$ using Algebra, Lemma 10.112.4 we conclude that $\dim (A) = 2$.

Let $f = \sum a_ i x^ i$ be a power series such that $k^ p(a_0, a_1, a_2, \ldots )$ has infinite degree over $k^ p$. Then $f \not\in A$ but $f^ p \in A$. We set

\[ B = A[f] \subset k[[x, y]] \]

Since $B$ is finite over $A$ we see that $B$ is Noetherian. Also, $B$ is complete with respect to the ideal generated by $xy$, see Algebra, Lemma 10.97.1. In fact $B$ is free over $A$ with basis $1, f, f^2, \ldots , f^{p - 1}$; we omit the proof.

We claim the ring

\[ (B_ y)^\wedge = (B[1/y])^\wedge = \mathop{\mathrm{lim}}\nolimits B[1/y]/(xy)^ n B[1/y] = \mathop{\mathrm{lim}}\nolimits B[1/y] / x^ n B[1/y] \]

is nonreduced. Namely, this ring is free over

\[ (A_ y)^\wedge = (A[1/y])^\wedge = \mathop{\mathrm{lim}}\nolimits A[1/y]/(xy)^ n A[1/y] = \mathop{\mathrm{lim}}\nolimits A[1/y] / x^ n A[1/y] \]

with basis $1, f, \ldots , f^{p - 1}$. However, there is an element $g \in (A_ y)^\wedge $ such that $f^ p = g^ p$. Namely, we can just take $g = \sum a_ i x^ i$ (the same expression as we used for $f$) which makes sense in $(A_ y)^\wedge $. Hence we see that

\[ (B_ y)^\wedge = (A_ y)^\wedge [f]/(f^ p - g^ p) \cong (A_ y)^\wedge [\tau ]/(\tau ^ p) \]

is nonreduced. In fact, this example shows slightly more. Namely, observe that $(A_ y)^\wedge $ is a dvr with uniformizer $x$ and residue field the fraction field of the dvr $D$ given above. Hence we see that even

\[ (B_ y)^\wedge [1/(xy)] = ((B_ y)^\wedge )_{xy} \]

is nonreduced. This produces an example of the following kind.

Lemma 109.17.1. There exists a local Noetherian $2$-dimensional domain $(B, \mathfrak m)$ complete with respect to a principal ideal $I = (b)$ and an element $f \in \mathfrak m$, $f \not\in I$ such that the $I$-adic completion $C = (B_ f)^\wedge $ of the principal localization $B_ f$ is nonreduced and even such that $C_ b = C[1/b] = (B_ f)^\wedge [1/b]$ is nonreduced.

Proof. See discussion above. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GHH. Beware of the difference between the letter 'O' and the digit '0'.