## 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. $\square$

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