108.16 Local rings with nonreduced completion

In Algebra, Example 10.118.5 we gave an example of a characteristic $p$ Noetherian local domain $R$ of dimension $1$ whose completion is nonreduced. In this section we present the example of [Proposition 3.1, Ferrand-Raynaud] which gives a similar ring in characteristic zero.

Let $\mathbf{C}\{ x\}$ be the ring of convergent power series over the field $\mathbf{C}$ of complex numbers. The ring of all power series $\mathbf{C}[[x]]$ is its completion. Let $K = \mathbf{C}\{ x\} [1/x]$ be the field of convergent Laurent series. The $K$-module $\Omega _{K/\mathbf{C}}$ of algebraic differentials of $K$ over $\mathbf{C}$ is an infinite dimensional $K$-vector space (proof omitted). We may choose $f_ n \in x\mathbf{C}\{ x\}$, $n \geq 1$ such that $\text{d}x, \text{d}f_1, \text{d}f_2, \ldots$ are part of a basis of $\Omega _{K/\mathbf{C}}$. Thus we can find a $\mathbf{C}$-derivation

$D : \mathbf{C}\{ x\} \longrightarrow \mathbf{C}((x))$

such that $D(x) = 0$ and $D(f_ i) = x^{-n}$. Let

$A = \{ f \in \mathbf{C}\{ x\} \mid D(f) \in \mathbf{C}[[x]]\}$

We claim that

1. $\mathbf{C}\{ x\}$ is integral over $A$,

2. $A$ is a local domain,

3. $\dim (A) = 1$,

4. the maximal ideal of $A$ is generated by $x$ and $xf_1$,

5. $A$ is Noetherian, and

6. the completion of $A$ is equal to the ring of dual numbers over $\mathbf{C}[[x]]$.

Since the dual numbers are nonreduced the ring $A$ gives the example.

Note that if $0 \not= f \in x\mathbf{C}\{ x\}$ then we may write $D(f) = h/f^ n$ for some $n \geq 0$ and $h \in \mathbf{C}[[x]]$. Hence $D(f^{n + 1}/(n + 1)) \in \mathbf{C}[[x]]$ and $D(f^{n + 2}/(n + 2)) \in \mathbf{C}[[x]]$. Thus we see $f^{n + 1}, f^{n + 2} \in A$! In particular we see (1) holds. We also conclude that the fraction field of $A$ is equal to the fraction field of $\mathbf{C}\{ x\}$. It also follows immediately that $A \cap x\mathbf{C}\{ x\}$ is the set of nonunits of $A$, hence $A$ is a local domain of dimension $1$. If we can show (4) then it will follow that $A$ is Noetherian (proof omitted). Suppose that $f \in A \cap x\mathbf{C}\{ x\}$. Write $D(f) = h$, $h \in \mathbf{C}[[x]]$. Write $h = c + xh'$ with $c \in \mathbf{C}$, $h' \in \mathbf{C}[[x]]$. Then $D(f - cxf_1) = c + xh' - c = xh'$. On the other hand $f - cxf_1 = xg$ with $g \in \mathbf{C}\{ x\}$, but by the computation above we have $D(g) = h' \in \mathbf{C}[[x]]$ and hence $g \in A$. Thus $f = cxf_1 + xg \in (x, xf_1)$ as desired.

Finally, why is the completion of $A$ nonreduced? Denote $\hat A$ the completion of $A$. Of course this maps surjectively to the completion $\mathbf{C}[[x]]$ of $\mathbf{C}\{ x\}$ because $x \in A$. Denote this map $\psi : \hat A \to \mathbf{C}[[x]]$. Above we saw that $\mathfrak m_ A = (x, xf_1)$ and hence $D(\mathfrak m_ A^ n) \subset (x^{n - 1})$ by an easy computation. Thus $D : A \to \mathbf{C}[[x]]$ is continuous and gives rise to a continuous derivation $\hat D : \hat A \to \mathbf{C}[[x]]$ over $\psi$. Hence we get a ring map

$\psi + \epsilon \hat D : \hat A \longrightarrow \mathbf{C}[[x]][\epsilon ].$

Since $\hat A$ is a one dimensional Noetherian complete local ring, if we can show this arrow is surjective then it will follow that $\hat A$ is nonreduced. Actually the map is an isomorphism but we omit the verification of this. The subring $\mathbf{C}[x]_{(x)} \subset A$ gives rise to a map $i : \mathbf{C}[[x]] \to \hat A$ on completions such that $i \circ \psi = \text{id}$ and such that $D \circ i = 0$ (as $D(x) = 0$ by construction). Consider the elements $x^ nf_ n \in A$. We have

$(\psi + \epsilon D)(x^ nf_ n) = x^ n f_ n + \epsilon$

for all $n \geq 1$. Surjectivity easily follows from these remarks.

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.

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 02JD. Beware of the difference between the letter 'O' and the digit '0'.