## 109.16 Local rings with nonreduced completion

In Algebra, Example 10.119.5 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

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

We claim that

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

$A$ is a local domain,

$\dim (A) = 1$,

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

$A$ is Noetherian, and

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

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

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

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