Lemma 108.42.2. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring of prime characteristic $p > 0$ such that $[\kappa : \kappa ^ p] < \infty $. Then the canonical map $A \to A^\wedge $ to the completion of $A$ is flat and formally unramified. However, if $A$ is regular but not excellent, then this map is not formally étale.

**Proof.**
Flatness of the completion is Algebra, Lemma 10.97.2. To show that the map is formally unramified, it suffices to show that $\Omega _{A^\wedge /A} = 0$, see Algebra, Lemma 10.148.2.

We sketch a proof. Choose $x_1, \ldots , x_ r \in A$ which map to a $p$-basis $\overline{x}_1, \ldots , \overline{x}_ r$ of $\kappa $, i.e., such that $\kappa $ is minimally generated by $\overline{x}_ i$ over $\kappa ^ p$. Choose a minimal set of generators $y_1, \ldots , y_ s$ of $\mathfrak m$. For each $n$ the elements $x_1, \ldots , x_ r, y_1, \ldots , y_ s$ generate $A/\mathfrak m^ n$ over $(A/\mathfrak m^ n)^ p$ by Frobenius. Some details omitted. We conclude that $F : A^\wedge \to A^\wedge $ is finite. Hence $\Omega _{A^\wedge /A}$ is a finite $A^\wedge $-module. On the other hand, for any $a \in A^\wedge $ and $n$ we can find $a_0 \in A$ such that $a - a_0 \in \mathfrak m^ nA^\wedge $. We conclude that $\text{d}(a) \in \bigcap \mathfrak m^ n \Omega _{A^\wedge /A}$ which implies that $\text{d}(a)$ is zero by Algebra, Lemma 10.51.4. Thus $\Omega _{A^\wedge /A} = 0$.

Suppose $A$ is regular. Then, using the Cohen structure theorem $x_1, \ldots , x_ r, y_1, \ldots , y_ s$ is a $p$-basis for the ring $A^\wedge $, i.e., we have

with $I = (i_1, \ldots , i_ r)$, $J = (j_1, \ldots , j_ s)$ and $0 \leq i_ a, j_ b \leq p - 1$. Details omitted. In particular, we see that $\Omega _{A^\wedge }$ is a free $A^\wedge $-module with basis $\text{d}(x_1), \ldots , \text{d}(x_ r), \text{d}(y_1), \ldots , \text{d}(y_ s)$.

Now if $A \to A^\wedge $ is formally étale or even just formally smooth, then we see that $\mathop{N\! L}\nolimits _{A^\wedge /A}$ has vanishing cohomology in degrees $-1, 0$ by Algebra, Proposition 10.138.8. It follows from the Jacobi-Zariski sequence (Algebra, Lemma 10.134.4) for the ring maps $\mathbf{F}_ p \to A \to A^\wedge $ that we get an isomorphism $\Omega _ A \otimes _ A A^\wedge \cong \Omega _{A^\wedge }$. Hence we find that $\Omega _ A$ is free on $\text{d}(x_1), \ldots , \text{d}(x_ r), \text{d}(y_1), \ldots , \text{d}(y_ s)$. Looking at fraction fields and using that $A$ is normal we conclude that $a \in A$ is a $p$th power if and only if its image in $A^\wedge $ is a $p$th power (details omitted; use Algebra, Lemma 10.158.2). A second consequence is that the operators $\partial /\partial x_ a$ and $\partial /\partial y_ b$ are defined on $A$.

We will show that the above lead to the conclusion that $A$ is finite over $A^ p$ with $p$-basis $x_1, \ldots , x_ r, y_1, \ldots , y_ s$. This will contradict the non-excellency of $A$ by a result of Kunz, see [Corollary 2.6, Kun76]. Namely, say $a \in A$ and write

with $a_{I, J} \in A^\wedge $. To finish the proof it suffices to show that $a_{I, J} \in A$. Applying the operator

to both sides we conclude that $a_{I, J}^ p \in A$ where $I = (p - 1, \ldots , p - 1)$ and $J = (p - 1, \ldots , p - 1)$. By our remark above, this also implies $a_{I, J} \in A$. After replacing $a$ by $a' = a - a_{I, J}^ p x^ I y^ J$ we can use a $1$-order lower differential operators to get another coefficient $a_{I, J}$ to be in $A$. Etc. $\square$

## 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.

## Comments (0)