The Stacks project

Lemma 109.43.2. 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

\[ A^\wedge = \bigoplus \nolimits _{I, J} (A^\wedge )^ p x_1^{i_1} \ldots x_ r^{i_ r} y_1^{j_1} \ldots y_ s^{j_ s} \]

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

\[ a = \sum \nolimits _{I, J} (a_{I, J})^ p x_1^{i_1} \ldots x_ r^{i_ r} y_1^{j_1} \ldots y_ s^{j_ s} \]

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

\[ (\partial /\partial x_1)^{p - 1} \ldots (\partial /\partial x_ r)^{p - 1} (\partial /\partial y_1)^{p - 1} \ldots (\partial /\partial y_ s)^{p - 1} \]

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. $\square$

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