Lemma 108.40.1. Let $A = \mathbb {F}_ p[T]$ be the polynomial ring in one variable over $\mathbb {F}_ p$. Let $A_{perf}$ denote the perfect closure of $A$. Then $A \rightarrow A_{perf}$ is flat and formally unramified, but not formally étale.

## 108.40 Flat and formally unramified is not formally étale

In More on Morphisms, Lemma 37.8.7 it is shown that an unramified flat morphism of schemes $X \to S$ is formally étale. The goal of this section is to give two examples that illustrate that we cannot replace ‘unramified' by ‘formally unramified'. The first example exploits special properties of perfect rings, while the second example shows the result fails even for maps of Noetherian regular rings.

**Proof.**
Note that under the Frobenius map $F_ A : A \to A$, the target copy of $A$ is a free-module over the domain with basis $\{ 1, T, \dots , T^{p - 1}\} $. Thus, $F_ A$ is faithfully flat, and consequently, so is $A \to A_{perf}$ since it is a colimit of faithfully flat maps. Since $A_{perf}$ is a perfect ring, the relative Frobenius $F_{A_{perf}/A}$ is a surjection. In other words, $A_{perf} = A[A_{perf}^ p]$, which readily implies $\Omega _{A_{perf}/A} = 0$. Then $A \rightarrow A_{perf}$ is formally unramified by More on Morphisms, Lemma 37.6.7

It suffices to show that $A \rightarrow A_{perf}$ is not formally smooth. Note that since $A$ is a smooth $\mathbb {F}_ p$-algebra, the cotangent complex $L_{A/\mathbb {F}_ P} \simeq \Omega _{A/\mathbb {F}_ p}[0]$ is concentrated in degree $0$, see Cotangent, Lemma 90.9.1. Moreover, $L_{A_{perf}/\mathbb {F}_ p} = 0$ in $D(A_{perf})$ by Cotangent, Lemma 90.10.3. Consider the distinguished triangle of cotangent complexes

in $D(A_{perf})$, see Cotangent, Section 90.7. We find $L_{A_{perf}/A} = \Omega _{A/\mathbb {F}_ p} \otimes _ A A_{perf}[1]$, that is, $L_{A_{perf}/A}$ is equal to a free rank $1$ $A_{perf}$ module placed in degree $-1$. Thus $A \rightarrow A_{perf}$ is not formally smooth by More on Morphisms, Lemma 37.13.5 and Cotangent, Lemma 90.11.3. $\square$

The next example also involves rings of prime characteristic, but is perhaps a little more surprising. The drawback is that it requires more knowledge of characteristic $p$ phenomena than the previous example. Recall that we say a ring $A$ of prime characteristic is $F$-finite if the Frobenius map on $A$ is finite.

Lemma 108.40.2. 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 10.96.2. To show that the map is formally unramified, it suffices to show that $\Omega _{A^\wedge /A} = 0$, see Algebra, Lemma 10.147.2.

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 10.50.4. 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 10.137.8. It follows from the Jacobi-Zariski sequence (Algebra, Lemma 10.133.4) 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 10.156.2). 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. $\square$

Remark 108.40.3. Non-excellent regular rings whose residue fields have a finite $p$-basis can be constructed even in the function field of $\mathbb {P}^2_ k$, over a characteristic $p$ field $k = \overline{k}$. See [$\mathsection 4.1$, DS18].

The proof of Lemma 108.40.2 actually shows a little more.

Lemma 108.40.4. Let $(A, \mathfrak m, \kappa )$ be a regular local ring of characteristic $p > 0$. Suppose $[\kappa : \kappa ^ p] < \infty $. Then $A$ is excellent if and only if $A \to A^\wedge $ is formally étale.

**Proof.**
The backward implication follows from Lemma 108.40.2. For the forward implication, note that we already know from Lemma 108.40.2 that $A \to A^\wedge $ is formally unramified or equivalently that $\Omega _{A^\wedge /A}$ is zero. Thus, it suffices to show that the completion map is formally smooth when $A$ is excellent. By Néron-Popescu desingularization $A \to A^\wedge $ can be written as a filtered colimit of smooth $A$-algebras (Smoothing Ring Maps, Theorem 16.12.1). Hence $\mathop{N\! L}\nolimits _{A^\wedge /A}$ has vanishing cohomology in degree $-1$. Thus $A \to A^\wedge $ is formally smooth by Algebra, Proposition 10.137.8.
$\square$

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