The Stacks project

Proof. Note that the flatness of $A \to \Lambda $ implies that $A \to \Lambda $ is injective, so the lemma really tells us that $\Lambda $ is a directed union of these types of subrings $A \subset \Lambda $. Let $n$ be the minimal integer such that $\mathfrak m^ n = 0$. We will prove this lemma by induction on $n$. The case $n = 1$ is clear as a field extension is a union of finitely generated field extensions.

Pick $\lambda _1, \ldots , \lambda _ d \in \mathfrak m$ which generate $\mathfrak m$. As $K$ is formally smooth over $\mathbf{F}_ p$ (see Algebra, Lemma 10.158.7) we can find a ring map $\sigma : K \to \Lambda $ which is a section of the quotient map $\Lambda \to K$. In general $\sigma $ is not a $k$-algebra map. Given $\sigma $ we define

using $\sigma $ on elements of $K$ and mapping $x_ i$ to $\lambda _ i$. Claim: there exists a $\sigma : K \to \Lambda $ and a subfield $k \subset F \subset K$ finitely generated over $k$ such that the image of $k$ in $\Lambda $ is contained in $\Psi _\sigma (F[x_1, \ldots , x_ d])$.

We will prove the claim by induction on the least integer $n$ such that $\mathfrak m^ n = 0$. It is clear for $n = 1$. If $n > 1$ set $I = \mathfrak m^{n - 1}$ and $\Lambda ' = \Lambda /I$. By induction we may assume given $\sigma ' : K \to \Lambda '$ and $k \subset F' \subset K$ finitely generated such that the image of $k \to \Lambda \to \Lambda '$ is contained in $A' = \Psi _{\sigma '}(F'[x_1, \ldots , x_ d])$. Denote $\tau ' : k \to A'$ the induced map. Choose a lift $\sigma : K \to \Lambda $ of $\sigma '$ (this is possible by the formal smoothness of $K/\mathbf{F}_ p$ we mentioned above). For later reference we note that we can change $\sigma $ to $\sigma + D$ for some derivation $D : K \to I$. Set $A = F[x_1, \ldots , x_ d]/(x_1, \ldots , x_ d)^ n$. Then $\Psi _\sigma $ induces a ring map $\Psi _\sigma : A \to \Lambda $. The composition with the quotient map $\Lambda \to \Lambda '$ induces a surjective map $A \to A'$ with nilpotent kernel. Choose a lift $\tau : k \to A$ of $\tau '$ (possible as $k/\mathbf{F}_ p$ is formally smooth). Thus we obtain two maps $k \to \Lambda $, namely $\Psi _\sigma \circ \tau : k \to \Lambda $ and the given map $i : k \to \Lambda $. These maps agree modulo $I$, whence the difference is a derivation $\theta = i - \Psi _\sigma \circ \tau : k \to I$. Note that if we change $\sigma $ into $\sigma + D$ then we change $\theta $ into $\theta - D|_ k$.

Choose a set of elements $\{ y_ j\} _{j \in J}$ of $k$ whose differentials $\text{d}y_ j$ form a basis of $\Omega _{k/\mathbf{F}_ p}$. The Jacobi-Zariski sequence for $\mathbf{F}_ p \subset k \subset K$ is

As $\dim H_1(L_{K/k}) < \infty $ we can find a finite subset $J_0 \subset J$ such that the image of the first map is contained in $\bigoplus _{j \in J_0} K\text{d}y_ j$. Hence the elements $\text{d}y_ j$, $j \in J \setminus J_0$ map to $K$-linearly independent elements of $\Omega _{K/\mathbf{F}_ p}$. Therefore we can choose a $D : K \to I$ such that $\theta - D|_ k = \xi \circ \text{d}$ where $\xi $ is a composition

Let $f_ j = \xi (\text{d}y_ j) \in I$ for $j \in J_0$. Change $\sigma $ into $\sigma + D$ as above. Then we see that $\theta (a) = \sum _{j \in J_0} a_ j f_ j$ for $a \in k$ where $\text{d}a = \sum a_ j \text{d}y_ j$ in $\Omega _{k/\mathbf{F}_ p}$. Note that $I$ is generated by the monomials $\lambda ^ E = \lambda _1^{e_1} \ldots \lambda _ d^{e_ d}$ of total degree $|E| = \sum e_ i = n - 1$ in $\lambda _1, \ldots , \lambda _ d$. Write $f_ j = \sum _ E c_{j, E} \lambda ^ E$ with $c_{j, E} \in K$. Replace $F'$ by $F = F'(c_{j, E})$. Then the claim holds.

Choose $\sigma $ and $F$ as in the claim. The kernel of $\Psi _\sigma $ is generated by finitely many polynomials $g_1, \ldots , g_ t \in K[x_1, \ldots , x_ d]$ and we may assume their coefficients are in $F$ after enlarging $F$ by adjoining finitely many elements. In this case it is clear that the map $A = F[x_1, \ldots , x_ d]/(g_1, \ldots , g_ t) \to K[x_1, \ldots , x_ d]/(g_1, \ldots , g_ t) = \Lambda $ is flat. By the claim $A$ is a $k$-subalgebra of $\Lambda $. It is clear that $\Lambda $ is the filtered colimit of these algebras, as $K$ is the filtered union of the subfields $F$. Finally, these algebras are essentially of finite type over $k$ by Algebra, Lemma 10.54.4. $\square$