The Stacks project

Proof. We could deduce this theorem from Theorem 16.13.1 using that the henselization $R^ h$ is a G-ring by More on Algebra, Lemma 15.50.8 and writing $R^ h$ as a directed colimit of étale extension $R'$. Instead we prove this by redoing the proof of the previous theorem in this case.

Let $c_ i \in R$ be an element such that $a_ i - c_ i \in \mathfrak m^ N$. Choose generators $\mathfrak m^ N = (d_1, \ldots , d_ M)$. Write $a_ i = c_ i + \sum a_{i, l} d_ l$. Consider the polynomial ring $R[x_{i, l}]$ and the elements

The system of equations $g_ j = 0$ has the solution $(a_{i, l})$. Suppose that we can show that $g_ j$ as a solution $(b_{i, l})$ in $R'$ for some étale ring map $R \to R'$ endowed with a maximal ideal $\mathfrak m'$ such that $\kappa (\mathfrak m) = \kappa (\mathfrak m')$. Then it follows that $b_ i = c_ i + \sum b_{i, l}d_ l$ is a solution of $f_ j = 0$ which is congruent to $a_ i$ modulo $(\mathfrak m')^ N$. Thus it suffices to show that solvability over $R^\wedge $ implies solvability over some étale ring extension which induces a trivial residue field extension at some prime over $\mathfrak m$.

Let $A \subset R^\wedge $ be the $R$-subalgebra generated by $a_1, \ldots , a_ n$. Since we've assumed $R$ is a G-ring, i.e., that $R \to R^\wedge $ is regular, we see that there exists a factorization

with $B$ smooth over $R$, see Theorem 16.12.1. Denote $\kappa = R/\mathfrak m$ the residue field. It is also the residue field of $R^\wedge $, so we get a commutative diagram

Since the vertical arrow is smooth, More on Algebra, Lemma 15.9.14 implies that there exists an étale ring map $R \to R'$ which induces an isomorphism $R/\mathfrak m \to R'/\mathfrak mR'$ and an $R$-algebra map $B \to R'$ making the diagram above commute. Let $b_ i \in R'$ be the image of $a_ i$ under the ring maps $A \to B \to R'$. Since all of these maps are $R$-algebra maps, we see that $(b_1, \ldots , b_ n)$ is a solution in $R'$. $\square$