Theorem 16.13.1. Let $R$ be a Noetherian local ring. Let $f_1, \ldots , f_ m \in R[x_1, \ldots , x_ n]$. Suppose that $(a_1, \ldots , a_ n) \in (R^\wedge )^ n$ is a solution in $R^\wedge $. If $R$ is a henselian G-ring, then for every integer $N$ there exists a solution $(b_1, \ldots , b_ n) \in R^ n$ in $R$ such that $a_ i - b_ i \in \mathfrak m^ NR^\wedge $.
Proof. 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
\[ g_ j = f_ j(c_1 + \sum x_{1, l} d_ l , \ldots , c_ n + \sum x_{n, l} d_{n, l}) \in R[x_{i, l}] \]
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 $R$. 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 $R$.
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
\[ A \to B \to R^\wedge \]
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
\[ \xymatrix{ B \ar[rd] \ar@{..>}[r] & R' \ar@{..>}[d] \\ R \ar[r] \ar[u] & \kappa } \]
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. Since $R$ is henselian we see that $R \to R'$ has a section, see Algebra, Lemma 10.153.3. Let $b_ i \in R$ be the image of $a_ i$ under the ring maps $A \to B \to R' \to R$. Since all of these maps are $R$-algebra maps, we see that $(b_1, \ldots , b_ n)$ is a solution in $R$. $\square$
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