Lemma 16.14.1. Let $(A, I)$ be a henselian pair with $A$ Noetherian. Let $A^\wedge $ be the $I$-adic completion of $A$. Assume at least one of the following conditions holds

$A \to A^\wedge $ is a regular ring map,

$A$ is a Noetherian G-ring, or

$(A, I)$ is the henselization (More on Algebra, Lemma 15.12.1) of a pair $(B, J)$ where $B$ is a Noetherian G-ring.

Given $f_1, \ldots , f_ m \in A[x_1, \ldots , x_ n]$ and $\hat{a}_1, \ldots , \hat{a}_ n \in A^\wedge $ such that $f_ j(\hat{a}_1, \ldots , \hat{a}_ n) = 0$ for $j = 1, \ldots , m$, for every $N \geq 1$ there exist $a_1, \ldots , a_ n \in A$ such that $\hat{a}_ i - a_ i \in I^ N$ and such that $f_ j(a_1, \ldots , a_ n) = 0$ for $j = 1, \ldots , m$.

**Proof.**
By More on Algebra, Lemma 15.50.15 we see that (3) implies (2). By More on Algebra, Lemma 15.50.14 we see that (2) implies (1). Thus it suffices to prove the lemma in case $A \to A^\wedge $ is a regular ring map.

Let $\hat{a}_1, \ldots , \hat{a}_ n$ be as in the statement of the lemma. By Theorem 16.12.1 we can find a factorization $A \to B \to A^\wedge $ with $A \to P$ smooth and $b_1, \ldots , b_ n \in B$ with $f_ j(b_1, \ldots , b_ n) = 0$ in $B$. Denote $\sigma : B \to A^\wedge \to A/I^ N$ the composition. By More on Algebra, Lemma 15.9.14 we can find an étale ring map $A \to A'$ which induces an isomorphism $A/I^ N \to A'/I^ NA'$ and an $A$-algebra map $\tilde\sigma : B \to A'$ lifting $\sigma $. Since $(A, I)$ is henselian, there is an $A$-algebra map $\chi : A' \to A$, see More on Algebra, Lemma 15.11.6. Then setting $a_ i = \chi (\tilde\sigma (b_ i))$ gives a solution.
$\square$

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