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The following form of Hensel's Lemma in Algebraic Geometry is well-documented in the literature:

$\textbf{Theorem 1}$: Let $R$ be an Henselian local ring with maximal ideal $\mathfrak{m}$, and let $X$ be a smooth $R$-scheme. Then $X(R)\to X(R/\mathfrak{m})$ is surjective.

One can for example refer to $[EGA_{IV}]$, Théorème $18.5.17$.

Now I would like to use the following variation:

$\textbf{Theorem 2}$: Let $R$ be a Henselian local ring with maximal ideal $\mathfrak{m}$, and let $X$ be a smooth $R$-scheme. Then $X(R/\mathfrak{m}^l)\to X(R/\mathfrak{m}^k)$ is surjective, for any integers $l\geq k>0$ (where $l=\infty$ is allowed, in which case $\mathfrak{m}^l = (0)$).

And I would really prefer to avoid saying that the proof is a straightforward adaptation of the proof of Theorem 1 (or to write it). Is there a reference for Theorem 2 ?

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    $\begingroup$ Since it suffices to treat $\ell = \infty$ (that doesn't need to be explained to the reader), you could say that the same proof applies verbatim upon making just one change: replace the reference to 18.5.11(b) with a reference to 18.5.4(b) (taking $S$ to be ${\rm{Spec}}(R)$ and $S_0$ to be ${\rm{Spec}}(R/\mathfrak{m}^k)$ in the notation of 18.5.4). $\endgroup$
    – nfdc23
    Commented Mar 29, 2016 at 3:28
  • $\begingroup$ Indeed, thanks very much ! In order to be able to use 18.5.4(b), I just have to mention that since ($\text{Spec}(R)$,$\text{Spec}(R/\mathfrak{m})$) is a henselian couple, so is ($\text{Spec}(R)$,$\text{Spec}(R/\mathfrak{m}^k)$). As mentioned below 18.5.5, this is a direct consequence of Definition 18.5.5 and of $[EGA_{I}]$, 5.1.8. $\endgroup$
    – thierry stulemeijer
    Commented Mar 29, 2016 at 8:38
  • $\begingroup$ Yes, though the reader doesn't need to be told about the reference 5.1.8 in EGA I (even though Grothendieck felt the need to mention it in that IV$_4$ 18.5 discussion). $\endgroup$
    – nfdc23
    Commented Mar 30, 2016 at 0:39

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