5
$\begingroup$

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 ?

$\endgroup$
  • 3
    $\begingroup$ 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 Mar 29 '16 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 Mar 29 '16 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 Mar 30 '16 at 0:39
1
$\begingroup$

The question has been answered in the comments

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.