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Let $A$ be a complete noetherian local ring and $\mathfrak{m}$ be its maximal ideal.

If we have several polynomials $f_i \in A[X_1, \dots, X_m]$ which have a common zero $x_n$ in $A/\mathfrak{m}^n$ for any $n$, then must they have a common zero in $A$? In other words, if we have a scheme $X$ finite type over $A$, does $X(A/\mathfrak{m}^n) \not= \varnothing \ \forall n$ imply $X(A) \not= \varnothing $?

This seems an easy question, but the main problem is compatibility for $x_n$. If $X$ is smooth we conclude by Hensel lemma.

  • If $k=A/\mathfrak{m}$ is finite then it's true because inverse limit of non-empty finite sets is non-empty.
  • If $A$ is a DVR then we know from a general Hensel lemma from "RATIONAL POINTS IN HENSELIAN DISCRETE VALUATION RINGS" by MARVIN J. GREENBERG.

  • If $k=A/\mathfrak{m}$ is algebraically closed and uncountable, it is also true by considering constructible topology and the inverse limit of non-empty compact Hausdorff spaces is non-empty.

    What about general case? Here we assume that $A$ is a complete noetherian local ring, but this also covers the excellent Henselian local ring case because of Artin approximation.

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    $\begingroup$ If you assume $X$ to be smooth over the punctured spectrum of $A$, then the answer is yes by Thm. 1 in Elkik's "Solutions d'équations a coefficients dans un anneau hensélien" (the role of smoothness is to ensure that the assumption which concerns the ideal $H_B$ there is met). A similar result in a non-Noetherian setting is Lemma 5.4.8 ff in Gabber, Ramero "Almost ring theory". Also, it would be be interesting to have a non-Noetherian result of this sort beyond principal ideals, e.g., for finitely generated weakly proregular ideals (this would, in particular, recover Elkik's result)... $\endgroup$
    – Kestutis Cesnavicius
    Commented Dec 11, 2018 at 9:01
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    $\begingroup$ This is true in general, by the "strong approximation property" due essentially to Popescu. See the survey arxiv.org/abs/1506.04717v4 by Guillaume Rond, or the Bourbaki talk by Teissier numdam.org/item/SB_1993-1994__36__259_0 $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 11, 2018 at 15:03
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    $\begingroup$ @LaurentMoret-Bailly Theorem $3.16$ in this good survey solves the problem, and your work shows Theorem $3.16$ implies $A$ is an excellent Henselian local ring, so basically that's all we can say. I didn't notice that strong arrpoximation property before, thank you. $\endgroup$
    – Zhiyu
    Commented Dec 11, 2018 at 16:14

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