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Let $C$ be a singular curve defined over a local field $K$. Let $\tilde{C}$ be its smooth compactification(maybe this is not normalization).

Why $\tilde{C}(K)\neq \emptyset$ implies ${C}(K)\neq \emptyset$ ?

First line of page $3$ of https://arxiv.org/pdf/2112.02470.pdf , this is because of 'implicit function theorem'. In this paper, we especially think the case $C:qy^2=x^4-p$($p,q$ are caprice integers)

But I'm stucking with what the statement of 'implicit function theorem' is here. I searched a lot, but I couldn't find any statement which fits in this context. I will be appreciated if you could tell me what the 'implicit function theorem' is here.

Reference or another approach(not using implicit function theorem) is also appreciated.

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    $\begingroup$ The argument in the preprint concerns a specific situation: an explicit curve $C$, with condition on $K$. I think you could mention that in the OP. $\endgroup$
    – François Brunault
    Commented Mar 27, 2023 at 6:41
  • $\begingroup$ More information about $C$ and $K$ has nothing to do with this question, but I added some comment regarding it. $\endgroup$
    – Duality
    Commented Mar 29, 2023 at 9:16
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    $\begingroup$ Probably “implicit function theorem” refers to Hensel’s Lemma. The more common way to refer to Hensel’s Lemma is “algebraic Newton’s method.” Since Newton’s method, the implicit function theorem, and the Picard-Lindelof Theorem are all corollaries of the Contraction Mapping Fixed Point Theorem, some people refer to them as if they were a single theorem $\endgroup$
    – Jason Starr
    Commented Mar 29, 2023 at 11:12
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    $\begingroup$ Local fields are the fraction fields of complete local rings. The sharp version of Hensel's Lemma does not just give a lift of a point over the residue field to the entire complete local ring (and thus also to the fraction field), it gives estimates on what kind of lifts we can find, e.g., their restrictions to quotients of the complete local ring by powers of the maximal ideal. It gives "so many" lifts, that they cannot all be contained in the complement $\overline{C}\setminus C$. So some of the lifts give points over the fraction field that are contained in $C$. $\endgroup$
    – Jason Starr
    Commented Mar 29, 2023 at 11:16

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