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Let $K$ be a number field and $v$ be its place. Let $p$ be a prime element of ring of integers of $K$. Let $K_v$ be completion of $K$ at $v$. I want to prove $C: y^2=x^4-p$ has $K_v$ rational point.

I tried to use Hensel lemma, but I'm stucking with which $x$ I should choose.

In the case $K=\Bbb{Q}$, $C$ has trivial real point $(p^{1/4},0)$, so we should consider $C(\Bbb{Q}_l)$ has rational point or not. As necessarily condition, we require Hilbert symbol satisfies $(1/p,1/p)_l=1$.

Thank you for your help.

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    $\begingroup$ Is there any relation between $v$ and $p$? Do you know that a number field has many different "places" , so saying "its place" is unclear? $\endgroup$
    – Torsten Schoeneberg
    Apr 4, 2023 at 18:12
  • $\begingroup$ $v$ is a fixed nontrivial places, $p$ and $v$ are independent. $\endgroup$
    – Pont
    Apr 4, 2023 at 18:41
  • $\begingroup$ For example, in the case of $K=\Bbb{Q}$, $v$ runs prime number $2,3,5,...,p ,...$ or real place. $\endgroup$
    – Pont
    Apr 4, 2023 at 19:46
  • $\begingroup$ I see, thanks for clarifying. Well then, do you see that the case of the real place is easy? So let's deal with finite places. Would you have an idea in case $K= \mathbb Q$ i.e. $K_v =\mathbb Q_\ell$? It would be very helpful to know how the subgroups of squares, and fourth powers, in (the multiplicative groups of) those fields look like. Do you know that? $\endgroup$
    – Torsten Schoeneberg
    Apr 5, 2023 at 0:59
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    $\begingroup$ If $v(x) < 0$ then $x^4-p= x^4(1-p/x^4)$ is a square. $\endgroup$
    – reuns
    Apr 5, 2023 at 13:32

1 Answer 1

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[Rewritten answer.]

A crucial thing to know is that all elements $z \in K_v$ such that

$$v(z-1) \ge 2v(2)+1$$

(which simplifies to $v(z-1) \ge 1$ if $v$ lies above an odd prime $\ell$) are squares.

One can see this with $\ell$-adic logarithms, or with expansion of the binomial series $(1+y)^{1/2}$ ($y:=z-1$). But it is also a straightforward application of Hensel's Lemma: For the case $\ell$ odd the standard version suffices; for the case that $v$ lies over $\ell=2$ one can use the general version of Hensel's Lemma from here, cf. section 4 and 8 here. (The condition $v(z-1) \ge 2v(2)+1$ is the additively rewritten $\lvert f(1) \rvert <\lvert f'(1)\rvert^2$ for $f(T) = T^2-z$.)

Now we introduce that element $p$ which at least satisfies $v(p)\ge 0$. We can certainly find many $x$ (namely, $v(x) < $ ? -- for most $v$, $\le -1$ will suffice here) such that $v(p/x^4) \ge 2v(2)+1$ and hence, per reuns' hint in the comment:

then $x^4-p = x^4(1-p/x^4)$ is a square

which should allow you to finish. It means that all such $x$ will give a $y$.

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  • $\begingroup$ For $p$:odd, let $f(T)=T^2-z$. Then $f(1)=1-z≡0modl$, $f'(1)=2\not\equiv 0modl$. From Hensel lemma, $z^{1/2}$ is in $\Bbb{Q}_l$. For $l=2$, what kind of modification is needed ? $\endgroup$
    – Pont
    Apr 6, 2023 at 8:37
  • $\begingroup$ Is that a different question? Is $p=l$? What is $z$? $\endgroup$
    – Torsten Schoeneberg
    Apr 6, 2023 at 14:24
  • $\begingroup$ No, this is about your second paragraph. Notation $l$, $z$ is the same as yours. $l$ is prime number, $p$ and $l$ are independent. $\endgroup$
    – Pont
    Apr 6, 2023 at 14:28
  • $\begingroup$ Ah I see. There is no $p$ there so far. Indeed one can show that with Hensel, with a slightly more general version in case $\ell=2$, see edit. $\endgroup$
    – Torsten Schoeneberg
    Apr 7, 2023 at 3:45
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    $\begingroup$ @TorstenSchoeneberg it looks good. More generally, for a local field $F$, $a \in F$, and $n$ a nonzero even number, $y^2 = x^n - a$ has a solution $x, y \in F$. To make Hensel's lemma for $f(T) = T^2 - (1 - a/x^n)$ applicable with approximate root $1$, choose $x \in F$ so that $|x|^n > |a/4|$, which is possible using large enough $x$ when $n > 0$ and small enough nonzero $x$ when $n < 0$. Then $|f(1)| < |f'(1)|^2$, so by Hensel's lemma there is $t \in F$ such that $f(t) = 0$, meaning $t^2 = 1 - a/x^n$, so $y^2 = x^n - a$, where $y = tx^{n/2}$. $\endgroup$
    – KCd
    Apr 7, 2023 at 15:39

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