Let $a \in \mathbb{Q}_3$ and suppose that $\vert a-1 \vert_3 \leq 3^{-2}$. Show that $a \in {\mathbb{Q}_3}^3$ using Hensel's Lemma.

My idea is the following: I consider $f(x) = x^3 - a$ and want to apply Hensel to deduce the claim. Observe that: $f'(x) = 3x^2$, $\vert f(1)\vert_3 \leq 3^{-2}$ and $\vert f'(1) \vert_3 = 3^{-1}$. However, the condition in Hensel's Lemma ($\vert f(1)\vert_3 < {\vert f'(1)\vert_3}^2$) does not hold. Could someone give me a hint on how to apply Hensel in this setup?

  • 1
    $\begingroup$ You can write $a=1+9b$ and use $x=1+3y$ for Hensel's lemma, then pick what $y$ to plug in that works. $\endgroup$
    – Merosity
    Commented Mar 27, 2022 at 22:11
  • $\begingroup$ Why can I write a=1+9b? $\endgroup$
    – Anton2107
    Commented Mar 27, 2022 at 23:00
  • 1
    $\begingroup$ Because of the hypothesis that $|a-1|_3\le3^{-2}$, i.e. that $\text{ord}_3(a-1)\ge2$. $\endgroup$
    – Lubin
    Commented Mar 27, 2022 at 23:12
  • $\begingroup$ ah yeah, makes sense. Thanks a lot! My solution proposal would be: Take $x=1-3b$. $f(1-3b)=3^3(b^2-3b^3)$, write $3^{\delta_1}z = b^2-3b^3$ for z coprime to p and $\delta_1 \geq 1$. $\vert f(1-3b) \vert_3 = 3^{-3} 3^{-\delta_1} \leq 3^{-3}$. $f'(1-3b) = 3(1-6b+3^2b^2)$, thus ${\vert f'(1-3b) \vert_3}^2 = 3^{-2}\vert 1-6b+3^2b^2\vert_3 = 3^{-2}$ since 1 and 3 are coprime and conclude using Hensel's Lemma. Are you d'accord? $\endgroup$
    – Anton2107
    Commented Mar 28, 2022 at 14:09
  • $\begingroup$ Theorem 4.5 in kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf is a more general result (replacing $3$ by an arbitrary odd prime). $\endgroup$
    – KCd
    Commented Apr 4, 2022 at 7:14


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