# Find 3rd root in $\mathbb{Q}_3$ using Hensels Lemma

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?

• You can write $a=1+9b$ and use $x=1+3y$ for Hensel's lemma, then pick what $y$ to plug in that works. Commented Mar 27, 2022 at 22:11
• Why can I write a=1+9b? Commented Mar 27, 2022 at 23:00
• Because of the hypothesis that $|a-1|_3\le3^{-2}$, i.e. that $\text{ord}_3(a-1)\ge2$. Commented Mar 27, 2022 at 23:12
• 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? Commented Mar 28, 2022 at 14:09
• Theorem 4.5 in kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf is a more general result (replacing $3$ by an arbitrary odd prime).
– KCd
Commented Apr 4, 2022 at 7:14