# Compute an approximation of some cubic polynomial over a cubic totally ramified extension over the 3-adics

Remark: Since I made a mistake with in this post, we have $$L=K$$ there which was not supposed to happen. In this post, I have changed the element $$\alpha$$ to obtain another field $$L=K(\alpha)$$.

Let $$K = \mathbb{Q}_3(\sqrt[4]{-3},i)$$ and $$L = K(\alpha)$$

where $$\alpha = \sqrt[3]{2} \frac{(1-\zeta_3)(1-\sqrt{-7})}{6}$$ and $$\zeta_3 \in K$$ is a primitive 3rd root of unity.

Furthermore, let $$v$$ be the valuation on $$L$$ with $$v(3)=1$$.

Question: Is there a unit $$\epsilon \in L^\times$$ satisfying the equation $$\epsilon^3 \equiv \frac{1}{4}$$ modulo an element of valuation $$\frac{9}{4}$$?

Ideas and Approaches:

• I tried to use Hensel's Lemma on the polynomial $$f(X)= 4X^3-1$$. However, since $$f'(X) = 12X^2$$ vanishes modulo $$3$$, it cannot be applied.
• By using Magma, my colleague found out that $$\alpha^3 \in K$$, i.e. the minimal polynomial of $$\alpha$$ over $$K$$ is $$x^3-\alpha^3$$.
• It is $$v(\alpha) = -\frac{1}{2}$$, so $$\tilde{\alpha} = (1-\zeta_3) \alpha$$ is a unit in $$L$$ since $$v(1-\zeta_3)=\frac{1}{2}$$. Maybe this can be used for constructing an appropriate $$\epsilon$$.

Since $$\zeta_3 = \frac{-1+\sqrt[4]{-3}^2}{2}$$ and $$\sqrt{-7} = i \sqrt{7}$$ are both contained in $$\mathbb{Q}_3(\sqrt[4]{-3}, i)$$ (Don't forget $$\sqrt{7}$$ is already in $$\mathbb{Q}_3$$), when you adjoin $$\alpha$$ we are only really gaining $$\sqrt[3]{2}$$ and can simply rearrange by ordinary algebra,
$$\epsilon = \left(\frac{(1-\zeta_3)(1-\sqrt{-7})}{6\alpha} \right)^2= \frac{1}{\sqrt[3]{2}^2}$$
This is true for all valuations, not just $$\frac{9}{4}$$ (which I assume originally was desired as that's the minimum requirement to guarantee lifting with the general Hensel lifting method).