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 $\epsilon$.

Now I ran out of ideas. Could you please help me with this problem? Thanks in advance!


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).

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