# Compute an approximation of some cubic polynomial over an extension of the $3$-adics

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

where $$\alpha = \frac{(1-\zeta_3)(1-\sqrt{-7})}{6}$$ 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 $$\epsilon$$.

Edit: After skimming through the answers, I noticed that I took the wrong element $$\alpha$$. In this post, I changed it to the one I intended to use. I will check the answer here more thoroughly though.

• The uniformizer of $$L$$ is $$(-3)^{1/4}$$

• If $$\epsilon^3 = \frac{1}{4}\bmod (-3)^{9/4}$$ for some $$\epsilon\in L$$ then $$\epsilon-1$$ is a root of $$(x+1)^3-1/4 =x^3+3 x^2+3x+3-9\bmod (-3)^{9/4}$$

This implies that $$v(\epsilon-1)=1/3$$ which is a contradiction since $$L/\Bbb{Q}_3$$ is tamely ramified with ramification degree $$4$$, no element of $$L$$ has valuation $$1/3$$

Also note that $$v{1/3\choose n}= v(\prod_{k=0}^{n-1} (1/3-k))-v(n!) = -n-\sum_{k\ge 1} \lfloor n/3^k \rfloor\le -n-n\frac{3^{-1}}{1-3^{-1}}$$ thus $$(1+x)^{1/3}=\sum_n {1/3\choose n}x^n$$ converges for $$v(x)>1+\frac1{3-1}$$ and hence $$\epsilon^3=1/4\bmod 3^2$$ gives that $$(4\epsilon^3)^{1/3}\in L, 4^{1/3}\in L$$ which is a contradiction since no element of $$L$$ has valuation $$1/3$$. The convergence of the binomial series can be stated in term of Hensel lift, when $$f'(a)=0\bmod \pi$$ we need to use the higher derivatives.

• Rather more efficient than my answer, as usual. – Lubin Feb 23 at 5:43
• May I ask how you obtain the equality $(x+1)^3 - \frac{1}{4} = x^3+3x^2+3x+1-\frac{1}{4} = x^3 +3x^2+3x-3-9$ modulo $(-3)^{9/4}$? And how can we derive $v(\epsilon^3-1/4) = 1/3$ from this equality? – Diglett Feb 24 at 14:47
• $1/(1+3)= 1-3+9 =1-3+9 \bmod 27$. It is immmediate that a root of $x^3+3x^2+3x+3+o(3)$ with valuation $\ge 0$ must have valuation $1/3$. – reuns Feb 24 at 15:00
• Thanks, I was able to understand the first answer. Though I cannot see yet why the second answer is immediate. Let $\alpha$ be a root of $f(x) = x^3+3x^2+3x+3$ with valuation $\geq 0$. Then the valuation of $f(\alpha)$ is $\geq 3 \min\{v(\alpha), 1/3\}$, depending on which one is less. If $v(\alpha) \neq 1/3$, then we have equality. But why can $v(f(\alpha))$ is neither $3 v(\alpha)$ nor $1$? – Diglett Feb 24 at 16:07
• If the valuation of $\alpha$ is $<1/3$ then $v(f(\alpha))=v(\alpha^3)<1$. If it is $> 1/3$ then $v(f(\alpha))=v(3)=1$. Thus for $v(f(\alpha))> 1$ we need $v(\alpha)=1/3$ @Diglett – reuns Feb 24 at 17:16

First, notice that $$\sqrt{-7}$$ is already in $$\Bbb Q_3(i)$$, since $$X^2+7\equiv X^2+1\pmod3$$: the latter factors in $$\Bbb Q_3(i)$$, so the former must as well. Thus $$L=K$$. This field has ramification index $$e=4$$ over $$\Bbb Q_3$$, and the residue-field extension degree is $$f=2$$. For a prime element, we can take $$\pi=\sqrt{-3}$$.

Second, since you’re talking about units, and $$2$$ is one such, finding an $$\epsilon$$ of the type you want is equivalent to finding a unit $$\delta$$ with such that $$v(\delta^3-2)\ge\frac94$$: just take $$\epsilon=\delta/2$$.

We may assume that such a $$\delta$$ will be $$\equiv-1\pmod\pi$$: otherwise multiply by a cube root of unity. Thus we write $$\delta=-1+\rho$$ with $$v(\rho)>0$$, and expand $$\delta^3-2=-1+3\rho-3\rho^2+\rho^3-2$$. Let’s look carefully: $$\delta^3-2=-3+3\rho-3\rho^2+\rho^3$$ In case $$v(\rho)<\frac13$$, the $$v$$-value of the whole sum is $$3v(\rho)<1$$, while if $$v(\rho)>\frac13$$, the $$v$$-value of the whole is $$1$$, the minimum valuation being taken on at the first term in the sum. Of course these two are the only possible cases, since there are no elements of $$v$$-value $$1/3$$. In other words, it’s not possible for $$v(\delta^3-2)$$ to take on the value $$9/4$$, similarly for $$v(\epsilon^3-\frac14)$$.