Existence of a square root of an non-uniformizer over an extension of $\mathbb{Q}_5$

Question

Let $L/\mathbb{Q}_5$ be an extension with ramification index $6$. Assume that there is an element $\alpha \in L$ with valuation $1/3$.

Question Is there a square root of $\alpha$ in L (i.e. an element $\beta \in L$ with $\beta^2 = \alpha$)?

My own effort:

• If there is such an element $\beta$, it would have valuation $1/6$, i.e. it would be a uniformizer.

• I cannot apply Hensel's Lemma to $f = x^2 - \alpha \in L[x]$ since its reduction is not separable.

Could you please help me to advance with my question? Thank you!

Answer 1

Not necessarily. In fact, if every element of valuation $1/3$ has a square root, then by taking ratios of these it follows that every element of valuation 0 has a square root as well. This is never true when $L/\mathbb Q_5$ is finite. (For example, in $\mathbb Q_5(5^{1/6})$, the element $5^{1/3}$ has a square root but $2 \cdot 5^{1/3}$ does not, since the ratio 2 isn't a square in $\mathcal O_L/\mathfrak m_L = \mathbb F_5$.)

What is true that if there exists $\beta$ such that $\beta^2 - \alpha$ has valuation greater than $1/3$ (i.e. "$\alpha$ has a square root up to first-order approximation"), then $\alpha$ has a square root. You can prove this by the strategy in reuns's answer.

Answer 2

Choose some uniformizer $\pi_L$, it is of valuation $1/6$, then apply Hensel lemma to $\alpha/\pi_L^2\in O_L^\times$