roots of unity in the maximal unramified extension and Kummer extensions

Let $K$ be a field complete with respect to a discrete valuation $v$ with residue field $k$ of positive characteristic $p$. Consider $$K \subset K_{ur} \subset K_s$$

with $K_s$ the separable closure of $K$, and $K_{ur}$ the maximal unramified su-extension of $K$ in $K_s$. In particular Serre in his ''Local Fields'' explains that we have a unique extension $w$ of the valuation $v$ on $K_{ur}$ and the residue field is the separable closure $k_s$ of $k$.

Can I use the Hensel lemma to say that $K_{ur}$ contains the $d$-th roots of unity with $\mathrm{gcd}(d,p)=1$?

I done the follow reasoning:

Let $d \geq 1$ be an integer such that $\mathrm{gcd}(d,p)=1$. Then the polynomial $f(x)=x^d -1$ is separable over $k$. Hence the roots of $f(x)$ belong to $k_s$. I would lift this roots to $K_{ur}$ and I thought that Hensel lemma it's a good way. However can apply the lemma to the valuation ring of $K_{ur}$?

Add question Take $\pi$ a uniformizer of $K_{ur}$ and consider the field $L=K_{ur}(\pi^{1/d})$. I would show that the extension $L/K_{ur}$ is tamely and of degree $d$.

It is enough show that the minimal monic irreducible polynomial over $K_{ur}$ of $\pi^{1/d}$ is $x^{d}-\pi$? Could I have an hint to show that $x^{d}-\pi$ is irreducible?

Maybe the fact that the minimal polynomial $f(x)$ of $\pi^{1/d}$ is a product of the form $$\prod_{k\in \Gamma} (x-\omega^k_d\pi^{1/d})$$ with $\Gamma \subset \{1,\ldots ,d\}$, $\omega_d$ a primitive $d$-th roots of unity, can it help?

I thought that: the extension $K_s/K_{ur}$ is totally ramified then it must be $L/K_{ur}$ since the inertia degree is multiplicative in towers.

What's useful about weakening the notion from completeness to Henselian'ness' is that any algebraic extension of henselian field is also henselian. Therefore you can use hensel's lemma on $K_{ur}$ as $K$ is complete (and hence henselian).
• I find it quite weird/possibly rude to deselect as accepted answer as you decided to ask another question. (would make sense to ask a new question) Well anyhow irreducibility of $x^d-\pi$ follows from Eisenstein's Criterion generalised slightly (can be found in wikipedia if you don't know). I'm not so sure about what you mean by the second part. Any extension of $K_ur$ is totally ramified. – Jack Yoon Aug 6 '15 at 11:47