Artin Schreier equation Let $K$ be the field obtained by adjoining to $\mathbb{Q}_p$ a root $\alpha$ of the polynomial $f(x)=x^p-x-\dfrac{1}{p}$. I should prove that $K \supset \mathbb{Q}_p$ is a Galois extension of degree $p$. I managed to prove that $K$ has degree $p$:taking $\dfrac{1}{\alpha}$ we see that is a root of $x^p-px^{p-1}-p$ which is an Eisenstein polynomial.
I do not know how to show that we have all roots of $f$.
 A: You are correct. One can use (the same ideas as in the proof of) Hensel's Lemma. Let's prove something slightly more general. Let $[K:\mathbf{Q}_p] < \infty$ be the base field with residue field $k$. Let $q$ divide the order of $k$, and let $\pi_K$ denote a uniformizer of $K$. Then any root of
$$f(x) = x^q - x - \pi^{-1}_K$$
defines a Galois extension of $K$ with abelian Galois group of exponent $p$ and order $q$. Take $K = \mathbf{Q}_p$ and $q = p = |\mathbf{F}_p|$ and $\pi_{\mathbf{Q}_p} = p$ and you get your specific question. 
As in your argument, the polynomial is a reciprocal of an Eisenstein polynomial and so is totally ramified of degree $q$. Let $\pi:=\pi_L$ denote a uniformizer of $L$ so that $\pi^{-1}$ is a root of $f(x)$ and $L = K(\pi)$. Normalize valuations so that $v(\pi_L) = 1$ and $v(\pi_K) = q$ and $v(p) \ge q$.
$$\begin{aligned}
f(x + t) = \ & (x+t)^q - (x+t) - 1/\pi_K \\
= \ & f(x) + t^q - t + p h(x,t),
\end{aligned}
$$
where
$$h(x,t) = t \sum_{k=1}^{q-1}  \frac{1}{p} \binom{q}{k} x^k t^{q-k-1}
\in t \mathbf{Z}[x,t]$$
is a polynomial of degree at most $q-1$ in $x$.
In particular, if $v(x) = -1$ and $v(t) \ge 0$, then
$$v(p h(x,t)) \ge v(t) + v(p) + v(x^{q-1}) \ge  v(t) + v(\pi_K) - (q-1) v(\pi) = v(t \pi).$$
In particular, with $v(x) = -1$ and $v(t) \ge 0$ we have
$$f(x+t) = f(x) + t^q - t \mod \pi t.$$
Let $S$ be any fixed choice of representatives in $\mathcal{O}_K$ for $k$ (say the Teichmuller lifts). Since $q$ divides the order of $k$, there is a subfield $\mathbf{F}_q \subset k$.
Claim: Let $a_0 \in S$ be an element which reduces to $\mathbf{F}_q \subset k$. There exists a (unique) sequence $a_i \in S$ (depending on $a_0$) such that, if
$$x_n = \pi^{-1} + a_0 + a_1 \pi + \ldots + a_{n-1} \pi^{n-1} \mod \pi^n,$$
then 
$$f(x_n) \equiv 0 \mod \pi^{n}.$$
We prove this by induction. For the case $n = 0$ (in which we have no choice), we get (noting that $f(\pi^{-1}) = 0$):
$$f(x_0) = f(\pi^{-1} + a_0) = a^q_0 - a_0 \pmod \pi = 0 \pmod \pi,$$
by construction of $a_0$ since $t^q - t$ vanishes over $\mathbf{F}_q \subset k$.  For $n > 0$,
$$f(x_{n-1} + a \pi^n) = f(x_{n-1}) - a \pi^n \pmod \pi^{n+1},$$
and so one can choose $a = a_n$ uniquely so that $f(x_{n}) \equiv 0 \pmod \pi^{n+1}$.
Of course, $x_{\infty}$ is then a root of $f(x)$ in $L$ such that $x_{\infty} = \pi^{-1} + a_0 + \ldots$,
and thus there are $q$ distinct roots in $L$ and hence $L/K$ is the splitting field of the polynomial $x^q - x - 1/\pi_K$. If $\sigma \in \mathrm{Gal}(L/K)$ is a Galois automorphism, it is determined by the action on $\pi^{-1}$. Hence $\sigma$ can be labelled by $a \in \mathbf{F}_q$ via the identification:
$$[a] \pi^{-1} = \pi^{-1} + a \mod \pi.$$
Clearly this defines an injective homomorphism
$$\mathrm{Gal}(L/K) \rightarrow \mathbf{F}_q,$$
and the claim is proved.
A: It's just Hensel's lemma.
First note that the $p$-adic valuation of $\alpha$ is $-1/p$: from the identity $\alpha^p-\alpha=1/p$, it follows that $v_p(\alpha) <0$, and then by triangle inequality $v_p(\alpha^p)=v_p(\alpha^p-\alpha)=-1$.
Putting $K = \mathbb{Q}_p(\alpha)$ and $y = x - \alpha$, we want to show that the polynomial $g(y) = (y + \alpha)^p - (y + \alpha) - \frac{1}{p} \in K[y]$ has $p$ roots in $K$.
Now write $g(y) = y^p + c_{p - 1}y^{p - 1} + \cdots + c_2y^2 + (c_1 - 1)y$, where every $c_i$ is in the maximal ideal of $\mathcal{O}_K$. In fact, every $c_i$ is of the form $p\cdot d_i \cdot \alpha^{p-i}$, where $d_i$ is an integer. We thus have $v_p(p\cdot \alpha^{p-i})= 1-(p-i)/p=i/p>0$.
In the residue field, we have $\overline{g}(\overline{y}) = \overline{y}^p - \overline{y}$, which has $p$ different roots, namely the elements of $\mathbb{F}_p$.
Thus by Hensel's lemma, for every element of $\mathbb{F}_p$ there is a unique root of $g$ in $K$ lifting that element.
