Algebraic extensions of p-adic closed fields I have been working with p-adically closed fields and there are two results that are used time and times again in what I am reading, but I cannot find any references where they are proved...
The first is that p-adically closed fields have a finite number of algebraic extensions of a given degree, the other that all algebraic extensions of a p-adically closed field are generated by an element that is algebraic over Q.
As there is (it seems to me) a lot more literature on p-adic fields than on p-adically closed fields, if anyone had an idea where these results might be proved for Qp, I would be very grateful (although if someone had a reference for p-adically closed fields directly I do not mind).
(added on the 17th of July)
I have been wondering lately if there is a way, when one has a finite extension $K$ of $Q_p$ of finding a generator $a$ of $K$ that is algebraic over $Q$ that is also a generator of the valuation ring of $K$ over $Z_p$?
There must be a straighforward reason why this is possible, but I can not figure it out...
 A: Both the results you ask about, in the case of $p$-adic fields, follow from Krasner's lemma.
See for example, Proposition 3 and 4 of Lang, Algebraic Number Theory, p.43,44. 
The main point
is that two irreducible polynomials which are sufficiently close in the $p$-adic topology have roots which generate the same extension. 
The second clearly follows from this by approximating a polynomial with coefficients in (say) $\mathbb{Q}_p$ by a polynomial with coefficients in $\mathbb{Q}$. To get the first result, one may assume the field is generated by a root of a monic irreducible polynomial with (say) $\mathbb{Z}_p$ coefficients and then use the compactness of $\mathbb{Z}_p$ (and approximation as above) to get finiteness.
A: For $\mathbb{Q}_p$, you can have a look at Laurent Berger's notes http://perso.ens-lyon.fr/laurent.berger/coursM2/Poly2010.pdf, especially theorem 15.3. They are written in French, but guessing from your name, I would not be surprised if you could read French.
Basically, the argument is as follows. Pick an extension $K$ of $\mathbb{Q}_p$ of finite degree d. Let me denote by e the ramification index and  by f the degree of the residual extension. We have d=ef. The maximal unramified subextension $L$ is obtained by adjoining a root of unity (of order $p^f-1$). Then you go from $L$ to $K$ by adjoining a root of an Eisenstein polynomial of degree e. Now, the coefficients of such a polynomial lives in the ring of integers of $L$, which is compact. By Krasner's lemma, the extension will not change if you change the Eisenstein polynomial by another one which is close enough. Hence you only have a finite number of extensions and they are generated by elements which are algebraic over $\mathbb{Q}$.
Unfortunately, I am not really familiar with general p-adically closed fields and cannot tell you if the previous argument still works. By the way, do you have any good reference for them?
A: Here is a completely elementary proof of the fact that a finite extension $F$ of $\mathbf{Q}_p$ has only finitely many extensions of any given degree $n>0$.  As $F$ has a unique unramified extension of any given degree, we are reduced to the case of totally ramified extensions. Next, it is easily seen that there are only finitely many (totally ramified) extensions of degree prime to $p$ (essentially because $F^\times/F^{\times m}$ is finite for any $m$ prime to $p$), so we are reduced to the case $n=p^r$.  Passing to the galoisian closure and using the fact that any $p$-group admits a filtration whose successive quotients are of order $p$, we are reduced to the case $n=p$. Now, degree-$p$ extensions of $F$ become cyclic when translated to $K$, the extension of $F$ obtained by adjoining to $F$ the $(p-1)$-th roots of everything in $F$; this follows from "Galois's last theorem" (MO24081), see for example arXiv:1005.2016v3.  Two degree-$p$ extensions of $F$ give rise to the same cyclic extension of $K$ if and only if they are conjugate over $F$, and the number of conjugates of a degree-$p$ extension $E$ is $1$ (if $E$ is cyclic over $F$) or exactly $p$ (otherwise).  As $K$ automatically contains a primitive $p$-th root of $1$, cyclic extensions of $K$ correspond to $\mathbf{F}_p$-lines in $K^\times/K^{\times p}$, which is finite.  In fact, the compositum of all degree-$p$ extensions of $F$ is the extension of $K$ obtained by adjoining the $p$-th roots of everything in $K$.  Done.  (It is at this point that the proof breaks down if $F$ had been a finite extension of $\mathbf{F}_p((t))$; such $F$ have infinitely many separable (indeed cyclic) extensions of degree $p$.)
