Minimal polynomial of cyclic $p^{n}$ extension of $\mathbb{Q}$? Excuse me if the following question is stupid:
When $p$ is an odd prime, we have $\operatorname{Gal}(\mathbb{Q}(\mu_{p^{n+1}})/\mathbb{Q})=\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}/p^{n}\mathbb{Z}$.
Therefore, there is an extension of $\mathbb{Q}$ of Galois group $\mathbb{Z}/p^{n}\mathbb{Z}$ lying in $\mathbb{Q}(\mu_{p^{n+1}})$.
Can its minimal polynomial be described in a generic expression?
(I know it is $x^3-3\cdot x+1$ if ${p}^{n}={3}$, for instance.)
 A: Maybe it’s not the best pedagogy to give you a quick rundown on the recipe and then spend a lot of time justifying it, but that’s what I’ll do.
Fixing $p$, we set $[p](X)=pX+X^p$, and $[p^n]=[p]\circ[p^{n-1}]$, so that $[p^n]=[p]^{\circ n}$, the $n$-fold iteration of $[p]$. Notice that $[p^n]$ is a (monic) $\Bbb Z$-polynomial, of degree $p^n$ since composition of polynomials multiplies the degrees. Also, let $g(X)=p+X^{p-1}$, an Eisenstein polynomial whose roots are the nonzero roots of $[p]$. This means that
$$
\frac{[p^n](X)}{[p^{n-1}](X)}=g\bigl([p^{n-1}](X)\bigr)=p+\bigl([p^{n-1}](X)\bigr)^{p-1}\,,
$$
a monic Eisenstein $\Bbb Z$-polynomial of degree $p^n-p^{n-1}$. But in fact when you look at this polynomial closely, you see that all nonzero monomials are in degrees divisible by $p-1$, so that it has the shape $h(X^{p-1})$, where now $h$ is monic Eisenstein of degree $p^{n-1}$. For the cyclic subextension of $\Bbb Q(\zeta_{p^n})$ of degree $p^{n-1}$, this $h(X)$ can be used to give your extension.
Before I justify all the above, let me work it out for your case $p=3$, $n=2$. Here $[3](X)=3X+X^3$, $g(X)=3+X^2$, and $h(X^2)=3+([3](X))^2=3+9X^2+6X^4+X^6$, so that $h(X)=3+9X+6X^2+X^3$. Now, $-h(-X-1)=X^3-3X^2+1$, whose roots’ reciprocals satisfy your polynomial.
Now for the justification:
As you may have seen, the open unit disc in $\Bbb C_p$ carries infinitely many essentially inequivalent $\Bbb Z_p$-analytic group structures. These are called the formal group laws over $\Bbb Z_p$. One of these is the multiplicative formal group law $\mathscr M(X,Y)=X+Y+XY$. It describes the action of taking elements $z,w\in\Bbb C_p$ with $|z|<1$, $|w|<1$, adding $1$ to both to get principal units, multiplying these, and then retranslating by subtracting $1$. The torsion elements of this structure are just the translates of the $p$-power roots of unity—you show that there’s no other torsion in this abelian group.
It turns out that $\Bbb Z_p$ is injected into the endomorphism ring of such a structure, and for $a\in\Bbb Z_p$. we write $[a](X)$ for the analytic expansion of this endomorphism. The coefficients will always be in $\Bbb Z_p$. In the case of $\mathcal M$, we have, for $m\in\Bbb Z$,
$[m]_{\mathcal M}(X)=(1+X)^m-1$, so that $[p]_{\mathcal M}(X)=pX+\frac{p(p-1)}2X^2+\dots+pX^{p-1}+X^p$, and if you divide by $X$ you get the translated version of the $p$-cylotomic polynomial $\Phi_p$.
In his answer, Nguen Quang Do mentioned the so-called Lubin-Tate theory, and I’m afraid now that I have to quote results from the paper of mine with John Tate (fifty years ago!) that exposes this theory. You can find it elsewhere, too, but the writeup (almost totally by Tate) is very clear. I’ll simplify and specialize the results to the specific case we’re interested in. Almost all of the results I quote are contained in Lemma 1 and Theorem 1 in the first few pages, and the comments following them.
The first important fact is that there is a $\Bbb Z_p$-formal group law $F_p(X,Y)$ whose $p$-endomorphism is $[p]_F(X)=pX+X^p$. The second is that there is a $\Bbb Z_p$-analytic isomorphism between this $F_p$ and the multiplicative formal group law $\mathcal M$, though we never need to write this power series out. A consequence is that over $\Bbb Q_p$, the torsion points of $\mathcal M$ and the torsion points of $F_p$ generate the same fields. The third fact I want to use is that a $\Bbb Z_p$-series commutes with $[p]_F(X)$ in the sense of composition of series if and only if it is an endomorphism of $F_p$. In particular, we have the $(p-1)$-th roots of unity in $\Bbb Z_p$, and if $\zeta$ is one of these, then $[\zeta](X)=\zeta X$ is an automorphism, and leaves the group of $p^n$-torsion elements stable.
I think we can look at my recipe now. The polynomial $h(X^{p-1})$, being $[p^n]_F(X)\big/[p^{n-1}]_F(X)$, has for its roots the primitive $p^n$-torsion points of $F_p$, and these generate $\Bbb Q_p(\zeta_{p^n})=L_n$. But the extension you’re looking for, call it $K_n$ for now, has $\text{Gal}(L_n|K_n)$cyclic of order $p-1$, and since the $p-1$-roots of unity are there, you can conclude that $K_n$ is generated by the $(p-1)$-th powers of $p^n$-torsion points of $F_p$, in other words by the roots of $h(X)$.
What have we achieved? We’ve found a $\Bbb Z$-Eisenstein polynomial $h(X)$ of degree $p^{n-1}$ whose roots generate a cyclic totally ramified extension of $\Bbb Q_p$. At the moment, I don’t see clearly that the $\Bbb Q$-splitting field of $h(X)$ is exactly the extension that you want, but at least locally at $p$, it does the trick.
A: I  think that you should not  expect  any "generic" expression . Of course I can't prove that such a thing does not exist, but take the much simpler example of the fields  $Q_p$(mu_${p^n}$) and consider their union K, generally called the cyclotomic tower of $Q_p$. For any a = $p^m$.u, u a p-adic unit, one of the problems of local class fieldetheory is to determine the automorphism $s_a$ of the maximal abelian extension of $Q_p$ which is attached to a by the local reciprocity law. Well, the restriction  of $s_a$ to K is known, but, according to Serre, chapter 6 of Cassels-Fröhlich's book "Algebraic Number Theory", this can  be proved locally only by "hard local methods (Dwork)" or by "Lubin-Tate theory" (not trivial). If a "generic expression" were known, things should not be so complicated. Again, these considerations are no proof.
