Proof that there exists an unramified extension of a particular defree.

Question

Let $F$ be a $p$-adic field. I am trying to understand the proof of the following statement, and I think I am just missing something from basic field theory: For any $n\geq 1$ there exists an extension $L/F$ which is unramified and verifies $[L:F]=n$.

So if we let $\mathbb{F}$ be the residual field for $F$, then $\mathbb{F}$ has an extension $\mathbb{L}/\mathbb{F}$ of degree $n$ which is unique up to isomorphism. Let $a\in\mathbb{L}$ be such that $\mathbb{L}=\mathbb{F}[a]$ and let $f$ be the minimal polynomial of $a$ over $\mathbb{F}$. We then have that $\deg f=n$. Choose a Polynomial $P\in\mathcal{O}_F[x]$ which is monic of degree $n$ such that $\overline{P}=f$. Finally, put $L=F[\alpha]$ where $\alpha$ is a root of $P$.

We have $[L:F]\leq n$ since $\deg(P)=n$. The next statement is what gives me trouble: We have $\alpha\in\mathcal{O}_L$ so $\overline{\alpha}$ is a root of $\overline{P}=f$ in the residual field $\mathbb{L}$, and it follows that $[\mathbb{L}:\mathbb{F}]\ge n$.

In particular, why does it follow that $[\mathbb{L}:\mathbb{F}]\ge n$?

Answer 1

Because $O_F$ is a complete DVR with finite residue field $$O_F/(\pi_F) = \Bbb{F}_{p^f}$$ then $$L=F(\zeta_{p^{fn}-1})$$ is the unique unramified extension of degree $$[L:F]=[O_L/(\pi_L):O_F/(\pi_F)]= [\Bbb{F}_{p^{fn}} :\Bbb{F}_{p^f} ]=n$$

Moreover for any $h\in O_F[x]$ monic of degree $n$ which is separable irreducbile $\bmod (\pi_F)$ then $L = F[x]/(h)$.

Proof : let $a_0,\ldots,a_r$ be representatives in $O_F$ of $O_F/(\pi_F)$ then $O_F = \{ \sum_{m \ge 0} c_m \pi_F^m, c_m\in a_0,\ldots,a_r\}$, doing the same for $O_L$ and taking $b_1,\ldots,b_n\in O_L$ a $O_F/(\pi_F)$-basis of $O_L/(\pi_L)$ we obtain $O_L = \sum_{j=1}^n b_j O_F$ so that $[L:F]=[O_L:O_F]=[O_L/(\pi_L):O_F/(\pi_F)]$.

Thus it is purely a problem of extension of residue field, there is only one field with $p^{fn}$ elements thus only one extension of degree $n$.

Moreover with $g \in O_L$ of order $p^{fn}-1$ modulo $(\pi_L)$ then $\lim_{k \to \infty}g^{p^{fnk}}$ converges in $O_L$ and it is a $p^{fn}-1$ root of unity.