# Example of fully ramified extension of local field

For $Q_p$, the unramified extension is adding roots of 1 of order prime to p. So what will be the fully ramified extension of $Q_p$?

Thanks!

• A partial answer is given by your first sentence. – franz lemmermeyer Sep 30 '17 at 19:39
• As @nguyenquangdo suggests in his excellent answer below, there probably is no “the” totally ramified extension of $\Bbb Q_p$. By adjoining one $m$-th root of $p$ for each $m>1$ in a clever way, you can get an infinite extension with no unramified part, but the fact that the subfields of this are hardly all totally ramified fields contributes to the difficulty and importance of “Higher Ramification Theory”. – Lubin Oct 10 '17 at 22:31

The answer to your question is the same for any local $p$-adic field $K$, i.e. any finite extension of $\mathbf Q_p$. The good behaviour of the unramified extensions is that the compositum of two of them is still unramified. So we can introduce the maximal unramified extension $K_{nr}$ of $K$, and then the property that you recalled can be stated as : " $K_{nr}$ is the extension of $K$ obtained by adding all the roots of unity of order prime to $p$ ". No such theorem exists for totally ramified extensions because the compositum of two such extensions can contain an unramified extension (*). So my answer will be threefold :
1) Generation of a totally ramified extension : an Eisenstein polynomial in $K[X]$ is a polynomial of the form $X^n + a_{n-1} X^{n-1} +...+ a_1 X + a_0$ with $v_K (a_i) \ge 1$ for $i=1,...,n-1$ and $v_K (a_0)=1$. The following theorem is classical : "An Eisenstein polynomial is irreducible. If $\pi$ is a root, then $L=K(\pi)$ is totally ramified and $v_L (\pi)=1$. Conversely, if $L/K$ is totally ramified and $v_L (\pi)=1$, the irreducible polynomial of $\pi$ over $K$ is Eisenstein."
2) Abhyankar's lemma : if $L$ and $M$ are two finite extensions of $K$ with ramification indices $e_L$ and $e_M$ s.t. $p$ does not divide $e_M$ (tame ramification) and $e_M$ divides $e_L$, then the compositum $L.M$ is unramified over $L$.
• You probably want the conclusion of Abhyankar's lemma to be that $LM/L$ is unramified, not that $LM/K$ is. – Ravi Oct 23 '17 at 6:26