For $\mathbb{Q}_p$, the unramified extension is adding roots of 1 of order prime to p. So what will be the fully ramified extension of $\mathbb{Q}_p$?
Thanks!
For $\mathbb{Q}_p$, the unramified extension is adding roots of 1 of order prime to p. So what will be the fully ramified extension of $\mathbb{Q}_p$?
Thanks!
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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."
Now let us illustrate the property (*) :
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 $L$.
3) An example in wild ramification : see e.g. the first part of https://math.stackexchange.com/a/2422663/300700