# Does totally ramified extension really exist?

The answer is certainly "Yes", but this is the problem I met in Algebraic Number Theory by Neukirch. I guess that I must be doing something wrong, since otherwise I will get the statement "There are no totally ramified extensions except the trival ones".

Let $K$ be Henselian field, $L/K$ be a finite, totally ramified extension. Let $\lambda$ and $\kappa$ be the residue field of $L$ and $K$ respectively. Because $L/K$ is totally ramified, $K$ is the maximal unramified subextension, so we have $\lambda=\kappa$. If $L\ne K$, let $a \in L-K$. Since the valuation is non-trival, by multiplying by an element in $K$, we can suppose $a \in O_L$, the valuation ring of $L$.

Because "the valuation ring of $L$ is the integral closure of the valuation ring of $K$ in $L$ "(P144, Chapt 2 Theorem (6.2) of Neukirch), let $f(x) \in O_K[x]$ be the minimal polynomial of $a$ in $O_K$, where $O_K$ is the valuation ring of $K$. (One can prove $f(x)$ is monic, and I guess it may differ from the minimal polynomial over $K$.) Let $\bar{f}(x)$ be the corresponding polynomial over the residue field $\kappa$. It must be the minimal polynomial of $\bar{a} \in \lambda$, because otherwise, by Hensel's lemma, $\bar{f}$ admits a factorization in $\kappa[x]$ implies $f$ admits a factorization in $O_K[x]$. But since $\lambda=\kappa$, we get $deg(\bar{f})=deg(f)=1$ ($f$ is monic). This means $a\in K$, a contradiction. This means $L=K$ !

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## 1 Answer

Henselian-ness means that any factorisation of $\overline{f}$ into coprime monic polynomials in $\kappa[X]$ lifts to $K[X]$. But the factorisation of $\overline{f}$ could be a power of an irreducible polynomial, and this is indeed what happens in practice: consider $K = \mathbb{Q}_p$ and $L = K[x]$ where $x^2 = p$. Then $\overline{f}$ is just $x^2$, which is obviously reducible but cannot be factorised nontrivially as a product of coprime factors.

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Great! Thank you very much! –  Li Zhan Jun 6 '11 at 9:58