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Unramified extension of local field is automatically galois because there is bijection between unramified extension of local field and extension of residue finite field, that is galois.

But, what about ramified extension of local field ? Does every ramified extension of local field is galois ?

Pont
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    Do you know any example of such extension ? – reuns Feb 18 '22 at 13:52
  • What do you mean by 'such' ? My question is asking whether there is an example of titled one. If you mean some extension of ramified extension of local field, then, $ \Bbb{Q_p}(p^{1/n},μ_{p^m-1})$ is ramified index is $m$ and residue index is $n$ over $ \Bbb{Q_p}$. – Pont Feb 18 '22 at 14:08
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    What about $\mathbb Q_p(\sqrt[p]p)$? – Mathmo123 Feb 18 '22 at 14:40
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    Also... every unramified extension of local fields is Galois. So if every ramified extension was Galois too, then all extensions would be Galois! – Mathmo123 Feb 18 '22 at 15:11
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    I think by "such extensions" reuns means ramified extensions, and what he means is that in fact, "most" ramified extensions are not Galois. That is, if you tried out a few of degree 3 or bigger, you should have stumbled upon a non-Galois one. In fact, generalizing @Mathmo123 's comment, it was shown in https://math.stackexchange.com/q/4256962/96384 that $\mathbb Q_p(\sqrt[n]{p})$ is Galois only if $n$ divides $p-1$, or $n=p=2$. – Torsten Schoeneberg Feb 19 '22 at 16:54
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    Just to sharpen my point a bit: every non-Galois extension of local fields is ramified. So just find any non-Galois extension of local fields. Radical extensions are a good place to start. – Mathmo123 Feb 19 '22 at 16:57

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Well, at least as a placeholder, when $\mathbb Q_p$ contains no $n$th root of unity, then "the" extension $\mathbb Q_p(p^{1/n})$ is not Galois. It is of degree $n$, by Eisenstein's criterion, but is not "normal", because it is missing $n$th roots of unity, etc.

I would think that more complicated extensions would tend to have this property, as well, so this is not any sort of pathology.

paul garrett
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