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Given $\mathbb{K}$ local non archimedean field, how can I find an example of two totally ramified extensions of $\mathbb{K}$ whose compositum is not totally ramified?

I know that every such extension is generated by a uniformizer, but I don't know how to start attacking the problem. Any hint?

Angelo Rendina
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1 Answers1

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Hint: Try extending $\Bbb{Q}_2$ by adjoining two distinct cube roots of two. If you adjoin both of them you also get something that gives an unramified extension.

Jyrki Lahtonen
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    Ok, it works! Thanks for the tip! – Angelo Rendina Dec 24 '14 at 16:23
  • Glad to hear that. Well done, Angelo! – Jyrki Lahtonen Dec 24 '14 at 16:30
  • Let $x,y$ be two distinct cube roots of $2$. Then $\mathbb{Q}_2(x)/\mathbb{Q}_2$ is totally ramified. But for $\mathbb{Q}_2(x,y)/\mathbb{Q}_2$, if I am not wrong. we have ramification index $e=3$, inertia index $f=2$. In particular, it is ramified extension. What do you mean by "something that gives an unramified extension"? – Doug Aug 17 '22 at 19:10
  • @DougLiu $\Bbb{Q}_2(x/y)/\Bbb{Q}_2$ is unramified. That's all I meant. – Jyrki Lahtonen Aug 17 '22 at 20:40
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    @DougLiu As per your calculation you saw that $\Bbb{Q}_2(x,y)$ is not totally ramified, which is all we need here. There will then automatically be an intermediate extension that is unramified. Also, when a Galois extension is not totally ramified, there will be an intermediate field that is unramified over the base field. – Jyrki Lahtonen Aug 19 '22 at 05:38