When two simple extensions of field are the same

Asked Aug 28 '24 at 19:05

Active Aug 28 '24 at 20:45

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Let $k$ be a field of $\operatorname{char}(k)=0$. Let $a, b\in k^{*}$ and $n\in \mathbb{N}$. Moreover, we assume $k$ contains all $n$-th root of unities. I want to show that $$k(\sqrt[n]{a})=k(\sqrt[n]{b}) \quad \text{if and only if} \quad (a/b)\in k^{*n}. $$ If $(a/b)\in k^{*n},$ then $a=bc^{n}$ for some $c$. Then $k(\sqrt[n]{a})=k(\sqrt[n]{b})$. But I do not know to how to prove the other way. Can you help me?

Thanks for the comments! I think this statement is not true. The correct statement should be related to Kummer theory which I did not know before.

edited Aug 28 '24 at 20:45

asked Aug 28 '24 at 19:05

Alexander

For $k=\Bbb Q$ this has been proved here. – Dietrich Burde Aug 28 '24 at 19:17
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Yes, I think you need that $\gcd(n,p)=1$. See also this case. – Dietrich Burde Aug 28 '24 at 19:23
@JyrkiLahtonen. Yeah, you give a counterexample, so what is the exact description? – Alexander Aug 28 '24 at 19:39
My guess would be that you need $a/b^i\in (k^*)^n$ for some integer $i$ with $\gcd(i,n)$. The case where $n$ is a prime was covered in the latter thread Dietrich found. I'm not totally sure, and it's too late an hour... – Jyrki Lahtonen Aug 28 '24 at 20:11
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Alexander, can you give a bit more context? Why are you talking about the algebraic closure $\overline{k}$, which is not used afterwards? Why do you need the result? In which context? Do you know about Kummer theory? – Dietrich Burde Aug 28 '24 at 20:41

When two simple extensions of field are the same

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