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Let $n$ be a natural number, $K$ a field of characteristic not dividing $n$. Let $L/K$ be the field extension of $K$ obtained by adjoining a primitive $n$th root of unity.

Can there be elements of $K$ which are an $n$th power in $L$, but not in $K$?

If $n$ is prime, then the statement follows by observing that if $a \in K$ such that $a$ is not an $n$th power in $K$, then adjoining an $n$th root of $a$ defines a degree $n$ extension, which cannot be contained in $L$, which is of degree dividing $n-1$.

What if e.g. $n=4$?

Bib-lost
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2 Answers2

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$-4=(1+i)^4$ became a fourth power when $i$, a primitive fourth root of unity, was adjoined to $\Bbb{Q}$.

Jyrki Lahtonen
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    Do take a look at this result. Other prime squares $p^2,p>2,$ are a bit more delicate in the sense that it seems to me that $a$ has to be a $p$th power in $K$. For otherwise $x^{p^2}-a$ is irreducible over $K$, and we run into the problem in the OP's argument. – Jyrki Lahtonen Oct 19 '19 at 19:00
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    Similarly, -27 becomes a sixth power when adjoining a 6th root of unity to $\mathbb{Q}$. My earlier claim that the statement was true for $n$ square free, was wrong. – Bib-lost Oct 19 '19 at 21:41
  • @Bib-lost A nice example as well. Any idea what happens with the ninth powers/roots of unity. I thought I had an example, but it was a late night half-drunken mirage :-/ – Jyrki Lahtonen Oct 20 '19 at 04:33
  • By your previous remark, the question reduces (?) to: "If $a \in K$ is not a third root of unity nor a third power, can it become a third power when adjoining a ninth root of unity?" I don't have an answer yet. – Bib-lost Oct 20 '19 at 21:22
  • I did think that this matter had something special to do with $n=2$. But I just had a cocktail at dinner, so I have to be a spectator, at least for a while. – Lubin Oct 21 '19 at 00:53
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    @JyrkiLahtonen The ninth cyclotomic field contains a (unique) three-dimensional subfield isomorphic to the root field of $X^3 - 3X + 1$ (generated by the sum of a ninth root of unity and its inverse). In the field $K = \mathbb{Q}(\sqrt{3}i)$ the element $\epsilon = \frac{1 + \sqrt{3}i}{2}$ is not a third power. But adjoining a cube root of $\epsilon$ is the same as adding a root of $X^3 - 3X + 1$. Hence, when adjoining a ninth root of unity to $K$, $\epsilon$ becomes a third power. – Bib-lost Nov 07 '19 at 17:30
  • @JyrkiLahtonen : Only a year later do a realize that the $\epsilon$ I gave satisfies $\epsilon^3 = -1$, which is a ninth power. So I am again unsure about the case where $n$ is odd. – Bib-lost Nov 20 '20 at 17:41
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Take $K=\mathbb R$ and $L=\mathbb C=K(i)$. Then all elements of $K$ become fourth powers in $L$. Note that $i$ is a primitive fourth root of unity. Only the positive real numbers are already fourth powers in $K$.

lhf
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  • True, thanks. As I was discussing with Jyrki Lahtonen in the comments of their answer, it might be the case that this same problem does not occur when $n$ is odd. Do you have any thoughts on that? – Bib-lost Oct 20 '19 at 22:18