Consider the following question asked in an assignment worksheet which I am solving by myself.
If n is an odd integer such that K contains a primitive nth root of unity and $ char K \neq 2 $, then K also contains a primitive 2n th root of unity. Here K is a field.
Let x be primitive n th root of unity. Then x can be written as $ x= y^{d/n}$ , (d,n)=1. n=2m+1 for some integer m. What I need to prove that $y^{d'/2n}$ , where (d', 2n)=1. I have proved that $(-y)^{d/n} \neq -1$ for all the d lying in {1,..., n-1} but unable to prove that $(-y)^{2n-1 /2n}\neq 1$ which will complete my proof as $(-x)^{2n}=1$ as $x^{n}=1$. I tried to prove that -x is a primitive root of unity.
So, I request your help.
As you noted we only need to prove that $(-x)^k\ne1$ for $0<k<2n$. If $(-x)^k=1$ for such a $k$ then $x^{2k}=1$, hence $n\mid 2k$ as $x$ is a primitive $n$-th root. As $n$ is odd we get $n\mid k$ and therefore $n=k$. But then $1=(-x)^k=(-1)^nx^n=-1\ne1$ as are not in characteristic $2$ and $n$ is odd. Hence $(-x)^k\ne1$ for $0<k<2n$, so it follows that $-k$ is a primitive $2n$-th root of unity.