Is $\omega^{n/2}=-1$ in any ring if $\omega$ is a primitive nth-root of unity and $n$ is an even number?

Question

I know this statement is true when we are working with the complex numbers, but I can't find out a way to prove it in any ring. I tried assuming by contradiction that $\omega^{n/2}=k\neq -1$ but can't get anywhere.

Thanks!

Edit: I added that $\omega$ is also a primitive (not only a nth root of unity) and that $n$ is even.

Answer 1

Consider what happens in the ring of $2\times 2$ matrices over $\mathbb{C}$.

Take, for example $\omega=\text{diag}(i,1)$. Then $\omega^4=\text{diag}(1,1)=I$, but $\omega^2=\text{diag}(-1,1)\not= I$.

Answer 2

I came to this: since $\omega$ is a primitive nth root of unity, then it generates all the nth roots of unity. Then, there must exist a $0\leq k\leq n-1$ such that $ \omega^k=-1$. By contradiction, lets supose that $k\neq n/2$. If $k<n/2$ then $\omega^{2k}=1$ so $\omega$ was not a primitive nth root of unity. Else if $k>n/2$ then $\omega^{2k}=\omega^{2k-n}\omega{n}=\omega^{2k-n}\cdot 1=1$, and because $2k-n<n$ then $\omega$ is not a primitive but only a root.

In both cases its a contradiction, so we conclude that $\omega^{n/2}=-1$.