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Fix a prime $p$. Let $\Bbb F_p$ be the finite field with $p$ elements with $\omega$ a primitive $n^{th}$ root of unity. From Galois theory of finite fields, the extension $\Bbb F_p[\omega]/\Bbb F_p$ is Galois with degree $$\left[\Bbb F_p[\omega]:\Bbb F_p\right]=\left|\operatorname{Gal}(\Bbb F_p[\omega]/\Bbb F_p)\right|=\left|\left<\sigma_p\right>\right|,$$ where $\sigma_p(\omega) = \omega^p$.

So that the degree of the extension $\Bbb F_p[\omega]/\Bbb F_p$ is exactly the order of $p$ in $(\Bbb Z/n\Bbb Z)^{\times}$. I am guessing that $$\omega,\omega^p,\omega^{p^2},...$$ is a set of basis. This has the right length as a basis, but I struggled to prove this spans $\Bbb F_p[\omega]$ over $\Bbb F_p$. How can one show it spans?

William Sun
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    Isn't this a simple extension? If degree of the extension is $k$, then $1, \omega, \omega^2,\dots, \omega^{k-1}$ forms a basis. – Yathi Nov 14 '23 at 14:19
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    The conjugates of $\omega$ need not form a basis. Possibly the simplest to describe case when this fails is $p=3$, $n=4$. Customarily the symbols $i$ is often used for a fourth root of unity, so I use it instead of $\omega$. It is in the quadratic extension of $\Bbb{F}_p$ as a root of $x^2+1$. However, its conjugate is $i^3=-i$, and surely ${i,-i}$ is linearly dependent :-) – Jyrki Lahtonen Nov 14 '23 at 15:56
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    (cont'd) A slightly more complicated example occurs when $p=2$, $n=7$. We have the modulo two factorization $$x^7-1=(x-1)(x^3+x+1)(x^3+x^2+1),$$ so the roots of $x^3+x+1$ are seventh roots of unity. If $\omega$ is one of them, by Galois theory, the other two are $\omega^2$ and $\omega^4$. So in the splitting field $$x^3+x+1=(x-\omega)(x-\omega^2)(x-\omega^4).$$ Expanding the right hand side and comparing the coefficients of the quadratic terms gives the linear dependency relation $$0=\omega+\omega^2+\omega^4.$$ – Jyrki Lahtonen Nov 14 '23 at 16:00
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    (cont'd) However, the roots of the other cubic factor do form a basis. To close I give you the buzzword normal basis. Feed that into a search engine (of the on-site search), and you will find a lot of hits such as this old answer of mine. So every extension of finite fields has a normal basis, but it's not immediately clear to me for which choices of $n$ such a basis exists. – Jyrki Lahtonen Nov 14 '23 at 16:03
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    A point worth bearing in mind is that every non-zero element of every finite field is a primitive root of unity of some order. In this sense roots of unity abound in the realm of finite fields :-) – Jyrki Lahtonen Nov 14 '23 at 16:10

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