# Cyclic extensions of degree $n$ which contain a primitive $n$-th root of unity

Let $$F$$ be a field which contains a primitive $$n$$-th root of unity $$\zeta$$. Furthermore, let $$E/F$$ be a cyclic extension of degree $$n$$, and consider a generator $$\sigma$$ of the Galois group of $$E/F$$. Assume there is an element $$\alpha \in E$$ with $$\sigma(\alpha) = \zeta \alpha$$.

Question Why is $$E = F(\alpha)$$?

The $$\supseteq$$-inclusion is obvious, so we only need to show the other direction. I tried to use the fact that $$\alpha^n \in F$$ and there is no smaller power of $$\alpha$$ lying in $$F$$. Maybe one can show that $$X^n - \alpha^n$$ is the minimal polynomial of $$\alpha$$ over $$F$$ but this is just a different type of problem which I cannot prove. Maybe I am also making the argument too difficult, I really don't know.

Say the minimal polynomial of $$\alpha$$ is some $$p(x)\in F[X]$$ dividing $$X^n-\alpha^n$$.

As all roots of $$p(x)$$ must be roots of $$X^n-\alpha^n$$, we can write

$$p(x)=\prod_{i\in I} (X-\zeta^i\alpha)$$

where $$I$$ is some nonempty subset of $$\{0,1,2\dots,n-1\}$$.

The trick now is just in defining an automorphism on $$E[X]$$ based on $$\sigma$$. Specifically, define $$\bar \sigma: E[X]\rightarrow E[X]$$ so that for any $$e(x)=e_0+e_1x+\dots+e_kx^k\in E[X]$$, we have $$\bar \sigma(e(x))=\sigma(e_0)+\sigma(e_1)x+\dots+\sigma(e_k)x^k$$ (I'll leave it to you to check that this an automorphism).

Since $$p(x)$$ is in $$F[X]$$, all of $$p(x)$$'s coefficients are in $$F$$, so (by the definition of $$\bar \sigma$$), we ought to have $$\bar \sigma(p(x))=p(x)$$. This means

$$\bar \sigma\left(\prod_{i\in I} (X-\zeta^i\alpha)\right)=\prod_{i\in I} (X-\zeta^i\alpha)$$

So

$$\prod_{i\in I} \bar\sigma(X-\zeta^i\alpha)=\prod_{i\in I} (X-\zeta^i\alpha)$$

$$\prod_{i\in I} (X-\zeta^i\sigma(\alpha))=\prod_{i\in I} (X-\zeta^i\alpha)$$

$$\prod_{i\in I} (X-\zeta^{i+1}\alpha)=\prod_{i\in I} (X-\zeta^i\alpha)$$

where above we have used the fact that $$\zeta\in F$$. We get that whenever $$i\in I$$, $$i+1$$ (or at least something congruent to $$i+1\;\textrm{mod}\; n$$) is also in $$I$$. It follows that $$I=\{0,1,2\dots,n-1\}$$, so that $$p(x)=X^n-\alpha^n$$, as desired.

Since, $$[F(\alpha):F]=\textrm{deg}(p(x))$$ and $$[E:F]=n$$, it follows directly that $$E=F(\alpha)$$.

(feel free to comment or edit for any corrections or suggestions)

Notice that $$\sigma^i(\alpha)=\zeta^i\alpha$$ for $$1\leq i\leq n$$. So the minimal polynomial of $$\alpha$$ over $$F$$ has $$n$$ distinct roots. Thus $$E=F (\alpha)$$.

• Could you please elaborate on the part about the distinct roots? It goes a bit too quick for me. Sep 3, 2020 at 3:19
• @Ribbity each $\zeta^i$ is distinct for $\zeta^i = \zeta^j$ so $\zeta^{i-j} = 1$ - hence $i-j \equiv 0 \pmod n$ ($\zeta$ is primitive) and so $i = j$. Sep 3, 2020 at 3:32
• @Mummytheturkey: What I meant was the implication "minimal polynomial of $\alpha$ over $F$ has $n$ distinct roots $\Rightarrow E = F(\alpha)$". Sep 3, 2020 at 3:48
• @Ribbity Since the minimal polynomial has $n$ roots, it is of degree $n$. So $[F(\alpha):F]=n=[E:F]$. Since $F(\alpha)\subseteq E$,....
– cqfd
Sep 3, 2020 at 4:28