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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.

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2 Answers 2

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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)

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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) $.

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

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