# If $K = \mathbb{F}_p(\alpha)$ where $\alpha^n \in \mathbb{F}_p$ and $n$ is the minimal such $n$. Does this imply that $[K : \mathbb{F}_p] = n$?

If $K = \mathbb{F}_p(\alpha)$ where $\alpha^n \in \mathbb{F}_p$ and $n$ is the minimal such $n$. Does this imply that $[K : \mathbb{F}_p] = n$?

If not, is there a condition on $\alpha$ where this is the case?

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No, it does not imply that $[K:\mathbb{F}_p]=n$.

Let $[K:\mathbb{F}_p]=d$. We can choose an $\alpha\in K$ such that $\alpha$ generates the multiplicative group $K^\times$, so that the smallest $s$ such that $\alpha^s=1$ is $s=p^d-1$, and therefore the smallest $n$ such that $\alpha^n\in\mathbb{F}_p$ has to be $n\geq \frac{p^d-1}{p-1}$, but in general we have that $$\frac{p^d-1}{p-1}> d$$

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Consider $\mathbb{F}_2$. Let $x^2 + x + 1 \in \mathbb{F}_2[x]$, a irreducible polynomial. Let $\theta$ be a root of this polynomial (in the algebraic closure). Then $\mathbb{F}_2[\theta]$ is a degree 2 extension. However $\theta^2 = \theta + 1 \notin \mathbb{F}_2$, since $\theta^2 + \theta + 1 = 0$.

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This is giving a counterexample for the opposite direction of implication to the one the question is about, isn't it? – huon Jun 8 '12 at 3:55
@dbaupp $\theta^3 = \theta(\theta^2) = \theta(\theta + 1) = \theta^2 + \theta = 1 \in \mathbb{F}_2$. $3$ is minimal but the extension is degree $2$. – William Jun 8 '12 at 4:00
It is also a counterexample for the converse. – William Jun 8 '12 at 4:03
Oh, yeah, of course. Thanks for explaining :) – huon Jun 8 '12 at 4:08