Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

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 $$\frac{p^d-1}{p-1}> d$$

share|improve this answer

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 $\theta^2 + \theta + 1 = 0$.

share|improve this answer
    
This is giving a counterexample for the opposite direction of implication to the one the question is about, isn't it? –  dbaupp 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 :) –  dbaupp Jun 8 '12 at 4:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.