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I need to prove this part of a theorem: given a field $K$ such that $|K| = p^n$, a subfield $H \subset K$, and $\xi$ a primitive element of $K$; i need to say that $H(\xi) \subseteq K$.

Of course $K$ contains all the polynomial in $H[ \xi ]$.

$\xi$ is a root of the polynomial $x^{|K| - 1} - 1 \in H[x]$, so $\xi$ is a root of a monic factor of $x^{|K| - 1} - 1 \in H[x]$ irreducible in $H[x]$.

Now, i think that i must require also that this monic factor has degree $n$. Is it right? If so, how to prove it?

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    @JyrkiLahtonen Since your comment appears to have served as an answer to the OP, perhaps you could post it as one (so as to clear this from the unanswered queue)?2013-05-22

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Here we actually have $H(\xi)=K$.

Let $F=\mathbb{F}_p$ be the common prime field of both $H$ and $K$. $K$ is a finite extension of $F$, so all of its elements, $\xi$ in particular, are algebraic over $F$. Therefore $H(\xi)=H[\xi]$.

Because $\xi$ is a primitive element of $K$, we have $K=F(\xi)=F[\xi]$. Therefore we have the following inclusions $ K=F[\xi]\subseteq H[\xi]\subseteq K[\xi]=K. $ From this chain of inclusions we can infer the claimed equality.

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    Sorry about leaving this hanging. I was half hoping the OP to type up an answer, but forgot to A) urge him, B) check this up afterwards.2013-05-23