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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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    No. The degree of the minimal polynomial of $\xi$ over $H$ will depend on what $H$ is, so the degree should be $[K:H]$. But an easier approach to your problem is to use the fact that $K$ is the smallest field containing $\xi$ (follows from the primitivity).2012-01-12
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    Probably this comment is wrong because it didn´t take in consideration the primitivity of $\xi$ nor the finiteness of $K$. But if $H[\xi]\subseteq K$ then by minimality of the fraction field also $H(\xi)\subseteq K$. What am I missing?2012-01-12
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    @JyrkiLahtonen: Okok, i agree about the fact that the degree should be $[K : H]$. But how to prove that it is really $[K : H]$ and not bigger? For the second part, if I undestand well, the fact that $K$ is the smallest field that contains $\xi$ implies that $K \subseteq H(\xi)$, but not the viceversa. Is it correct?2012-01-12
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    @Aslan986: You are right in that $K\subseteq H(\xi)$ for the reason that you stated. But also ... $H\subseteq K$ and $\xi\in K$, so what algebraic operations would introduce elements of $H(\xi)$ that are not in $K$?2012-01-12
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    @JyrkiLahtonen: oh ok! Finally I understood. Thank you.2012-01-12
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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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