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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). –  Jyrki Lahtonen Jan 12 '12 at 11:10
    
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? –  Giovanni De Gaetano Jan 12 '12 at 11:14
    
@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? –  Aslan986 Jan 12 '12 at 11:43
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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 $K$? –  Jyrki Lahtonen Jan 12 '12 at 12:35
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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)? –  Lord_Farin May 22 '13 at 14:10

1 Answer 1

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. –  Jyrki Lahtonen May 23 '13 at 7:11

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