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I have to show : $GF(2^9)$ contains $GF(2^3)$ as only proper intermediate field.

MJD
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ana
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    And what is your question? – MJD Mar 24 '15 at 21:28
  • Sorry for being slow, but what is $GF(2^{9})$? You mean $GL(F_{2^{9}})$ or something? I have never seen this notation before. – Bombyx mori Mar 24 '15 at 21:30
  • @Bombyxmori $GF(p^n)$ is standard notation for $F_{p^n}$. – John Gowers Mar 24 '15 at 21:32
  • Hint: $3$ divides $9$. – Lullaby Mar 24 '15 at 21:33
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    Hi Ana! I think you've misunderstood the point of the site. It exists to help you find solutions to your problems, not to answer them for you. You're more likely to get a useful answer if you tell us exactly what your question is, why you're interested in, what you've tried and where exactly you've got stuck. – John Gowers Mar 24 '15 at 21:33
  • Ok, Ana. What do you know about extensions of finite fields? Does the following result ring a bell? If $GF(q)$ is a subfield of $GF(Q)$, then $Q=q^n$ for some positive integer $n$. This task can be completed using that result, and not really without it, so... – Jyrki Lahtonen Mar 24 '15 at 21:34
  • But the others have a point. You should add some extra "context" to the question. Like a related example done in class? Did you try something, but it didn't work? Such pieces of information are needed, because only then we can explain to you how to make progress and understand what you are supposed to do here. – Jyrki Lahtonen Mar 24 '15 at 21:36
  • @Bombyxmori: $GF(q)$ is a common alternative notation for $\Bbb{F}_q$. For example coding theorists use it. Even in relatively recent texts. Our local number theorists also used it in the past, but I'm not sure about the current generation :-) – Jyrki Lahtonen Mar 24 '15 at 21:39
  • @JyrkiLahtonen: I see. May I ask what does $GF$ means at here? I have never seen it before. – Bombyx mori Mar 24 '15 at 21:42
  • @Bombyxmori: It is short for Galois Field. – Jyrki Lahtonen Mar 24 '15 at 21:43
  • @JyrkiLahtonen: Thanks! I learned something new. – Bombyx mori Mar 24 '15 at 21:55

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Hint: put $L=\mathbb{F}_{2^9}$, the field of $512$ elements. Then $[L:\mathbb{F}_2]=9$. If $\mathbb{F}_2 \subset K \subset L$ is an intermediate field, then $[L:K] \mid [L:\mathbb{F}_2]$, so ...

Nicky Hekster
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