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How to prove that order of a field $\mathbb{GF}(q)$ is a power of its characteristic $\lambda$?

What I can say is that every field has a subfield of order $\lambda$ if it helps

Mahdi Khosravi
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  • Your last sentence does help (+1 for that). But this question has already been answered here and here, so I vote for this question being a duplicate (or to be merged). +1 to the correct answers by relatively new members here, of course. – Jyrki Lahtonen Sep 29 '13 at 06:17

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Hint 1: call $K$ the subfield of order $\lambda$. Then $\mathbb{GF}(q)$ becomes a vector space over $K$.
Hint 2: if $x_1,\dots,x_m$ is a basis, every element can be written uniquely as $\sum_{i=1}^m\alpha_i x_i$ with $\alpha_i\in K$. How many $m$-uples $(\alpha_1,\dots,\alpha_m)$ are there?

Mizar
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  • Thanks. I am a beginner. It takes time for me to see what you mean. It would be better if you answered it completely but thank you again – Mahdi Khosravi Sep 28 '13 at 12:11
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    @MahdiKhosravi , carizio did answer completely your question. That he didn't write down the explicit answer is because you must do some work on your own question...BTW, + 1 – DonAntonio Sep 28 '13 at 12:17
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${\mathbb{GF}}(q)$ is a vectorspace over ${\mathbb{GF}}(\lambda)$ [follows trivially from the definition of vectorspace]. Therefore it is, as a vectorspace, isomorphic to ${\mathbb{GF}}(\lambda)^d$ for some $d$ [by picking a basis]. So it has $\lambda^d$ elements.

Magdiragdag
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