Suppose one is asked to find $[GF(2^{12}) : GF(2^3)]$, where $GF(p^n)$ denotes the unique finite field of order $p^n$. How do I go about solving this sort of problem?
How to find degree of field of extension of finite field
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abstract-algebra
3 Answers
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Recall that the degree $[L:K]$ of a field extension $L/K$ is just the dimension of $L$ as a $K$-vector space. If $n = [L:K] = \dim_K(L)$ is finite, then $L$ is isomorphic to $K^n$ as a $K$-vector space. If $K$ is finite, then $\#L = (\#K)^n$.
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If you have $\Bbb F\subseteq \Bbb K\subseteq \Bbb L$, with $\Bbb F,\Bbb K,\Bbb L$ fields, $\Bbb L$ an extension of $\Bbb K$ and $\Bbb K$ an extension of $\Bbb F$ then the degrees formula tells you that $|\Bbb L :\Bbb F|=|\Bbb L: \Bbb K|\cdot|\Bbb K:\Bbb F|$.
In your particular case you have that $|GF(2^{12}):GF(2^{3})|=\frac{|GF(2^{12}):GF(2)|}{|GF(2^{3}):GF(2)|}=4$.
From here you can immediately prove the more general result $|GF(p^m):GF(p^n)|=\frac mn$ if $n| m$
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0Thanks so much for the response. I'm quite rusty on my field theory, however. What result leads to $\vert GF(p^m) : GF(p) \vert = m$? – 2017-01-04
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0Actually, I think I see the solution. The elements of $GF(p^n)$ look like $$a_0 + a_1u +a_2u^2 + \cdots + a_{n-1}u^{n-1},$$ where $a_i \in \mathbb{Z}_p$. Clearly then, $\left \{1, u, u^2, \cdots, u^{n-1} \right \}$ is a basis for $GF(p^n)$, the vector space of $GF(p^n)$ over $GF(p)$ has dimension $n$. – 2017-01-04
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0yeah, that seems good, it should be an immediate consequence of your construction of $GF(p^n)$, but there are different ways to do so @JaneDoe – 2017-01-04
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we have $[GF(2^{12}):GF(2)]=[GF(2^{12}):GF(2^3)][GF(2^{3}):GF(2)]$ then $12=[GF(2^{12}):GF(2^3)]\times 3$ therfore $[GF(2^{12}):GF(2^3)]=4$