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We know that the binary splitting field of the equation $x^n-1$ is the $GF(2^q)$ where $q$ is the least positive number such that $n\mid 2^q-1$.

My question: what is the binary splitting field of the equation $x^n-\alpha$ such that $\alpha$ is the element of $GF(2^k)$.

In fact, I want to ask: Is there a relation between the binary splitting field of the equations $x^n-1$ and $x^n-\alpha$ such that $\alpha \neq 1$.

Edit(1): Edit question by comment of @ ancient mathematician.

user0410
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    I think you need to replace "if" in your first line by "where $q$ is the least positive number such that". – ancient mathematician Mar 16 '17 at 12:32
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    If $a \neq 0$ (I think we assume this anyway), the splitting field of $x^n-a$ contains the splitting field of $x^n-1$. Apart from this, not much can be said, because it depends on $a$. For instance look at $x^3-a$ over $\mathbb F_7$. If $a=-1$ it already splits over the base field. If $a=2$ it is irreducible, i.e. the splitting field is $\mathbb F_{7^3}$. – MooS Mar 16 '17 at 12:53
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    Look at $K(α^{1/p^e})$ for each prime divisor of $n$. If $r$ is the largest exponent such that $α^{1/p^r} \in K$, I'd say that $[K(\alpha^{1/p^{e}}):K] =[K(\alpha^{1/p^r},(\alpha^{1/p^r})^{1/p^{e-r}})):K(\alpha^{1/p^{r}})] = \varphi(p^{e-r})$ so that $[K(\alpha^{1/n}):K] = \varphi(m)$ where $m = \prod_i p_i^{e_i-r_i}$ and $\mathbb{F}q(\alpha^{1/n}) = \mathbb{F}{q^{\varphi(m)}}$ – reuns Mar 16 '17 at 12:55
  • I also ought to have clarified: you do mean $n$ odd, don't you? – ancient mathematician Mar 16 '17 at 13:35
  • @ancientmathematician yes $n$ has to be odd. Thanks for comment. – user0410 Mar 16 '17 at 13:55
  • @user1952009 Is it possible to ask you to explain your method with an example as a answer of this question. Your method is a bit complicated for me to understand. Thanks. – user0410 Mar 16 '17 at 14:14
  • @MooS If $n$ in the equation $x^n-\alpha$ dose not divide $2^k-1$, can we say that the equation $x^n-\alpha$ has only one solution in the $GF(2^k)$ that this solution is the cube root of $\alpha$ that exist because $GF(2^q)^*$ is cyclic.Thanks – user0410 Mar 16 '17 at 14:15
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    I think what I said is wrong (see there) and I can only say $[K(\alpha^{1/p^{e}}):K]$ divides $\varphi(p^{e-r})$. I'd like to know how to compute it, i.e. find the Galois group of $K(\alpha^{1/p^{e}})$ – reuns Mar 16 '17 at 16:04
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    There is an answer http://math.stackexchange.com/questions/801846/galois-group-of-k-sqrtpa-over-k when $\mu_p \in K$. Not sure how to use it here. – reuns Mar 16 '17 at 18:37
  • @user1952009 Thanks for useful comment. – user0410 Mar 16 '17 at 20:39

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