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I need to show that the Galois group of $ x^{p^n} − 2 $ over is $F_p$ is a cyclic group of order n.

I need rigorous proof for this.

I know that the Galois group of a polynomial over F is Galois group of its splitting field.

Result1 I know that $F_p$ has precisely $p^n$ elements and the Galois group of polynomial $ x^{p^n} − x $ over $F_p$ is isomorphic to Galois group of order n generated by the frobenius automorphism.

Result 2 Let F be a field of charecteristic not dividing n which contains the nth roots of unity. Then the extension $F(a^\frac{1}{n})$ for $a \in $ F is cyclic over F of degree n. $F(a^\frac{1}{n})$ is the splitting field of $x^n-a$, also $F(a^\frac{1}{n})$ is Galois over F with Gal($F(a^\frac{1}{n})$ /F) isomorphic to the cyclic group of $U_n$ , the group of nth roots of unity of order n."

I guess result 2) can be used here but I am just unable to compare things. Here a=2, what about characteristic of field $F_p$ , what is n here?

Please help . I am getting confused .

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    $\begingroup$ In characteristic $p$ : $x^{p^n}-2 = x^{p^n}-2^{p^n} = (x-2)^{p^n}$ $\endgroup$ – reuns Nov 11 '17 at 13:46
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    $\begingroup$ The Galois group of $f \in K[x]$ with splitting field $L= K(\alpha_1,\ldots,\alpha_m)$ (with $\alpha_j$ the roots of $f$) is the Galois group $\text{Aut}(L/K)$ represented as a subgroup of the permutation group $S_m$. When $f$ is non-separable $L/K$ is only a normal extension. $\endgroup$ – reuns Nov 11 '17 at 13:54
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    $\begingroup$ Are you sure you have written the problem correctly? As written, it is not true unless $n=1$... $\endgroup$ – Eric Wofsey Nov 14 '17 at 18:28
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    $\begingroup$ @KavitaSahu To be clear : the splitting field of $x^{p^n}-2 =(x-2)^{p^n} \in \mathbf{F}_p[x]$ is of course $\mathbf{F}_p$ itself, so the extension and the Galois group is trivial (cyclic of order $1$..) $\endgroup$ – reuns Nov 14 '17 at 18:33
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    $\begingroup$ Clear ? Then what can you say of the splitting field of $x^{p^n}-x\in \mathbf{F}_p[x]$ ? $\endgroup$ – reuns Nov 14 '17 at 18:37
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This is not true unless $n=1$. Indeed, since $(a+b)^p=a^p+b^p$ in characteristic $p$, $(x-2)^{p^n}=(x-2)^{p^n}$. So this polynomial splits already in $\mathbb{F}_p$, and $\mathbb{F}_p$ is the splitting field, with trivial Galois group. So the Galois group is always cyclic of order $1$, no matter what $n$ is.

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