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Let $K$ be the splitting field of $f(x)=x^5-x+1=(x^2+x+3)(x^3-x^2-2x-2)$ over $F=\mathbb{F}_7$. I want to find $\text{Gal}(K/F)$. Let $\alpha_1,\alpha_2$ be the roots of $x^2+x+3$ and $\beta_1,\beta_2,\beta_3$ the roots of $x^3-x^2-2x-2$. Then the elements of $\text{Gal}(K/F)$ are the permutations of the roots $(\alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3)$ that fix $F$ (and so $\text{Gal}(K/F)$ embeds into $S_5$).

By Vieta's formulas we have relations $\alpha_2=-1-\alpha_1$, $\beta_1+\beta_2+\beta_3=1$, and $\beta_1\beta_2\beta_3=2$. From the latter two we can obtain $\beta_2,\beta_3=\frac{1}{2}((1-\beta_1)\pm \sqrt{(1-\beta_1)^2-8/\beta_1})=4((1-\beta_1)\pm \sqrt{(1-\beta_1)^2-1/\beta_1})$. Letting $\gamma^2=(1-\beta_1)^2-1/\beta_1$, we can express the roots as:

$$(\alpha_1,\alpha_2,\beta_1,\beta_2,\beta_3)=(\alpha_1,-1-\alpha_1,\beta_1,4(1-\beta_1+\gamma),4(1-\beta_1-\gamma)).$$

So $K=F(\alpha_1,\beta_1,\gamma)$. Now consider the following automorphisms of $K$.

$$\phi_1(\alpha_1)=-1-\alpha_1\text{ and fixes }\beta_1,\gamma.$$ $$\phi_2(\gamma)=-\gamma\text{ and fixes }\alpha_1,\beta_1.$$ $$\phi_3(\beta_1)=4(1-\beta_1+\gamma),\, \phi_3(\gamma)=4(1-3\beta_1-\gamma)\text{ and fixes }\alpha_1.$$

Then $\phi_1$ generates $S_2\subset S_5$ that permute only the $(\alpha_i)$ and $\phi_2,\phi_3$ generate $S_3\subset S_5$ that permute only the $(\beta_i)$, so $\text{Gal}(G/F)$ should equal $S_2\times S_3\subset S_5$. But I've done something wrong since any Galois extension of a finite field has a cyclic Galois group.

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    $\begingroup$ Your two separate Galois groups are cyclic of order $2$ and $3$, respectively. Certainly $C_2\times C_3$, cyclic of order $6$, embeds nicely into $S_5$. $\endgroup$ – Lubin Mar 29 '15 at 20:47
  • $\begingroup$ That doesn't seem right. $\phi_2$ is the permutation $(45)$ and $\phi_3$ is $(345)$, so together they generate $S_3$. $\endgroup$ – jcai Mar 30 '15 at 4:18
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    $\begingroup$ No, the two subfields are (linearly) disjoint, so there’s a nontrivial autom of the quadratic field leaving the cubic field elementwise fixed, and vice versa. Let me put it into an answer. $\endgroup$ – Lubin Mar 30 '15 at 12:36
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    $\begingroup$ I guess your error is to think that there’s something leaving your $\alpha_1$ and $\beta_1$ fixed, but moving something else. Maybe this is an implicit assumption that the extension $\Bbb F_7(\alpha_1,\beta_1)$ is not already Galois. But your $\gamma$ is already in the field $\Bbb F_7(\alpha_1,\beta_1)$. $\endgroup$ – Lubin Mar 30 '15 at 18:52
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You have made things much too difficult for yourself. Note that $7^2=49$, $7^3=343$, and $7^4=2401$. The nontrivial automorphism of the quadratic extension $k_2$ is given by $a\mapsto a^7$, but this formula also moves the elements of the cubic extension $k_3$. But $a\mapsto a^{343}$ has the same effect on $k_2$, and leaves $k_3$ elementwise fixed. On the other hand $a\mapsto a^{2401}$ leaves $k_2$ fixed, but has the effect $a\mapsto a^7$ on $k_3$. The two commute with each other.

Alternatively, if you want to call $A$ one root of $x^2 + x + 3$, the other root being $-1-A$, then $A^7=-1-A$; and if you call $B$ one of the roots of $x^3-x^2-2x-2$, then $B^7 = 2B+2$, and $B^{49} = 4B-1$. Your formulas should have found these as conjugates of the roots of your irreducible factors.

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