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To find the Galois group, I first begun by finding the splitting field:

I substituted $u=x^2$ and found the roots of the polynomial to be $\sqrt{3}\cdot e^{i\frac{\pi(2k+1)}{4}}$ for $k=\{0,1,2,3\}$. So the splitting field is generated by a single root, when $k=0$.

Applying Eisenstein to $(x+1)^4+9=x^4+4x^3+6x^2+4x+10$, with $p=2$, we see that it is irreducible. Hence the extension is of degree $4$. Now, the Galois group must have order $4$ too, and since the polynomial is separable and irreducible, the Galois group must be a transitive subgroup of $S_4$. There are two transitive subgroups of $S_4$, one is isomorphic to $Z_2\times Z_2$, and the other to $Z_4$.

To choose between those two, I computed the discriminant which was $4^4\cdot 3^6$ which is a perfect square in $\mathbb{Q}$. Hence the Galois group is a subgroup of $A_4$, and it must thus be isomorphic to $Z_2\times Z_2$.

Is my proof correct? Is there a better way to go through this problem? I would have preferred to see how the roots interact, but I couldn't manage to do it.

Dude111
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    "Since the roots are all complex, the polynomial is irreducible." Not true. $(x^2+1)(x^2+2)$ has only complex roots, but is reducible. – Arthur Jul 29 '19 at 09:00
  • Right, I forgot about that. Thanks. Back to square one for now. – Dude111 Jul 29 '19 at 09:01
  • @Arthur would this argument work: Since $(x^2-i3)(x^2+i3)$ is a factorization of $x^4+9$ in $\mathbb{C}[x]$ and $\mathbb{C}[x]$ is a UFD, then this factorization must be unique. Since $\mathbb{Q}$ is a subfield not containing $i$, then that means it is irreducible over $\mathbb{Q}$. – Dude111 Jul 29 '19 at 09:05
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    @Dude111 No, not really. My example polynomial may be factored as $(x^2 - (1+\sqrt2)ix -\sqrt2)(x^2 +(1+\sqrt2)ix-\sqrt2)$. It's still reducible over $\Bbb Q$. UFD only says something about irreducible factoring, so over $\Bbb C$, that means there is only one way to fector a polynomial into linear factors. – Arthur Jul 29 '19 at 09:09
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    I think I fixed it by applying eisenstein to $(x+1)^4+9$. – Dude111 Jul 29 '19 at 09:37
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    @Dude111 Also $\mathbb{R}$ is a subfield not containing $i$, but the polynomial $x^4+9$ is reducible in $\mathbb{R}[x]$, namely as $(x^2-x\sqrt{6}+3)(x^2+x\sqrt{6}+3)$. This proves it is irreducible over $\mathbb{Q}$. – egreg Jul 29 '19 at 10:24
  • @egreg I got to that part when trying to factor $x^4+9$ into quadratics, but I couldn't find an easy way to show that $\sqrt{6}$ was not rational, except with the rational roots theorem but I found it computationally hard. I am assuming that is the reason it proves irreducibility? – Dude111 Jul 29 '19 at 10:41
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    @Dude111 No squarefree integer has a rational square root: apply Eisenstein to $x^2-6$. – egreg Jul 29 '19 at 10:42
  • Ah yes, sorry I was confusing the wrong coefficient I had gotten $\sqrt{2\cdot\sqrt{3}}$ with $\sqrt{6}$. – Dude111 Jul 29 '19 at 10:48

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I don´t see anything wrong with what you've done, but you say you're interested in a different approach.

Recall that the primitive $8$-th root of unity is $\frac{1+i}{\sqrt{2}}$, so that your four roots are $\sqrt{\frac{3}{2}}(\pm 1 \pm i)$. A little manipulation shows that the splitting field is just $\mathbb{Q}[\sqrt{\frac{3}{2}}, i]$. The automorphisms must now swap or fix $\pm \sqrt{\frac{3}{2}}$ and swap or fix $\pm i$, so the group is indeed the fours-group. The action on the four roots of $X^4+9$ is easy to compute.

ancient mathematician
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  • Yes, this was what I had in mind, thanks. Just to make sure, to show that the $\sqrt{3/2}$ is not in $\mathbb{Q}$, we can use the rational roots theorem on $2x^2-3$ and see that there is no rational solution to that, right? – Dude111 Jul 29 '19 at 10:37
  • Yes, or just prove it by hand. – ancient mathematician Jul 29 '19 at 10:39
  • Got it thanks. Also for reference if anyone sees this, Eisenstein $p=3$ works too for irreducibility. – Dude111 Jul 29 '19 at 10:44