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I want to determine the splitting field, galois group and intermediate fields of the polynomial $f(X)=(X^2+12)(X^3-5)\in\mathbb Q[X]$.

I want to obtain the splitting field by adjoining the roots of the polynomial to $\mathbb Q$ which is $\sqrt[3]{5}$ for $(X^3-5)$ but I don't understand how to determine the splitting field for $(X^2+12)$. It should be something like $\mathbb Q(i\sqrt{12})$ but I have also seen problems where roots of unity were used in similar situations, so I'm not sure.

Since I have the polynomial already written as irreducible factors it is reducible and because it does not split into linear factors its Galoid group is $\mathbb Z/2\mathbb Z$.

To determine the intermediate fields of the splitting field $/\mathbb Q$ I think I have to apply the main theorem of Galois theory and reduce it to a group problem but I have never seen how that is done.

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The splitting field of $x^2+12$ is the same as the splitting field of $x^2+3$, i.e. the cyclotomic field $\mathbf Q(\zeta)$ generated by the cube roots of unity. It's a quadratic extension, with Galois group isomorphic to $\mathbf Z/2\mathbf Z$.

On $\mathbf Q(\zeta)$, $x^3-5\,$ is irreducible (one can check it has no roots, or one can apply Eisentein's criterion on the ring of integers of $\mathbf Q(\zeta)$ : the prime $5$ remains prime in this euclidean ring). The polynomial $x^3-5$ splits completely in the field $\mathbf Q(\zeta,\sqrt[3]5)$. Hence $\mathbf Q(\zeta,\sqrt[3]5)$ is the splitting field of $(x^2+12)(x^3-5)$.

$$[\mathbf Q(\zeta,\sqrt[3]5):\mathbf Q]=$[\mathbf Q(\zeta,\sqrt[3]5):\mathbf Q(\zeta)]\cdot [\mathbf Q(\zeta):\mathbf Q]=6. $$

Its Galois group is thus of order $6$. There remains to know if it is isomorphic with the cyclic group of order $6$ or with the group $S_3$.

The Galois group of $\mathbf Q(\zeta)/\mathbf Q$ is generated by conjugation. Also the Galois group of $\mathbf Q(\zeta,\sqrt[3]5)/\mathbf Q$ sends $\sqrt[3]5$ to itself, or to $\zeta\sqrt[3]5$ or to $\bar\zeta\sqrt[3]5$. You can check computation doesn't commute with, say, $\sqrt[3]5 \mapsto\zeta\sqrt[3]5$, so you can conclude that $\,\,\operatorname{Gal}(\mathbf Q(\zeta,\sqrt[3]5)/\mathbf Q)\simeq S_3$.

It has only four non-trivial subgroups, three isomorphic to $\mathbf Z/2\mathbf Z$ and one to $\mathbf Z/3\mathbf Z$.

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  • $\begingroup$ The group $S_3$ has four nontrivial proper subgroups, not two. Even if two different subgroups are isomorphic, they are different. $\endgroup$ – KCd Jan 15 '15 at 1:57
  • $\begingroup$ @KCd: I wrote without re-reading myself – it's late here, too… Thanks for pointing the error, it's fixed now. $\endgroup$ – Bernard Jan 15 '15 at 2:16
  • $\begingroup$ I don't understand why the splitting filed of $x^2+12$ is the same as the one of $x^2+3$ and why it is a cube-root of unity ($e^{\frac{2\pi i}{3}}$) then? $\endgroup$ – sj134 Jan 15 '15 at 2:46
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    $\begingroup$ @sj134 $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$ and $e^{2\pi i/3} = \frac{-1 + i\sqrt{3}}{2}$. $\endgroup$ – André 3000 Jan 15 '15 at 2:53
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    $\begingroup$ It's the same because $12=4\times 3$ and $4$ is a square in $\mathbf Q$. The cube root of unity comes from the algebraic value: $\dfrac{-1+\mathrm i\sqrt 3}{2}$: it amounts to the same knowing this cube root ot $1$ or knowing $ \mathrm i\sqrt3 $. $\endgroup$ – Bernard Jan 15 '15 at 2:56
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  • Since $x^2+12$ is a polynomial of degree 2 that doesn't have root in $\mathbb{Q}$ then it's irreducible over $\mathbb{Q}$ (it's easy to see why). The same line of reasoning applies to $x^3-5$.

  • From the proof of existence of the splitting field, we adjoin a root of $f(x)$ to $\mathbb{Q}$ then test $f$ on $\mathbb{Q}$ if it has another root such that doesn't belong to the new field. Here is the process:

$x^2+12=(x+i\sqrt{12})(x-i\sqrt{12}) $ since it's irreducible from first extension theorem (I am not sure about the name) $\mathbb{Q}[x]/(x^2 + 12)$ is a field, from the second extension theorem $1,i\sqrt{12} $ is a basis of $\mathbb{Q}/(f)=\mathbb{Q}[i\sqrt{12}]= \{ \lambda_1, \lambda_2 i\sqrt{12}: \lambda_1, \lambda_2 \in \mathbb{Q} \} $ it's obvious the other root is $-i\sqrt{12} \in \mathbb{Q}[i\sqrt{12}]$


Apply the same process for $f(x)= (x^2+12))(x^3-5)=(x-i\sqrt{12})(x+i\sqrt{12})(x^3-5)$ on $\mathbb{Q}[i\sqrt{12}]$ Note: $x^3-5 = (x-\sqrt[3]{5})(x^2+\sqrt[3]{5}x+\sqrt[3]{5}^2)$

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