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I am trying to find the Galois group $\operatorname{Gal}(E/\mathbf{Q})$ where $E$ is the splitting field for $(X^2-2)(X^3-3)$. First the splitting field for $X^2-2$ and $X^3-3$ over $\mathbf{Q}$ are $\mathbf{Q}[\sqrt2]$ and $\mathbf{Q}[3^{1/3},w]$ respectively, where $w$ is the primitive root of unity. Also I know that $\operatorname{Gal}(\mathbf{Q}[\sqrt2]/\mathbf{Q})$ is $C_2$ and $\operatorname{Gal}(\mathbf{Q}[3^{1/3},w]/\mathbf{Q})$ is $S_3$. Since $\mathbf{Q}[\sqrt2] \cap \mathbf{Q}[3^{1/3},w] = \mathbf{Q} $, then the Galois group $\operatorname{Gal}(E/ \mathbf{Q})$ is the direct product $C_2 \times S_3$. I think something should be wrong with my reasoning because if it is correct, then we can generalize this idea to any given polynomial $(X^2-p)(X^3-q)$ where $q$ and $p$ are (not necessarily distinct) primes, but I could not find such a generalization on similar questions.

I want to ask what is wrong with my reasoning.

Rócherz
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CCCC
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    Why should it be wrong? $\omega$ is a primitive third root of unity, not "the" qubic root. – Dietrich Burde Mar 23 '25 at 19:34
  • Yes, I fixed my question accordingly. I am not sure, there are discussions on similar questions like https://math.stackexchange.com/questions/1883839/galois-group-of-x3-2x2-2-over-mathbbq?rq=1 and I did not see my straightforward idea somewhere like there. Simply, I am not sure because it is just my solution – CCCC Mar 23 '25 at 19:44
  • In fact I just saw in the link I sent, the second idea is the same as mine. But still I did not see somewhere that we can generalize this to any polynomial of the form $(X^2-p)(X^3-q)$ – CCCC Mar 23 '25 at 19:47
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    As long as both polynomials are irreducible ... – Dietrich Burde Mar 23 '25 at 19:56
  • Do we really need to indicate that the polynomials are irreducible? They are irreducible by Eisenstein since $p$ and $q$ are primes. Thank you by the way, I think you verify my idea – CCCC Mar 23 '25 at 20:03
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    This was only referring to your previous comment, any polynomial of the form $(X^2-p)(X^3-q)$. This is not true, take $p=4$. Of course, in the post it is $p,q$ prime, but you need this (so you need irreducible). – Dietrich Burde Mar 23 '25 at 20:20
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    Do observe that the claim is false, if you allow negative primes. The troublemaker is then $x^2+3$ that splits itself in the splitting fields of all $x^3-q$, $q$ a prime (positive or negative). – Jyrki Lahtonen Mar 23 '25 at 21:24
  • Because the roots of $x^2+3$, which are $\pm i\sqrt{3}$ are contained in $Q[w] = Q[\frac{-1+ i\sqrt{3}}{2}]$, right? But we are safe when the primes are positive. – CCCC Mar 23 '25 at 21:37

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