Q: Let K be the splitting field for $(x^{5}-1)(x^{3}-2)$ over $\mathbb{Q}$. Compute the cardinality of the Galois group $G$ for $\mathbb{Q} \subset K$, and show that G is not abelian.

So first I computed the splitting fields and Galois groups for the two factors. For the first factor it is just $\mathbb{Q}(\zeta_{5})$ and the group is $Z^{\times}_{5}$, and for the other it is $\mathbb{Q}(\zeta_{3}, \sqrt[3]{2})$ and the Galois group is $S_{3}$.

The splitting field of the polynomial is then the join of the two splitting fields, since it contains all the roots, and at the same time is the smallest field containing $\mathbb{Q}(\zeta_{3}, \sqrt[3]{2})$ and $\mathbb{Q}(\zeta_{5})$.

Now, I would like to say $\mathbb{Q}(\zeta_{3}, \sqrt[3]{2})\cap\mathbb{Q}(\zeta_{5})=\mathbb{Q}$, in which case $Gal(Q/K)=Z^{\times}_{5}\times S_{3}$ which is then non-abelian of order $24$, but I'm not sure how to say this. Intuitively, since $\sqrt[3]{2}$ lies in a degree 3 extension, it can't be in a degree 4 extension, hence not in $\mathbb{Q}(\zeta_{5})$, so then we are left with considering $\mathbb{Q}(\zeta_{3})\cap\mathbb{Q}(\zeta_{5})=\mathbb{Q}$, which is true since 3 and 5 are coprime, but is this correct?

Is there a more formal way of showing the last statement I made is true (if it is true at all)?

  • $\begingroup$ Looks quite good to me. The intersection $L=\Bbb{Q}(\root3\of2,\zeta_3)\cap\Bbb{Q}(\zeta_5)$ has to be of degree 1 or two, because those two fields have degrees 6 and 4 respectively and $\gcd(6,4)=2$. Presumably you have shown earlier that 1) if two Galois extensions intersect trivially, then their compositum is Galois, and the Galois group of the compositum is the direct product, and 2) the intersection of $\Bbb{Q}(\zeta_3)\cap\Bbb{Q}(\zeta_5)$ is trivial? $\endgroup$ Commented Dec 6, 2014 at 20:43
  • $\begingroup$ Yes, those two statements I have at my disposal, and I see that $L$ has to have degree 1 or 2 by drawing a diagram. But what is a good way to argue it is not 2? $\endgroup$ Commented Dec 6, 2014 at 20:55
  • $\begingroup$ Good. To your question: Because $S_3$ has only one subgroup of index two, namely $A_3$, the field $\Bbb{Q}(\root3\of2,\zeta_3)$ has only one subfield of degree two, namely $\Bbb{Q}(\zeta_3)$. Sorry about forgetting to include that bit. $\endgroup$ Commented Dec 6, 2014 at 21:00
  • $\begingroup$ Ah, yes I see now. Thank you! :) $\endgroup$ Commented Dec 6, 2014 at 21:11

1 Answer 1


(Moving comment to an answer) A simple way of completing your argument is to use known bits about the subgroups of $S_3$. It has four non-trivial subgroups: $A_3$ generated by either 3-cycle and three subgroups of order two generated by a single 2-cycle each. Therefore there is only one subgroup of index two. By Galois correspondence this means that $\Bbb{Q}(\root3\of2,\zeta_3)$ has only one subfield that is quadratic over the rationals, namely $\Bbb{Q}(\zeta_3)$.

The OP is clearly in possession of results that carried him to the destination from this point on.


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