Given irreducible quartic $f(x) \in F[x]$ with roots $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ and Galois group $G = S_4$, what is the degree of the extension $E = F(\alpha_1+\alpha_2)$ over $F$? Find all subfields of E.

I began by trying to find the subgroup $H$ of $S_4$ that corresponds to $E$. I believe $H$ would need to fix the sum of the first two roots, which would automatically fix the sum of the remaining two roots. That is, $F(\alpha_1+\alpha_2) = F(\alpha_1+\alpha_2,\alpha_3+\alpha_4)$. Thus $H$ can permute each of the pairs of roots as well as the order of the pairs: $$H = \{ (),(12),(34),(12)(34),(13)(24),(14)(23),(1423),(1324) \} \simeq D_8. $$

This would mean that $[E:F] = 24/8 = 3$. And there are no subfields since $D_8$ is maximal in $S_4$.

Is this correct? It seems to make sense to me, but I wonder since the question asks for the subfields and there do not appear to be any.

  • $\begingroup$ This seems reasonable to me. $F$ is always a subfield, so maybe note that. $\endgroup$ – Ian Coley Dec 10 '13 at 22:37
  • 1
    $\begingroup$ Why are $(1423)$ and $(1324)$ elements of $H$? They map $\alpha_1 + \alpha_2$ and $\alpha_3 + \alpha_4$ to each other, but elements of $H$ should leave them fixed, shouldn't they? $\endgroup$ – Magdiragdag Dec 10 '13 at 22:42
  • 1
    $\begingroup$ Also, $(13)(24)$ and $(14)(13)$ should not be there for the same reason. $\endgroup$ – Berci Dec 10 '13 at 22:46
  • $\begingroup$ @Berci indeed, same for those two. $\endgroup$ – Magdiragdag Dec 10 '13 at 22:47
  • $\begingroup$ If you consider the depressed quartic, I believe $\alpha_1+\alpha_2 = -(\alpha_3+\alpha_4)$. Hmmm... I'll need to think about this. $\endgroup$ – Steve Dec 10 '13 at 22:49

The corresponding subgroup $H$ fixes each element of $E$, in particular, fixes $\alpha_1+\alpha_2$.

We cannot have the identity $\alpha_1+\alpha_2=\alpha_3+\alpha_4$ between them because then this would be satisfied with any permutation of indices, in particular $\alpha_1+\alpha_3=\alpha_2+\alpha_4$, leading to $\alpha_2-\alpha_3=\alpha_3-\alpha_2$ so $\alpha_2=\alpha_3$ -- unless the characteristic is $2$.

For weaker identity, such as $\alpha_1+\alpha_2=-\alpha_3-\alpha_4$ observe that e.g. $(13)(24)$ doesn't fix $\alpha_1+\alpha_2$ but takes it to $\alpha_3+\alpha_4=-(\alpha_1+\alpha_2)$.

Otherwise the thought works, but in $H$ we only have $\{(),(12),(34),(12)(34)\}$. So that, $\alpha_1+\alpha_2$ has order $24/4=6$, and $E$ will indeed have subfields. Can you find them?

  • $\begingroup$ Alright, that makes sense. We need to completely fix the value. I guess I tripped myself up with that minus sign. As I said in the other comment, the group I gave would contain the correct $H$, and the corresponding field could be written as $F((\alpha_1+\alpha_2)*(\alpha_3+\alpha_4))$. With the extra multiplication, we can now flip the two pairs without disrupting the 'fixed' value. $\endgroup$ – Steve Dec 10 '13 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.