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
    Commented Dec 10, 2013 at 22:37
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    $\begingroup$ 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? $\endgroup$ Commented Dec 10, 2013 at 22:42
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    $\begingroup$ Also, $(13)(24)$ and $(14)(13)$ should not be there for the same reason. $\endgroup$
    – Berci
    Commented Dec 10, 2013 at 22:46
  • $\begingroup$ @Berci indeed, same for those two. $\endgroup$ Commented Dec 10, 2013 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
    Commented Dec 10, 2013 at 22:49

1 Answer 1


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
    Commented Dec 10, 2013 at 23:23

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