Let $K$ be a splitting field in $\mathbb C$ of the polynomial $f(X)=X^4-X^3-5X+5$ over $\mathbb Q$.

  • Construct the splitting field $K$ and find the degree of the extension $K:\mathbb Q$.

$f(X)=(X-1)(X^3-5)$, hence we have roots $\sqrt[3]{5},\sqrt[3]{5}\zeta_3$ and $\sqrt[3]{5}\zeta_3^2$, where $\zeta_3$ is the primitive 3rd root of unity. Thus our field extension is $\mathbb{Q}(\sqrt[3]{5},\zeta_3)$. By the Tower Law $[\mathbb{Q}(\sqrt[3]{5},\zeta_3)]=[\mathbb{Q}(\sqrt[3]{5},\zeta_3):\mathbb Q(\sqrt[3]{5})]\cdot[\mathbb{Q}(\sqrt[3]{5}):\mathbb Q]=6$.

Have I missed anything important out?

  • Find the order and structure of $Gal(K:\mathbb Q)$.

The order of $Gal(K:\mathbb Q)$ is also 6 because the extension is normal and separable. I believe the six automorphisms are:

$id: \sqrt[3]{5} \mapsto \sqrt[3]{5} $ , $\zeta_3 \mapsto \zeta_3$

$\alpha: \sqrt[3]{5} \mapsto \zeta_3\sqrt[3]{5} $ , $\zeta_3 \mapsto \zeta_3$

$\alpha: \sqrt[3]{5} \mapsto \zeta_3^2\sqrt[3]{5} $ , $\zeta_3 \mapsto \zeta_3$

$\beta: \sqrt[3]{5} \mapsto \sqrt[3]{5} $ , $\zeta_3 \mapsto \zeta_3^2$

$\beta: \sqrt[3]{5} \mapsto \zeta_3^2\sqrt[3]{5} $ , $\zeta_3 \mapsto \zeta_3^2$

which is isomorphic to the symmetric group $S_3$?

  • Find all subfields of $K$ via Galois correspondence.

I'm trying to get my head around fixed fields and Galois correspondence, could anyone show me clearly how this part is done?

  • Find all constructible numbers in $K$.

I assume this leads on from the previous part, I know constructible numbers must be of a degree which is a power of 2? So would it be all the elements of $K$ with such an order?

Hope my attempts weren't too hard to follow, any help would be great!


1 Answer 1


First some review on what you have already done. You found the field and degree nicely, so well done there. You say there are six automorphisms, but only list five (you're missing $\sqrt[3]{5}\mapsto\zeta_3\sqrt[3]{5}, \zeta_3\mapsto\zeta_3^2$). Also, you have called two of the automorphisms $\alpha$ and two of them $\beta$, which makes it a bit difficult to talk about any specific one.

As for fixed fields and subgroups, let's look at the subgroups, shall we? One subgroup is just the identity. It fixes all of $K$, so that's that one. Then there are three subgroup of order $2$ and of order $3$.

One of the order $2$ groups is generated by $\zeta_3\mapsto \zeta_3^2$, which is basically just complex conjugation. It clearly keeps $\sqrt[3]{5}$ fixed, so it keeps $\Bbb Q(\sqrt[3]{5})$ fixed.

The next order $2$ group is generated by $\sqrt[3]{5}\mapsto\zeta_3\sqrt[3]{5}, \zeta_3\mapsto\zeta_3^2$. This time it's not so immediate to see the fixed subfield, but it is $\Bbb Q(\zeta_3^2\sqrt[3]{5})$, since the automorphism sends $\zeta_3^2\sqrt[3]{5}$ to $$(\zeta_3^2)^2\cdot \zeta_3\sqrt[3]{5} = \zeta_3^5\sqrt[3]{5} = \zeta_3^2\sqrt[3]{5}$$

I guess you can handle the third subgroup of order $2$ by yourself now. As for the subgroup of order $3$, it is generated by $\sqrt[3]{5}\mapsto \zeta_3\sqrt[3]{5}$, and the corresponding fixed field is $\Bbb Q(\zeta_3)$.

Finally, for the last part, a number is constructible iff it has degree a power of $2$ over $\Bbb Q$. What elements of $K$ have degree $1$ or $2$? Are there elements of degree $4$? What about higher powers of $2$?

  • $\begingroup$ Thank you so much for your answer, it's extremely helpful. May I check that the last order 2 subgroup is generated by $\sqrt[3]{5}\mapsto\zeta_3\sqrt[3]{5}, \zeta_3\mapsto\zeta_3^2$ and fixes $\mathbb Q(\zeta_3 \sqrt[3]{5})$? Can I ask you to expand on the last part? Obviously the degree of the extension is not a power of 2 but the first part of it's tower law $[\mathbb{Q}(\sqrt[3]{5},\zeta_3):\mathbb Q(\sqrt[3]{5})]$ is, is this the correct approach? $\endgroup$
    – user484410
    Commented Nov 21, 2017 at 12:47
  • 1
    $\begingroup$ You are right about the last element and subfield. As for constructibility, the number $\alpha$ is constructible iff $[\Bbb Q(\alpha):\Bbb Q]$ is a power of $2$. The degree $[K:\Bbb Q(\alpha)]$ is not directly relevant, although it might help you with some calculations. $\endgroup$
    – Arthur
    Commented Nov 21, 2017 at 12:52

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