# Determining all intermediate fields of $X^3-3$

I have to establish all subgroups of $\text{Gal}(L/\mathbb{Q})$ and all intermediate fields $L\geq M\geq \mathbb{Q}$, where $L$ is splitting field of the polynomial $X^3-3 \in \mathbb{Q}[X]$.

Now I know what the Cayley table of $\text{Gal}(L/\mathbb{Q})$ looks like (its isomorphic to $S_3$) so I can easily find all subgroups. But I don't have any idea how to find the splitting fields. Please bear with me, since I'm a very beginner, so I hope for answer, that really explains (or give detailed hints) where the connections lies - not just one-liner comments, that help only someone who already is familiar with Galois theory.

Hint: the roots of $\,x^3-3\,$ over $\,\Bbb Q\,$ are $\,\sqrt[3] 3\,,\,w\sqrt[3] 3\,,\,w^2\sqrt[3] 3\,\,\,,\,\,w=e^{2\pi i/3}\,$, thus you have the following subfields:

$$\Bbb Q\subset \Bbb Q(\sqrt[3] 3)\,,\,\Bbb Q(w)\,,\,\Bbb Q(w\sqrt[3] 3)\,...etc.$$

Try to take it from here (Check the automorphism, what each of them fixes, etc.)

$$\Bbb Q\subset\Bbb Q(\sqrt[3] 3)\subset\Bbb Q(\sqrt[3] 3\,,\,w)=:L\,\,,\,[L:\Bbb Q]=6$$

and since $\,\left\{1,3^{1/3}3^{2/3}\right\}\,$ is a basis for $\,\Bbb Q(\sqrt[3]3)/\Bbb Q\,$ and $\,\{1,w\}\,$ is a basis for $\,L/\Bbb Q(\sqrt[3]3)\,$ (why?) , then $\,\left\{1,\,3^{1/3},\,3^{2/3},\,w,\,3^{1/3}w,\,3^{2/3}w\right\}\,$ is a basis for $\,L/\Bbb Q\,$.

So let us check the action of each automorphism in $\,G:=Gal(L/\Bbb Q)\cong S_3\,$ on the above basis, for example: naming the roots of $\,x^3-3\,$ as $\,a_1=\sqrt[3]3\,,\,a_2=\sqrt[3]3\,\,w\,,\,a_3=\sqrt[3]3\,\,w^2\,$ , and taking the automorphism $\,\sigma\in G\,$ that we identify with the involution $\,(1,2)\in S_3\,$ we get:

$$\sigma a_1=a_2\,\,\,,\,\,,\sigma a_2=a_1\,\,\,,\,\,\sigma a_3=a_3$$ so

$$\sigma\left(\begin{cases}1\\{}\\3^{1/3}\\{}\\3^{2/3}\\{}\\w\\{}\\3^{1/3}\,\,w\\{}\\3^{2/3}\,\,w\end{cases}\right)=\begin{cases}1\\{}\\3^{1/3}\,\,w\\{}\\-3^{2/3}\,\,w-3^{2/3}=3^{2/3}\,\,w^2\\{}\\w^2\\{}\\3^{1/3}\\{}\\-3^{2/3}-3^{2/3}\,\,w^2=3^{2/3}\,\,w\end{cases}$$

We can see that $\,\sigma\Bbb Q(3^{1/3}\,w)=\Bbb Q(3^{1/3}\,w)\,$ , and we can even check by means of the Fundamental Theorem of Galois Theory:

$$3=[\Bbb Q(3^{1/3}\,w):\Bbb Q]=[G:\langle \sigma \rangle]$$

and, btw, $\,\langle\sigma\rangle\ntriangleleft G\,\Longleftrightarrow \Bbb Q(3^{1/3}\,w)/\Bbb Q$ not a normal extension.

• But what is the (general) connection between the automorphism and the subfields ? Shouldn't I just adjoint every possible combination of $\sqrt{3},\sqrt[3]{3},w$ and $w^2$ and those are already all my possible subfields ?
– user47574
Dec 9, 2012 at 17:11
• @user47574 , I added some stuff to my answer that will hopefully help. Dec 9, 2012 at 19:16
• Subfields correspond to subgroups of the Galois group $S_3$. So you should be looking for 3 quadratic subfields, and one cubic subfield. Dec 10, 2012 at 16:16
• @DonAntonio: Do you mean $\sigma\mathbb Q(3^{2/3}w) = \mathbb Q(3^{2/3}w)$?
– user59083
May 3, 2014 at 20:40