Finding a Galois extension of $\Bbb Q$ of degree $3$

Solution 1:

Consider the cyclotomic polynomial $\Phi_7(x) = x^6+x^5+\ldots + x + 1$ which is irreducible and generates an extension of $\Bbb Q$ of degree $6$ which is abelian (i.e. it is Galois with abelian Galois group). Then if $\zeta_7$ is a primitive $7^{th}$ root of $1$, $F=\Bbb Q(\zeta_7)$ is the extension. The element $\zeta_7+\zeta_7^{-1}$ is fixed by complex conjugation (an element of order $2$) and no other automorphism (you can check directly by noting $\zeta_7\mapsto \zeta_7^{k}, 1\le k\le 6$ are the automorphisms of $F$ and that any other automorphism besides $k=6$ gives a different element.

But then $K= \Bbb Q(\zeta_7+\zeta_7^{-1})\subseteq F$ is an extension of degree $3$, because that is the index of the fixing Galois group generated by complex conjugation. Hence $K/\Bbb Q$ is the desired extension. You can even describe it explicitly as $K=\Bbb Q\left((\cos\left({2\pi\over 7}\right)\right)$.

Working out the details you can see it is generated by the polynomial

$$p(x) = x^3+x^2-2x-1.$$

Solution 2:

To get an Galois extension of degree $3$, you need all the roots to be real since otherwise complex conjugation is an automorphism of order $2$.

Whilst you're right that Euler's totient function doesn't take the value $3$, we can tweak this idea. Now let $\zeta_7$ be a primitive $7$th root of unity. Then $[\mathbb{Q}(\zeta_7):\mathbb{Q}]=6$ and it has cyclic Galois group. As I said above, complex conjugation creates an order 2 element so we shall take the fixed field corresponding to this.

Now the complex conjugate $\zeta_7$ is $\zeta_7^{-1}$ and it turns out that $[\mathbb{Q}(\zeta_7+\zeta_7^{-1}):\mathbb{Q}]=3$. Moreover, this extension is Galois since $\mathbb{Q}(\zeta_7)/\mathbb{Q}$ had an abelian Galois group.

In fact, all degree $3$ Galois extensions of $\mathbb{Q}$ will arise like this as being the subfield of some other cyclotomic field.

Solution 3:

Hint: Consider the cyclotomic field $\mathbb{Q}(\zeta_7)$. where $\zeta_7=e^{2\pi i/7}$. Its Galois group is $\mathrm{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong\mathbb{Z}/6\mathbb{Z}$, which is abelian and therefore all of its subgroups are normal. Can you think of a subgroup $H\subset \mathbb{Z}/6\mathbb{Z}$ such that the fixed subfield $\mathbb{Q}(\zeta_7)^H$ is a degree $3$ Galois extension over $\mathbb{Q}$? Use the Fundamental Theorem of Galois Theory.