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I want to find a Galois extension $K/\mathbb{Q}$ such that $[K:\mathbb{Q}]=3$. I thought about this for a while, but haven't been able to come up with one yet.

What I tried so far: (i) Taking a separable polynomial $f\in\mathbb{Q}[x]$ of degree three and considering its splitting field. (ii) Looking at the splitting fields of primitive roots of unity.

The second one doesn't work because the splitting field over such a root has as degree a value in the range of Euler's totient function, and this doesn't contain three.

The first approach also didn't work. I tried polynomials of the form $(x-\sqrt{p})(x-\sqrt{q})(x-\sqrt{r})$ for primes, but those have degree $8$. I then tried 'third roots' $\alpha$, but the minimal polynomials of those have complex as well as real roots, so the simple extensions $K(\alpha)$ aren't normal unless they're trivial.

Could anyone please give me a hint on what else to try.

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  • $\begingroup$ A useful discussion can be found here $\endgroup$
    – lulu
    Jan 24, 2017 at 20:21
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    $\begingroup$ My post here gives an easy-to-check necessary and sufficient condition on an irreducible cubic $f$ for $\text{Gal}(f/\mathbb{Q}) \cong \mathbb{Z}_3$ (warning: heavy spoilers). math.stackexchange.com/questions/2017810/… $\endgroup$
    – Kaj Hansen
    Jan 24, 2017 at 20:36

3 Answers 3

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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

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

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  • $\begingroup$ @see you do not need all of the subgroups, you know that complex conjugation generates some subgroup of order two, and that implies the index is three. The other subgroups are not relevant for the purposes of this question. $\endgroup$
    – Adam Hughes
    Jan 24, 2017 at 23:04
  • $\begingroup$ @see No, not quite, $\zeta_7\mapsto \zeta_7^0$ is not an automorphism as it sends something not $1$ to $1$, you mean $1\le k\le 6$. Complex conjugation is clearly a field automorphism because $F\subseteq\Bbb C$ is a subfield. It is exactly the map $\zeta_7\mapsto\zeta_7^6=\zeta_7^{-1}$, but I chose not to bother looking at it from that point-of-view because why complicate matters when "complex conjugation" is a much more familiar way to phrase it. If it helps you to think of it the other way you can, they're the same thing, but remember, you don't need to know the full Galois group (contd) $\endgroup$
    – Adam Hughes
    Jan 25, 2017 at 0:46
  • $\begingroup$ @see all you need is its order and the order of complex conjugation, since you know by Lagrange's theorem that a subgroup of order two in a group of order six has index $3$, and the FTGT tells you that the extension degree of the fixed field is this index--i.e. $3$--which is all the problem asks for. $\endgroup$
    – Adam Hughes
    Jan 25, 2017 at 0:48
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    $\begingroup$ @see yes, that's one approach, I opted for the faster route, since--as you'll find as you progress--reinventing the wheel every time takes a while, so we learn to use alternative ways with more general approaches. You will not want to work out entire Galois groups for every problem unless the problem asks you to find every subfield, and exploiting complex conjugation is one of the best tricks to add to your toolbox to avoid the excess work. $\endgroup$
    – Adam Hughes
    Jan 25, 2017 at 1:42
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    $\begingroup$ The minimal polynomial for any algebraic number is the product of linear terms in the conjugates of that number, form $(x-\alpha)$. Since you know the conjugates are all $\zeta^j+\zeta^{-j}$ it's fairly straightforward. $\endgroup$
    – Adam Hughes
    Jul 31, 2019 at 22:49
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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.

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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.

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