Spring 2016 - Problem 10

Galois theory

Determine the Galois group of the polynomial $X^4 - 2$ over $\Q$ , as a subgroup of a permutation group. Also, give generators and relations for this group.
Determine the Galois group of the polynomial $X^3 - 3X - 1$ over $\Q$ . (Hint: for polynomials of the form $X^3 + aX + b$ , the quantity $\Delta = -4a^3 - 27b^2$ , known as the discriminant, plays a key theoretical role.) Explain your answer.

Over $\C$ , the roots of $f = X^4 - 2$ are $2^{1/4}, i2^{1/4}, -2^{1/4}$ , and $-i2^{1/4}$ , so the splitting field of $f$ is $\Q\p{2^{1/4}, i}$ . Consider the tower
$\Q \subseteq \Q\p{2^{1/4}} \subseteq \Q\p{2^{1/4}, i}.$
Note that $\br{\Q\p{2^{1/4}} : \Q} = 4$ since $X^4 - 2$ is irreducible by Eisenstein with $p = 2$ . For the second inclusion, note that $X^2 + 1$ is still irreducible over $\Q\p{2^{1/4}}$ , since $i \notin \Q\p{2^{1/4}} \subseteq \R$ , so $\br{\Q\p{2^{1/4}, i} : \Q} = 8$ .

The maps $\sigma\colon 2^{1/4} \mapsto i2^{1/4}$ and $\tau\colon i \mapsto -i$ are all field automorphisms fixing $\Q$ . Moreover, these satisfy the relation $\sigma^4 = \tau^2 = \sigma\tau\sigma\tau = \id$ . Thus, $\gen{\sigma, \tau} = D_4$ in the Galois group, but because $\Q\p{2^{1/4}, i}/\Q$ is Galois, it follows that $\Gal\p{\Q\p{2^{1/4}, i}} \simeq D_4 \leq S_4$ .
Notice that by the rational roots theorem, the only possible roots in $\Q$ are $1$ and $-1$ , which are not roots. Thus, $f = X^3 - 3X - 1$ has no roots in $\Q$ , hence irreducible as any reducible cubic polynomial will have a linear factor.

The discriminant of $f$ is $\Delta = -4 \cdot \p{-3}^3 - 27 \cdot \p{-1}^2 = 81 = 9^2$ . Thus, any $\sigma$ in the Galois group $G$ of $f$ fixes $\sqrt{\Delta} = 9$ , so $G \leq A_3$ . We deduce $\abs{G} \leq 3$ , but because $f$ is irreducible, we also have $\abs{G} \geq 3$ , so $\abs{G} = 3$ , hence $G \simeq C_3$ .