Example 3

Suppose that $K = \mathbf{Q}$ and take $f = X^3 - 2$. Using the decomposition of integers into products of prime factors (I, p. 51), we easily see that 2 is not the cube of an element of $\mathbf{Q}$. Hence the polynomial $f$ is irreducible, for if not, it would have a root in $\mathbf{Q}$. Let $A = \{ x_1, x_2, x_3 \}$ be the set of roots of $f$ in $\Omega$ and $\Gamma$ the Galois group of $f$. It acts transitively on $A$; its order is therefore divisible by three. On the other hand, the quotient $j = \frac{x_2}{x_1}$ is different from 1 and we have $j^3 = 1$. Hence $j$ satisfies the relation $j^2 + j + 1 = 0$; now the polynomial $T^2 + T + 1 = \left( T + \frac{1}{2} \right)^2 + \frac{3}{4}$ has no roots in $\mathbf{Q}$, which shows (Example 1) that $[\mathbf{Q}(j) : \mathbf{Q}] = 2$. Thus $[\mathbf{N} : \mathbf{Q}]$ is divisible by 2, and it follows that the order of $\Gamma$ is divisible by 6. Since $\Gamma$ is contained in the group $\mathfrak{S}_A$ of order 6, we have $\Gamma = \mathfrak{S}_A$.