Question

If $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\in }\mathbf{\mathbb{Q}}\mathbf{\left[}\mathit{x}\mathbf{\right]}$is irreducible of prime degree p and $\mathit{f}\left(x\right)$has exactly two nonreal roots, prove that the Galois group of $\mathit{f}\left(x\right)$is ${\mathit{S}}_{\mathbf{p}}$. [Example 5 is essentially the case p=5.]

Short answer

It is proved that Galois group of$f\left(x\right)$ is $S_p$.

Step-by-step solution

Determine the splitting field

The shortest field extension of the extension field over which a minimum polynomial break or dissociates into linear components is known as the splitting field.

Determine the proof

Suppose that the splitting fieldF of $f\left(x\right)$over $\mathrm{\mathbb{Q}}$. Given that $f\left(x\right)\in \mathrm{\mathbb{Q}}\left[x\right]$is irreducible of prime degreep and splitting field has exactly two nonreal roots.

Now, if it is assumed an order 2 automorphism represented because a single pair of complex conjugate root contained by the splitting field. This means $G=Gal(F/\mathrm{\mathbb{Q}})$.

One say that $[F:\mathrm{\mathbb{Q}}]=p$because $S_5$, a subgroup of order 5, consists an order 2 subgroup which is generated by single 2 cycle.

Now, the order of the group is divided byp by the Galois theory and ap cycle is consisted inG by the Cauchy’s Theorem.

By the above calculations, we can say that $S_p$is generated by ap cycle and a 2 cycle.

Thus, the Galois group of the given splitting field$f\left(x\right)$ is $S_p$.