Question

Construct a polynomial in $\mathbf{\mathbb{Q}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ of degree 7 whose Galois group is S₇.

Short answer

The polynomial$p\left(x\right)=x^7+x^6+12x^5-7x^4+28x^{3}14x^2-9x+1$ has its Galois group S₇.

Step-by-step solution

Galois Group

If K is a finite dimension, normal, separable extension field of the field F so K is a Galois extension of F. In other words, K is Galois over F.

Find the polynomial

Suppose K is a fixed field over $\mathrm{\mathbb{Q}}$ and 29^th root of unity is $\varsigma =exp\left(\frac{2\mathrm{\pi i}}{29}\right)$. Then the Galois group of the field localid="1659536564738" $\mathbb{Q}\left(\zeta \right)/\mathbb{Q}$is:

localid="1659536581864" $\begin{array}{rcl}& & \begin{array}{c}G={(\mathbb{Z}/29\mathbb{Z})}^x\\ \simeq (\mathbb{Z}/28\mathbb{Z})\\ \simeq (\mathbb{Z}/4\mathbb{Z})\times (\mathbb{Z}/7\mathbb{Z})\end{array}\end{array}$

Here, the root of mod 29 is 2, so, the automorphism localid="1659449995592" $\zeta \to {\zeta }^2$generates the Galois group. Now, the automorphism $\zeta \to {\zeta }^{2^7}={\zeta }^{12}$can be used to determine the element of four order by 7^th power.

Suppose K be the normal intermediate field and also the subgroup of order 4 extension field. Then,localid="1659536599460" $G/H\simeq \mathbb{Z}/7\mathbb{Z}$becomes the Galois field of the fixed field. The minimal polynomial can be obtained by using the series given below:

$\begin{array}{rcl}\zeta & =& \sum _{\sigma \in H}\sigma \left(\zeta \right)\\ & =& \zeta +{\zeta }^{12}+{\zeta }^{28}+{\zeta }^7\end{array}$

The minimal polynomial that contains the power of zetas in the above series and that has a degree of 7 is:

$p\left(x\right)=x^7+x^6+12x^5-7x^4+28x^3+14x^2-9x+1$

Thus, the polynomial$p\left(x\right)=x^7+x^6+12x^5-7x^4+28x^3+14x^2-9x+1$ has its Galois group S₇.