Villaretq0

2022-06-24

What is the Galois group of $27{x}^{8}-72{x}^{4}-16$ over the rationals?

Eli Shaffer

Expert

The roots are easily found to be

with a,b being positive real numbers.
The given polynomial is irreducible over $\mathbb{Q}$ and hence $\left[\mathbb{Q}\left(a\right):\mathbb{Q}\right]=8$. The splitting field L of the polynomial contains $a,ai$ and hence $i\in L$. But $i\notin \mathbb{Q}\left(a\right)$ and hence $\left[\mathbb{Q}\left(a,i\right):\mathbb{Q}\right]=16$
It can be proved with some effort that $L=\mathbb{Q}\left(a,i\right)$ is the splitting field of the polynomial and hence the Galois group is of order 16.
We can note that

and next we show that $c\in L$. Since c is real we have in fact $c\in \mathbb{Q}\left(a\right)$. We can observe that
${c}^{2}=2\sqrt{3}=\frac{9{a}^{4}-12}{4}\in \mathbb{Q}\left(a\right)$
and
$\begin{array}{rl}9{a}^{4}& =4{c}^{2}+12\\ & =4{c}^{2}+{c}^{4}\\ & ={c}^{2}\left({c}^{2}+4\right)\\ & ={c}^{2}\left(4+2\sqrt{3}\right)\\ & ={c}^{2}\left(1+\sqrt{3}{\right)}^{2}\end{array}$
This implies
$3{a}^{2}=c\left(1+\sqrt{3}\right)$
ie
$c=\frac{6{a}^{2}}{2+{c}^{2}}\in \mathbb{Q}\left(a\right)$
It now follows that $b=c/\left(3a\right)\in \mathbb{Q}\left(a\right)\subset L$ and L is the desired splitting field.
The Galois group is ${D}_{16}$ (dihedral group of order 16) as explained in another answer.

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