Andreasihf

2022-09-12

First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: $\mathbb{Q}\left(\sqrt[3]{2},\gamma \right)$, where ${\gamma }^{3}=1$ and ${\gamma }^{2}=1-\gamma$. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.
I know that my Galois group will consist of the following automorphisms:
${\sigma }_{i}$ = identity
${\sigma }_{1}:\sqrt[3]{2}↦\sqrt[3]{2}\gamma$ and $\gamma$ is fixed.
${\sigma }_{2}:\sqrt[3]{1}↦\sqrt[3]{2}{\gamma }^{2}$ and $\gamma$ is fixed.
${\sigma }_{3}:\gamma ↦{\gamma }^{2}$ and $\sqrt[3]{2}$ is fixed.
${\sigma }_{4}:\sqrt[3]{2}↦\sqrt[3]{2}\gamma$ and $\gamma ↦{\gamma }^{2}$.
${\sigma }_{5}:\sqrt[3]{1}↦\sqrt[3]{2}{\gamma }^{2}$ and $\gamma ↦{\gamma }^{2}$.
If I didn't miss something or make some error.
My specific question is, when constructing my automorphisms, why exactly is it that I can't send $\sqrt[3]{2}$ to something like $\sqrt[3]{4}$ or $\gamma$ or some other random thing? Why are these automorphisms so strictly defined?

Isaiah Haynes

Explanation:
By the Fundamental Theorem of Galois Theorem, if the Galois group were abelian, then all subfields would be normal extensions. But this cannot be, as $\mathbb{Q}\left(\sqrt[3]{2}\right)/\mathbb{Q}$ is not a normal extension.

sooxicyiy

Step 1
let $j=\frac{1}{2}\left(-1+i\sqrt{3}\right)$. Clearly $\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}},\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}j\phantom{\rule{thinmathspace}{0ex}}$ and $\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}{j}^{2}$ are roots of $f\left(x\right)={x}^{3}-2$. We claim $E=\mathbb{Q}\left(\sqrt[3]{2},j\right)$ is splitting field f over $\mathbb{Q}$ because
$E=\mathbb{Q}\left(\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}},\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}j\phantom{\rule{thinmathspace}{0ex}},{j}^{2}\sqrt[3]{2}\right)\subset \mathbb{Q}\left(\sqrt[3]{2},j\right)$
and
$j=\frac{1}{2}\sqrt[3]{2}\left(\sqrt[3]{2}{\right)}^{2}j\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathbb{Q}\left(\sqrt[3]{2},j\right)\subset E=\mathbb{Q}\left(\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}},\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}j\phantom{\rule{thinmathspace}{0ex}},{j}^{2}\sqrt[3]{2}\right)$
Step 2
The polynomial f is not irreducible over Q as a result minimal polynomial of $\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}$ is f and $\left[\mathbb{Q}\left(\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}\right):\mathbb{Q}\right]=3$ on the other hand
$\left[\mathbb{Q}\left(\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}},j\phantom{\rule{thinmathspace}{0ex}}\right):\mathbb{Q}\right]=\left[\mathbb{Q}\left(\sqrt[3]{2},\phantom{\rule{thinmathspace}{0ex}}j\right):\mathbb{Q}\left(\sqrt[3]{2}\right)\right]\left[\mathbb{Q}\left(\sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}\right):\mathbb{Q}\right]=2×3=6$
in the other words
$\left[E:\mathbb{Q}\right]=6$
so the Galois group is order 6. let
A: The metamorphism which fixes j and maps $j\to {j}^{2}$
B: The metamorphism which fixes $\sqrt[3]{2}$ and maps $\sqrt[3]{2}\to \sqrt[3]{2}\phantom{\rule{thinmathspace}{0ex}}j$
The Galois group is thus:
$⟨A,B\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{A}^{2},\phantom{\rule{thinmathspace}{0ex}}{B}^{3},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}AB={B}^{-1}A⟩\cong {D}_{6}$

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