Can a Single Complex Number Generate the Splitting Field of \(x^3-2\)?

[mathmari]

Hey!

Let $E$ the splitting field of $x^3-2 \in \mathbb{Q}[x]$.

Applying the algorithm of the proof of the Primitive element theorem, find a complex $c$ with $E=\mathbb{Q}(c)$.

I have done the following:

The splitting field is $E=\mathbb{Q}(\sqrt[3]{2}, \omega_3)$.

Since $\sqrt[3]{2}, \omega_3$ are algebraic over $\mathbb{Q}$, we have that $\mathbb{Q} \leq \mathbb{Q}(\sqrt[3]{2}, \omega_3)$ is finite, that means that $\mathbb{Q} \leq E$ is finite.

Since $\mathbb{Q}$ is a perfect field, we have that the finite expansion $\mathbb{Q} \leq E$ is seperable.

So, from the Primitive element theorem, $\exists c \in E$ with $E = \mathbb{Q}(c)$.

We have that $E=\mathbb{Q}(\sqrt[3]{2}, \omega_3 )$ with $\omega_3$ separable over $\mathbb{Q}$.

Let $f$ and $g$ the minimum polynomials of $\sqrt[3]{2}$ and $\omega_3$ over $\mathbb{Q}$, and let $L$ be a splitting field for $fg$ containing $E$.

Let $a_1=\sqrt[3]{2}, a_2, \dots , a_s$ be the roots of $f$ in $L$, and let $b=\omega_3 , b_2, \dots , b_t$ be the roots of $g$.

For $j \neq 1, b_j \neq b=\omega_3$ so the equation $a_ix+xb_j=a+xb \Rightarrow a_i+xb_j=\sqrt[3]{2}+x \omega_3$ has exactly one solution, $x=\frac{a_i-\sqrt[3]{2}}{\omega_3-b_j}$.

We choose $\gamma \in \mathbb{Q}$ differeent from any of the roots, then $a_i+\gamma b_j \neq \sqrt[3]{2}+\gamma \omega_3$, unless $i=1=j$.

Let $c=\sqrt[3]{2}+\gamma \omega_3$.

We claim that $\mathbb{Q}(\sqrt[3]{2}, \omega_3)=\mathbb{Q}(c)$.

The polynomkials $g(x)$ and $f(c-\gamma x)$ have coefficients in $\mathbb{Q}(c)$ and have $\omega_3$ as a root:
$g(\omega_3)=0, f(c-\gamma \omega_3)=f(\sqrt[3]{2})=0$

Indeed, $\omega_3$ is their only common solution, since we chose $\gamma$ so that $c-\gamma b_j \neq a_i$, unless $i=1=j$.

So, $gcd(g(x), f(c-\gamma x))=x-\omega_3$.

We computed the $gcd$ in $L[x]$, but it is equal to the $gcd$ in $\mathbb{Q}[x]$.

So, $\omega_3 \in \mathbb{Q}(c) \Rightarrow \sqrt[3]{2}=c-\gamma \omega_3 \in \mathbb{Q}(c)$.Is this correct?? Or should I have found a specific value for $c$ ?? (Wondering)

 

[jvicens]

Your proof looks correct! You have correctly applied the algorithm of the Primitive element theorem, and you have shown that $c=\sqrt[3]{2}+\gamma \omega_3$ satisfies the conditions for being the primitive element. Therefore, your choice of $c$ is valid. Great job!