Showing $K(\sqrt[3]{t}) / K$ is a normal extension.

Question

Let $K = \mathbb{F}_3(t)$. Consider the degree 3 extension $K(\sqrt[3]{t}) / K$.

$\sqrt[3]{t}$ has minimal polynomial $m(x) = x^3 - t$ over $K$.

I would like to show that this is a normal extension directly from the definition, rather than using that this is the splitting field of $m$.

Given $a \in K(\sqrt[3]{t})$, if the minimum polynomial of $a$ is degree 1 or 2 then all roots are contained in $K(\sqrt[3]{t})$. The case I am having trouble with is when this has degree 3.

Any help is appreciated. Thanks.

Answer 1

Let $\alpha=\sqrt[3]{t}$ .

Then $X^3-t=X^3-\alpha^3=(X-\alpha)^3,$ since $K$ has characteristic $3$. Hence your extension is the extension generated by the root(s) of $X^3-t$, hence it is a normal extension.