Is $\mathbf F_3(t,t^{1/3})/\mathbf F_3(t)$ a normal extension? Is it separable?

Question

I'm used to checking this sort of thing for Archimedian fields, but I'm not sure what to do in this case.

I know that $[\mathbf F_3(t,t^{1/3}):\mathbf F_3(t)]=3$ since $x^3-t$ is irreducible over $\mathbf F_3(t)$ by Eisenstein with prime $t$. I believe that this extension is not separable, since $x^3-t=(x-t^{1/3})^3$ so we have repeated factors. How do I go about checking whether or not this extension is normal?

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Edit: To clarify, the definition of normal I am using is the following: $L/K$ is normal if for all $\alpha\in L$, the minimal polynomial $m_{\alpha,K}$ has all its roots in $L$.

Answer 1

I showed it this way: Let $K=\mathbf F(t)$, and $L=K(t^{1/3})$, so $\{1,t^{1/3},t^{2/3}\}$ is a $K$-basis for $L$. Consider a generic element $\alpha=a+bt^{1/3}+ct^{2/3}\in L$. Then the minimal polynomial $m_{\alpha,K}$ of $\alpha$ divides $$f:=x^3-\alpha^3=x^3-a^3-bt-ct^2 \in K[x].$$ But $f=(x-\alpha)^3$, so clearly $f$ splits into linear factors over $L$ (and therefore so does $m_{\alpha,K}$). Thus $L/K$ is normal.

Answer 2

Well, the definition of normality that I like best says that $L\supset K$ is normal if: whenever $\Omega$ is an algebraically closed field containing $L$ and $f:L\to\Omega$ is a $K$-homomorphism, then $f(L)\subset L$.

Under this definition, I believe you should be able to show that your extension is normal.