 Solution

Take $p=11$. Write $Y=X^5$, so the extension isThe polynomial $t^5-Y$ is irreducible over ${\mathbb{F}}_{11}(Y)$, for example by Eisenstein's criterion at $Y$ in ${\mathbb{F}}_{11}[Y]$, and has $X$ as a root. Thus the extension has degree five.

The element $3\in {\mathbb{F}}_{11}^{\times }$ has order five. Hence all roots of $t^5-Y$ areand already lie in ${\mathbb{F}}_{11}(X)$. The extension is therefore the splitting field of $t^5-Y$ and is normal. Its derivative is $5t^4\ne 0$ in characteristic eleven, so the polynomial is separable. By part (c),