Exercise 5.3.8

Proof. Here $k$ is a field of characteristic 3.

Let $F=k(t,u)$, where $t,u$ are two variables, $f=(x^2-t)(x^3-u)$, and $\alpha ,\beta$ in a splitting field $L$ of $f$ such that ${\alpha }^2=t,{\beta }^3=u$.

(a)

The Exercise 4.2.9, applied to the field $k(u)$, shows that $x^2-t$ has no root in $F=k(t,u)=k(u)(t)$, so it is irreducible over $F$. So $x^2-t$ is the minimal polynomial of $\alpha$ over $F$.

In $L$, $x^2-t=(x-\alpha )(x+\alpha )$, and $\alpha \ne -\alpha$, otherwise $2\alpha =0$, with $2=-1\ne 0$ in $k$, and $\alpha \ne 0$ since ${\alpha }^2=t\ne 0$.

Thus the minimal polynomial $x^2-u\in F[x]$ of $\alpha$ over $F$ is separable, so $\alpha$ is separable.

(b)

Similarly, $x^3-u$ has no root in $F=k(t,u)=k(t)(u)$, and its degree is 3, thus it is irreducible over $F$: $x^3-u$ is the minimal polynomial of $\beta$ over $F$.

As the characteristic is 3, $x^3-u={(x-\beta )}^3$, so this polynomial is not separable : $\beta$ is not separable.

So $F\subset L$ is not a separable extension, and is not a purely inseparable extension. □