Exercise 5.4.8

Proof. Here $F=k(t,u)\subset L$, where the characteristic of $k$ is 3, $L$ is the splitting field of $(x^2-t)(x^3-u)$, and $\alpha ,\beta \in L$ are such that ${\alpha }^2=t,{\beta }^3=u$.

$\alpha$ is separable, but not $\beta$ (cf Exercise 5.3.8).

We know (by Exercise 5.3.8) that

are the respective minimal polynomials of $\beta$ and $\alpha$ over $F$.

The two polynomials

vanish at $\alpha$, since $g(\alpha )=0,f(\beta )=0$, and they are both in $F(\alpha +\beta )[x]$.

Thus $x-\alpha \mid h(x)=\gcd (g(x),f(\alpha +\beta -x))$.

$1\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)\le 2$. If $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=2$, as $h\mid g$, we would have $h=g=(x-\alpha )(x+\alpha )$, and then $x+\alpha \mid h\mid f(\alpha +\beta -x)$.

As $f(x)={(x-\beta )}^3$, then $f(\alpha +\beta -x)=-{(x-\alpha )}^3$, which is not divisible by $x+\alpha$, since $-\alpha \ne \alpha$.

Therefore $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=1$, and $h(x)=\gcd (g(x),f(\alpha +\beta -x))=x-\alpha$.

Thus there exists a Bézout’s relation

This proves that $\alpha \in F(\alpha +\beta )$, thus also $\beta =(\alpha +\beta )-\alpha \in F(\alpha +\beta )$, which implies that $L=F(\alpha ,\beta )=F(\alpha +\beta )$ : $\alpha +\beta$ is a primitive element of $L∕F$.

We compute explicitly the gcd of the polynomials $f(\alpha +\beta -x),g(x)$ :

The first Euclidean division of $f(\alpha +\beta -x)$ by $g(x)$ gives

We must then have

We compute a direct proof of this equality :

This equality proves also that $\alpha +\beta$ is a primitive element of $F\subset L$ □