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Set $K_r=K({\alpha }_1,\dots ,{\alpha }_r)$, beginning with $K_0=K$. Suppose $j:K_{r-1}\to L$ extends $i$. The minimal polynomial of ${\alpha }_r$ over $K_{r-1}$ divides its minimal polynomial over $K$. Its image under $j$ therefore divides $i(m_{{\alpha }_r})$, which splits into distinct linear factors in $L$. It consequently has exactly $[K_r:K_{r-1}]$ distinct roots in $L$. Part (i) supplies exactly this many extensions of $j$ to $K_r$.

Starting with the unique prescribed map on $K$, multiply these branching counts:For the final assertion, let $M$ be a splitting field of a separable polynomial and embed it in an algebraic closure $\Omega$. It is generated by roots whose minimal polynomials are separable and split in $\Omega$. Applying the just-proved count with $L=\Omega$ gives $[M:K]$ embeddings. Were $M/K$ inseparable, part (a) would give strictly fewer. Thus the splitting extension is separable. This works for the factorwise convention as well, since the generators' minimal polynomials divide the separable irreducible factors.

In field characteristic $p$, the intermediate binomial coefficients in the $p$th power vanish. HenceRepeated translation gives all $p$ roots $\alpha +j$, $j\in {\mathbb{F}}_p$. They are distinct because the prime field embeds in $F$. They exhaust the roots by degree; also the formal derivative of $X^p-X-a$ is $-1$, so this is a separable polynomial. If one root belongs to $F$, they all do, andThis is the root-translation property of an Artin–Schreier polynomial.

For an algebraic element $\alpha$ over a field $F$ of characteristic $p>0$, suppose first that $\alpha$ is a separable algebraic element. Its minimal polynomial over $E=F({\alpha }^p)$ divides its original separable polynomial and also $T^p-{\alpha }^p=(T-\alpha {)}^p$. It is therefore separable with only one distinct root, so has degree one and $\alpha \in E$. Conversely, if the irreducible minimal polynomial $m$ is inseparable, its formal derivative vanishes and $m(T)=g(T^p)$ for some $g$ of degree $\mathrm{deg}(m)/p$. Thus $[F({\alpha }^p):F]\le \mathrm{deg}(m)/p<[F(\alpha ):F]$, proving the converse.

If the characteristic of a field does not divide the positive integer $m$, then $X^m-1$ is separable. Its formal derivative is $m{X}^{m-1}$, and a common root would have to be both nonzero and zero.

Let a separable polynomial $f$ of degree four be irreducible over a field $F$. If all three pair sums $r_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$, $r_2={\alpha }_1{\alpha }_3+{\alpha }_2{\alpha }_4$, and $r_3={\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3$ lie in $F$, its Galois group is the Klein four-group. Indeed $r_1-r_2=({\alpha }_1-{\alpha }_4)({\alpha }_2-{\alpha }_3)\ne 0$, and similarly the other pair sums differ. Every Galois automorphism therefore fixes each of the three partitions into pairs. The kernel of this action of $S_4$ on three partitions is $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Irreducibility gives a transitive group action on the four roots, so the orbit-stabilizer theorem forces $4$ to divide the group order. Hence the Galois group equals $V_4$. The argument also works in characteristic two.

For a separable polynomial $f\in F[T]$ in characteristic two, the Vandermonde determinant of its roots is fixed by every Galois automorphism: root permutations change it only by a sign, and $-1=1$ in $F$. It therefore lies in $F$. Its square, multiplied by the square of the appropriate leading-coefficient power for a nonmonic polynomial, is the polynomial discriminant. Thus being a square does not detect containment of the Galois group in the alternating group. For example $T^2+T+1$ over ${\mathbb{F}}_2$ has discriminant one and Galois group $C_2$, while $A_2$ is trivial. The discriminant criterion for an alternating Galois group requires characteristic different from two.