$f$ is separable if and only if for every irreducible factor $g ∈ F[t]$ of $f$, the formal derivative $D(g)$ of $g$ is not zero. Since $f$ is irreducible over $F$ and $\deg D(f)<\deg f$, the polynomials $f$ and $D(f)$ are coprime in $F[t]$. Since $F[t]$ is a PID, there exist $p, q ∈ F[t]$ such that $p D(f)+q f=1$. Suppose that $K$ is a field extension in which $f$ has a repeated root, $α$ say. Then $f=(t-α)^2 g$ for some $g ∈ K[t]$. Hence $D(f)=2(t-α) g+(t-α)^2 D(g)$ also vanishes at $t=α$. Substituting $t=α$ into $p D(f)+q f=1$ then gives a contradiction.

Assume from now on that the field $F$ has characteristic $p$.
Since $f$ is irreducible, we have to show that $D(f)=0$ if and only if $f ∈ F[t^p]$. If $f=\sum_{i=0}^n a_i t^i$, then $D(f)=\sum_{i=1}^n i a_i t^{i-1}$. Hence $D(f)=0$ if and only if $a_i=0$ whenever $p ∤ i$. This is equivalent to $f ∈ F[t^p]$.

Show that the following are equivalent:

- $α$ is separable over $F$,
- $F(α)=F(α^p)$.

Suppose that $α$ is separable over $F$. Then $m:=m_{F, α}$ is separable, so $D(m) ≠ 0$.

Now, $α$ is a root of $t^p-α^p ∈ F(α^p)[t]$. Hence $m ∣ t^p-α^p$. Viewing this divisibility in the larger ring $F(α)[t]$, we see that $m ∣ t^p-α^p=(t-α)^p$, hence $m=(t-α)^r$ for some $1 ⩽ r ⩽ p$. Since $m$ has no repeated roots in $F(α)$ by part (a), $r=1$. Hence $t-α ∈ F(α^p)$ and $α ∈ F(α^p)$. Hence $F(α)=F(α^p)$.

Conversely, suppose that $α$ is not separable over $F$. Since $m=m_{F, α}$ is irreducible, $m ∈ F[t^p]$ by part (b), so there exists $g ∈ F[t]$ such that $m=g(t^p)$. Then $\deg m=p \deg g$, so $\deg g<\deg m$. Since $α^p$ is a root of $g$, we see that $[F(α^p): F] ⩽ \deg g<\deg m=[F(α): F]$. Hence $F(α^p) ≠ F(α)$.

Now, $α$ is a root of $t^p-α^p ∈ F(α^p)[t]$. Hence $m ∣ t^p-α^p$. Viewing this divisibility in the larger ring $F(α)[t]$, we see that $m ∣ t^p-α^p=(t-α)^p$, hence $m=(t-α)^r$ for some $1 ⩽ r ⩽ p$. Since $m$ has no repeated roots in $F(α)$ by part (a), $r=1$. Hence $t-α ∈ F(α^p)$ and $α ∈ F(α^p)$. Hence $F(α)=F(α^p)$.

Conversely, suppose that $α$ is not separable over $F$. Since $m=m_{F, α}$ is irreducible, $m ∈ F[t^p]$ by part (b), so there exists $g ∈ F[t]$ such that $m=g(t^p)$. Then $\deg m=p \deg g$, so $\deg g<\deg m$. Since $α^p$ is a root of $g$, we see that $[F(α^p): F] ⩽ \deg g<\deg m=[F(α): F]$. Hence $F(α^p) ≠ F(α)$.

Consider the following chains of fields:
\[
F ⊆ F(α+β) ⊆ F(α+β, β) \text{ and } F ⊆ F(α^p+β^p) ⊆ F(α^p+β^p, β^p) .
\]
Since $α, β$ are separable over $F$, part (c) implies that
\[
F(α^p+β^p, β^p)=F(α^p, β^p)=F(α, β)=F(α+β, β) .
\]
Hence by the Tower Law, we have
\[\tag1
[F(α+β): F] ⋅[F(α+β, β): F(α+β)]=[F(α^p+β^p): F] ⋅[F(α^p+β^p, β^p): F(α^p+β^p)] .
\]
Now $F(α^p+β^p)=F((α+β)^p) ⊆ F(α+β)$, so
\[\tag2
[F(α^p+β^p): F] ⩽[F(α+β): F] .
\]
If $g$ is the minimal polynomial of $β$ over $F(α+β)$, then $0=g(β)^p=h(β^p)$ for some monic $h ∈ F((α+β)^p)[t]=F(α^p+β^p)[t]$, with $\deg h=\deg g: h$ is the polynomial obtained from $g$ by raising each of its coefficients to the power of $p$. Hence
\[\tag3
[F(α^p+β^p, β^p): F(α^p+β^p)] ⩽ \deg h=\deg g=[F(α+β, β): F(α+β)] .
\]
Equation (1) then implies that we have equalities in (2) and (3). Hence $F(α+β)=F((α+β)^p)$, so $α+β$ is separable over $F$ by part (c).