 Solution

An algebraic element is a separable algebraic element over $K$ if its minimal polynomial of an algebraic element has distinct roots in a splitting field. A field extension is a separable extension if all its elements are separable algebraic elements.

The formal derivative is additive and satisfies $D(X^m)=m{X}^{m-1}$. For monomials, $D(X^r{X}^s)=(r+s)X^{r+s-1}=X^{r}D(X^s)+X^{s}D(X^r)$. Distributing over the finitely many coefficients proves the product rule $D(fg)=fDg+gDf$ over any field. If $f=(X-\alpha {)}^{2}h$ in an extension, the rule gives $Df=2(X-\alpha )h+(X-\alpha {)}^{2}Dh$, so $Df(\alpha )=0$, including in characteristic two. If $f,Df$ were relatively prime in $K[X]$, the Bézout's identity for polynomials would give $uf+vDf=1$, whose evaluation at $\alpha$ is impossible. Hence they are not relatively prime.

In characteristic zero, the formal derivative of a positive-degree irreducible polynomial $m$ is nonzero and has smaller degree. Thus $\gcd (m,Dm)=1$, since an irreducible polynomial has no proper nonconstant factor. By the previous argument, $m$ has no repeated root. Every element of a finite field extension is algebraic, so the extension is a separable extension.

For $K={\mathbb{F}}_p(t)$, let ${\alpha }^p=t$. The characteristic-$p$ identity gives $X^p-t=(X-\alpha {)}^p$, so $K(\alpha )$ is already a splitting field. It is a proper extension by the allowed no-root fact. More precisely, the minimal polynomial of an algebraic element $\alpha$ divides $X^p-t$ and, over $K(\alpha )$, equals $(X-\alpha {)}^r$ for some $1\le r\le p$. If $r<p$, its next-to-leading coefficient is $-r\alpha \in K$, forcing $\alpha \in K$ because $r\ne 0$ in ${\mathbb{F}}_p$, a contradiction. Hence its minimal polynomial of an algebraic element is $X^p-t$, with its one root repeated $p$ times. The extension is therefore inseparable, of degree $p$.

By contrast, every finite field extension $L/{\mathbb{F}}_p$ is a separable extension. If $[L:{\mathbb{F}}_p]=n$, then $|L|=p^n$. Lagrange's theorem in its nonzero multiplicative group gives $a^{p^n}=a$ for all $a\in L$. Thus every element is a root of $X^{p^n}-X$, whose formal derivative is $-1$; it has no repeated roots. Every minimal polynomial of an algebraic element divides this separable polynomial, proving the assertion.