Let $K/k$ be an algebraic field extension. Recall that an element $a \in K$ is called separable over $k$ if the minimal polynomial of $a$ over $k$ has no repeated root (in its splitting field). Otherwise, $a$ is called inseparable. If every element of $K$ is separable over $k,$ then $K$ is called a separable extension of $k.$ If every element of $K \setminus k$ is inseparable over $k,$ then $K$ is called a purely inseparable extension of $k.$ Recall also the following facts from basic field theory.

Fact 1. If $\text{char}(k)=0,$ then $K/k$ is separable. If $\text{char}(k) = p > 0,$ then an element $a \in K$ is separable over $k$ if and only if the minimal polynomial of $a$ over $k$ is not in $k[x^p].$

Fact 2. If $\text{char}(k) = p > 0$ and $K \neq k,$ then $K/k$ is purely inseparable if and only if for every $a \in K$ there exists some integer $m$ such that $a^{p^m} \in k.$

Theorem. (Jacobson, Noether) Let $D$ be a noncommutative division algebra with the center $k.$ If $D$ is algebraic over $k,$ then there exists an element of $D \setminus k$ which is separable over $k.$

Proof. If $\text{char}(k)=0,$ let $a \in D \setminus k$ and put $K=k(a).$ Then, by Fact 1, $K/k$ is separable and we are done. Suppose now that $\text{char}(k)=p > 0$ and every element of $D \setminus k$ is inseparable over $k.$ Then for every $a \in K \setminus k,$ the field $K=k(a)$ is purely inseparable over $k.$ So, by Fact 2, $a^{p^m} \in k$ for some integer $m \geq 1.$ Now, fix an element $a \in D \setminus k$ and an integer $m \geq 1$ such that $a^{p^m} \in k.$ Define the map $\delta: D \longrightarrow D$ by $\delta(x)= ax - xa$ for all $x \in D.$ Clearly $\delta = f - g,$ where $f,g : D \longrightarrow D$ are the $k$-linear maps defined by $f(x)=ax$ and $g(x)=xa$ for all $x \in D.$ Since $fg=gf$ and $\text{char}(k)=p,$ we have

$\delta^{p^m} = (f - g)^{p^m} = f^{p^m} - g^{p^m}.$

Therefore $\delta^{p^m}(x)=a^{p^m}x - xa^{p^m} = 0,$ for all $x \in D,$ because $a^{p^m} \in k.$ Hence

$\delta^n = 0, \ \forall n \geq p^m. \ \ \ \ \ \ (*)$

Now, since $a \notin k,$ we have $\delta \neq 0$ and so $\delta(b) \neq 0$ for some $b \in D.$ Thus, by $(*),$ there exists an integer $n \geq 1$ which is maximal with respect to the property $\delta^n(b) \neq 0.$ Hence $\delta^{n+1}(b)=0$ and so if we let $c = \delta^n(b),$ then $c \neq 0$ and $\delta(c) = 0,$ i.e. $c$ is invertible and it commutes with $a.$ Let $d = c^{-1}a \delta^{n-1}(b).$ See that $ad - da = a$ and so $d = 1 + a^{-1}da.$ By Fact 2, there is an integer $r \geq 1$ such that $d^{p^r} \in k.$ Then

$d^{p^{r}} = (1+a^{-1}da)^{p^r}=1 + (a^{-1}da)^{p^r}=1+a^{-1}d^{p^r}a = 1 + d^{p^r},$

which is non-sense. $\Box$