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


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