The Stacks project

Proof. If the characteristic of $k$ is zero, then $k' = k$ is the unique choice. Assume the characteristic of $k$ is $p > 0$. For every $n > 0$ there exists a unique algebraic extension $k \subset k^{1/p^ n}$ such that (a) every element $\lambda \in k$ has a $p^ n$th root in $k^{1/p^ n}$ and (b) for every element $\mu \in k^{1/p^ n}$ we have $\mu ^{p^ n} \in k$. Namely, consider the ring map $k \to k^{1/p^ n} = k$, $x \mapsto x^{p^ n}$. This is injective and satisfies (a) and (b). It is clear that $k^{1/p^ n} \subset k^{1/p^{n + 1}}$ as extensions of $k$ via the map $y \mapsto y^ p$. Then we can take $k' = \bigcup k^{1/p^ n}$. Some details omitted. $\square$