
Every field has a unique perfect closure.

Lemma 10.44.4. For every field $k$ there exists a purely inseparable extension $k \subset k'$ such that $k'$ is perfect. The field extension $k \subset k'$ is unique up to unique isomorphism.

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. $\square$

Comment #1366 by Herman Rohrbach on

Suggested slogan: Every field has a unique unique purely inseparable perfect extension.

Comment #1371 by jojo on

Maybe a different suggestion : Every field has a unique perfect closure.

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