Definition 10.44.1. Let $k$ be a field. We say $k$ is *perfect* if every field extension of $k$ is separable over $k$.

## 10.44 Perfect fields

Here is the definition.

Lemma 10.44.2. A field $k$ is perfect if and only if it is a field of characteristic $0$ or a field of characteristic $p > 0$ such that every element has a $p$th root.

**Proof.**
The characteristic zero case is clear. Assume the characteristic of $k$ is $p > 0$. If $k$ is perfect, then all the field extensions where we adjoin a $p$th root of an element of $k$ have to be trivial, hence every element of $k$ has a $p$th root. Conversely if every element has a $p$th root, then $k = k^{1/p}$ and every field extension of $k$ is separable by Lemma 10.43.1.
$\square$

Lemma 10.44.3. Let $k \subset K$ be a finitely generated field extension. There exists a diagram

where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is a separable field extension. In this situation we can assume that $K' = k'K$ is the compositum, and also that $K' = (k' \otimes _ k K)_{red}$.

**Proof.**
By Lemma 10.41.4 we can find such a diagram with $k' \subset K'$ separably generated. By Lemma 10.43.2 this implies that $K'$ is separable over $k'$. The compositum $k'K$ is a subextension of $k' \subset K'$ and hence $k' \subset k'K$ is separable by Lemma 10.41.2. The ring $(k' \otimes _ k K)_{red}$ is a domain as for some $n \gg 0$ the map $x \mapsto x^{p^ n}$ maps it into $K$. Hence it is a field by Lemma 10.35.19. Thus $(k' \otimes _ k K)_{red} \to K'$ maps it isomorphically onto $k'K$.
$\square$

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.
$\square$

Definition 10.44.5. Let $k$ be a field. The field extension $k \subset k'$ of Lemma 10.44.4 is called the *perfect closure* of $k$. Notation $k \subset k^{perf}$.

Note that if $k \subset k'$ is any algebraic purely inseparable extension, then $k' \subset k^{perf}$. Namely, $(k')^{perf}$ is isomorphic to $k^{perf}$ by the uniqueness of Lemma 10.44.4.

Lemma 10.44.6. Let $k$ be a perfect field. Any reduced $k$ algebra is geometrically reduced over $k$. Let $R$, $S$ be $k$-algebras. Assume both $R$ and $S$ are reduced. Then the $k$-algebra $R \otimes _ k S$ is reduced.

**Proof.**
The first statement follows from Lemma 10.43.3. For the second statement use the first statement and Lemma 10.42.5.
$\square$

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