Definition 10.45.1. Let k be a field. We say k is perfect if every field extension of k is separable over k.
10.45 Perfect fields
Here is the definition.
Lemma 10.45.2. 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 pth 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 pth root of an element of k have to be trivial, hence every element of k has a pth root. Conversely if every element has a pth root, then k = k^{1/p} and every field extension of k is separable by Lemma 10.44.2. \square
Lemma 10.45.3. Let K/k be a finitely generated field extension. There exists a diagram
\xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] }
where k'/k, K'/K are finite purely inseparable field extensions such that K'/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 10.42.4 we can find such a diagram with K'/k' separably generated. By Lemma 10.44.3 this implies that K' is separable over k'. The compositum k'K is a subextension of K'/k' and hence k' \subset k'K is separable by Lemma 10.42.2. 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 10.36.19. Thus (k' \otimes _ k K)_{red} \to K' maps it isomorphically onto k'K. \square
Lemma 10.45.4.slogan For every field k there exists a purely inseparable extension k'/k such that k' is perfect. The field extension k'/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^ nth 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 10.45.5. Let k be a field. The field extension k'/k of Lemma 10.45.4 is called the perfect closure of k. Notation k^{perf}/k.
Note that if k'/k is any algebraic purely inseparable extension, then k' is a subextension of k^{perf}, i.e., k^{perf}/k'/k. Namely, (k')^{perf} is isomorphic to k^{perf} by the uniqueness of Lemma 10.45.4.
Lemma 10.45.6. 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 10.44.4. For the second statement use the first statement and Lemma 10.43.5. \square
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