In this section we continue the discussion started in Section 10.42.
Lemma 10.44.1. Let k be a field of characteristic p > 1. Let K/k be a field extension generated by x_1, \ldots , x_{n + 1} \in K such that
\{ x_1, \ldots , x_ n\} is a transcendence base of K/k,
for every k-linearly independent subset \{ a_1, \ldots , a_ m\} of K the set \{ a^ p_1, \ldots , a_ m^ p\} is k-linearly independent.
Then there is 1 \leq j \leq n+1 such that \{ x_1, \ldots , \widehat{x}_ j, \ldots , x_{n+1}\} is a separating transcendence base for K / k.
Proof. By assumption x_{n + 1} is algebraic over k(x_1, \ldots , x_ n) so there exists a non-zero polynomial F \in k[X_1, \ldots , X_{n + 1}] such that F(x_1, \ldots , x_{n+1}) = 0. Choose F of minimal total degree. Then F is irreducible, because at least one irreducible factor must also have the same property.
We claim that, for some i, not all powers of X_ i appearing in F are multiples of p. Suppose for a contradiction that all the exponents appearing in F were multiples of p, then the set
is k-linearly dependent where \lambda _\alpha are the coefficients of F. By assumption (2) we conclude the set
is also k-linearly dependent, contradicting minimality of \deg (F).
Choose i for which a non-pth power of X_ i appears in F. Then we see that x_ i is algebraic over L = k(x_1, \ldots , x_{i - 1}, x_{i + 1}, \ldots , x_{n+1}). By Fields, Lemma 9.26.3 we see that x_1, \ldots , x_{i - 1}, x_{i + 1}, \ldots , x_{n+1} is a transcendence base of K/k. Thus L is the fraction field of the polynomial ring over k in x_1, \ldots , x_{i - 1}, x_{i + 1}, \ldots , x_{n + 1}. By Gauss' Lemma we conclude that
is irreducible. By construction P(T) is not contained in L[T^ p]. Hence K/L is separable as required. \square
Let p be a prime number and let k be a field of characteristic p. In this case we write k^{1/p} for the extension of k gotten by adjoining pth roots of all the elements of k to k. (In other words it is the subfield of an algebraic closure of k generated by the pth roots of elements of k.)
Lemma 10.44.2. Let k be a field of characteristic p > 0. Let K/k be a field extension. The following are equivalent:
K is separable over k,
for every k-linearly independent subset \{ a_1, \ldots , a_ m\} of K the set \{ a^ p_1, \ldots , a_ m^ p\} is k-linearly independent,
the ring K \otimes _ k k^{1/p} is reduced, and
K is geometrically reduced over k.
Proof. The implication (1) \Rightarrow (4) follows from Lemma 10.43.6. The implication (4) \Rightarrow (3) is immediate.
Assume (3). Consider the ring homomorphism m : K \otimes _ k k^{1/p} \rightarrow K given by
Note that x^ p = m(x) \otimes 1 for all x \in K \otimes _ k k^{1/p}. Since K \otimes _ k k^{1/p} is reduced we see m is injective. If \{ a_1, \ldots , a_ m\} \subset K is k-linearly independent, then \{ a_1 \otimes 1, \ldots , a_ m \otimes 1\} is k^{1/p}-linearly independent. By injectivity of m we deduce that no nontrivial k-linear combination of a_1^ p, \ldots , a_ m^ p is is zero. Hence (3) implies (2).
Assume (2). To prove (1) we may assume that K is finitely generated over k and we have to prove that K is separably generated over k. Let \{ x_1, \ldots , x_ d\} be a transcendence base of K/k. By Fields, Lemma 9.8.6 we have [K : K'] < \infty where K' = k(x_1, \ldots , x_ d). Choose the transcendence base such that the degree of inseparability [K : K']_ i is minimal. If K / K' is separable then we win. Assume this is not the case to get a contradiction. Then there exists x_{d + 1} \in K which is not separable over K', and in particular [K'(x_{d+1}) : K']_ i > 1. Then by Lemma 10.44.1 there is 1 \leq j \leq n + 1 such that K'' = k(x_1, \ldots , \widehat{x}_ j, \ldots , x_{d+1}) satisfies [K'(x_{d+1}) : K'']_ i = 1. By multiplicativity [K : K'']_ i < [K : K']_ i and we obtain the contradiction. \square
Lemma 10.44.3. A separably generated field extension is separable.
In the following lemma we will use the notion of the perfect closure which is defined in Definition 10.45.5.
Lemma 10.44.4. Let k be a field. Let S be a k-algebra. The following are equivalent:
k' \otimes _ k S is reduced for every finite purely inseparable extension k' of k,
k^{1/p} \otimes _ k S is reduced,
k^{perf} \otimes _ k S is reduced, where k^{perf} is the perfect closure of k,
\overline{k} \otimes _ k S is reduced, where \overline{k} is the algebraic closure of k, and
S is geometrically reduced over k.
Proof. Note that any finite purely inseparable extension k'/k embeds in k^{perf}. Moreover, k^{1/p} embeds into k^{perf} which embeds into \overline{k}. Thus it is clear that (5) \Rightarrow (4) \Rightarrow (3) \Rightarrow (2) and that (3) \Rightarrow (1).
We prove that (1) \Rightarrow (5). Assume k' \otimes _ k S is reduced for every finite purely inseparable extension k' of k. Let K/k be an extension of fields. We have to show that K \otimes _ k S is reduced. By Lemma 10.43.4 we reduce to the case where K/k is a finitely generated field extension. Choose a diagram
as in Lemma 10.42.4. By assumption k' \otimes _ k S is reduced. By Lemma 10.43.6 it follows that K' \otimes _ k S is reduced. Hence we conclude that K \otimes _ k S is reduced as desired.
Finally we prove that (2) \Rightarrow (5). Assume k^{1/p} \otimes _ k S is reduced. Then S is reduced. Moreover, for each localization S_{\mathfrak p} at a minimal prime \mathfrak p, the ring k^{1/p}\otimes _ k S_{\mathfrak p} is a localization of k^{1/p} \otimes _ k S hence is reduced. But S_{\mathfrak p} is a field by Lemma 10.25.1, hence S_{\mathfrak p} is geometrically reduced by Lemma 10.44.2. It follows from Lemma 10.43.7 that S is geometrically reduced. \square
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