In this section we talk about separability for nonalgebraic field extensions. This is closely related to the concept of geometrically reduced algebras, see Definition 10.43.1.
Definition 10.42.1. Let $K/k$ be a field extension.
We say $K$ is separably generated over $k$ if there exists a transcendence basis $\{ x_ i; i \in I\} $ of $K/k$ such that the extension $K/k(x_ i; i \in I)$ is a separable algebraic extension.
We say $K$ is separable over $k$ if for every subextension $k \subset K' \subset K$ with $K'$ finitely generated over $k$, the extension $K'/k$ is separably generated.
With this awkward definition it is not clear that a separably generated field extension is itself separable. It will turn out that this is the case, see Lemma 10.44.3.
Lemma 10.42.2. Let $K/k$ be a separable field extension. For any subextension $K/K'/k$ the field extension $K'/k$ is separable.
Proof. This is direct from the definition. $\square$
Lemma 10.42.3. Let $K/k$ be a separably generated, and finitely generated field extension. Set $r = \text{trdeg}_ k(K)$. Then there exist elements $x_1, \ldots , x_{r + 1}$ of $K$ such that
$x_1, \ldots , x_ r$ is a transcendence basis of $K$ over $k$,
$K = k(x_1, \ldots , x_{r + 1})$, and
$x_{r + 1}$ is separable over $k(x_1, \ldots , x_ r)$.
Proof. Combine the definition with Fields, Lemma 9.19.1. $\square$
Lemma 10.42.4. Let $K/k$ be a finitely generated field extension. There exists a diagram
where $k'/k$, $K'/K$ are finite purely inseparable field extensions such that $K'/k'$ is a separably generated field extension.
Proof. This lemma is only interesting when the characteristic of $k$ is $p > 0$. Choose $x_1, \ldots , x_ r$ a transcendence basis of $K$ over $k$. As $K$ is finitely generated over $k$ the extension $k(x_1, \ldots , x_ r) \subset K$ is finite. Let $K/K_{sep}/k(x_1, \ldots , x_ r)$ be the subextension found in Fields, Lemma 9.14.6. If $K = K_{sep}$ then we are done. We will use induction on $d = [K : K_{sep}]$.
Assume that $d > 1$. Choose a $\beta \in K$ with $\alpha = \beta ^ p \in K_{sep}$ and $\beta \not\in K_{sep}$. Let $P = T^ n + a_1T^{n - 1} + \ldots + a_ n$ be the minimal polynomial of $\alpha $ over $k(x_1, \ldots , x_ r)$. Let $k'/k$ be a finite purely inseparable extension obtained by adjoining $p$th roots such that each $a_ i$ is a $p$th power in $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. Such an extension exists; details omitted. Let $L$ be a field fitting into the diagram
We may and do assume $L$ is the compositum of $K$ and $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. Let $L/L_{sep}/k'(x_1^{1/p}, \ldots , x_ r^{1/p})$ be the subextension found in Fields, Lemma 9.14.6. Then $L_{sep}$ is the compositum of $K_{sep}$ and $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. The element $\alpha \in L_{sep}$ is a zero of the polynomial $P$ all of whose coefficients are $p$th powers in $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$ and whose roots are pairwise distinct. By Fields, Lemma 9.28.2 we see that $\alpha = (\alpha ')^ p$ for some $\alpha ' \in L_{sep}$. Clearly, this means that $\beta $ maps to $\alpha ' \in L_{sep}$. In other words, we get the tower of fields
Thus this construction leads to a new situation with $[L : L_{sep}] < [K : K_{sep}]$. By induction we can find $k' \subset k''$ and $L \subset L'$ as in the lemma for the extension $L/k'$. Then the extensions $k''/k$ and $L'/K$ work for the extension $K/k$. This proves the lemma. $\square$
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