In this section we talk about separability for nonalgebraic field extensions. This is closely related to the concept of geometrically reduced algebras, see Definition 10.42.1.

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 10.43.2.

Lemma 10.41.4. Let $k \subset K$ be a finitely generated field extension. There exists a diagram

\[ \xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] } \]

where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ 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 $k(x_1, \ldots , x_ r) \subset K_{sep} \subset K$ 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 $P = T^ d + a_1T^{d - 1} + \ldots + a_ d$ be the minimal polynomial of $\alpha $ over $k(x_1, \ldots , x_ r)$. Let $k \subset k'$ 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

\[ \xymatrix{ K \ar[r] & L \\ k(x_1, \ldots , x_ r) \ar[u] \ar[r] & k'(x_1^{1/p}, \ldots , x_ r^{1/p}) \ar[u] } \]

We may and do assume $L$ is the compositum of $K$ and $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. Let $k'(x_1^{1/p}, \ldots , x_ r^{1/p}) \subset L_{sep} \subset L$ 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

\[ \xymatrix{ K \ar[r] & L \\ K_{sep}(\beta ) \ar[r] \ar[u] & L_{sep} \ar[u] \\ K_{sep} \ar[r] \ar[u] & L_{sep} \ar@{=}[u] \\ k(x_1, \ldots , x_ r) \ar[u] \ar[r] & k'(x_1^{1/p}, \ldots , x_ r^{1/p}) \ar[u] \\ k \ar[r] \ar[u] & k' \ar[u] } \]

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 $k' \subset L$. Then the extensions $k \subset k''$ and $K \subset L'$ work for the extension $k \subset K$. This proves the lemma.
$\square$

## Comments (2)

Comment #406 by Keenan on

Comment #408 by Johan on