Section 10.42 (030I): Separable extensions

10.42 Separable extensions

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

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

where $k'/k$ , $K'/K$ are finite purely inseparable field extensions such that $K'/k'$ is a 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

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

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

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

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

Comments (4)

Comment #406 by Keenan on December 26, 2013 at 20:45

I could be mistaken, but it seems the application of 031V in the proof of 04KM is not completely justified. The minimal polynomial of $\alpha$ over $k(x_1,\ldots,x_r)$ , $P$ , has all its coefficients $p$ -th powers in $k^\prime(x_1^{1/p},\ldots,x_r^{1/p})$ . But 031V involves the minimal polynomial of $\alpha$ over $k^\prime(x_1^{1/p},\ldots,x_r^{1/p})$ , whose relationship with $P$ is not completely clear to me (other than the obvious divisibility). With that said, it seems to me that the proof of 031V shows that irreducibility of $P$ over the relevant field ( $k^\prime(x_1^{1/p},\ldots,x_r^{1/p})$ in this case) doesn't actually matter; all that is needed is the separability. In fact, the divisibility that is deduced in the proof of 031V takes place in the extension $L$ , where the polynomial $Q$ of that proof is not necessarily irreducible (all that one needs is the irreducibility of $T^p-\alpha$ over $L$ there).

Comment #408 by Johan on December 27, 2013 at 16:02

Thanks a bunch for finding this mistake as well as pointing out how to fix it! The corresponding edits are here.

Comment #5878 by Paolo on January 11, 2021 at 08:55

I think that in Lemma 04KM it would be better to use two different letters for $d=[K:K_{sep}]$ and $d=\mathrm{deg}(P)$ in order to avoid confusion (they are different things).

Comment #6088 by Johan on April 29, 2021 at 18:24

Yep! Thanks very much and fixed here.

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10.42 Separable extensions

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