## 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 $k \subset K$ be a field extension.

1. 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 $k(x_ i; i\in I) \subset K$ is a separable algebraic extension.

2. We say $K$ is separable over $k$ if for every subextension $k \subset K' \subset K$ with $K'$ finitely generated over $k$, the extension $k \subset K'$ 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 10.44.2.

Lemma 10.42.2. Let $k \subset K$ be a separable field extension. For any subextension $k \subset K' \subset K$ the field extension $k \subset K'$ is separable.

Proof. This is direct from the definition. $\square$

Lemma 10.42.3. Let $k \subset K$ 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

1. $x_1, \ldots , x_ r$ is a transcendence basis of $K$ over $k$,

2. $K = k(x_1, \ldots , x_{r + 1})$, and

3. $x_{r + 1}$ 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 $k \subset K$ 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 $P = T^ n + a_1T^{n - 1} + \ldots + a_ n$ 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. $\square$

Comment #406 by Keenan on

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 on

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

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).

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