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.

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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (1)

Comment #738 by Keenan Kidwell on

There are also: