
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 (1)

Comment #738 by Keenan Kidwell on

The $L^{\prime\prime}$ in the last line should be $L^\prime$.

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• 2 comment(s) on Section 10.41: Separable extensions

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