
## 10.43 Separable extensions, continued

In this section we continue the discussion started in Section 10.41. Let $p$ be a prime number and let $k$ be a field of characteristic $p$. In this case we write $k^{1/p}$ for the extension of $k$ gotten by adjoining $p$th roots of all the elements of $k$ to $k$. (In other words it is the subfield of an algebraic closure of $k$ generated by the $p$th roots of elements of $k$.)

Lemma 10.43.1. Let $k$ be a field of characteristic $p > 0$. Let $k \subset K$ be a field extension. The following are equivalent:

1. $K$ is separable over $k$,

2. the ring $K \otimes _ k k^{1/p}$ is reduced, and

3. $K$ is geometrically reduced over $k$.

Proof. The implication (1) $\Rightarrow$ (3) follows from Lemma 10.42.6. The implication (3) $\Rightarrow$ (2) is immediate.

Assume (2). Let $k \subset L \subset K$ be a subextension such that $L$ is a finitely generated field extension of $k$. We have to show that we can find a separating transcendence basis of $L$. The assumption implies that $L \otimes _ k k^{1/p}$ is reduced. Let $x_1, \ldots , x_ r$ be a transcendence basis of $L$ over $k$ such that the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$ is minimal. If $L$ is separable over $k(x_1, \ldots , x_ r)$ then we win. Assume this is not the case to get a contradiction. Then there exists an element $\alpha \in L$ which is not separable over $k(x_1, \ldots , x_ r)$. Let $P(T) \in k(x_1, \ldots , x_ r)[T]$ be the minimal polynomial of $\alpha$ over $k(x_1, \ldots , x_ r)$. After replacing $\alpha$ by $f \alpha$ for some nonzero $f \in k[x_1, \ldots , x_ r]$ we may and do assume that $P$ lies in $k[x_1, \ldots , x_ r, T]$. Because $\alpha$ is not separable $P$ is a polynomial in $T^ p$, see Fields, Lemma 9.12.1. Let $dp$ be the degree of $P$ as a polynomial in $T$. Since $P$ is the minimal polynomial of $\alpha$ the monomials

$x_1^{e_1} \ldots x_ r^{e_ r} \alpha ^ e$

for $e < dp$ are linearly independent over $k$ in $L$. We claim that the element $\partial P/\partial x_ i \in k[x_1, \ldots , x_ r, T]$ is not zero for at least one $i$. Namely, if this was not the case, then $P$ is actually a polynomial in $x_1^ p, \ldots , x_ r^ p, T^ p$. In that case we can consider $P^{1/p} \in k^{1/p}[x_1, \ldots , x_ r, T]$. This would map to $P^{1/p}(x_1, \ldots , x_ r, \alpha )$ which is a nilpotent element of $k^{1/p} \otimes _ k L$ and hence zero. On the other hand, $P^{1/p}(x_1, \ldots , x_ r, \alpha )$ is a $k^{1/p}$-linear combination the monomials listed above, hence nonzero in $k^{1/p} \otimes _ k L$. This is a contradiction which proves our claim.

Thus, after renumbering, we may assume that $\partial P/\partial x_1$ is not zero. As $P$ is an irreducible polynomial in $T$ over $k(x_1, \ldots , x_ r)$ it is irreducible as a polynomial in $x_1, \ldots , x_ r, T$, hence by Gauss's lemma it is irreducible as a polynomial in $x_1$ over $k(x_2, \ldots , x_ r, T)$. Since the transcendence degree of $L$ is $r$ we see that $x_2, \ldots , x_ r, \alpha$ are algebraically independent. Hence $P(X, x_2, \ldots , x_ r, \alpha ) \in k(x_2, \ldots , x_ r, \alpha )[X]$ is irreducible. It follows that $x_1$ is separably algebraic over $k(x_2, \ldots , x_ r, \alpha )$. This means that the degree of inseparability of the finite extension $k(x_2, \ldots , x_ r, \alpha ) \subset L$ is less than the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$, which is a contradiction. $\square$

In the following lemma we will use the notion of the perfect closure which is defined in Definition 10.44.5.

Lemma 10.43.3. Let $k$ be a field. Let $S$ be a $k$-algebra. The following are equivalent:

1. $k' \otimes _ k S$ is reduced for every finite purely inseparable extension $k'$ of $k$,

2. $k^{1/p} \otimes _ k S$ is reduced,

3. $k^{perf} \otimes _ k S$ is reduced, where $k^{perf}$ is the perfect closure of $k$,

4. $\overline{k} \otimes _ k S$ is reduced, where $\overline{k}$ is the algebraic closure of $k$, and

5. $S$ is geometrically reduced over $k$.

Proof. Note that any finite purely inseparable extension $k \subset k'$ embeds in $k^{perf}$. Moreover, $k^{1/p}$ embeds into $k^{perf}$ which embeds into $\overline{k}$. Thus it is clear that (5) $\Rightarrow$ (4) $\Rightarrow$ (3) $\Rightarrow$ (2) and that (3) $\Rightarrow$ (1).

We prove that (1) $\Rightarrow$ (5). Assume $k' \otimes _ k S$ is reduced for every finite purely inseparable extension $k'$ of $k$. Let $k \subset K$ be an extension of fields. We have to show that $K \otimes _ k S$ is reduced. By Lemma 10.42.4 we reduce to the case where $k \subset K$ is a finitely generated field extension. Choose a diagram

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

as in Lemma 10.41.4. By assumption $k' \otimes _ k S$ is reduced. By Lemma 10.42.6 it follows that $K' \otimes _ k S$ is reduced. Hence we conclude that $K \otimes _ k S$ is reduced as desired.

Finally we prove that (2) $\Rightarrow$ (5). Assume $k^{1/p} \otimes _ k S$ is reduced. Then $S$ is reduced. Moreover, for each localization $S_{\mathfrak p}$ at a minimal prime $\mathfrak p$, the ring $k^{1/p}\otimes _ k S_{\mathfrak p}$ is a localization of $k^{1/p} \otimes _ k S$ hence is reduced. But $S_{\mathfrak p}$ is a field by Lemma 10.24.1, hence $S_{\mathfrak p}$ is geometrically reduced by Lemma 10.43.1. It follows from Lemma 10.42.7 that $S$ is geometrically reduced. $\square$

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