In this section we continue the discussion started in Section 10.42.
Lemma 10.44.1. Let $k$ be a field of characteristic $p > 1$. Let $K/k$ be a field extension generated by $x_1, \ldots , x_{n + 1} \in K$ such that
$\{ x_1, \ldots , x_ n\} $ is a transcendence base of $K/k$,
for every $k$-linearly independent subset $\{ a_1, \ldots , a_ m\} $ of $K$ the set $\{ a^ p_1, \ldots , a_ m^ p\} $ is $k$-linearly independent.
Then there is $1 \leq j \leq n+1$ such that $\{ x_1, \ldots , \widehat{x}_ j, \ldots , x_{n+1}\} $ is a separating transcendence base for $K / k$.
Proof. By assumption $x_{n + 1}$ is algebraic over $k(x_1, \ldots , x_ n)$ so there exists a non-zero polynomial $F \in k[X_1, \ldots , X_{n + 1}]$ such that $F(x_1, \ldots , x_{n+1}) = 0$. Choose $F$ of minimal total degree. Then $F$ is irreducible, because at least one irreducible factor must also have the same property.
We claim that, for some $i$, not all powers of $X_ i$ appearing in $F$ are multiples of $p$. Suppose for a contradiction that all the exponents appearing in $F$ were multiples of $p$, then the set
is $k$-linearly dependent where $\lambda _\alpha $ are the coefficients of $F$. By assumption (2) we conclude the set
is also $k$-linearly dependent, contradicting minimality of $\deg (F)$.
Choose $i$ for which a non-$p$th power of $X_ i$ appears in $F$. Then we see that $x_ i$ is algebraic over $L = k(x_1, \ldots , x_{i - 1}, x_{i + 1}, \ldots , x_{n+1})$. By Fields, Lemma 9.26.3 we see that $x_1, \ldots , x_{i - 1}, x_{i + 1}, \ldots , x_{n+1}$ is a transcendence base of $K/k$. Thus $L$ is the fraction field of the polynomial ring over $k$ in $x_1, \ldots , x_{i - 1}, x_{i + 1}, \ldots , x_{n + 1}$. By Gauss' Lemma we conclude that
is irreducible. By construction $P(T)$ is not contained in $L[T^ p]$. Hence $K/L$ is separable as required. $\square$
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 10.44.2. Let $k$ be a field of characteristic $p > 0$. Let $K/k$ be a field extension. The following are equivalent:
$K$ is separable over $k$,
for every $k$-linearly independent subset $\{ a_1, \ldots , a_ m\} $ of $K$ the set $\{ a^ p_1, \ldots , a_ m^ p\} $ is $k$-linearly independent,
the ring $K \otimes _ k k^{1/p}$ is reduced, and
$K$ is geometrically reduced over $k$.
Proof. The implication (1) $\Rightarrow $ (4) follows from Lemma 10.43.6. The implication (4) $\Rightarrow $ (3) is immediate.
Assume (3). Consider the ring homomorphism $m : K \otimes _ k k^{1/p} \rightarrow K$ given by
Note that $x^ p = m(x) \otimes 1$ for all $x \in K \otimes _ k k^{1/p}$. Since $K \otimes _ k k^{1/p}$ is reduced we see $m$ is injective. If $\{ a_1, \ldots , a_ m\} \subset K$ is $k$-linearly independent, then $\{ a_1 \otimes 1, \ldots , a_ m \otimes 1\} $ is $k^{1/p}$-linearly independent. By injectivity of $m$ we deduce that no nontrivial $k$-linear combination of $a_1^ p, \ldots , a_ m^ p$ is is zero. Hence (3) implies (2).
Assume (2). To prove (1) we may assume that $K$ is finitely generated over $k$ and we have to prove that $K$ is separably generated over $k$. Let $\{ x_1, \ldots , x_ d\} $ be a transcendence base of $K/k$. By Fields, Lemma 9.8.6 we have $[K : K'] < \infty $ where $K' = k(x_1, \ldots , x_ d)$. Choose the transcendence base such that the degree of inseparability $[K : K']_ i$ is minimal. If $K / K'$ is separable then we win. Assume this is not the case to get a contradiction. Then there exists $x_{d + 1} \in K$ which is not separable over $K'$, and in particular $[K'(x_{d+1}) : K']_ i > 1$. Then by Lemma 10.44.1 there is $1 \leq j \leq n + 1$ such that $K'' = k(x_1, \ldots , \widehat{x}_ j, \ldots , x_{d+1})$ satisfies $[K'(x_{d+1}) : K'']_ i = 1$. By multiplicativity $[K : K'']_ i < [K : K']_ i$ and we obtain the contradiction. $\square$
Lemma 10.44.3. A separably generated field extension is separable.
In the following lemma we will use the notion of the perfect closure which is defined in Definition 10.45.5.
Lemma 10.44.4. Let $k$ be a field. Let $S$ be a $k$-algebra. The following are equivalent:
$k' \otimes _ k S$ is reduced for every finite purely inseparable extension $k'$ of $k$,
$k^{1/p} \otimes _ k S$ is reduced,
$k^{perf} \otimes _ k S$ is reduced, where $k^{perf}$ is the perfect closure of $k$,
$\overline{k} \otimes _ k S$ is reduced, where $\overline{k}$ is the algebraic closure of $k$, and
$S$ is geometrically reduced over $k$.
Proof. Note that any finite purely inseparable extension $k'/k$ 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 $K/k$ be an extension of fields. We have to show that $K \otimes _ k S$ is reduced. By Lemma 10.43.4 we reduce to the case where $K/k$ is a finitely generated field extension. Choose a diagram
as in Lemma 10.42.4. By assumption $k' \otimes _ k S$ is reduced. By Lemma 10.43.6 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 10.25.1, hence $S_{\mathfrak p}$ is geometrically reduced by Lemma 10.44.2. It follows from Lemma 10.43.7 that $S$ is geometrically reduced. $\square$
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