## 15.46 Field extensions, revisited

In this section we study some peculiarities of field extensions in characteristic $p > 0$.

Definition 15.46.1. Let $p$ be a prime number. Let $k \to K$ be an extension of fields of characteristic $p$. Denote $kK^ p$ the compositum of $k$ and $K^ p$ in $K$.

1. A subset $\{ x_ i\} \subset K$ is called p-independent over $k$ if the elements $x^ E = \prod x_ i^{e_ i}$ where $0 \leq e_ i < p$ are linearly independent over $kK^ p$.

2. A subset $\{ x_ i\}$ of $K$ is called a p-basis of $K$ over $k$ if the elements $x^ E$ form a basis of $K$ over $kK^ p$.

This is related to the notion of a $p$-basis of a $\mathbf{F}_ p$-algebra which we will discuss later (insert future reference here).

Lemma 15.46.2. Let $k \subset K$ be a field extension. Assume $k$ has characteristic $p > 0$. Let $\{ x_ i\}$ be a subset of $K$. The following are equivalent

1. the elements $\{ x_ i\}$ are $p$-independent over $k$, and

2. the elements $\text{d}x_ i$ are $K$-linearly independent in $\Omega _{K/k}$.

Any $p$-independent collection can be extended to a $p$-basis of $K$ over $k$. In particular, the field $K$ has a $p$-basis over $k$. Moreover, the following are equivalent:

1. $\{ x_ i\}$ is a $p$-basis of $K$ over $k$, and

2. $\text{d}x_ i$ is a basis of the $K$-vector space $\Omega _{K/k}$.

Proof. Assume (2) and suppose that $\sum a_ E x^ E = 0$ is a linear relation with $a_ E \in k K^ p$. Let $\theta _ i : K \to K$ be a $k$-derivation such that $\theta _ i(x_ j) = \delta _{ij}$ (Kronecker delta). Note that any $k$-derivation of $K$ annihilates $kK^ p$. Applying $\theta _ i$ to the given relation we obtain new relations

$\sum \nolimits _{E, e_ i > 0} e_ i a_ E x_1^{e_1}\ldots x_ i^{e_ i - 1} \ldots x_ n^{e_ n} = 0$

Hence if we pick $\sum a_ E x^ E$ as the relation with minimal total degree $|E| = \sum e_ i$ for some $a_ E \not= 0$, then we get a contradiction. Hence (1) holds.

If $\{ x_ i\}$ is a $p$-basis for $K$ over $k$, then $K \cong kK^ p[X_ i]/(X_ i^ p - x_ i^ p)$. Hence we see that $\text{d}x_ i$ forms a basis for $\Omega _{K/k}$ over $K$. Thus (a) implies (b).

Let $\{ x_ i\}$ be a $p$-independent subset of $K$ over $k$. An application of Zorn's lemma shows that we can enlarge this to a maximal $p$-independent subset of $K$ over $k$. We claim that any maximal $p$-independent subset $\{ x_ i\}$ of $K$ is a $p$-basis of $K$ over $k$. The claim will imply that (1) implies (2) and establish the existence of $p$-bases. To prove the claim let $L$ be the subfield of $K$ generated by $kK^ p$ and the $x_ i$. We have to show that $L = K$. If $x \in K$ but $x \not\in L$, then $x^ p \in L$ and $L(x) \cong L[z]/(z^ p - x)$. Hence $\{ x_ i\} \cup \{ x\}$ is $p$-independent over $k$, a contradiction.

Finally, we have to show that (b) implies (a). By the equivalence of (1) and (2) we see that $\{ x_ i\}$ is a maximal $p$-independent subset of $K$ over $k$. Hence by the claim above it is a $p$-basis. $\square$

Lemma 15.46.3. Let $k \subset K$ be a field extension. Let $\{ K_\alpha \} _{\alpha \in A}$ be a collection of subfields of $K$ with the following properties

1. $k \subset K_\alpha$ for all $\alpha \in A$,

2. $k = \bigcap _{\alpha \in A} K_\alpha$,

3. for $\alpha , \alpha ' \in A$ there exists an $\alpha '' \in A$ such that $K_{\alpha ''} \subset K_\alpha \cap K_{\alpha '}$.

Then for $n \geq 1$ and $V \subset K^{\oplus n}$ a $K$-vector space we have $V \cap k^{\oplus n} \not= 0$ if and only if $V \cap K_\alpha ^{\oplus n} \not= 0$ for all $\alpha \in A$.

Proof. By induction on $n$. The case $n = 1$ follows from the assumptions. Assume the result proven for subspaces of $K^{\oplus n - 1}$. Assume that $V \subset K^{\oplus n}$ has nonzero intersection with $K_\alpha ^{\oplus n}$ for all $\alpha \in A$. If $V \cap 0 \oplus k^{\oplus n - 1}$ is nonzero then we win. Hence we may assume this is not the case. By induction hypothesis we can find an $\alpha$ such that $V \cap 0 \oplus K_\alpha ^{\oplus n - 1}$ is zero. Let $v = (x_1, \ldots , x_ n) \in V \cap K_\alpha$ be a nonzero element. By our choice of $\alpha$ we see that $x_1$ is not zero. Replace $v$ by $x_1^{-1}v$ so that $v = (1, x_2, \ldots , x_ n)$. Note that if $v' = (x_1', \ldots , x'_ n) \in V \cap K_\alpha$, then $v' - x_1'v = 0$ by our choice of $\alpha$. Hence we see that $V \cap K_\alpha ^{\oplus n} = K_\alpha v$. If we choose some $\alpha '$ such that $K_{\alpha '} \subset K_\alpha$, then we see that necessarily $v \in V \cap K_{\alpha '}^{\oplus n}$ (by the same arguments applied to $\alpha '$). Hence

$x_2, \ldots , x_ n \in \bigcap \nolimits _{\alpha ' \in A, K_{\alpha '} \subset K_\alpha } K_{\alpha '}$

which equals $k$ by (2) and (3). $\square$

Lemma 15.46.4. Let $K$ be a field of characteristic $p$. Let $\{ K_\alpha \} _{\alpha \in A}$ be a collection of subfields of $K$ with the following properties

1. $K^ p \subset K_\alpha$ for all $\alpha \in A$,

2. $K^ p = \bigcap _{\alpha \in A} K_\alpha$,

3. for $\alpha , \alpha ' \in A$ there exists an $\alpha '' \in A$ such that $K_{\alpha ''} \subset K_\alpha \cap K_{\alpha '}$.

Then

1. the intersection of the kernels of the maps $\Omega _{K/\mathbf{F}_ p} \to \Omega _{K/K_\alpha }$ is zero,

2. for any finite extension $K \subset L$ we have $L^ p = \bigcap _{\alpha \in A} L^ pK_\alpha$.

Proof. Proof of (1). Choose a $p$-basis $\{ x_ i\}$ for $K$ over $\mathbf{F}_ p$. Suppose that $\eta = \sum _{i \in I'} y_ i \text{d}x_ i$ maps to zero in $\Omega _{K/K_\alpha }$ for every $\alpha \in A$. Here the index set $I'$ is finite. By Lemma 15.46.2 this means that for every $\alpha$ there exists a relation

$\sum \nolimits _ E a_{E, \alpha } x^ E,\quad a_{E, \alpha } \in K_\alpha$

where $E$ runs over multi-indices $E = (e_ i)_{i \in I'}$ with $0 \leq e_ i < p$. On the other hand, Lemma 15.46.2 guarantees there is no such relation $\sum a_ E x^ E = 0$ with $a_ E \in K^ p$. This is a contradiction by Lemma 15.46.3.

Proof of (2). Suppose that we have a tower $K \subset M \subset L$ of finite extensions of fields. Set $M_\alpha = M^ p K_\alpha$ and $L_\alpha = L^ p K_\alpha = L^ p M_\alpha$. Then we can first prove that $M^ p = \bigcap _{\alpha \in A} M_\alpha$, and after that prove that $L^ p = \bigcap _{\alpha \in A} L_\alpha$. Hence it suffices to prove (2) for primitive field extensions having no nontrivial subfields. First, assume that $L = K(\theta )$ is separable over $K$. Then $L$ is generated by $\theta ^ p$ over $K$, hence we may assume that $\theta \in L^ p$. In this case we see that

$L^ p = K^ p \oplus K^ p\theta \oplus \ldots K^ p\theta ^{d - 1} \quad \text{and}\quad L^ pK_\alpha = K_\alpha \oplus K_\alpha \theta \oplus \ldots K_\alpha \theta ^{d - 1}$

where $d = [L : K]$. Thus the conclusion is clear in this case. The other case is where $L = K(\theta )$ with $\theta ^ p = t \in K$, $t \not\in K^ p$. In this case we have

$L^ p = K^ p \oplus K^ pt \oplus \ldots K^ pt^{p - 1} \quad \text{and}\quad L^ pK_\alpha = K_\alpha \oplus K_\alpha t \oplus \ldots K_\alpha t^{p - 1}$

Again the result is clear. $\square$

Lemma 15.46.5. Let $k$ be a field of characteristic $p > 0$. Let $n, m \geq 0$. Let $K$ be the fraction field of $k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$. As $k'$ ranges through all subfields $k/k'/k^ p$ with $[k : k'] < \infty$ the subfields

$\text{fraction field of } k'[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ m^ p] \subset K$

form a family of subfields as in Lemma 15.46.4. Moreover, each of the ring extensions $k'[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ m^ p] \subset k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$ is finite.

Proof. Write $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ m]$ and $A' = k'[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ m^ p]$. We also denote $K'$ the fraction field of $A'$. The ring extension $k'[[x_1^ p, \ldots , x_ d^ p]] \subset k[[x_1, \ldots , x_ d]]$ is finite by Algebra, Lemma 10.97.7 which implies that $A \to A'$ is finite. For $f \in A$ we see that $f^ p \in A'$. Hence $K^ p \subset K'$. Any element of $K'$ can be written as $a/b^ p$ with $a \in A'$ and $b \in A$ nonzero. Suppose that $f/g^ p \in K$, $f, g \in A$, $g \not= 0$ is contained in $K'$ for every choice of $k'$. Fix a choice of $k'$ for the moment. By the above we see $f/g^ p = a/b^ p$ for some $a \in A'$ and some nonzero $b \in A$. Hence $b^ p f \in A'$. For any $A'$-derivation $D : A \to A$ we see that $0 = D(b^ pf) = b^ p D(f)$ hence $D(f) = 0$ as $A$ is a domain. Taking $D = \partial _{x_ i}$ and $D = \partial _{y_ j}$ we conclude that $f \in k[[x_1^ p, \ldots , x_ n^ p]][y_1^ p, \ldots , y_ d^ p]$. Applying a $k'$-derivation $\theta : k \to k$ we similarly conclude that all coefficients of $f$ are in $k'$, i.e., $f \in A'$. Since it is clear that $A^ p = \bigcap \nolimits _{k'} A'$ where $k'$ ranges over all subfields as in the lemma we win. $\square$

Comment #6541 by Yijin Wang on

In the first paragraph of the first lemma,the last line,'hence(2)holds' should be 'hence (1) holds'.

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