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

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

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:

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

$\text{d}x_ i$ 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.
$\square$

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