**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 $L/M/K$ 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.
$\square$

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