Lemma 10.158.3. Let $k$ be a field of characteristic $p > 0$. Let $a_1, \ldots , a_ n \in k$ be elements such that $\text{d}a_1, \ldots , \text{d}a_ n$ are linearly independent in $\Omega _{k/\mathbf{F}_ p}$. Then the field extension $k(a_1^{1/p}, \ldots , a_ n^{1/p})$ has degree $p^ n$ over $k$.
Proof. By induction on $n$. If $n = 1$ the result is Lemma 10.158.2. For the induction step, suppose that $k(a_1^{1/p}, \ldots , a_{n - 1}^{1/p})$ has degree $p^{n - 1}$ over $k$. We have to show that $a_ n$ does not map to a $p$th power in $k(a_1^{1/p}, \ldots , a_{n - 1}^{1/p})$. If it does then we can write
\begin{align*} a_ n & = \left(\sum \nolimits _{I = (i_1, \ldots , i_{n - 1}),\ 0 \leq i_ j \leq p - 1} \lambda _ I a_1^{i_1/p} \ldots a_{n - 1}^{i_{n - 1}/p}\right)^ p \\ & = \sum \nolimits _{I = (i_1, \ldots , i_{n - 1}),\ 0 \leq i_ j \leq p - 1} \lambda _ I^ p a_1^{i_1} \ldots a_{n - 1}^{i_{n - 1}} \end{align*}
Applying $\text{d}$ we see that $\text{d}a_ n$ is linearly dependent on $\text{d}a_ i$, $i < n$. This is a contradiction. $\square$
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