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

1. $\{ x_1, \ldots , x_ n\}$ is a transcendence base of $K/k$,

2. 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

$\{ x_1^{\alpha _1} \ldots x^{\alpha _{n+1}}_{n+1} \mid \lambda _\alpha \neq 0\} \subset K$

is $k$-linearly dependent where $\lambda _\alpha$ are the coefficients of $F$. By assumption (2) we conclude the set

$\{ x_1^{\alpha _1 / p} \ldots x^{\alpha _{n+1} / p}_{n+1} \mid \lambda _\alpha \neq 0 \}$

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

$P(T) = F(x_1, \ldots , x_{i - 1}, T, x_{i + 1}, \ldots , x_{n + 1}) \in L[T]$

is irreducible. By construction $P(T)$ is not contained in $L[T^ p]$. Hence $K/L$ is separable as required. $\square$

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