If $x_1,\dots,x_n$ purely inseparable and linearly independent over $K$, then they are linearly independent over a separable extension $E$ of $K$.

Question

Let $K$ be a field and $L$ be an algebraic closure of $K$. Let $E,F$ be subfields of $L$ containing $F$ such that $E/K$ is separable and $F/K$ is purely inseparable. Let $x_1,\cdots,x_n$ be some elements in $F$ such that they are $K$-linearly independent. Prove that $x_1,\cdots,x_n$ are $E$-linearly independent.

Answer 1

Suppose you have $\sum_{i\in I}\lambda_i x_i = 0$ with $\lambda_i\in E$. Then we can assume that the $\lambda_i$ all belong to some finite subextention $E'\subset E$. Since $E'/K$ is separable, $E'= K(\theta)$, and we write $\mu_\theta \in K[X]$ for the minimal polynomial of $\theta$, of degree $d = [E':K]$.

Then if $\lambda_i = \sum_{k=0}^{d-1}a_{i,k}\theta^k$ with $a_{i,k}\in K$, we get $\sum_{k=0}^{d-1}(\sum_{i\in I}a_{i,k}x_i)\theta^k = 0$, and we want to conclude that all $a_{i,k}$ are zero.

Let $y_k = \sum_{i\in I}a_{i,k}x_i$, and $P = \sum_{k=0}^{d-1}y_k X^k\in F[X]$ ; then $P(\theta) = 0$. Let's show that $\mu_\theta$ is the minimal polynomial of $\theta$ over $F$ ; this is enough to conclude since then $P=0$ because $\deg P< \deg \mu_\theta$, so $y_k=0$ for all $k$, and then $a_{i,k}=0$ because of the independence of the $x_i$ over $K$.

So let $Q\in F[X]$ be the minimal polynomial of $\theta$ over $F$. Then since $F/K$ is purely inseparable, there is $r\in \mathbb{N}$ such that $Q^{p^r}\in K[X]$ ; thus $\mu_\theta | Q^{p^r}$. But since $Q$ is irreducible in $F[X]$, this implies that $\mu_\theta = Q^m$ for some $m\in \mathbb{N}$. Now $\mu_\theta$ is separable, so $m=1$, and $Q=\mu_\theta$ as expected.