abstract algebra - Proving that a field $K$ can be generated by algebraically independent elements and an separable element

Let $k$ be a perfect field (either $k$ has characteristic $0$, or characteristic $p > 0$ and every element has a $p$th root), and let $K$ be a finitely generated extension field.

I have a question about a step in the proof of the following statement. This can be found in the Vol. 1 of Shafarevich (Appendix 5) or "Introduction to Algebraic Geometry and Algebraic Groups" by Geck (exercise 1.8.15).

Let $d$ be the transcendence degree of $K/k$. The claim is that then there exist elements $z_1, \ldots, z_{d+1}$ such that:

1. $K = k(z_1, \ldots, z_{d+1})$.


2. $z_1, \ldots, z_d$ are algebraically independent.



3. $z_{d+1}$ is separable algebraic over $k(z_1, \ldots, z_d)$.

The proof proceeds as follows. Now there exist $a_i$ such that $K = k(a_1, \ldots, a_n)$, with $d \leq n$, and $a_1, \ldots, a_d$ are algebraically independent. The case $n = d$ is easy, so assume $n > d$ and proceed by induction on $n$.

First: $\{a_1, \ldots, a_{d+1}\}$ is not algebraically independent, so there exists a nonzero, nonconstant irreducible polynomial $F \in k[X_1, \ldots, X_{d+1}]$ such that $F(a_1, \ldots, a_{d+1}) = 0$.

Since $k$ is perfect it follows that for some $i$ the partial derivative of $F$ with respect to $X_i$ is nonzero.

So far this makes sense. The next claim is that $a_i$ is separable algebraic over $L = k(a_1, \ldots, a_{i-1}, a_{i+1}, \ldots, a_{d+1})$, and this is the only thing in the proof I have a problem with. Why is this true? In the proof they claim we can use $F$ because $X_i$ appears in it, but I fail to see how this follows. Couldn't $F(a_1, \ldots, a_{i-1}, X, a_{i+1}, \ldots, a_{d+1})$ be the zero polynomial in $L[X]$?