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 $L[X]$?

Without loss of generality, assume $F$ is irreducible and of minimal multi-degree so that $F(a_1,\dots,a_{d+1})=0$. By multi-degree, I mean the maximal sum of exponents of a monomial appearing in $F$ with nonzero coefficient. Write
$$F = \sum_{j=0}^h f_j X_i^j,$$ for $f_j \in k[X_1,\dots,X_{i-1},X_{i+1},\dots,X_{d+1}]$. Note that
$$\frac{\partial F}{\partial X_i} = \sum_{j=1}^h j \cdot f_j X_i^{j-1}. \,\, (*)$$
If $F(a_1,\dots,a_{i-1},X,a_{i+1},\dots,a_{d+1})\equiv 0$ in $L[X]$, then all of the coefficient polynomials $f_j$ vanish at $(a_1,\dots,a_{i-1},a_{i+1},\dots,a_{d+1})$. Since $\frac{\partial F}{\partial X_i} \neq 0$, this contradicts our minimality assumption on the degree of $F$.
• Thanks, this seems to work. It confuses me that no explanation is given in the proof in Shafarevich. Just adding to the proof that "choose $F$ of minimal degree" would probably have been enough for me to figure this out.
• @spin It is strange. I was trying to see if something in the proof more directly implied $F(a_1,\dots,a_{i-1},X,a_{i+1},\dots,a_{d+1}) \neq 0$, but that doesn't seem to be the case. Jul 29, 2014 at 17:11