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 $L[X]$?


1 Answer 1


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 $F$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – spin
    Jul 29, 2014 at 16:37
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 29, 2014 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.