Algebraic solutions of a set of simultaneous polynomial equations, with nonzero Jacobian.

Question

Let $f_1, \ldots, f_n \in \mathbb{Q}[x_1, \ldots, x_n]$ and let $a \in \mathbb{C}^n$ be a point such that $f_1(a) = \cdots = f_n(a) = 0$, with Jacobian determinant $J(f_1, \ldots, f_n)(a) \neq 0$. Does this imply that every coordinate $a_i$ of $a = (a_1, \ldots a_n)$ is algebraic over $\mathbb{Q}$?

Answer 1

We work over $K=\overline{\mathbb Q}$, the algebraic closure of $\mathbb Q$. The kernel of the evaluation map at $(a_1,\dots,a_n)$, say $P$, is a prime ideal of $K[x_1,\dots,x_n]$. We have $f_i\in P$ and $J(f_1,\dots,f_n)\notin P$. Then, by Hilbert's Nullstellensatz, there is a maximal ideal $M=(x_1-b_1,\dots,x_n-b_n)$, $b_i\in K$, such that $P\subseteq M$ and $J(f_1,\dots,f_n)(b)\ne0$. Now use Taylor expansion and get $P+M^2=M$. By Nakayama $P=M$, so $a_i=b_i$ for all $i$.