Question

I'm getting started in Real Algebraic Geometry (from a model-theory perspective), and a paper makes the following assertion (here $K\subset L$ are real-closed fields):

Suppose $Q\in L^n$, $f_1, \ldots, f_n\in K[x_1, \ldots, x_n]$, $f_1(Q)=\cdots=f_n(Q)=0$, and $\dfrac{\partial f_1, \ldots, f_n}{\partial x_1, \ldots, x_n}(Q)\not=0$. Then every coordinate of $Q$ is algebraic over $K$.

By real-closure of $K$ or model-completeness or whatever you like, this implies $Q\in K$. But I don't see how to prove it. The author declares it "elementary algebra," but I guess I don't see it.

I know it's trivial in the $n=1$ case, and I know the determinant condition is necessary (otherwise you can just under-specify the coordinates, and they'll be algebraic over each other but not over $K$, e.g. $n=2$ and $f_1=f_n=x_1-x_2$). But that's all I've got.

Answer 1

If $Q$ is transcendental, then viewing a transcendental coordinate as a parameter (i.e. view $Q$ as a point defined in an extension of $K(t)$, you conclude that the variety $X:f_1=\ldots=f_n=0$ which is defined over (the algebraic closure of) $K$ contains a curve, this contradicts the Jacobian determinant condition which implies that $X$ is zero-dimensional.