Nonsingular zeroes are algebraic?

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.

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This is a fact of algebraic geometry for any field $K$: for $J = (f_1,\dots,f_n) \subset K[x] := K[x_1,\dots,x_n]$ and $d = \det(\partial f_i/\partial x_j) \in K[x]$, if we let $A= (K[x]/J)[1/d]$ (so $K$-algebra maps from $A$ into an extension field $L/K$ correspond to $Q$ you care about) then $A$ is finite-dimensional as a $K$-vector space (and even is an etale $K$-algebra: a finite product of finite separable extensions of $K$). Intuitively, $f:\{d\ne 0\}\rightarrow \mathbf{A}^n$ with components $f_i$ satisfies an "algebraic inverse function theorem", so it has finite geometric fibers. –  grp Oct 5 '12 at 15:28
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.