MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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