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I have been told that:

Let $f_1,\dots,f_n,g\in F[x_1,\dots,x_m]$ be polynomials in $m$ variables with coefficients in the algebraically closed field $F$. Then if the system: $$\left\{\begin{array}{@{}l@{}} f_1(u)=0 \\ f_2(u)=0 \\ \vdots \\ f_n(u)=0 \\ g(u)\neq0 \end{array}\right.$$ has a solution in some extension $\overline{F}\geq F$, it has solution in $F$ as well.

I haven't been given any proof of this however, and I really don't know how to proceed, also because I've been told it is «a consequence of a classical result of unknown quantity elimination», which hasn't been stated to me. So how do I prove this?

MickG
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    Looks like this follows from applying Hilbert's Nullstellensatz to the ideal $\langle f_1,f_2,\ldots,f_n,1-x_{m+1}g\rangle\subseteq F[x_1,x_2,\ldots,x_m,x_{m+1}]$. – Jyrki Lahtonen Jul 21 '14 at 16:08
  • Uh… This is an ingredient for a proof of the Nullstellensatz so… vicious circle detected :). – MickG Jul 21 '14 at 16:14
  • Let $K$ be an algebraic closure of $\overline{F}$. Nullstellensatz (over $K$) says that the radical of the above ideal (in $K[x_1,\ldots,x_{m+1}]$) is not the entire polynomial algebra. In particular $1$ is not in there. Therefore $1$ is not in the corresponding ideal over $F$ either. Therefore (by Nullstellensatz over $F$) the system has a solution in $F$ as well. – Jyrki Lahtonen Jul 21 '14 at 16:23
  • Yes. This is more or less equivalent to Nullstellensatz. The trick to deal with the extra inequality is bread and butter. All I'm saying is that you look up Nullstellensatz :-) – Jyrki Lahtonen Jul 21 '14 at 16:24
  • "quantity elimination". I think you mean "quantifier elimination". The theory of algebraically closed fields has this nice property (just google a little bit - the proof is not so hard). – Martin Brandenburg Jul 21 '14 at 19:59
  • Maybe. I am translating «eliminazione di incognite», i.e. the elimination of unknown variables in an equations, or at least that's what is suggested but the name, as «incognita» is the unknown quantity/ies in an equation. How do you call that in English? – MickG Jul 22 '14 at 12:48

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As noted in comments, this is essentially the statement of the Nullstellensatz.

There are several proofs, which you will find many accounts of with a bit of googling. (If you search on MathOveflow, there are a couple of threads there comparing some of the different proofs.)

I think it's reasonable to say that the proof is non-trivial, and not likely something you will discover yourself, or get your head around in five minutes (once you start to read it).

user160609
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