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Given a system of rational polynomials in some number of variables with at least one real solution, I want to prove that there exists a solution that is a tuple of algebraic numbers. I feel like this should be easy to prove, but I can't determine how to. Could anyone give me any help?

I have spent a while thinking about this problem, and I can't think of any cases where it should be false. However I have absolutely no idea how to begin to show that it is true, and I can't stop thinking about it. Are there any simple properties of the algebraic numbers that would imply this? I don't want someone to prove it for me, I just need someone to point me in the direction of a proof, or show me how to find a counterexample if it is actually false (which would be very surprising and somewhat enlightening). If anyone knows anything about this problem I would be thankful if they could give me a little bit of help.

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  • $\begingroup$ By 'all variables algebraic' you mean one solution will be a tuple of algebraic numbers? $\endgroup$ Commented Jul 4, 2013 at 15:46
  • $\begingroup$ Yes, I will edit my post to specify that. $\endgroup$
    – Parakee
    Commented Jul 4, 2013 at 16:22
  • $\begingroup$ Does no one have any idea how to answer this? $\endgroup$
    – Parakee
    Commented Jul 4, 2013 at 23:14
  • $\begingroup$ Hmm. Take that one point you have. Its coordinates generate an extension field of the rationals. Among them you can find a transcendence basis of that extension. I would guess that you can treat the elements of the transcendence basis as variables, and solve the others in terms of them. So you should be able to assign the elements of transcendental basis arbitrary values, and be able to solve for the others. After all that you would have another point with all coordinates algebraic. A leaky argument, indeed, and needs a lot of results about transcendence bases and such. Cannot fill it in :-( $\endgroup$ Commented Jul 7, 2013 at 12:11

2 Answers 2

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Here's a thought. Let's look at the simplest non-trivial case. Let $P(x,y)$ be a polynomial in two variables with rational (equivalently, for our purposes, integer) coefficients, and a real zero.

If that zero is isolated, then $P$ is never negative (or never positive) in some neighborhood of that zero, so the graph of $z=P(x,y)$ is tangent to the plane $z=0$, so the partial derivatives $P_x$ and $P_y$ vanish at the zero, so if you eliminate $x$ from the partials (by, say, taking the resultant) you get a one-variable polynomial that vanishes at the $y$-value, so the $y$-value must be algebraic, so the $x$-value must be algebraic.

If the zero is not isolated, then $P$ vanishes at some nearby point with at least one algebraic (indeed, rational) coordinate, but that point must then have both coordinates algebraic.

Looking at the general case, many polynomials in many variables, you ought to be able to use resultants to get down to a single polynomial, and then do an induction on the number of variables --- we've already done the base case.

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This follows from the fact that the real algebraic numbers $\mathbb{R} \cap \overline{\mathbb{Q}}$ are elementary equivalent to $\mathbb{R}$ (see real closed field) and that the condition that a system of polynomial equations over $\mathbb{Q}$ have a solution is a first-order statement in the theory of fields, so is true over $\mathbb{R}$ iff it's true over any real closed field.

With a bit more effort this elementary equivalence proves the harder and more interesting fact that if such a system has a real solution then it has an algebraic solution arbitrarily close to the real solution; in other words, the algebraic solutions are dense. But this argument really uses that we're working over $\mathbb{Q}$ and I don't see how to generalize it to a bigger subfield of $\mathbb{R}$.

I came across this old question while trying to google something relevant to this question.

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