# Algebraic Solutions to Systems of Polynomial Equations

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.

• By 'all variables algebraic' you mean one solution will be a tuple of algebraic numbers? – Gregor Botero Jul 4 '13 at 15:46
• Yes, I will edit my post to specify that. – Parakee Jul 4 '13 at 16:22
• Does no one have any idea how to answer this? – Parakee Jul 4 '13 at 23:14
• 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 :-( – Jyrki Lahtonen Jul 7 '13 at 12:11

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.