roots of multi-variable polynomials and extension fields I am teaching a course in (standard single-variable) Galois theory and the following, presumably naive, question occurred to me:
Given a finite collection of polynomial equations in a finite number of variables with coefficients in some field $K$ such that they have a finite number of solutions, are the coordinates of the solutions necessarily algebraic over $K$?

Since googling multi-variable Galois theory brought up very little except this MO question, I assume the answer is no, but I would love to know something about it,
 A: Let me state your question more precisely (since you don't say where these solutions are, if they aren't in an algebraic extension of $K$):

Suppose $f_1, \dots f_m$ are polynomials in $K[x_1, \dots x_n]$ such that the system $f_1 = \dots = f_m = 0$ has finitely many solutions in any field $L$ containing $K$. Are these solutions always algebraic over $K$?

The answer is yes, which you can see as follows. By Noether normalization, the ring $R = K[x_1, \dots x_n]/(f_1, \dots f_m)$ is a finite integral extension of some free subring $S = K[y_1, \dots y_d]$. If $L$ is any infinite and algebraically closed field containing $K$, it follows that there are infinitely many $L$-points (homomorphisms $R \to L$ of $K$-algebras, or solutions over $L$ of our system of equations) unless $d = 0$. 
If $d = 0$, then $R$ is finite-dimensional over $K$, from which it follows that there are at most finitely many $L$-points for any $L$, whose coordinates must be algebraic. This is because, since $R$ is finite-dimensional, for any $x_i$ there is some linear dependence among the monomials $\{ 1, x_i, x_i^2, \dots \}$, so each $x_i$ satisfies a polynomial over $K$.
In general, there is a subject one might call "multivariable Galois theory," but people usually call it "algebraic geometry" (over non-algebraically closed fields), or more specifically "Galois descent." 

Edit: Here's a lower-tech but longer argument using only transcendence degree. We'll prove the contrapositive: if the system has a solution over some field $L$ whose coordinates are not algebraic over $K$, then it has infinitely many solutions over some other field. 
Suppose $(\ell_1, \dots \ell_n) \in L^n$ is a transcendental solution over $L$, and let $D$ be the $K$-subalgebra of $L$ generated by the coordinates. ($D$ is a quotient of $R$ above, but unlike $R$ it is an integral domain.) We'll attempt to construct infinitely many new solutions over some other field $L''$ by writing down infinitely many $K$-algebra homomorphisms $D \to L''$.
Let $L'$ be the fraction field of $D$. By hypothesis, $L'$ has positive transcendence degree over $K$. Hence $L'$ is an algebraic extension of a transcendental extension of $K$. Since $D$ is finitely generated as a $K$-algebra, $L'$ is a finite algebraic extension of a finite transcendental extension $K(t_1, \dots t_d)$ of $K$. Let $\alpha_1, \dots \alpha_k$ be the finitely many elements generating $L'$ as an algebraic extension of $K(t_1, \dots t_d)$.
By construction, we can write each $\ell_i$ as a polynomial in the $\alpha_i$ with coefficients in $K(t_1, \dots t_d)$. We will now attempt to write down infinitely many $K$-algebra homomorphisms $g : D \to L''$ for some third field $L''$ containing $K$. We will do this by picking values for $g(\alpha_i), g(t_i)$ such that each $\ell_i$ continues to be well-defined; equivalently, we just need to arrange for the denominators of each term of each $\ell_i$ to be nonzero. An easy way to do this is to take $L''$ to be the algebraic closure of $K(s_1, s_2, \dots)$. Then we can send $g(t_i)$ to any $s_j$, and send $g(\alpha_i)$ to various roots of various polynomials, which exist since $L''$ is algebraically closed. If there are infinitely many $s_j$ then there are infinitely many such homomorphisms, as desired. 
(Other arguments are possible at this last step if $K = \mathbb{Q}, L = \mathbb{C}$. Then we can take $L'' = \mathbb{C}$. We need to avoid the zero sets of a finite set of polynomials in the $t_i$, but there are many ways to see that there are infinitely many ways to do this: for example, the zero sets of polynomials have measure zero.)
(Note how much work we have to do to cobble together a substitute for Noether normalization!) 
