TITLE: Finitely many complex solutions imply all solutions algebraic
QUESTION [4 upvotes]: I came across this rather interesting problem:
Suppose that $f_1(x_1, \dots , x_n), \dots , f_m(x_1, \dots , x_n)$ are $m$ polynomials with integer coefficients, and that these have a finite number of common roots in $\mathbb{C}^n$. Prove that each of these solutions lie in $\bar{\mathbb{Q}}^n$, where $\bar{\mathbb{Q}}$ denotes the field of algebraic numbers.
I believe that this should involve Zariski's lemma or the (weak) Nullstellensatz, but so far all I can prove is that if there is a solution in $\mathbb{C}^n$ then there is a solution in $\bar{\mathbb{Q}}^n$. But this doesn't show all the solutions lie in $\bar{\mathbb{Q}}^n$. I haven't used the finiteness condition yet, nor the condition that all the coefficients in $f_1(x_1, \dots , x_n), \dots , f_m(x_1, \dots , x_n)$ are integers, but I can't seem to see how to use them.

REPLY [10 votes]: Let $K\subset L$ be a field extension. Let $R=K[X_1,\dots,X_n]$, and $f_1,\dots,f_m\in R$ with the property that the system of equations $f_1(x)=0,\dots,f_m(x)=0$ has (only) finitely many solutions in $L^n$. Then any solution $(a_1,\dots,a_n)\in L^n$ is algebraic over $K$, that is, $a_i$ is algebraic over $K$ for all $i=1,\dots,n$. 

Let $I=(f_1,\dots,f_m)$ be the ideal of $R$ generated by $f_1,\dots,f_m$. Set $S=L[X_1,\dots,X_n]$. The hypothesis tells us that the ideal $J=IS$ of $S$ is zero-dimensional, that is, $\dim_LS/IS<\infty$. But $S/IS=R/I\otimes_KL$, and then $\dim_KR/I<\infty$. This shows that the ideal $I$ of $R$ is zero-dimensional. Then $I\cap K[X_i]\ne(0)$ for all $i=1,\dots,n$. Now choose a non-zero polynomial $g_i\in I\cap K[X_i]$, and notice that $g_i(a_i)=0$ hence $a_i$ is algebraic over $K$.
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