Finitely many complex solutions imply all solutions algebraic 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.
 A: We use a little model theory.
Suppose that the polynomials have exactly $k$ common roots in $\mathbb{C}^n$.
Let theory $T$ be the (first-order) theory of algebraically closed fields of characteristic $0$, over the usual language $L$ that has constant symbols $0$ and $1$, and two binary function symbols (addition and multiplication). 
Let $\varphi$ be the sentence that says that our specific polynomials have exactly $k$ common roots. This can be written as a sentence of $L$.
The sentence $\varphi$ is true in the complex numbers. But all models of $T$ are elementarily equivalent (the theory $T$ is complete).
Thus $\varphi$ is true over the algebraic numbers. It follows that the polynomials have exactly $k$ common roots in $\bar{\mathbb{Q}}^n$. There are therefore no additional common roots in $\mathbb{C}^n$,  and therefore all the common roots are in $\bar{\mathbb{Q}}^n$.
A: Let $X = V(I) = Spec(\mathbb{Z}[x_1,\dots,x_n]/I)$ with $I=(f_1,\dots,f_m)$. The starting point is that $X(\mathbb{C})$ is finite. This implies that $X_\mathbb{C}$ is zero-dimensional, thus so is $X_\mathbb{Q}$, so it is discrete. 
The question Closed points are dense in $\operatorname{Spec} A$ recalls why closed points are dense, which is a consequence of the Nullstellensatz. But since $X_\mathbb{Q}$ is a discrete space, it means that all points are closed.
Now a closed point is always defined over a finite extension, so in particular it is defined over $\overline{\mathbb{Q}}$.

I will break down this reasoning in a proof as elementary as possible. In particular, I make use of the analytical topology of $\mathbb{C}^n$ to simplify an argument.
Let $A = \mathbb{Q}[x_1,\cdots,x_n]/I$, and $V(K) = \{ a\in K^n\,|\ \forall i,\, f_i(a)=0\}$ for all fields $K$ of characteristic $0$. Then there is a canonical bijection between $V$ and the ring morphisms $A\to K$ : each $\varphi: A\to K$ defines $(\varphi(x_1),\dots,\varphi(x_n))\in V(K)$, and for each $a\in V$ there is a unique $\varphi:A\to K$ such that $\varphi(x_i)=a_i$. 
Our starting point is that $V(\mathbb{C})$ is finite, and we want $V(\mathbb{C}) = V(\overline{\mathbb{Q}})$
First we show that every prime ideal of $A$ is maximal. Indeed, suppose we have $\mathfrak{p}\subset A$ prime but not maximal. Then there is $\mathfrak{q}\subset \mathbb{Q}[x_1,\dots,x_n]$ prime but not maximal with $I\subset \mathfrak{q}$ (according to the description of the prime ideals of a quotient ring). Now the irreducible variety in $\mathbb{C}^n$ it defines must be pathwise connected so it is infinite and for each $a$ in this variety you have $\mathfrak{q}\subset \mathfrak{m}_a\subset \mathbb{C}[x_1,\dots,x_n]$. In particular each of these $\mathfrak{m}_a$ contains $I$ and this each of these $a\in \mathbb{C}^n$ are actually in $V(\mathbb{C})$, which is a contradiction.
Now let $a\in V(\mathbb{C})$, and let $\varphi: A\to \mathbb{C}$ be the corresponding morphism. Then $\varphi(A)\subset \mathbb{C}$ so $\varphi(A)$ is an integral domain, and in particular $\mathfrak{p}= \ker(\varphi)$ is prime. Then it must be maximal, so $\varphi(A) = A/\mathfrak{p}$ is a finitely generated algebra over $\mathbb{Q}$ which is a field, so by Zariski's lemma, it is a finite extension of $\mathbb{Q}$. Hence $\varphi(A)\subset \overline{\mathbb{Q}}$, so $a\in V(\overline{\mathbb{Q}})$.
If you want the $f_i$ to have coefficients in $\overline{\mathbb{Q}}$ you can just replace $\mathbb{Q}$ by $\overline{\mathbb{Q}}$ everywhere in the proof.
A: 
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 $K$. 
