Another proof for complex solutions and solutions in finite fields 
Let $S = \{f_1 , ... , f_r\} \subset \mathbb{Z}[x_1,...,x_n]$
Prove that there exists $x \in \mathbb{C}$ such that $\forall i, f_i(x) = 0 \iff $ for infinitely many $p$, $\exists x_p \in F_p$ such that   $\forall i, f_i(x_p) = 0$ where $F_p$ is some field of characteristic $p$.

I have been stuck on this question for a while now. I was looking for a proof and I found a SE post here
Relationship between a solution in $\mathbb{C}$ to a finite set of polynomials over $\mathbb{Z}$ and solutions in finite fields
but I wanted to know if there was a purely algebraic way to prove this?
Any help would be appreciated!
 A: The purely algebraic way to prove this is by using Nullstellensatz.
If $f_1, \dotsc, f_r$ don't have a common zero in $\mathbb C$, then by Nullstellensatz, the ideal generated by these polynomials over $\mathbb C[x_1, \dotsc, x_n]$ contains the constant polynomial $1$. Hence there exist polynomials $g_1, \dotsc, g_r \in \mathbb C[x_1, \dotsc, x_n]$ such that $f_1g_1 + \dotsc + f_rg_r = 1$ as an identity of polynomials.
Now you should convince yourself that, by replacing the coefficients of the $g_i$'s with indeterminates, the identity $f_1g_1 + \dotsc + f_rg_r = 1$ becomes a system of linear equations in these indeterminates, with integral coefficients.
The existence of $g_1, \dotsc, g_r$ then means that this system of linear equations has a solution in $\mathbb C$. But since all the coefficients are in $\mathbb Z$, hence in $\mathbb Q$, we know that there must already be a solution in $\mathbb Q$.
Therefore, there exist polynomials $g_1, \dotsc, g_r\in \mathbb Q[x_1, \dots, x_n]$, such that $f_1g_1 + \dotsc + f_rg_r = 1$. Clearing the denominators, we get polynomials $h_1, \dotsc, h_r\in\mathbb Z[x_1, \dots, x_n]$, such that $f_1h_1 + \dotsc + f_rh_r = d$ for some non-zero integer $d$. Hence for any $p$ coprime to $d$, the polynomials $f_1, \dotsc, f_r$ cannot have a common zero in any field of characteristic $p$.

The other direction is similar. Assume that $f_1, \dotsc, f_r$ have a common zero in $\mathbb C$.
Let $\overline{\mathbb Q}$ be the algebraic closure of $\mathbb Q$. By Nullstellensatz, they already have a common zero $(y_1, \dotsc, y_n)$ in $\overline{\mathbb Q}$ (otherwise the ideal they generate in $\overline{\mathbb Q}[x_1, \dotsc, x_n]$ would contain the constant polynomial $1$, hence no common zero in $\mathbb C$).
Let $E$ be the number field generated by the algebraic numbers $y_1, \dotsc, y_n$ and let $\mathcal{O}_E$ be the ring of integers of $E$. Choose a non-zero integer $d$ such that each $dy_i$ are algebraic integers.
Now for any prime number $p \nmid d$, choose a prime ideal $\mathcal P$ of $\mathcal{O}_E$ above $p$, so that the quotient $F_p = \mathcal{O}_E/\mathcal P$ is a finite field of characteristic $p$.
It is then clear that, if we put $z_i = d^{-1}((dy_i)\mod {\mathcal P}) \in F_p$, then $(z_1, \dotsc, z_n)$ is a common zero of $f_1, \dotsc, f_r$ in $F_p$.
