Suppose $f \in \mathbb{Q}[x_1,x_2,\dotsc,x_i]$. How do I prove the following:
There exists $x_1 \in \mathbb{C}$ such that for all $x_2,\dotsc,x_i \in \mathbb{C}$, $$f(x_1,x_2,\dots,x_i) = 0$$ if and only if $f$ is divisible by some $g \in \mathbb{Q}[x_1]$.
(I suspect this is simply down to me not knowing proper definitions...)
In general, for any given constant $c\in{\mathbb C}$, an arbitrary polynomial $h$ can be Taylor-decomposed with respect to $x_1$ as
$$ f(x_1,x_2,x_3,\ldots ,x_i)=\sum_{k=0}^d h_k(x_2,x_3,\ldots ,x_i) (x_1-c)^k $$
It is easy then to see that $f(c,x_2,x_3,\ldots, x_i)=0$ for any $x_2,x_3,\ldots, x_i$ iff $x_1-c$ divides $f$.
If $c$ is transcendental, then $f=0$ (exercise : why?). If $c$ is algebraic and $Q$ is the minimal polynomial of $c$ over $\mathbb Q$, show that $Q(x_1)$ divides $f$ iff $x_1-c$ divides $f$.
UPDATE : Suppose that $c$ is transcendental. Decompose $f$ as
$$ f=\sum_{k=0}^d g_k(x_2,x_3,\ldots ,x_i) x_1^k $$
where $g_k$ are polynomials with rational coefficients, not involving $x_1$. Since $c$ is transcendental, the numbers $1,c,c^2,c^3, \ldots, c^d$ are linearly indepedent over $\mathbb Q$. So the hypothesis $f(c,\ldots)=0$ entails that all the $g_k$ are zero, so $f=0$.
Suppose that $c$ is algebraic, with minimal polynomail $Q$ (over $\mathbb Q$). Denote by $c_1=c,c_2, \ldots, c_r$ the conjugates of $c$ (over $\mathbb Q$). Since $x_1-c$ divides $f$, by conjugation $x_1-c_j$ also divides $f$, for any $j$. So $Q$ is divsible by the product of the $x_1-c_j$, which is none other than $Q(x_1)$.
