We consider a polynomial $F \in \mathbb{C}[X_1, \ldots , X_n, Y_1, \ldots , Y_m]$ in $n+m$ variables. Define the set $A_F = \{ x \in \mathbb{C}^n \mid F(x_1, \ldots , x_n, Y_1, \ldots , Y_m) = 0\}$ where equality is meant to be understood in the polynomial ring$\mathbb{C}[ Y_1, \ldots , Y_m]$. My question: Is $A_F$ an algebraic variety?
I would need to find a finite number of polynomials whose common roots are exactly $A_F$. It is easy to find a set of defing polynomials for $A_F$, namely $\{ F(X_1, \ldots , X_n, y_1, \ldots , y_m) \mid y \in \mathbb{C}^n \}$. This is true because for any $f \in \mathbb{C}[ Y_1, \ldots , Y_m]$ we have $f=0 \iff \forall y \in \mathbb{C}^n ~ f(y) = 0$. However these are infinitely many polynomials. Intuitively I should be able to pick a finite number of those, but I don't see a convincing argument why this suffices.
And assuming that you can pick out finitely many of these polynomials, how many (if chosen randomly) would suffice to describe $A_F$ almost surely?
Any help on this would be appreciated.
Upgrading my hint in to an answer: Write $F=\sum_\alpha p_\alpha(X_1,\cdots,X_n)Y^\alpha$ as a polynomial in the $Y_i$ with coefficients in the $X_j$ where $\alpha$ is a multi-index to save us some notation. Then the set you describe is exactly the zero locus of all of the $p_\alpha$, of which there are finitely many because a polynomial has only finitely many nonzero coefficients.