"Zero set of finitely many polynomials." but what polynomials are we talking about? This is a follow-up question of Zero set of finitely many polynomials.
When it says "As the zero set of finitely many polynomials,  is a closed subset of ×"
I understand we are saying that determinant of submatrix is all zero. But I'm not sure how it is associated with polynomials. It just looks like a bunch of computations. What polynomials are we talking about? Could you give me any concrete example or formula for this? I would really appreciate the answers.
 A: Let $(a, b, c)$ and $(d, e, f)$ be two vectors in $\mathbb{R}^3$: there is a linear dependence between these vectors if and only if all of the 2x2 subdeterminants of the matrix $M$ vanish, where
$$ M = \begin{pmatrix} a & d \\ b & e \\ c &  f \end{pmatrix}.$$
There are three subdeterminants:
$$ M_{1, 2} = ae - bd, \quad M_{1, 3} = af-cd, \quad M_{2, 3} = bf-ce.$$
So to check that $(a, b, c)$ and $(d, e, f)$ are colinear, we just need to check that $M_{1, 2}$, $M_{1, 3}$, and $M_{2, 3}$ are all zero. You can find many proofs of this fact (or facts like this) around the place.
The connection with polynomials is that each $M_{i, j}$ can be treated as a polynomial in the six variables $a, b, c, d, e, f$, for example
$$ M_{1, 2}(a, b, c, d, e, f) = ae - bd.$$
We can then imagine $M_{1, 2}$ as a function on the space $\mathbb{R}^6$, and we can look at the set
$$ V(M_{1, 2}) = \{(a, b, c, d, e, f) \in \mathbb{R}^6 \mid M_{1, 2}(a, b, c, d, e, f) = 0\} \subseteq \mathbb{R}^6.$$
This set $V(M_{1, 2})$ is called the vanishing of the polynomial $M_{1, 2}$. In order to find colinear vectors, we are looking at the simultaneous vanishing $V(M_{1, 2}) \cap V(M_{1, 3}) \cap V(M_{2, 3})$ of three polynomials (finitely many polynomials).
