I have two questions regarding algebraic sets. Let $k$ be a field.
Given a finite set of points $S\subset k^n$, can we always find a set of polynomials $T\subset k[x_1,x_2,\dots,x_n]$ such that $Z(T)=S$?
Let $a=(a_1,a_2,\dots,a_n)$ be a point in $k^n$. We know that $Z(z_1-a_1,x_2-a_2,\dots,x_n-a_n)$ is $a$. However, is $I(a)=(z_1-a_1,x_2-a_2,\dots,x_n-a_n)$? Or can it contain some other element outside of $(z_1-a_1,x_2-a_2,\dots,x_n-a_n)$?
An affirmative answer to (1) follows from the fact that algebraic sets are the closed sets of the Zariski topology on $k^n$. In any topological space, a finite union of closed sets is closed. Since points are Zariski closed in $k^n$, any finite set in $k^n$ is a Zariski closed set. This means that it is of the form $Z(T)$ for some $T \subset k[x_1,\dots,x_n]$.
For (2), recall that the ideal $I(a)$ is defined as the set of a polynomials which vanish at the point $a$. It's easy to see that $f(a)=0$ implies that $f$ can be written as a polynomial in $(x_i-a_i)$ with zero constant term (to check this, change coordinates so that $a_i=0$ for all $i$). In particular, $f(a)=0 \implies f \in \langle x_1-a_1,x_2-a_2,\dots,x_n-a_n \rangle$.
If you haven't already, I'd suggest you learn about Hilbert's Nullstellensatz and understand its proof. This technical result is needed to make more sophisticated statements about the relationship between algebraic sets and their ideals. While the theorem is for $k$ algebraically closed, understanding its proof will help you to become more comfortable with these ideas in general.