Please explain when the Zariski topology is $T_1$ (depending on $k \ge 1$).
Let $X$ be a topological space and let $x$ and $y$ be points in $X$. We say that $x$ and $y$ are separated if each lies in a neighbourhood that does not contain the other point.
$X$ is called a $T_1$ space if any two distinct points in $X$ are separated.
Definition of Zariski topology on any field $\mathbb{K}^k$: has as closed sets $V(J)$, where we hvae that $J \subseteq \mathbb{K}[x_1,...,x_k]$ an ideal.
I don't know when the Zariski topology is $T_1$ (depending on $k \ge 1$).
There are two possible answers to your question: the classical answer, and the more modern Grothendieck answer.
First, the classical answer: the Zariski topology is $T_1$. A point is always closed, because $\{(a_1,\ldots,a_k)\}$ is the zero set of the ideal $(x_1-a_1,\ldots,x_k-a_k)$. $T_1$ is equivalent to saying all points are closed, so the space is $T_1$.
Grothendieck and his collaborators reformulated algebraic geometry; this is sometimes called the "scheme revolution". Part of this recasting involved generalizing the notion of "point" to include all irreducible components. This may seem strange, but I'll suggest briefly the motivation below. The closure of a "point" now becomes the set of all irreducible components contained in it. With this definition, the answer is typically no. Two lines (regarded as "points") that intersect in a classical point cannot be separated, because their closures intersect in the classical point.
Why this expanded notion of "point"? Well, even classically we have the correspondence between ideals and algebraic sets. Recall that this is order-reversing. Let's restrict to the classical case where the ground field $\mathbb{K}$ is algebraically closed. The space of all classical points corresponds to MaxSpec, the maximal spectrum, which is the collection of all maximal ideals. The closed sets in the Zariski topology on MaxSpec are of the form $\{M\in\text{MaxSpec}\;|\;M\supseteq I\}$, where $I$ is an ideal.
It turns out the prime spectrum Spec has nicer properties than the maximal spectrum. This is especially true when we want to generalize beyond the classical case of an algebraically closed ground field. We define the Zariski topology on Spec just the same way: the closed sets are things of the form $\{P\in\text{Spec}\;|\;P\supseteq I\}$, where $I$ is an ideal.
Translating back using the order-reversing correspondence, the switch from MaxSpec to Spec means we should consider all irreducible components to be "points" of the space.
Caveat: If you glance at the Wikipedia entry for "schemes", you will see I've barely hinted at all the complexities of the Grothedieck approach. I've tried to say just enough to answer your question.