I always find it difficult to motivate Zariski topology on $\text{Spec}(k[X_1,...,X_n])$, both the underlying set and the topology. As far as I understand, Zariski topology isn't really that essential for proving many classical theorems like Bezout's (for example Fischer doesn't mention Zariski topology once in his Plane Algebraic Curves); applications of Zariski topology where it does play an essential role are probably too advanced for me (although counterexamples are much appreciated).
Then I wonder if I can at least motivate Zariski topology intrinsically. I can somehow agree with the naturalness of Zariski topology on $k^n$: points should be closed and polynomials should be continuous; this is a bit like weak topology in functional analysis. But why do we want to consider prime ideals, i.e., to put points and irreducible varieties on the same footing? A usual answer is that preimage of maximal ideal is in general not maximal, but it is maximal in the case of finitely generated $k$-algebras, which seems good enough (and why is it so important to take preimage?). Also, why should we think of the zero ideal as the "generic point"? For me it's hard to relate that to those classical generic point arguments.
But now I think I finally found a way to motivate Zariski topology. Please help me check if the math is correct.
Let $K\supseteq k$ be a larger algrabraically closed field of infinite transcendence degree, e.g., if $k=Q^{\text{alg}}$ then we may take $K=\mathbb{C}$. For each $a=(a_1,...,a_n)\in K^n$, define $\pi(a)=\{f\in k[X_1,...,X_n]: f(a)=0\}$. This is a prime ideal.
We claim that $\pi:K^n\rightarrow\text{Spec}(k[X_1,...,X_n])$ is surjective. For any prime ideal $\mathfrak{p}$, the image of $X_1,...,X_n$ in $k[X_1,...,X_n]/\mathfrak{p}$ is a tuple that satisfies exactly those polynomials in $\mathfrak{p}$. The algebraic closure of $k[X_1,...,X_n]/\mathfrak{p}$ has transcendence degree at most $n$ over $k$, and hence embeds into $K$.
We claim that with both $K^n$ and $\text{Spec}(k[X_1,...,X_n])$ endowed with Zariski topology, $\pi$ is a closed map, hence a quotient map. If $f(a)=0$ then $\pi(a)\ni f$; conversely if $\mathfrak{p}\ni f$, choose any $a$ that maps to $\mathfrak{p}$, and we must have $f(a)=0$. This shows the image of a basic closed set in $K^n$ is a basic closed set in $\text{Spec}(k[X_1,...,X_n])$; the general case is similar.
So the Zariski topology on $\text{Spec}(k[X_1,...,X_n])$ is just the quotient of the (in my opinion) more intuitive topology on $K^n$. As for generic point, if $\pi(a)=\mathfrak{p}$ then the tuple $a$ satisfies those polynomials in $\mathfrak{p}$ and nothing else, as with "most" $K$-points on the variety determined by $\mathfrak{p}$; in particular $\pi(a)=(0)$ iff $a=(a_1,...,a_n)$ are algebraically independent. This makes more sense to me than "$(0)$ is a point that lies everywhere". As for the reason of going to a larger field $K$: truly generic point doesn't exist if we only look at $k$. All algebraically closed fields (of a fixed characteristic) are more or less the same, so the structure of varieties don't really change when we move to a larger field.
Is this a useful way to picture Zariski topology? Perhaps it's too early for me, but is this related to the the functor of points?