Zariski topology on $\mathbb{C}[X, Y]$ For a commutative ring $A$, let Spec$(A)$ be the set of prime ideals. A topology on Spec$(A)$ is defined by the closed sets
$$
\mathcal{V}(T) = \lbrace \mathbb{p} \in \text{Spec}(A) \vert T \subseteq \mathbb{p} \rbrace
$$
for some $T \subseteq A$, called the Zariski topology.
We studied some basic properties of this topology in class (e.g. it is $\mathtt{T}_0$ and sober) and worked out the rather trivial examples where $A$ is a field or a principal ideal domain. More complicated examples, like $\mathbb{Z}[T]$ or $\mathbb{C}[X, Y]$, were left as an exercise.
I have no idea how to begin to answer the question 'describe the Zariski topology on Spec$(A)$' for these more complicated rings. I understand the theory, but what can be said about this topology in these cases? As a more general question: how to analyse the Zariski topology of a given ring?
This is not a homework assignment, I'm just trying to get a better grip on what I have to study. I have gotten no further than a few remarks:


*

*Both rings are domains, hence $\lbrace 0 \rbrace$ is the minimum of Spec$(A)$ in both cases.

*By the Nullstellensatz, we know that the maximal ideals of $\mathbb{C}[X, Y]$ are of the form $(X - a, Y - b)$. Also the prime ideals of height one must be principal ideals, generated by an irreducible polynomial.

*In $\mathbb{Z}[T]$, the prime ideals of height one are principal ideals, generated by an irreducible polynomial or a prime in $\mathbb{Z}$.

 A: If $A$ is a Noetherian ring, there is a fairly easy way to get a grip on the Zariski topology on $\operatorname{Spec}(A)$.  Every closed set $C$ can be written as a union of finitely many irreducible closed sets $C=C_1\cup\dots\cup C_n$ (The proof is by Noetherian induction: if every proper closed subset of $C$ has this property, then either $C$ is irreducible so we can let $n=1$ and $C_1=C$, or else we can write $C=A\cup B$ where $A$ and $B$ are smaller closed sets, and by induction we have such a decomposition for $A$ and $B$ and we can combine them to get one for $C$.)  Furthermore, this decomposition is unique if we assume it is irredundant, in the sense that no $C_i$ is contained in another $C_j$.  Since $\operatorname{Spec}(A)$ is sober, each irreducible closed set is the closure of a unique point.  So if we know all of the prime ideals of $A$ and how they are ordered by inclusion, we know the closed subsets of $\operatorname{Spec}(A)$: they are just finite unions of sets of the form $\overline{\{P\}}=\{Q:P\subseteq Q\}$ for prime ideals $P$.
Let's see how this works in practice, say for $A=\mathbb{C}[X,Y]$.  As you note, there are three kinds of primes in $A$.  There are the maximal ideals, which are all of the form $(X-a,X-b)$.  We can identify such a maximal ideal with the point $(a,b)\in\mathbb{C}^2$, so we have a copy of  $\mathbb{C}^2$ in $\operatorname{Spec}(A)$ (though not with the usual topology).  There are the height one primes, which are of the form $(f(X,Y))$ for some irreducible polynomial $f(X,Y)$ (note that contrary to what you say, $f$ need not be linear--for instance, $f(X,Y)=X^2+Y^3$ is irreducible).  The closure of such a point in $\operatorname{Spec}(A)$ contains all of the maximal ideals $(X-a,Y-b)$ such that $f(a,b)=0$: that is, it is the curve in $\mathbb{C}^2$ defined by the equation $f(a,b)=0$, together with the one additional "generic" point $(f)$.  Finally, there is the ideal $0$, whose closure is all of $\operatorname{Spec}(A)$.
So if a closed subset of $\operatorname{Spec}(A)$ is not the entire space, we can write it as a union of finitely many single points of $\mathbb{C}^2$ and finitely many curves in $\mathbb{C}^2$ defined by irreducible polynomials, where for each such curve we also throw in an additional "generic point" of the curve.  This decomposition is unique if we additionally stipulate that none of our finitely many points should lie on any of the curves.
