February 23rd
Today I learned about the Zariski topology. I think the first definition came from algebraic varieties. If you have an affine space (say, over $\CC$) which we'll name $\AA^n,$ then the Zariski topology is defined by its closed sets\[V(S)=\left\{x\in\AA^n:f\in S\implies f(x)=0\right\},\]where $S$ is a set of polynomials in $n$ variables. Note that if $f$ and $g$ are polynomials, then $f(x)=0$ and $g(x)=0$ is equivalent to $af(x)+bg(x)=0$ for all $a,b\in\AA.$ Thus, the closed set generated by $S$ is equal to the closed set generated by the ideal generated by $S.$ So we consider ideals $I$ of polynomials in $n$ variables from here on.
We can quickly check that this is indeed a topology. For the easy stuff, note that $S=\emp$ gives $V(S)=\AA^n,$ and $S=\{1\}$ gives $V(S)=\emp.$ So we see that $\emp$ and $\AA^n$ are closed sets.
The intersection $V(I)\cap V(J)$ is extracting points which are $0$ on everyone in $I$ and everyone in $J,$ so we claim\[V(I)\cap V(J)=V(I+J).\]Indeed, $x\in V(I+J)$ if and only if $x$ is $0$ on every polynomial $af(x)+bg(x)$ for $f,g\in I,J$ and $a,b\in\AA.$ Setting $a=0$ and then $b=0$ tells us this occurs if and only if $x$ is $0$ on every polynomial $f\in I$ and every polynomial $g\in J,$ which is what we wanted. Generalizing, the same argument tells us that for a collection $\{I_\alpha\}_{\alpha\in\lambda}$\[\bigcap_{\alpha\in\lambda}V(I_\alpha)=V\left(\sum_{\alpha\in\lambda}I_\alpha\right).\]So our closed sets are closed under arbitrary intersection.
Similarly, the union $V(I)\cup V(J)$ is extracting points which are $0$ on either everyone in $I$ or everyone in $J.$ This is a bit fuzzier, but we claim\[V(I)\cup V(J)=V(IJ).\]If $x\in V(I)\cup V(J),$ then $x\in V(IJ),$ for $IJ$ is generated by products $fg$ with $f\in I$ and $g\in J,$ and we must have $(fg)(x)=0.$ Conversely, if $x\notin V(I)\cup V(J),$ then there exists a polynomial $f\in I$ and a polynomial $g\in J$ such that\[f(x)\ne0\quad\text{and}\quad g(x)\ne0.\]Thus, $(fg)(x)\ne0$ with $fg\in IJ,$ so $x\notin V(IJ).$ So the claim follows. Inducting, we see that for a collection $\{I_n\}_{n\in\NN}$\[\bigcup_{n\in\NN}V(I_n)=V\left(\prod_{n\in\NN}I_n\right).\]Note that this argument had to use a contrapositive to get the second inclusion, which I think is why we don't have arbitrary union.
Of course, the reason why I care about the Zariski topolgy is the spectrum of a ring. With this in mind, we state explicitly that we are looking towards a ring-theoretic Zariski topology. I think the motivation here comes from looking at the case $\AA^n=\CC^1.$ (I think we should be looking at $\CC P^1,$ but I don't know enough algebraic geometry to be able to tell the difference.) In this case, our closed sets are\[V(S)=\{\alpha\in\CC:f\in S\implies f(\alpha)=0\}\]for $S\subseteq\CC[t].$ Now, we remark that all polynomials over $\CC[t]$ factor, so for $f\in\CC[t],$ we see $V(f)$ is just the zeroes of $f.$ So our topology is "really'' generated closed sets consisting of finitely many points. Of course, simple factorization in $\CC[t]$ doesn't generalize well, but ideal factorization does, where we can still write\[(f(t))=\prod_{f(\alpha)=0}(t-\alpha).\]Here $(t-\alpha)$ are the prime ideals of $\CC[t],$ and they both associate with $\CC$ and encode our closed sets. At this point, writing\[V(I)=\{(t-\alpha):I\subseteq(t-\alpha)\}\]turns our definition into an entirely ring-theoretic one. Note that this is pretty much the same Zariski topology over $\CC,$ but now we're encoding points $\alpha\in\CC$ with their ideals $(t-\alpha).$ With this in mind, here's our definition.
Definition. For a commutative ring $R,$ we define the spectrum of a ring $\op{Spec}R$ with points the prime ideals of $R$ and a topology where \[V(I)=\{\mf p\subseteq R:I\subseteq\mf p\}\] is a closed set for each ideal $I\subseteq R.$
We provide the check that this is a topology to hammer home that this really deserve the name "Zariski topology.'' Notably, these proofs are pretty much identical. Again, $I=(0)$ means $\op{Spec}R$ is closed, and $I=R$ means $\emp$ is closed.
For intersections, we still have, for a collection $\{I_\alpha\}_{\alpha\in\lambda},$ that\[\bigcap_{\alpha\in\lambda}V(I_\alpha)=V\left(\sum_{\alpha\in\lambda}I_\alpha\right).\]Now, $\mf p$ is in the intersection is and only if each $I_\bullet\subseteq\mf p.$ But the sum $\sum I_\bullet$ is generated by (finite) linear combinations of elements in the $I_\bullet,$ so indeed $\mf p$ is $V\left(\sum I_\bullet\right).$ And then because $I_\bullet\subseteq V\left(\sum I_\bullet\right),$ we see $\mf p\in V\left(\sum I_\bullet\right)$ implies\[I_\bullet\subseteq\mf p\]for each $I_\bullet.$ So $\mf p$ is also in the intersection. So we have arbitrary intersection.
For unions, we again still have $V(I)\cup V(J)=V(IJ).$ Here, we see $\mf p\in V(I)\cup V(J)$ if and only if $I,J\subseteq\mf p.$ So because $IJ$ is generated by products $ab$ with $a\in I$ and $b\in J,$ we see $IJ\subseteq\mf p$ because each $ab\in\mf p.$ Conversely, if $\mf p\notin V(I)\cup V(J),$ then there exists $a\in V(I)$ and $b\in V(J)$ with\[a\notin\mf p\quad\text{and}\quad b\notin\mf p\]But then $ab\notin\mf p$ because $\mf p$ is prime (!), so $\mf p\notin V(IJ).$ Inducting from here gives finite unions.