Let $X = \text{Spec }A$ be an affine scheme. We always have that $X$ is quasi-compact, although in general it is certainly not true that any open set $U \subseteq X$ is quasi-compact.
I was recently trying to show that if $Z \subseteq X$ is a closed subset, then $Z$ can be covered by a finite number of affine open sets of the form $Z \cap D(f_{i})$ with $f_{i} \in A$. In the process, I seem to have "proved" the above claim, which I know is false. Although I am not actually able to find where my mistake is, so I am clearly not understanding something properly.
Let $\mathcal{I} \subseteq A$ be the ideal defining the closed set $Z$. That is, let $Z = V(\mathcal{I})$. Without loss of generality, we can take $\mathcal{I}$ to be radical. The points of $V(\mathcal{I})$ correspond to the prime ideals of $A$ containing $\mathcal{I}$. That is, prime ideals of $A / \mathcal{I}$. Also, $A / \mathcal{I}$ is a reduced ring, and we can choose a finite number of non-unit, non-nilpotent elements which generate the unit ideal,
$$
A / \mathcal{I} = \langle \bar{g_{1}}, \bar{g_{2}}, \ldots , \bar{g_{m}} \rangle.
$$
But any prime $\mathfrak{p} \subseteq A$ containing $\mathcal{I}$ must
fail to contain at least one of the $g_{j}$, since otherwise it would
be sent to the unit ideal in $A / \mathcal{I}$. So the distinguished
open sets $\{ D(g_{j}) \}_{j}$ cover $Z$.
Now let $U$ be the complement of $Z$ in $X$. Since $U$ is open, it may be covered by a collection of distinguished open sets $\{D(f_{i}) \}_{i}$. Taking the collection of all the $D(f_{i})$ and $D(g_{j})$ together then gives a cover for $X$ with the former collection contained in $U$ and the latter collection contained in $Z$. By quasi-compactness, we can choose a finite subcollection of this combined collection, and so there must be only finitely many of the $D(f_{i})$ necessary to cover $U$.
Where exactly does this argument fail? The only place I can see it possibly may is the part I have highlighted above. Is it really true that $\mathfrak{p}$ being sent to the unit ideal means that it itself must not be proper?
Best Answer
Note that the $D(g_j)$ are not contained in $Z$ - when you're covering $Z$ by affine open subsets, you're taking intersections of affine open subsets of $X$ with $Z$.
Per the comments, this question comes from a Hartshorne problem that sets out to show that every subscheme structure on $Z$ has the form $\operatorname{Spec}(A/\mathfrak a)$ with $\sqrt {\mathfrak{ a}} = \mathcal I$. Note that one cannot give meaning to the phrase "affine open subspace of $Z$" without a specified scheme structure on $Z$, but one can say that $Z$ is quasi-compact as a topological space.