Compactness of $\operatorname{Spec}(A)$

So after all the input from Pierre and Dylan, I have decided to post my answer here:

Suppose that $X$ is covered by $\bigcup_{i\in I} X_{f_i}$. Our goal is to show that $X$ can also be covered by $\bigcup_{i \in J} X_{f_i}$ where $J$ is some finite subset of $I$.

This is equivalent to proving (as in Dylan's comment) that $\emptyset = \bigcap_{i \in J} V(f_i)$. Supposing that this is non-empty, we have a prime ideal $\mathfrak{p}$ that contains each $f_i$ for all $i \in J$. Now by the reasoning in my post above we know that the ideal $\sum_{i \in I} (f_i) = (1)$. But then by definition of the sum of ideals, the ideal $\sum_{i \in I} (f_i)$ consists of elements of the form $\sum x_i$ where $x_i \in (f_i)$ and almost all of the $x_i$ (i.e. all but a finite set) are zero.

This means that we have a finite subset J of I such that $\sum_{i \in J} (f_i) = (1)$. Recall by assumption that we have a prime ideal $\mathfrak{p}$ that contains each $f_i$ for $i \in J$. However $\mathfrak{p}$ necessarily contains all linear combinations of the $f_i's$. In particular there exists a linear combination of the $f_i's$ that gives us $1$. But then $1 \in \mathfrak{p}$ which is a contradiction. Hence this finite intersection is empty. Since our initial open cover for $X$ was arbitrary, we are done.

$\hspace{6in} \square$

Here is a way to prove that $X$ is quasi-compact without invoking the fact the $X_f:=X\setminus V(f)$ generate the topology of $X:=\text{Spec}A$.

Let $(\mathfrak a_i)_{i\in I}$ be a family of ideals satisfying $$ \bigcap_{i\in I}V(\mathfrak a_i)=\varnothing, $$ and observe successively

$\bullet\quad\displaystyle V\left(\sum_{i\in I}\mathfrak a_i\right)=\bigcap_{i\in I}V(\mathfrak a_i)=\varnothing,$

$\bullet\quad\displaystyle\sum_{i\in I}\mathfrak a_i=(1),$

$\bullet\quad\displaystyle\sum_{i\in F}\mathfrak a_i=(1)$ for some finite subset $F$ of $I$,

$\bullet\quad\displaystyle\bigcap_{i\in F}V(\mathfrak a_i)=V\left(\sum_{i\in F}\mathfrak a_i\right)=V(1)=\varnothing$.

EDIT. The purpose of this edit is (a) to state and prove Proposition I.$1.1.4$ page $195$ in the Springer version of EGA I (see reference below), and (b) to perform the mental experiment consisting in defining the Zariski topology on the prime spectrum of a commutative ring in terms of open (instead of closed) subsets.

Precise reference: Éléments de Géométrie Algébrique I, Volume $166$ of Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen mit besonderer Berücksichtigung der Anwendungsgebiete, A. Grothendieck, Jean Alexandre Dieudonné, Springer-Verlag, $1971$.

Let $A$ be a commutative ring and $X$ the set of its prime ideals. For any subset $M$ of $A$ we write $$ U(M) $$ for the set of those prime ideals of $A$ which do not contain $M$.

If $\mathfrak a$ is the ideal generated by $M$, then $U(M)=U(\mathfrak a)=U(r(\mathfrak a))$.

We have the nice formulas $$ M\subset N\implies U(M)\subset U(M), $$ $$ U\left(\bigcup_{i\in I}\ M_i\right)=\bigcup_{i\in I}\ U(M_i), $$ and, for ideals $\mathfrak a$ and $\mathfrak b$, $$ U(\mathfrak a\cap\mathfrak b)=U(\mathfrak a)\cap U(\mathfrak b). $$ More generally we have $$ U(0)=\varnothing,\quad U(1)=X, $$ $$ U\left(\bigcup_{i\in I}M_i\right)=U\left(\sum_{i\in I}M_i\right)=\bigcup_{i\in I}\ U(M_i), $$ $$ U(\mathfrak a\cap\mathfrak b)=U(\mathfrak a\mathfrak b)=U(\mathfrak a)\cap U(\mathfrak b), $$ which shows that the $U(M)$ form a topology. Note $$ U(\mathfrak a)\subset U(\mathfrak b)\iff r(\mathfrak a)\subset \mathfrak b. $$ The equality $$ U(\mathfrak a)=\bigcup_{f\in\mathfrak a}\ U(f)\qquad(*) $$ shows that the $U(f),f\in A$, form a basis for our topology.

Proposition I.1.1.4 of the Springer version of EGA I says

$U(\mathfrak a)$ is quasi-compact $\iff$ $U(\mathfrak a)=U(f_1,\dots,f_n)$ for some $f_1,\dots,f_n$ in $A$.

Proof.

$\Longrightarrow:\ $ This follows immediately from $(*)$.

$\Longleftarrow:\ $ As $U(f_1,\dots,f_n)$ is the union of the $U(f_j)$, it suffices to prove that $U(f)$ is quasi-compact. If $$ U(f)\subset\bigcup_{i\in I}\ U(\mathfrak a_i), $$ then some power $f^k$ of $f$ is in $\sum_{i\in I}\mathfrak a_i$. But then $f^k$ is in $\sum_{i\in F}\mathfrak a_i$ for some finite subset $F$ of $I$, implying $$ U(f)\subset\bigcup_{i\in F}\ U(\mathfrak a_i). $$