In an exercise in Atiyah-Macdonald it asks to prove that the prime spectrum $\operatorname{Spec}(A)$ of a commutative ring $A$ as a topological space $X$ (with the Zariski Topology) is compact.
Now because the basic open sets $X_f = \{\mathfrak{p} \in \operatorname{Spec} (A) : \{f\} \not\subseteq \mathfrak{p} \}$ form a basis for the Zariski Topology it suffices to consider the case when
$$X = \bigcup_{i \in I} X_{f_i}$$
where $I$ is some index set. Then taking the complement on both sides we get that
$$\emptyset = \bigcap_{i \in I} X_{f_i}^c$$
so there is no prime ideal $\mathfrak{p}$ of $A$ such that all the $f_i$'s are in $\mathfrak{p}$. Now from here I am able to show that the ideal generated by the $f_i$'s is the whole ring as follows. Since there is no prime ideal $\mathfrak{p}$ such that all the $f_i \in \mathfrak{p}$, it is clear that there is no $\mathfrak{p}$ such that $(f_i) \subseteq \mathfrak{p}$ for all $i \in I.$ Taking a sum over all the $i$ then gives $$\sum_{i \in I} (f_i) = (1).$$
Now here's the problem:
How do I show from here that there is an equation of the form $1 = \sum_{i \in J} f_ig_i,$ where $g_i \in A$ and $J$ some finite subset of $I$?
This part has been giving me a headache. I am not sure if the finiteness bit has to do with algebra, topology or the fact that we are dealing with prime ideals.
This is not a homework problem but rather for self-study.
$\textbf{Edit:}$ I have posted my answer below after the discussion with Dylan and Pierre.
Best Answer
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.
$\hspace{6in} \square$