Vtgyjfy

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 $X_f = {mathfrak{p} in operatorname{Spec} (A) : {f} notsubseteq mathfrak{p} }$ 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.

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

Suppose that $X$ is covered by $bigcup_{iin I} X_{f_i}$. 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 $X_f:=Xsetminus V(f)$ generate the topology of $X:=text{Spec}A$.

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

$bulletquaddisplaystyle Vleft(sum_{iin I}mathfrak a_iright)=bigcap_{iin I}V(mathfrak a_i)=varnothing,$

$bulletquaddisplaystylesum_{iin I}mathfrak a_i=(1),$

$bulletquaddisplaystylesum_{iin F}mathfrak a_i=(1)$ for some finite subset $F$ of $I$,

$bulletquaddisplaystylebigcap_{iin F}V(mathfrak a_i)=Vleft(sum_{iin F}mathfrak a_iright)=V(1)=varnothing$.

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
$$
Msubset Nimplies U(M)subset U(M),
$$
$$
Uleft(bigcup_{iin I} M_iright)=bigcup_{iin I} U(M_i),
$$
and, for ideals $mathfrak a$ and $mathfrak b$,
$$
U(mathfrak acapmathfrak b)=U(mathfrak a)cap U(mathfrak b).
$$
More generally we have
$$
U(0)=varnothing,quad U(1)=X,
$$
$$
Uleft(bigcup_{iin I}M_iright)=Uleft(sum_{iin I}M_iright)=bigcup_{iin I} U(M_i),
$$
$$
U(mathfrak acapmathfrak b)=U(mathfrak amathfrak b)=U(mathfrak a)cap U(mathfrak b),
$$
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
$$
U(mathfrak a)=bigcup_{finmathfrak a} U(f)qquad(*)
$$
shows that the $U(f),fin A$, 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
$$
U(f)subsetbigcup_{iin I} U(mathfrak a_i),
$$
then some power $f^k$ of $f$ is in $sum_{iin I}mathfrak a_i$. But then $f^k$ is in $sum_{iin F}mathfrak a_i$ for some finite subset $F$ of $I$, implying
$$
U(f)subsetbigcup_{iin F} U(mathfrak a_i).
$$