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

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 $f_i's \in \mathfrak{p}$, 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 $i's$ 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.

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It will be hard to not post a complete resolution of this, since you've done the hard work involved in the problem. You've shown that the collection $\{f_i\}_{i \in I}$ is not contained in any prime ideal. This implies that the smallest ideal containing the $\{f_i\}_{i \in I}$ is the unit ideal. You can check that the ideal generated by a family of elements (infinite or not) $\{f_i\}_{i \in I}$ consists of sums, for $J \subset I$ finite and $a_i \in A$ for each $i \in J$, $\sum_{i \in J} a_if_i$. – Dylan Moreland Jan 31 '12 at 14:44
Benjamin: Because of the definition you gave! (More precisely, because of the condition "almost all of the $x_i$ are zero".) (+1 for asking such a basic - and thus fundamental - question!) – Pierre-Yves Gaillard Jan 31 '12 at 14:47
@BenjaminLim As you say, when writing down elements you are always dealing with some finite subset of $I$. We don't really have a way of making sense of an infinite sum of non-zero elements (until Chapter 9!). So when you write $1 = \sum_{i \in I} a_if_i$, the $a_i$ are non-zero only for $i$ in a finite subset of $I$. – Dylan Moreland Jan 31 '12 at 14:49
@BenjaminLim Regarding finishing the problem, I would try to play around with $(\bigcup_{i \in J} X_{f_i})^c = \bigcap_{i \in J} (X_{f_i})^c = \bigcap_{i \in J} V(f_i)$. In the end, you want to show that this is $\varnothing$ and there are some helpful facts in some of the nearby exercises, IIRC. Let me know if I should say more. – Dylan Moreland Jan 31 '12 at 15:19
@DylanMoreland Dead easy. If $\bigcap_{i\in J} V(f_i)$ is not empty, there is a prime ideal that contains each of the $f_i's$. But then it contains any linear combination of them so that it must contain $1$ contradicting the assumption that a prime ideal must be a proper subset of the whole ring. – user38268 Jan 31 '12 at 15:30

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.

$\hspace{6in} \square$

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I very much enjoy it when the site works like this. – Dylan Moreland Jan 31 '12 at 19:50
@DylanMoreland You mean when people answer their own questions after some discussion? – user38268 Feb 1 '12 at 1:49
@BenjaminLim Thank you for the explanation. :) – Samuel Reid Feb 1 '12 at 10:02
@SamuelReid Thanks just wanted to show an example of compactness outside of the setting of metric spaces as in Rudin :D – user38268 Feb 1 '12 at 10:16

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 $$U(f)\subset\bigcup_{i\in F}\ U(\mathfrak a_i).$$

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It may just be that people avoid the "finite intersection property" formulation of compactness. A few of the hints in A-M are suboptimal in this way, too. Cool argument, though! – Dylan Moreland Feb 2 '12 at 23:37
@Pierre-YvesGaillard True but the thing with dealing with the basis sets seemed to me like the first "follow your nose" way of simplifying the problem. E.g. If we write an open cover of $X$, then each of the open sets be written as a union of basis elements, and then from there proceed to solve the problem. – user38268 Feb 3 '12 at 1:36