Commutative Algebra (VIII): Spectrum of a Ring

08 Mar 2019

• commutative algebra

We have arrived at one of the central objects of algebraic geometry, the spectrum of a ring.

Motivation: algebraic geometry

First, we cover some motivation for the definition of the spectrum from algebraic geometry. The reader may skip this introductory section and proceed straight to the definition of the spectrum.

We will get a lot of mileage out of the following philosophy:

To understand a space, study the functions on the space.

For instance, in the context of algebraic geometry, suppose we want to study $\bb C^n$ and its algebraic subvarieties. (In this setting we usually write $\bb C^n$ as $\bb A^n$, the affine $n$-space.) Then the appropriate function space to look at is the polynomial ring $\bb C[X_1,\ldots,X_n]$. Every ideal $\mf a\subseteq\bb C[X_1,\ldots,X_n]$ defines an affine algebraic variety in $\bb A^n$, which is its vanishing locus:

$$\bb V(\mf a)\coloneqq\{(x_1,\ldots,x_n)\in\bb A^n\,:\,f(x_1,\ldots,x_n)=0\text{ for all }f\in\mf a\}.$$

It can be shown that the collection of all algebraic varieties in $\bb A^n$ form the closed sets for a topology, known as the Zariski topology on $\bb A^n$.

More generally, for any affine algebraic variety $V\subseteq\bb A^n$, the corresponding function space is the ring of polynomial functions on $V$, called its coordinate ring and denoted $\bb C[V]$. Note that

$$\bb C[V]=\bb C[X_1,\ldots,X_n]/\bb I(V),$$

where $\bb I(V)$ is the ideal of polynomials vanishing on $V$. Again, it is possible to define the Zariski topology on $V$.

The geometry of affine algebraic varieties has a natural effect on the coordinate rings, as follows: any morphism of affine algebraic varieties $F:V\to W$ induces a pullback map on the coordinate rings

$$\begin{aligned} F^\#:\bb C[W]&\to\bb C[V],\\ g\quad&\mapsto g\circ F. \end{aligned}$$

The most important link between $V$ and $\bb C[V]$ is given by Hilbert’s Nullstellensatz, which implies that the points of $V$ correspond to maximal ideals of $\bb C[V]$, ie. elements of the maximal spectrum

$$\maxSpec(\bb C[V])\coloneqq\{\mf m\subseteq\bb C[V]\,:\,\mf m\text{ maximal ideal}\}.$$

Moreover, we can define the Zariski topology on $\maxSpec(\bb C[V])$ such that the correspondence is in fact a homeomorphism $V\to\maxSpec(\bb C[V])$.

The point of the above construction is this: we have recovered the geometric object $V$ from algebraic data in $\bb C[V]$.

The above discussion works for rings of the form $\bb C[V]$, or perhaps $k[V]$ for algebraically closed fields $k$. What if we try to extend the construction to arbitrary (commutative unital) rings? Here the philosophy is that every ring $A$ is the “function ring” of some geometrical object, called the spectrum $\Spec(A)$.

From the discussion above, we want $\Spec(A)$ to have the following nice properties:

1. $\Spec(A)$ has a topology, defined using purely algebraic constructions from $A$;
2. Every ring homomorphism $A\to B$ induces a continuous map $\Spec(B)\to\Spec(A)$;
3. In fact, we want this to be functorial, ie. $\Spec$ is a contravariant functor from the category of (commutative unital) rings to the category of topological spaces.

We note that $\maxSpec$ actually does not work, because it fails to be functorial: this is essentially because the inverse image of a maximal ideal under a ring homomorphism might not be maximal. However, by replacing “maximal” with “prime” we circumvent this problem, and this turns out to be the right generalisation.

Spectrum of a ring

Note: Since the definition of the spectrum is motivated by the specific case of coordinate rings in algebraic geometry, it will be very helpful to keep the examples $A=\bb C[X_1,\ldots,X_n]$ or $A=\bb C[V]$ in mind from this point on.

Let $A$ be a (commutative unital) ring. The (prime) spectrum of $A$, denoted $\Spec(A)$, is the set of all prime ideals of $A$.

The first order of business is to define a topology on $\Spec(A)$, which is well-behaved with respect to the ring structure. For each subset $E\subseteq A$, let $V(E)$ denote the set of all prime ideals which contain $E$:

$$V(E)=\{\mf p\in\Spec(A)\,:\,E\subseteq\mf p\}.$$

We record some easy consequences of the definition:

1. $V(0)=\Spec(A)$ and $V(A)=\emptyset$;
2. $V$ is inclusion-reversing: if $E_1\subseteq E_2$, then $V(E_1)\supseteq V(E_2)$;
3. If $\mf a=AE$ is the ideal generated by $E$, then $V(E)=V(\mf a)$.

Here are more properties of $V$.

Proposition: Let $\mf a\subseteq A$ be an ideal. Then $V(\mf a)=V(\rad(\mf a))$.

Proof: ($\supseteq$) Clear since $\mf a\subseteq\rad(\mf a)$.

($\subseteq$) If $\mf p\in V(\mf a)$, then $\mf a\subseteq\mf p$, so $\rad(\mf a)\subseteq\rad(\mf p)=\mf p$, ie. $\mf p\in V(\rad(\mf a))$. $\qed$

Proposition: Let $(E_i)_ {i\in I}$ be a family of subsets of $A$. Then

$$V\left(\bigcup_{i\in I}E_i\right)=\bigcap_{i\in I}V(E_i).$$

Proof: Note that

$$\begin{aligned} \mf p\in V\left(\bigcup_{i\in I}E_i\right)&\iff E_i\subseteq\mf p\text{ for all }i\\ &\iff\mf p\in V(E_i)\text{ for all }i\\ &\iff\mf p\in\bigcap_{i\in I}V(E_i).\ \qed \end{aligned}$$

Proposition: Let $\mf a,\mf b\subseteq A$ be ideals. Then

$$V(\mf a\cap\mf b)=V(\mf a\mf b)=V(\mf a)\cup V(\mf b).$$

Proof: $V(\mf a\cap\mf b)=V(\mf a\mf b)$: Note that $\subseteq$ is clear, since $\mf a\cap\mf b\supseteq\mf a\mf b$.

In the other direction, suppose that $\mf p\not\in V(\mf a\cap\mf b)$. Then $\mf a\cap\mf b\not\subseteq\mf p$, so there exists $x\in\mf a\cap\mf b$ with $x\not\in\mf p$. But then $x^2\in\mf a\mf b$ and $x^2\not\in\mf p$. Hence $\mf a\mf b\not\subseteq\mf p$, so $\mf p\not\in V(\mf a\mf b)$.

$V(\mf a\mf b)=V(\mf a)\cup V(\mf b)$: Note that $\supseteq$ is clear, since $\mf a\mf b\subseteq\mf a$ and $\mf a\mf b\subseteq\mf b$.

In the other direction, suppose that $\mf p\in V(\mf a\mf b)$, so $\mf a\mf b\subseteq\mf p$. By definition of a prime ideal, this implies either $\mf a\subseteq\mf p$ or $\mf b\subseteq\mf p$, so $\mf p\in V(\mf a)\cup V(\mf b)$. $\qed$

Hence we have shown that the family of sets

$$\{V(E)\,:\,E\subseteq A\}\subseteq\mc P(\Spec(A))$$

contains $\emptyset$ and $\Spec A$, and is closed under arbitrary intersection and finite union. Thus this family of sets are the closed sets for a topology on $\Spec(A)$, known as the Zariski topology.

Proposition: Let $\mf a\subseteq A$ be an ideal. Then $\Spec(A/\mf a)$ is homeomorphic to $V(\mf a)$ (as a subspace of $\Spec(A)$).

Proof: Note that closed sets in $V(\mf a)$ are exactly those of the form $V(\mf a)\cap V(E)=V(\mf a\cup E)=V(A(\mf a\cup E))$, ie. exactly those of the form $V(\mf b)$ for ideals $J$ with $\mf a\subseteq\mf b\subseteq A$.

Now the lattice isomorphism theorem gives a correspondence

$$\begin{aligned} \{\text{ideals of }A/\mf a\}&\leftrightarrow\{\text{ideals of }A\text{ containing }\mf a\}\\ \underline{\mf c}\qquad&\mapsto\quad q^{-1}(\underline{\mf c})\\ q(\mf b)\quad&\gets\qquad\mf b, \end{aligned}$$

which preserves inclusion. It is easy to check that the maps in both directions send prime ideals to prime ideals. Hence this map sends $V(\underline{\mf c})\mapsto V(q^{-1}(\underline{\mf c}))$ and $V(q(\mf b))\gets V(\mf b)$, so it is the desired homeomorphism. $\qed$

Proposition: Let $\mf p\in\Spec(A)$. Then $\overline{\{\mf p\}}=V(\mf p)$.

Proof: We have

$$\begin{aligned} \overline{\{\mf p\}}&=\bigcap\{C\subseteq\Spec(A)\text{ closed}\,:\,\{\mf p\}\subseteq C\}\\ &=\bigcap\{V(E)\,:\,\mf p\in V(E)\}\\ &=V\left(\bigcup\{E\,:\,E\subseteq\mf p\}\right)\\ &=V(\mf p).\ \qed \end{aligned}$$

Corollary: $\mf p$ is a closed point in $\Spec(A)$ (ie. $\{\mf p\}$ is closed) if and only if $\mf p$ is maximal. $\qed$

Basic open sets

In this section, we further study the spectrum as a topological space under the Zariski topology. As such, we write $X=\Spec(A)$.

For each $f\in A$, let $X_f=X\backslash V(f)$. Note that this is a basis of open sets for the Zariski topology: any open set in $X$ is of the form $U=X\backslash V(\mf a)$ for some ideal $\mf a\subseteq A$, and

$$\begin{aligned} \mf p\in V(\mf a)&\iff\mf a\subseteq\mf p\\ &\iff f\in\mf p\text{ for all }f\in\mf a\\ &\iff(f)\subseteq\mf p\text{ for all }f\in\mf a\\ &\iff\mf p\in\bigcap_{f\in\mf a}V(f), \end{aligned}$$

so $U=\bigcup_{f\in\mf a}X_f$.

Here we collect some simple properties of the $X_f$:

Proposition: For all $f,g\in A$, we have:

1. $X_f\cap X_g=X_{fg}$;
2. $X_f=\emptyset\iff f$ is nilpotent;
3. $X_f=X\iff f$ is a unit;
4. $V(f)\subseteq V(g)\iff\rad(f)\supseteq\rad(g)$;
5. $X_f=X_g\iff\rad(f)=\rad(g)$.

Proof: (1) We have $fg\not\in\mf p$ if and only if $f\not\in\mf p$ and $g\not\in\mf p$.

(2) Both statements hold if and only if $f$ is in the intersection of all prime ideals of $A$, which is the nilradical of $A$.

(3) If $X_f=X$ then $f$ is not contained in any maximal ideal of $A$. Thus $(f)$ is not a proper ideal, ie. $(f)=A$, so $f$ is a unit.

Conversely, if $f$ is a unit then $(f)=A$, so $f$ is not contained in any proper ideal (hence any prime ideal) of $A$.

(4) Note that $V(f)=V(\rad(f))$ and $V(g)=V(\rad(g))$, so ($\Leftarrow$) is clear since $V$ is order-reversing, and ($\Rightarrow$) follows since $\rad(f)\in V(\rad(f))\subseteq V(\rad(g))$ implies $\rad(g)\subseteq\rad(f)$ by definition.

(5) This is clear from (4). $\qed$

Following the convention in algebraic geometry, we say that a topological space is quasi-compact if every open cover has a finite subcover, and it is compact if it is quasi-compact and Hausdorff.

Proposition: $X$ is quasi-compact.

Proof: Let $(U_i)_ {i\in I}$ be an open covering of $X$; without loss of generality, we may assume that the $U_i$ are basic open sets, so $U_i=X_{f_i}$ for some $f_i\in A$. Now if $\mf a$ is the ideal generated by all the $f_i$, then

$$V(\mf a)=\bigcap_{i\in I}V(f_i)=\emptyset,$$

so $\mf a$ is not contained in any maximal ideal, ie. $\mf a=A$. In particular, $1\in\mf a$ implies that there exists $i_1,\ldots,i_n\in I$ and $a_1,\ldots,a_n\in A$ such that

$$1=a_1f_{i_1}+\cdots+a_nf_{i_n},$$

so $A=Af_{i_1}+\cdots+Af_{i_n}$ implies $V(f_{i_1})\cap\cdots\cap V(f_{i_n})=\emptyset$. Hence $X_{f_1},\ldots,X_{f_n}$ form a finite subcover of $X$. $\qed$

Corollary: For any $f\in A$, the subspace $X_f\subseteq X$ is quasi-compact. $\qed$