prime spectrum

1 Spec as a set

Let $R$ be any commutative ring with identity. The primespectrum $\mathrm{Spec}(R)$ of $R$ is defined to be the set

$$\{P⊊R\mid P\text{ is a prime ideal of }R\}.$$

For any subset $A$ of $R$, we define the variety of $A$ to bethe set

$$V(A):=\{P\in \mathrm{Spec}(R)\mid A\subset P\}\subset \mathrm{Spec}(R)$$

It is enough to restrict attention to subsets of $R$ which are ideals,since, for any subset $A$ of $R$, we have $V(A)=V(I)$ where $I$ isthe ideal generated by $A$. In fact, even more is true: $V(I)=V(\sqrt{I})$ where $\sqrt{I}$ denotes the radical of the ideal $I$.

2 Spec as a topological space

We impose a topology on $\mathrm{Spec}(R)$ by defining the sets $V(A)$ to bethe collection of closed subsets of $\mathrm{Spec}(R)$ (that is, a subset of$\mathrm{Spec}(R)$ is open if and only if it equals the complement of $V(A)$for some subset $A$). The equations

${\displaystyle \bigcap _{\alpha }}V(I_{\alpha })$$=$$V\left({\displaystyle \bigcup _{\alpha }}{I}_{\alpha }\right)$

${\displaystyle \bigcup _{i=1}^n}V(I_i)$$=$$V\left({\displaystyle \bigcap _{i=1}^n}{I}_i\right),$

for any ideals $I_{\alpha }$, $I_i$ of $R$, establish that this collection does constitute a topology on$\mathrm{Spec}(R)$. This topology is called the Zariski topology inlight of its relationship to the Zariski topology on an algebraicvariety (see Section 4 below). Note that a point $P\in \mathrm{Spec}(R)$ is closed if and only if $P\subset R$ is a maximal ideal.

A distinguished open set of $\mathrm{Spec}(R)$ is defined to be an openset of the form

$$\mathrm{Spec}{(R)}_f:=\{P\in \mathrm{Spec}(R)\mid f\notin P\}=\mathrm{Spec}(R)\setminus V(\{f\}),$$

for any element $f\in R$. The collection of distinguished open setsforms a topological basis for the open sets of $\mathrm{Spec}(R)$. In fact, wehave

$$\mathrm{Spec}(R)\setminus V(A)=\bigcup _{f\in A}\mathrm{Spec}{(R)}_f.$$

The topological space $\mathrm{Spec}(R)$ has the following additionalproperties:

• $\mathrm{Spec}(R)$ is compact (but almost never Hausdorff).

• A subset of $\mathrm{Spec}(R)$ is an irreducible closed set if and only if it equals $V(P)$ for some prime ideal $P$ of $R$.

• For $f\in R$, let $R_f$ denote the localization of $R$ at$f$. Then the topological spaces $\mathrm{Spec}{(R)}_f$ and $\mathrm{Spec}(R_f)$ arenaturally homeomorphic, via the correspondence sending a prime idealof $R$ not containing $f$ to the induced prime ideal in $R_f$.

• For $P\in \mathrm{Spec}(R)$, let $R_P$ denote the localization of $R$at the prime ideal $P$. Then the topological spaces $V(P)\subset \mathrm{Spec}(R)$ and $\mathrm{Spec}(R_P)$ are naturally homeomorphic, via thecorrespondence sending a prime ideal of $R$ contained in $P$ to theinduced prime ideal in $R_P$.

3 Spec as a sheaf

For convenience, we adopt the usual convention of writing $X$ for$\mathrm{Spec}(R)$. For any $f\in R$ and $P\in X_f$, let ${\iota }_{f,P}:R_f⟶R_P$ be the natural inclusion map. Define a presheaf of rings${𝒪}_X$ on $X$ by setting

$${𝒪}_X(U):=\left\{(s_P)\in \prod _{P\in U}{R}_P\right|\},$$

for each open set $U\subset X$. The restriction map ${\mathrm{res}}_{U,V}:{𝒪}_X(U)⟶{𝒪}_X(V)$ is the map induced by the projection map

$$\prod _{P\in U}{R}_P⟶\prod _{P\in V}{R}_P,$$

for each open subset $V\subset U$. The presheaf ${𝒪}_X$ satisfies thefollowing properties:

1. ${𝒪}_X$ is a sheaf.

2. ${𝒪}_X(X_f)=R_f$ for every $f\in R$.

3. The stalk ${({𝒪}_X)}_P$ is equal to $R_P$ for every $P\in X$. (In particular, $X$ is a locally ringed space.)

4. The restriction sheaf of ${𝒪}_X$ to $X_f$ is isomorphic as asheaf to ${𝒪}_{\mathrm{Spec}(R_f)}$.

4 Relationship to algebraic varieties

$\mathrm{Spec}(R)$ is sometimes called an affine scheme because of theclose relationship between affine varieties in ${𝔸}_k^n$ and the$\mathrm{Spec}$ of their corresponding coordinate rings. In fact, thecorrespondence between the two is an equivalence of categories,although a complete statement of this equivalence requires the notionof morphisms of schemes and will not be givenhere. Nevertheless, we explain what we can of this correspondencebelow.

Let $k$ be a field and write as usual ${𝔸}_k^n$ for the vector space$k^n$. Recall that an affine variety $V$ in ${𝔸}_k^n$ is the set ofcommon zeros of some prime ideal $I\subset k[X_1,\mathrm{\dots },X_n]$. The coordinate ring of $V$ is defined to bethe ring $R:=k[X_1,\mathrm{\dots },X_n]/I$, and there is an embedding $i:V\hookrightarrow \mathrm{Spec}(R)$ given by

$$i(a_1,\mathrm{\dots },a_n):=(X_1-a_1,\mathrm{\dots },X_n-a_n)\in \mathrm{Spec}(R).$$

The function $i$ is not a homeomorphism, because it is not abijection (its image is contained inside the set of maximal ideals of $R$). However,the map $i$ does define an order preserving bijection between the opensets of $V$ and the open sets of $\mathrm{Spec}(R)$ in the Zariski topology.This isomorphism between these two lattices of open sets can be usedto equate the sheaf $\mathrm{Spec}(R)$ with the structure sheaf of the variety$V$, showing that the two objects are identical in every respectexcept for the minor detail of $\mathrm{Spec}(R)$ having more points than$V$.

The additional points of $\mathrm{Spec}(R)$ are valuable in many situationsand a systematic study of them leads to the general notion ofschemes. As just one example, the classical Bezout’s theorem is onlyvalid for algebraically closed fields, but admits a scheme–theoreticgeneralization which holds over non–algebraically closed fields aswell. We will not attempt to explain the theory of schemes in detail,instead referring the interested reader to the references below.

Remark. The spectrum $\mathrm{Spec}(R)$ of a ring $R$ may be generalized to the case when $R$ is not commutative, as long as $R$ contains the multiplicative identity. For a ring $R$ with $1$, the $\mathrm{Spec}(R)$, like above, is the set of all proper prime ideals of $R$. This definition is used to develop the noncommutative version of Hilbert’s Nullstellensatz.