Spectrums of a commutative ring

Let $A$ be a commutative ring. (Recall that we always assume ring to be unital associative.) Amazingly enough(!?), any element $z$ in $A$ is central. As we have seen in the Schur's Lemma, for any ``finite dimensional'' irreducible representation $\rho$ of $A$ , $\rho(z)$ should be a scalar. Thus we see that any irreducible ``finite dimensional'' irreducible representation of $A$ should be one dimensional. Though this argument does not make sense when $A$ has no restriction such as ``$A$ is finitely generated over a field'', we may begin by considering a one-dimensional representation of $A$ . That means, a ring homomorphism

$$\rho: A \to K$$

where $K$ is a field. One knows that

1. $A/\operatorname{Ker}(\rho)$ is an integral domain. That means, it has no zero-divisor other than zero. (In this sense, $\operatorname{Ker}(\rho)$ is said to be a prime ideal of $A$ .)
2. $\rho$ is decomposed in the following way.
   $$A \to A/\operatorname{Ker}(\rho)\to Q(A/\operatorname{Ker}(\rho))\to K$$
   where $Q(B)$ is the field of fractions of a ring $B$ .

With a suitable definition of ``equivalence" of such representations, we may identify equivalence class of representation with the kernel $\operatorname{Ker}(\rho)$ .

In other words, we are interested in prime ideals.

DEFINITION 1.1   Let $A$ be a commutative ring. Then we define the set $\operatorname{Spec}(A)$ of spectrum of $A$ as the set of prime ideals of $A$ .

We note that for any $\mathfrak{p}\in \operatorname{Spec}(A)$ , we have a ring homomorphism (``representation associated to $\mathfrak{p}$ '') $\rho_\mathfrak{p}$ defined by

$$\rho_\mathfrak{p}:A \to A/\mathfrak{p}\to Q(A/\mathfrak{p}).$$

Since $A/\mathfrak{p}\to Q(A/\mathfrak{p})$ is an inclusion, we may say, by abuse of language, that the value of an element $f\in A$ under the representation $\rho_\mathfrak{p}$ is equal to $f \pmod \mathfrak{p}$ . We note further that

$$\mathfrak{p}=\{f \in A; \rho_\mathfrak{p}(f)=0 \}$$

holds.

Let us now define a topology on $\operatorname{Spec}(A)$ .

DEFINITION 1.2   Let $A$ be a commutative ring. For any $f\in A$ , we define a subset $O_f$ of $\operatorname{Spec}(A)$ defined by

$$O_f=\{\mathfrak{p}\in \operatorname{Spec}(A); \rho_\mathfrak{p}(f)\neq 0\}.$$

LEMMA 1.3   Let $A$ be a commutative ring. Then we have

$$O_f \cap O_g=O_{fg}$$

for any $f,g\in A$ . $\{O_f; f\in A\}$ . Thus we may introduce a topology on $\operatorname{Spec}(A)$ whose open sets are unions of various $O_f$ .

PROOF..

$$O_f \cap O_g =\{ \mathfrak{p}; \rho_\mathfrak{p}(f)\neq 0$$$$\text { and }\rho_\mathfrak{p}(g)\neq 0\} =\{ \mathfrak{p}; \rho_\mathfrak{p}(f g)\neq 0 \}$$

$\qedsymbol$

DEFINITION 1.4   The topology defined in the preceding Lemma is called the Zariski topology of $\operatorname{Spec}(A)$ .

In Part II, we always equip $\operatorname{Spec}(A)$ with the Zariski topology. Thus for any commutative ring $A$ , we may always associate a topological space $\operatorname{Spec}(A)$ .