$\operatorname {Spec}A$

DEFINITION 0.3   An ideal $I$ of a ring $A$ is said to be

1. a prime ideal if $A/I$ is an integral domain.
2. a maximal ideal if $A/I$ is a field.

DEFINITION 0.4   Let $A$ be a ring. Then we define its affine spectrum as

$$\operatorname{Spec}(A)=\{ \mathfrak{p}\subset A ; \mathfrak{p}$$ is a prime ideal of $A$$$\}.$$

DEFINITION 0.5   Let $A$ be a ring. For any $\mathfrak{p}\in \operatorname{Spec}(A)$ we define “evaluation map” $\operatorname{eval}_\mathfrak{p}$ as follows:

$$\operatorname{eval}_\mathfrak{p}: A \to A/\mathfrak{p}$$

Note that $A/\mathfrak{p}$ is a subring of a field $Q(A/\mathfrak{p})$, the field of fractions of the integral domain $A/\mathfrak{p}$. We interpret each element $f$ of $A$ as a something of a “fuction”, whose value at a point $\mathfrak{p}$ is given by $\operatorname{eval}_\mathfrak{p}(f)$.

We introduce a topology on $\operatorname{Spec}(A)$. We basically mimic the following Lemma:

LEMMA 0.6   Let $X$ be a topological space. then for any continuous function $f: X \to \mathbb{C}$, its zero points $\{ x \in X; f(x)=0\}$ is a closed subset of $X$. Furthermore, for any family $\{f_\lambda\}$ of continous $\mathbb{C}$-valued functions, its common zeros $\{x \in X; f_\lambda(x)=0 \quad (\forall \lambda)\}$ is a closed subset of $X$.

DEFINITION 0.7   Let $A$ be a ring. Let $S$ be a subset of $A$, then we define the common zero of $S$ as

$$V(S)=\{ \mathfrak{p}\in \operatorname{Spec}(A) ; \operatorname{eval}_\mathfrak{p}(f)=0 \qquad(\forall f\in S\}.$$

For any subset $S$ of $A$, let us denote by $\langle S \rangle_{A}$ the ideal of $A$ generated by $S$. Then we may soon see that we have $V(S)=V(\langle S\rangle_A)$. So when thinking of $V(S)$ we may in most cases assume that $S$ is an ideal of $A$.

LEMMA 0.8   Let $A$ be a ring. Then:

1. $V(0)=\operatorname{Spec}(A), \quad V(\{1\})(=V(A))=\emptyset.$
2. For any family $\{I_\lambda\}$ of ideals of $A$, we have $\cap_{\lambda} V(I_\lambda) = V(\sum_\lambda I_{\lambda}).$
3. For any ideals $I,J$ of $A$, we have $V(I)\cup V(J)=V(I \cdot J)$.

PROPOSITION 0.9   Let $A$ be a ring. $\{ V(I); I$    is an ideal of $A\}$ satisfies the axiom of closed sets of $\operatorname{Spec}(A)$. We call this the Zariski topology of $\operatorname{Spec}(A)$.

PROBLEM 0.10   Prove Lemma 3.8.

Algebraic geometry and Ring theory

Yoshifumi Tsuchimoto