Further properties of $\operatorname {Spec}$

LEMMA 0.11   Let $A$ be a ring. Then:

1. For any $f\in A$, $D(f)=\{\mathfrak{p}\in \operatorname{Spec}(A); f \notin \mathfrak{p}\}$ is an open set of $\operatorname{Spec}(A)$.
2. Given a point $\mathfrak{p}$ of $\operatorname{Spec}(A)$ and an open set $U$ which contains $\mathfrak{p}$, we may always find an element $f\in A$ such that $\mathfrak{p}\in D(f)\subset U$. (In other words, $\{D(f)\}$ forms an open base of the Zariski topology.

THEOREM 0.12   For any ring $A$, $\operatorname{Spec}(A)$ is compact. (But it is not Hausdorff in most of the case.)

DEFINITION 0.13   Let $X$ be a topological space. A closed set $F$ of $X$ is said to be reducible if there exist closed sets $F_1$ and $F_2$ such that

$$F=F_1 \cup F_2,\quad F_1\neq F, F_2 \neq F$$

holds. $F$ is said to be irreducible if it is not reducible.

Recall that we have defined, for any ring $A$ and for any ideal $I$, a closed subset $V(I)$ of $\operatorname{Spec}(A)$ by

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

We define:

DEFINITION 0.14   Let $A$ be a ring. Let $X$ be a subset of $\operatorname{Spec}(A)$. Then we define

$$I(X)= \{f \in A; \operatorname{eval}_{\mathfrak{p}}(f)=0 \quad ( \forall \mathfrak{p}\in I)\}.$$

LEMMA 0.15   Let $A$ be a ring. Then:

1. For any subset $X$ of $\operatorname{Spec}(A)$, $I(X)$ is an ideal of $A$.
2. (For any subset $S$ of $A$, $V(S)$ is a closed subset of $\operatorname{Spec}(A)$.)
3. For any subsets $X_1 \subset X_2$ of $\operatorname{Spec}(A)$, we have $I(X_1 )\supset I(X_2)$.
4. For any subsets $S_1\subset S_2$ of $A$, we have $V(S_1) \supset V(S_2)$.
5. For any subset $X$ of $\operatorname{Spec}(A)$, we have $V(I(X))\subset X$.
6. For any subset $S$ of $A$, we have $I(V(S))\subset S$.

COROLLARY 0.16   Let $A$ be a ring. Then:

1. For any subset $X$ of $\operatorname{Spec}(A)$, we have $I(V(I(X)))= I(X)$.
2. For any subset $S$ of $A$, we have $V(I(V(S)))=V(S)$.

DEFINITION 0.17   Let $I$ be an ideal of a ring $A$. Then we define its radical to be

$$\sqrt{I}=\{ x \in A; \exists N\in \mathbb{Z}_{>0}$$    such that $$x^N \in I\}.$$

PROPOSITION 0.18   Let $A$ be a ring. Then;

1. For any ideal $I$ of $A$, we have $V(I)=V(\sqrt{I})$.
2. For two ideals $I$, $J$ of $A$, $V(I)=V(J)$ holds if and only if $\sqrt{I}=\sqrt{J}$.
3. For an ideal $I$ of $A$, $V(I)$ is irreducible if and only if $\sqrt{I}$ is a prime ideal.