Zariski Topology on Prime Spectra

Fact [L1] gives exactly the four closed-set properties listed in the statement for the family $\{V(I)\}$ of subsets of $\mathrm{Spec}(R)$.

By [A1], any family with those four properties is the family of closed sets of a topology on the underlying set. Therefore the subsets $V(I)$ define a topology on $\mathrm{Spec}(R)$.

The vanishing sets are precisely the closed sets of the Zariski topology on $\mathrm{Spec}(R)$.

Choose $J$ with $Z=V(J)$ as in [A1]. Then $I(Z)={\bigcap }_{\mathfrak{p}\in V(J)}\mathfrak{p}=\sqrt{J}$ by [L1]. In particular, $I(Z)$ is radical.

Since step 1.1 gives $I(Z)=\sqrt{J}$, fact [L2] yields $V(I(Z))=V(\sqrt{J})=V(J)=Z$.

If also $Z=V(K)$ for an ideal $K$, then $V(K)=V(J)$, so [L2] gives $\sqrt{K}=\sqrt{J}=I(Z)$. Thus any radical ideal defining $Z$ equals $I(Z)$.

Therefore every Zariski-closed subset has the unique radical defining ideal $I(Z)$.

Let $\mathfrak{p}\in U$. Since $\mathfrak{p}\mathrm{\notin }V(I)$, the ideal $I$ is not contained in $\mathfrak{p}$. Choose $f\in I\setminus \mathfrak{p}$. Then $\mathfrak{p}\in D(f)$ by [L1], so $U\subseteq {\bigcup }_{f\in I}D(f)\mathrm{.}$

Conversely, if $\mathfrak{p}\in D(f)$ for some $f\in I$, then $f\mathrm{\notin }\mathfrak{p}$, so certainly $I⊈\mathfrak{p}$. Hence $\mathfrak{p}\mathrm{\notin }V(I)$ and therefore $\mathfrak{p}\in U$. Thus ${\bigcup }_{f\in I}D(f)\subseteq U\mathrm{.}$

Steps 1.1 and 1.2 prove the displayed equality, so every Zariski-open subset is a union of distinguished opens.

By [L1], the open set $U$ is a union of distinguished opens. Since $\mathfrak{p}\in U$, there exists $f\in R$ with $\mathfrak{p}\in D(f)$ and $D(f)\subseteq U$.

The chosen $D(f)$ is therefore a distinguished-open neighbourhood of $\mathfrak{p}$ contained in $U$.

Hence every point of a Zariski-open set has a distinguished-open neighbourhood inside that open set.

Let $Z\subseteq \mathrm{Spec}(R)$ be closed. By [L2], $Z=V(I)$ for some ideal $I⊴R$. Then [L1] gives $\mathrm{Spec}(\phi {)}^{-1}(Z)=\mathrm{Spec}(\phi {)}^{-1}(V(I))=V(IA)$, which is closed in $\mathrm{Spec}(A)$ by [L2]. Therefore $\mathrm{Spec}(\phi )$ is continuous.

For a prime ideal $\mathfrak{p}\subseteq R$ one has $\mathrm{Spec}({\mathrm{id}}_R)(\mathfrak{p})={\mathrm{id}}_R^{-1}(\mathfrak{p})=\mathfrak{p}$, so $\mathrm{Spec}({\mathrm{id}}_R)={\mathrm{id}}_{\mathrm{Spec}(R)}$.

For a prime ideal $\mathfrak{q}\subseteq B$, one has $\mathrm{Spec}(\psi \circ \phi )(\mathfrak{q})=(\psi \circ \phi {)}^{-1}(\mathfrak{q})={\phi }^{-1}({\psi }^{-1}(\mathfrak{q}))=\mathrm{Spec}(\phi )(\mathrm{Spec}(\psi )(\mathfrak{q}))$. Hence $\mathrm{Spec}(\psi \circ \phi )=\mathrm{Spec}(\phi )\circ \mathrm{Spec}(\psi )$.

Steps 1.1, 1.2, and 1.3 prove that $\mathrm{Spec}$ is a contravariant functor to topological spaces.

By [L1], contraction gives a bijection $c:\mathrm{Spec}(R\mathrm{/}I)\to V(I)$.

Let $J$ be an ideal of $R$ containing $I$. If $\mathfrak{q}\in \mathrm{Spec}(R\mathrm{/}I)$ has contraction $\mathfrak{p}={\pi }^{-1}(\mathfrak{q})$, then $\mathfrak{q}\supseteq J\mathrm{/}I⟺\mathfrak{p}\supseteq J\mathrm{.}$ Therefore $c(V_{R\mathrm{/}I}(J\mathrm{/}I))=V_R(J)\mathrm{.}$ Since $V_R(J)\subseteq V_R(I)$, this is closed in the subspace $V(I)$.

By [L2], every closed subset of $\mathrm{Spec}(R\mathrm{/}I)$ has the form $V_{R\mathrm{/}I}(K)$ for some ideal $K⊴R\mathrm{/}I$. Writing $J={\pi }^{-1}(K)$, one has $K=J\mathrm{/}I$, so step 1.2 shows that $c$ sends every closed subset of $\mathrm{Spec}(R\mathrm{/}I)$ to a closed subset of $V(I)$.

Conversely, let $C\subseteq V(I)$ be closed. By [A1], $C=V(I+J)$ for some ideal $J⊴R$. Since $I+J$ contains $I$, step 1.2 gives $C=V_R(I+J)=c(V_{R\mathrm{/}I}((I+J)\mathrm{/}I)),$ so $c^{-1}$ also carries closed sets to closed sets.

The bijection $c$ and its inverse both preserve closed sets, so $c$ is a homeomorphism from $\mathrm{Spec}(R\mathrm{/}I)$ onto the closed subspace $V(I)$.

By [L1], contraction gives a bijection $c:\mathrm{Spec}(S^{-1}R)\to X$.

Let $I⊴R$. If $\mathfrak{q}\in \mathrm{Spec}(S^{-1}R)$ has contraction $\mathfrak{p}$, then the inverse description in [L1] gives $\mathfrak{q}=S^{-1}\mathfrak{p}$. Therefore $\mathfrak{q}\supseteq S^{-1}I⟺\mathfrak{p}\supseteq I,$ and hence $c(V_{S^{-1}R}(S^{-1}I))=X\cap V_R(I)\mathrm{.}$

By [L2], every closed subset of $\mathrm{Spec}(S^{-1}R)$ has the form $V_{S^{-1}R}(K)$ for some ideal $K⊴S^{-1}R$. With $I={\lambda }^{-1}(K)$, assumption [A1] gives $K=S^{-1}I$, so step 1.2 shows that $c$ sends every closed subset of $\mathrm{Spec}(S^{-1}R)$ to a closed subset of $X$.

Conversely, if $C\subseteq X$ is closed, then [A2] gives $C=X\cap V_R(I)$ for some ideal $I⊴R$. Step 1.2 then yields $C=c(V_{S^{-1}R}(S^{-1}I)),$ so $c^{-1}$ also preserves closed sets.

The bijection $c$ and its inverse both preserve closed sets, so $c$ is a homeomorphism from $\mathrm{Spec}(S^{-1}R)$ onto the subspace $X$.

Apply [L1] with $S=\{1,f,f^2,\dots \}$. Its image consists of the prime ideals $\mathfrak{p}$ such that $\mathfrak{p}\cap S=\mathrm{\varnothing }$.

For a prime ideal $\mathfrak{p}$, the condition $\mathfrak{p}\cap \{1,f,f^2,\dots \}=\mathrm{\varnothing }$ is equivalent to $f\mathrm{\notin }\mathfrak{p}$, because $f\in \mathfrak{p}$ implies $f^n\in \mathfrak{p}$ for every $n\ge 1$, while $f^n\in \mathfrak{p}$ implies $f\in \mathfrak{p}$ by primality. By [L2], this image is exactly $D(f)$.

Therefore $\mathrm{Spec}(R_f)$ is homeomorphic to the distinguished open subset $D(f)$.

Let $J$ be the ideal generated by $\{f_{\lambda }{\}}_{\lambda \in \mathrm{\Lambda }}$. If $J\mathrm{\ne }R$, then $J$ is a proper ideal, so $R$ is nonzero and [L1] gives a maximal ideal $\mathfrak{m}$ containing $J$.

For every $\lambda$, one has $f_{\lambda }\in J\subseteq \mathfrak{m}$, so $\mathfrak{m}\mathrm{\notin }D(f_{\lambda })$. Hence $\mathfrak{m}\mathrm{\notin }{\bigcup }_{\lambda \in \mathrm{\Lambda }}D(f_{\lambda })$, contradicting the assumed cover of $\mathrm{Spec}(R)$.

The contradiction shows that $J$ cannot be proper. Therefore $J=R$.

Let $\mathfrak{p}\in \mathrm{Spec}(R)$. If $\mathfrak{p}$ were outside $D(f_1)\cup \cdots \cup D(f_n)$, then [L1] would give $f_i\in \mathfrak{p}$ for every $i$. Because $\mathfrak{p}$ is an ideal, it would then contain the sum $a_1{f}_1+\cdots +a_n{f}_n=1$, impossible for a prime ideal.

Therefore every prime ideal lies in at least one $D(f_i)$, so $\mathrm{Spec}(R)=D(f_1)\cup \cdots \cup D(f_n)$.

The displayed unit expression yields a finite distinguished-open subcover of the spectrum.

If $\mathrm{Spec}(R)=\mathrm{\varnothing }$, then the empty subfamily of $\mathcal{U}$ already covers it. By [L1], the spectrum is compact in this case.

Assume now that $\mathrm{Spec}(R)\mathrm{\ne }\mathrm{\varnothing }$. For each $\mathfrak{p}\in \mathrm{Spec}(R)$, choose $U_{\mathfrak{p}}\in \mathcal{U}$ with $\mathfrak{p}\in U_{\mathfrak{p}}$. By [L2], choose $f_{\mathfrak{p}}\in R$ with $\mathfrak{p}\in D(f_{\mathfrak{p}})\subseteq U_{\mathfrak{p}}$. Then $\{D(f_{\mathfrak{p}}){\}}_{\mathfrak{p}\in \mathrm{Spec}(R)}$ is a distinguished-open cover of the spectrum.

By [L3], the ideal generated by the family $\{f_{\mathfrak{p}}\}$ is $R$. Hence there exist finitely many primes ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n\in \mathrm{Spec}(R)$ and coefficients $a_1,\dots ,a_n\in R$ such that $1=a_1{f}_{{\mathfrak{p}}_1}+\cdots +a_n{f}_{{\mathfrak{p}}_n}$.

Applying [L4] to this identity gives $\mathrm{Spec}(R)=D(f_{{\mathfrak{p}}_1})\cup \cdots \cup D(f_{{\mathfrak{p}}_n})\subseteq U_{{\mathfrak{p}}_1}\cup \cdots \cup U_{{\mathfrak{p}}_n}$. Thus $\mathcal{U}$ has a finite subcover.

Steps 1.1 and 3.1 show that every open cover of $\mathrm{Spec}(R)$ has a finite subcover. Therefore $\mathrm{Spec}(R)$ is compact by [L1].

By [L1], there is a homeomorphism $h:\mathrm{Spec}(R_f)\to D(f)$. By [L2], the domain $\mathrm{Spec}(R_f)$ is compact.

Apply [A1] to the open cover $\mathcal{U}$ of $D(f)$. The inverse images $h^{-1}(U)$ with $U\in \mathcal{U}$ form an open cover of $\mathrm{Spec}(R_f)$, so compactness from step 1.1 gives finitely many $U_1,\dots ,U_n\in \mathcal{U}$ whose inverse images cover $\mathrm{Spec}(R_f)$. Then $U_1,\dots ,U_n$ cover $D(f)$.

Thus every open cover of $D(f)$ has a finite subcover, so $D(f)$ is compact by [L3].

Since $\mathfrak{p}\supseteq \mathfrak{p}$, the point $\mathfrak{p}$ lies in $V(\mathfrak{p})$ by [L2]. Because $V(\mathfrak{p})$ is closed, the closure $\overline{\{\mathfrak{p}\}}$ is contained in $V(\mathfrak{p})$.

Let $Z$ be a closed subset containing $\mathfrak{p}$. By [L1], write $Z=V(I)$ for its radical defining ideal $I$. Since $\mathfrak{p}\in Z$, fact [L2] gives $I\subseteq \mathfrak{p}$. Therefore every prime ideal containing $\mathfrak{p}$ also contains $I$, so $V(\mathfrak{p})\subseteq V(I)=Z$.

Step 1.2 shows that every closed set containing $\mathfrak{p}$ also contains $V(\mathfrak{p})$. Hence $V(\mathfrak{p})$ is the smallest closed set containing $\mathfrak{p}$, that is, $\overline{\{\mathfrak{p}\}}=V(\mathfrak{p})$.

By [L1] and [L2], $\mathfrak{q}$ is a specialisation of $\mathfrak{p}$ exactly when $\mathfrak{q}\in V(\mathfrak{p})$.

By the definition of $V(\mathfrak{p})$, the condition $\mathfrak{q}\in V(\mathfrak{p})$ is exactly $\mathfrak{p}\subseteq \mathfrak{q}$.

Therefore specialisation in $\mathrm{Spec}(R)$ is reverse inclusion of prime ideals.

If $\overline{\{\mathfrak{p}\}}=\overline{\{\mathfrak{q}\}}$, then each point lies in the closure of the other. Hence each is a specialisation of the other, so [L1] gives both $\mathfrak{p}\subseteq \mathfrak{q}$ and $\mathfrak{q}\subseteq \mathfrak{p}$. Therefore $\mathfrak{p}=\mathfrak{q}$, contrary to the hypothesis.

Thus distinct prime ideals have distinct closures. By [A1], this is exactly the $T_0$ property.

Therefore $\mathrm{Spec}(R)$ is $T_0$.

The point $\mathfrak{p}$ is closed exactly when $\overline{\{\mathfrak{p}\}}=\{\mathfrak{p}\}$. By [L1], this is equivalent to $V(\mathfrak{p})=\{\mathfrak{p}\}$.

If $\mathfrak{p}$ is maximal, then every prime ideal containing $\mathfrak{p}$ equals $\mathfrak{p}$ by [L2]. Hence $V(\mathfrak{p})=\{\mathfrak{p}\}$, so $\mathfrak{p}$ is a closed point.

Conversely, if $\{\mathfrak{p}\}$ is closed, then step 1.1 gives $V(\mathfrak{p})=\{\mathfrak{p}\}$. If $\mathfrak{p}\subseteq \mathfrak{q}$ for a prime ideal $\mathfrak{q}$, then $\mathfrak{q}\in V(\mathfrak{p})$ and therefore $\mathfrak{q}=\mathfrak{p}$. Thus no strictly larger proper ideal can contain $\mathfrak{p}$, so [L2] shows that $\mathfrak{p}$ is maximal.

Therefore the closed points of $\mathrm{Spec}(R)$ are exactly the maximal ideals.

Suppose first that $Z$ is irreducible. By [L1], write $Z=V(\mathfrak{a})$ with $\mathfrak{a}$ radical. Let $xy\in \mathfrak{a}$. Then $V(\mathfrak{a})\subseteq V((xy))=V((x))\cup V((y))$ by [L2]. Since $Z$ is irreducible, $Z$ is contained in $V((x))$ or in $V((y))$. By [L1], this means $x\in I(Z)=\mathfrak{a}$ or $y\in I(Z)=\mathfrak{a}$. Therefore $\mathfrak{a}$ is prime.

Suppose conversely that $\mathfrak{a}$ is prime. If $Z=F_1\cup F_2$ with $F_1,F_2$ closed subsets of $Z$, then each $F_i$ is closed in the ambient spectrum, so [L1] gives radical ideals ${\mathfrak{b}}_i$ with $F_i=V({\mathfrak{b}}_i)$. Using $Z=V(\mathfrak{a})$ and [L2], one has $V(\mathfrak{a})=V({\mathfrak{b}}_1)\cup V({\mathfrak{b}}_2)=V({\mathfrak{b}}_1{\mathfrak{b}}_2)$. Since [L1] identifies $\mathfrak{a}$ as the radical defining ideal of $Z$, one has ${\mathfrak{b}}_1{\mathfrak{b}}_2\subseteq \mathfrak{a}$. Primality of $\mathfrak{a}$ then gives ${\mathfrak{b}}_1\subseteq \mathfrak{a}$ or ${\mathfrak{b}}_2\subseteq \mathfrak{a}$, so $V(\mathfrak{a})\subseteq V({\mathfrak{b}}_1)$ or $V(\mathfrak{a})\subseteq V({\mathfrak{b}}_2)$. Hence $Z=F_1$ or $Z=F_2$, and $Z$ is irreducible.

Under either condition, steps 1.1 and 1.2 show that $\mathfrak{a}$ is prime and $Z=V(\mathfrak{a})$. By [L3], the closure of $\{\mathfrak{a}\}$ is $V(\mathfrak{a})=Z$, so $\mathfrak{a}$ is a generic point of $Z$ in the sense of Specialisations, generalisations, and generic points.

If $\eta$ is another generic point of $Z$, then [L3] gives $\overline{\{\eta \}}=Z=\overline{\{\mathfrak{a}\}}$. Fact [L4] now forces $\eta =\mathfrak{a}$. Thus the generic point is unique.

Therefore a nonempty Zariski-closed subset is irreducible exactly when its radical defining ideal is prime, and in that case it has the unique generic point $\mathfrak{a}$.

Let $\mathfrak{p}$ be a minimal prime ideal. Since $\mathfrak{p}$ is prime, fact [L1] shows that $V(\mathfrak{p})$ is irreducible.

If $V(\mathfrak{p})\subseteq Y$ for an irreducible closed subset $Y$, then [L1] gives $Y=V(\mathfrak{q})$ for a prime ideal $\mathfrak{q}$. The inclusion $V(\mathfrak{p})\subseteq V(\mathfrak{q})$ means $\mathfrak{q}\subseteq \mathfrak{p}$ by [L2]. Minimality of $\mathfrak{p}$ forces $\mathfrak{q}=\mathfrak{p}$, so $Y=V(\mathfrak{p})$. Hence $V(\mathfrak{p})$ is maximal among irreducible closed subsets, that is, an irreducible component.

Let $Y$ be an irreducible component. By [L1], $Y=V(\mathfrak{p})$ for a prime ideal $\mathfrak{p}$. If $\mathfrak{q}\subseteq \mathfrak{p}$ is another prime, then $V(\mathfrak{p})\subseteq V(\mathfrak{q})$ by [L2]. Since $V(\mathfrak{q})$ is irreducible by [L1], maximality of the component $Y$ forces $V(\mathfrak{p})=V(\mathfrak{q})$, and [L1] then gives $\mathfrak{p}=\mathfrak{q}$. Thus $\mathfrak{p}$ is minimal.

The prime ideal in step 1.3 is unique because [L1] gives a unique generic point for every irreducible component.

Steps 1.1, 1.2, 1.3, and 2.1 give the claimed correspondence between irreducible components and minimal prime ideals.

Let $Z_0\supseteq Z_1\supseteq Z_2\supseteq \cdots$ be a descending chain of closed subsets of $\mathrm{Spec}(R)$. By [L2], each $Z_n$ has a unique radical defining ideal ${\mathfrak{a}}_n$ with $Z_n=V({\mathfrak{a}}_n)$.

Since $Z_{n+1}\subseteq Z_n$, every prime ideal in $Z_{n+1}$ also lies in $Z_n$. Therefore ${\mathfrak{a}}_n={\bigcap }_{\mathfrak{p}\in Z_n}\mathfrak{p}\subseteq {\bigcap }_{\mathfrak{p}\in Z_{n+1}}\mathfrak{p}={\mathfrak{a}}_{n+1}\mathrm{.}$ Thus ${\mathfrak{a}}_0\subseteq {\mathfrak{a}}_1\subseteq {\mathfrak{a}}_2\subseteq \cdots$ is an ascending chain of ideals.

By [L3], the ideal chain from step 2.1 stabilizes: there is $N$ such that ${\mathfrak{a}}_n={\mathfrak{a}}_N$ for all $n\ge N$. Then $Z_n=V({\mathfrak{a}}_n)=V({\mathfrak{a}}_N)=Z_N$ for all $n\ge N$. Hence the closed chain stabilizes.

By [L1], stabilization of every descending closed chain means that $\mathrm{Spec}(R)$ is Noetherian.

By [L2], the ring $R$ has only finitely many minimal prime ideals.

By [L1], each irreducible component is $V(\mathfrak{p})$ for a unique minimal prime $\mathfrak{p}$, and each minimal prime gives an irreducible component. Therefore the set of irreducible components is finite.

Hence $\mathrm{Spec}(R)$ has only finitely many irreducible components.

By [L1], there are radical ideals $I,J⊴R$ such that $C=V(I)$ and $\mathrm{Spec}(R)\setminus C=V(J)$. Since $C$ and its complement are disjoint and cover the spectrum, [A1] gives $V(I+J)=\mathrm{\varnothing }$ and $V(IJ)=\mathrm{Spec}(R)$. If $I+J$ were proper, then either $R$ would be the zero ring, in which case $I+J=R$ anyway, or [L4] would place $I+J$ inside a maximal ideal, hence inside a prime ideal, contradicting $V(I+J)=\mathrm{\varnothing }$. Therefore $I+J=R$, and [L3] gives $I\cap J=IJ$.

Every element of $IJ$ therefore lies in every prime ideal of $R$. By [L2], $IJ\subseteq \mathrm{Nil}(R)$.

Choose $x\in I$ and $y\in J$ with $x+y=1$. Then $x(1-x)=xy\in IJ\subseteq \mathrm{Nil}(R)$ by step 2.1, so $x^n(1-x{)}^n=0$ for some $n\ge 1$. Expanding $1=(x+(1-x){)}^{2n-1}$ shows that every term is divisible by $x^n$ or by $(1-x{)}^n$, so there exist $a,b\in R$ with $1=a{x}^n+b(1-x{)}^n$. Put $e=a{x}^n$. Then $e\in I$, $1-e=b(1-x{)}^n\in J$, and $e(1-e)=ab{x}^n(1-x{)}^n=0$, so $e^2=e$.