Definition

Let $A$ be commutative ring with unit. $\text{Spec}(A)$ is the collection of prime ideals of $A$.

$V(E)$ is the collection of prime ideals containing $E \subseteq A$. $\{V(E); E \subseteq A\}$ is a collection of closed subsets of $\text{Spec} (A)$, since

$V(1)=\emptyset, V(0) =X$. Hence Axiom of Closed Set I is satisfied.
$\cap_{i \in I} V(E_{i}) =V(\cup_{i \in I} E_{i})$. Hence Axiom of Closed Set II is satisfied.
To verify axiom III, Consider the ideal $\mathsf{a}_{1}$ generated by $E_{1}$ and the ideal $\mathsf{a}_{2}$ generated by $E_{2}$. Note that $V(\mathsf{a}_{1}) \cup V(\mathsf{a}_{2}) \subseteq V(\mathsf{a}_{1} \cap \mathsf{a}_{2})$. Then, for any prime ideal $\mathsf{p}$ containing $\mathsf{a}_{1} \cap \mathsf{a}_{2}$, we have $\mathsf{a}_{1} \subseteq \mathsf{p}$ or $\mathsf{a}_{2} \subseteq \mathsf{p}$, since

\[ x_{1},x_{2} \notin \mathsf{p} \Rightarrow x_{1}x_{2} \notin \mathsf{p} \text{ while }\,\, x_{1}\in \mathsf{a}_{1}, x_{2}\in \mathsf{a}_{2} \Rightarrow x_{1}x_{2} \in \mathsf{a}_{1} \cap \mathsf{a}_{2}\]

This proves $V(\mathsf{a}_{1} \cap \mathsf{a}_{2}) \subseteq V(\mathsf{a}_{1}) \cup V(\mathsf{a}_{2})$. Therefore, these sets equal to each other $V(\mathsf{a}_{1} \cap \mathsf{a}_{2}) = V(\mathsf{a}_{1}) \cup V(\mathsf{a}_{2})$, which verifies Axiom of Closed Set III. $\square$

Proporsition 1 [Topological basis]

Given an element $f$ in $A$, we define $X_{f}$ to be the complement of $V(f)$ in $X=\text{Spec}(A)$, so \[ \mathscr{U}^{\prime} = \{X_{f}; f\in A\}\]

is a topological basis for \[\mathscr{U} = \{ X\backslash V(E); E \subseteq A\}\]

To see more clearly, Choose an open set $O \in \mathscr{U}$, it must be the complement of a closed set $V(E)$ generated by some subset $E$ of ring $A$, that is, $O = X\backslash V(E)$. Decompose $E$ into the union of its components $\left\{f\right\}$ we have $E = \cup_{f\in E}\{f\}$, by utilize axiom of closed set III

\[ O =X\backslash \left(\cap_{f \in E} V(f)\right) = \cup_{f\in E} \left(X \backslash V(f)\right) =\cup_{f\in E} X_{f} \]

That is to say $\mathscr{U}^{\prime}$ is a topological basis for $\mathscr{U}$. $\square$

Proporsition 1 [Countability Axiom, not $C_{1}$]

Example

Let $A$ be UFD. $I$ is an index set, $|I| \geq \aleph_{1}$. For each index $i\in I$ associate with a variable $x_{i}$, Consider polynomial ring with coeffiecients in $A$ \[ A[I] := \{\sum a_{i_{1},i_{2},\cdots,i_{k}} x_{i_{i}}^{n_{1}}x_{i_{2}}^{n_{2}} \cdots x_{i_{k}}^{n_{k}} : a_{i_{1},i_{2},\cdots,i_{k}}\in A \}\] . Since $A[I]$ in integral domain, $(0)$ is a prime ideal. The Zariski topological basis $\{X_{f}; {f \in A(I)}\}$ for ring $A(I)$ with emptyset exception, that is $f\neq 0$, is a neighborhood basis at $(0)$ for $\text{Spec}(A[I])$ . Since $A[I]$ is either UFD, $A[I]$ contains uncoutable many prime idels $(x_{i}), {i \in I}$, which implies $A[I]$ dose not satisfies first-countable axiom $C_{1}$.

For topological space, we are also interested in its separation, compactness and connectedness etc.

Proporsition 2 [Separation Axiom $T_{0}$]

A topological space $(X,\mathscr{U})$ satisfies separtion axiom $T_{0}$ separation, if for any $x,y \in X$, $x\neq y$, there exists open set $U \in \mathscr{U}$, such that $x\in U, y \notin U$ of if $y \in U, x \notin U$.

Let $x,y$ be distinct two points in $X=\text{Spec}(A)$, i.e. two distinct prime ideals $\mathsf{p}_{x},\mathsf{p}_{y}$ of ring $A$:

If $\mathsf{p}_{x} \subseteq \mathsf{p}_{y}$, then $V(\mathsf{p}_{y}) \subseteq V(\mathsf{p}_{x})$. Taking $U= X \backslash V(\mathsf{p}_{y})$ we have $x\in U, y \notin U$.
If $\mathsf{p}_{x} \subsetneq \mathsf{p}_{y}$, thereby $\mathsf{p}_{y} \notin V(\mathsf{p}_{x})$. Taking $U = X \backslash V(\mathsf{p}_{x})$ we have $y \in U, x \notin U$.

This together impies that $X=\text{Spec}(A)$ satisfies separation axiom $T_{0}$. $\square$

Propersition 3 [Compactness]

We next prove $X = \text{Spec} (A)$ is (compact)[]. It only suffices to consider covering from topological basis $\{X_{f}; {f\in E}\}$, where $E\neq \emptyset$ is a subset of ring $A$. Accrording to (De Morgan law)[] and axiom of closed set III, nota that \[ X \backslash \left(\cup_{f\in E} X_{f}\right) = \cap_{f\in E} \left(X \backslash X_{f} \right) = V (\cup_{f\in E} \{f\}) \] hence $\cup_{f\in E} X_{f} = X \Leftrightarrow V(E) =\emptyset$. Let the ideal generated by $E$ be $\mathsf{a}$, we show that $= (1) $:

Let $\Sigma$ consist of non-trivial ideals of $A$. Since the non-decreasing chain of ideals are also an ideal, by Zorn lemma, if $\Sigma \neq \emptyset$, $\Sigma$ contains a maximal element $\mathsf{a}_{0}$ with respect to $\subseteq$-partial ordering; it is indeed an maximal ideal. (Otherwise, if there exist another $\mathsf{a}_{0}^{\prime}$ such that $\mathsf{a}_{0} \subseteq \mathsf{a}_{0}^{\prime}$, $\mathsf{a}_{0}^{\prime} = \mathsf{a}_{0} \cup \mathsf{a}_{0}^{\prime}$ is maximal with respect to inclusion relation.) It also can be proved that if $\mathsf{a} \neq (1)$, $\mathsf{a}$ is contained in a maximal ideal. Hence if $\mathsf{a} \neq (1)$, $V(\mathsf{a})$ contains at least an element. Contradiction. $\square$ [See Zorn lemma]

Therefore $1$ can be represented by a finite sum of products of elements from $E$ and $A$: \[1 = g_{1}f_{1} + \cdots + g_{n}f_{n}, \quad g_{i}\in A, f_{i} \in E, 1 \leq i \leq n\] Take $F= \{f_{1}, \cdots, f_{n}\}$, thereby $V(F) = \emptyset$, which implies $\{X_{f}; {f \in F}\}$ covers $X$. $\square$

Definition [Nilradical and Radical]

The ideal $\mathfrak{R}$ of all nipotents in a commutative ring $A$ with unit $1$ is the insection of all prime ideals of $A$, that is, \[ \mathfrak{R} = \bigcap_{\mathsf{p} \text{ is a prime ideal of $A$}} \mathsf{p} \]

Of course $\mathfrak{R}$ is an ideal closed under addition and multiplicatio. We split the proof in two steps:

If $\mathsf{p}$ is a prime ideal, then $\mathfrak{R}\subseteq \mathsf{p}$. Since $0 \in \mathsf{p}$, and we know for any $f \in \mathfrak{R}$, there exist an integer $n>0$ such that \[f^{n} \in \mathsf{p} \Rightarrow \cdots \Rightarrow f \in \mathsf{p}\]
If $f$ in not nilpotent, then there is a prime ideal $\mathsf{p} \cap \{f\} = \emptyset$. Let \[\Sigma = \left\{ \mathsf{a} \text{ is an ideal of $A$}: \text{for any $n \in \mathbb{N}_{*}$, }f^{n} \notin \mathsf{a} \right\}\]

Since $(0) \in \Sigma$. By Zorn lemma, $\Sigma$ contains maximal element $\mathsf{p}$ with respect to $\subseteq$-inclusion partial ordering. This $\mathsf{p}$ is indeed a prime ideal, because: For any $x,y \notin \mathsf{p}$, thanks to the maximal proporty of $\mathsf{p}$ we have $\mathsf{p} + (x)$, $\mathsf{p} + (y)$ $\notin \Sigma$; however, \[\begin{aligned} f^{m} \in\mathsf{p} + (x) &, f^{n} \in \mathsf{p} +(y) \Rightarrow f^{m+n} \in \mathsf{p} + (xy) \end{aligned}\] implies $\mathsf{p} + (xy) \notin \Sigma$; therefore $xy \notin \mathsf{p}$. $\square$

Denote \[r(\mathsf{a}): = \{f\in A: \text{there exist integer } n >0 \text{ such that } f^{n} \in \mathsf{a}\}\] which is the intersection of all prime ideals containing $\mathsf{a}$ by correspondence theorem.

Definition [Irreducibility]

A topological space $(X,\mathscr{U})$ is said to be irreducible if $X\neq \emptyset$ and $X$ can't be union of any two non-empty open sets $X = X_{1} \cup X_{2}$, or equivalently, if $U_{1}, U_{2} \neq \emptyset$ are any two non-empty open sets, then $U_{1} \cap U_{2} \neq \emptyset$.

Remark: According to the definition, irreducibility if finer than connectness, because two open sets $X_{1},X_{2}$ can be intersected.

Theorem 1[Irreducibility]

$\text{Spec}(A)$ is irreducible if and only if its nilradical $\mathfrak{R}$ of $A$ is a prime ideal.

For any two non-empty open sets $\cup_{f \in F} X_{f}$ and $\cup_{g \in G} X_{g}$ in $X=\text{Spec}(A)$. \[ \begin{aligned} \cup_{f \in F} X_{f} \neq \emptyset &\Leftrightarrow \cap_{f \in E} V(f) \neq X \Leftrightarrow \exists f_{0} \notin \mathfrak{R}\\ \left(\cup_{f \in F} X_{f}\right) \cap \left(\cup_{g \in G} X_{g}\right) \neq \emptyset &\Leftrightarrow \cup_{f \in F, g\in G} \left(X_{fg}\right) \neq \emptyset \Leftrightarrow \exists f_{0}\in F, g_{0} \in G, f_{0}g_{0} \notin \mathfrak{R} \end{aligned} \]

Remark 1: Singletons $\{x\}, x\in X$ are irreducible subspaces; the closure of irreducible subspaces in $X$ are irreducible subspaces.

Remark 2: Each singleton $\{x\}, x\in X$ contained in a maximal irreducible subspace of $X$, which is called irreducible component. The irrducible components cover $X$.

Remark 3: In an irreducible Hausdorff space $X$, the irrducible components of $X$ are singletons. However, the conected components, for instance, Euclidean space, are itself.

By taking quotient $A/\mathsf{p}$ and noting that there is a one-to-one correspondence $h:\mathsf{a} \mapsto \mathsf{a}/ \mathsf{p}$ between $V(\mathsf{p})$ and $\text{Spec}(A/\mathsf{p})$ that preserves prime ideals, there is a one-to-one correspondence between $r(\mathsf{p}) = \cap_{\mathsf{p}_{x} \in V(\mathsf{p})} \mathsf{p}_{x}$ and $\mathfrak{R} (A/\mathsf{p}) = \cap_{x\in \text{Spec}(A/\mathsf{p})} x$, as is shown in the diagram

\[\begin{align} \require{AMScd} \begin{CD} V(\mathsf{p}) @>>h> \text{Spec} (A/\mathsf{p})\\ @VV{\tiny \text{cap}}V @VV{\tiny \text{cap}}V\\ r(\mathsf{p}) @>>h> \mathfrak{R} (A/\mathsf{p}) \end{CD} \end{align}\]

According to above theorem and

Non-zero ring $A$ has minimal prime ideals with respect to set inclusion. By Zorn's lemma, it only suffices to prove that the intersection $\cap_{n}\mathsf{p}_{n}$ of a non-increasing sequence of prime ideals $\cdots \supseteq \mathsf{p}_{n} \supseteq \mathsf{p}_{n+1}\supseteq \cdots$ is also a prime ideal. By considering $x,y \in A, xy \in \cap_{n}\mathsf{p}_{n}$, we have for each $n$, either $x$ or $y$ belongs to $\mathsf{p}_{n}$. Then one of $x$ and $y$ occurs in $\mathsf{p}_{n}$ for infinitely many times, say, for example, $x \in \cap_{n}\mathsf{p}_{n}$. $\square$

we only need to show that $h$ is a homeomorphism, which is indeed true since $h$ maps $V(\mathsf{a}) \cap V(\mathsf{p}) = V(\mathsf{a} + \mathsf{p})$ to $V((\mathsf{a} + \mathsf{p})/ \mathsf{p})$ and vice versa, where $\mathsf{a}$ is an ideal of $A$. Similarily, the closed subspace $V(\mathsf{p})$ of $\text{Spec}(A)$ is irreducible if and only if $r(\mathsf{p})$ is a prime ideal, where $\mathsf{p}$ is an ideal of $A$. $\square$

Property 4 [Homomorphism]

Let $\phi: A \to B$ be homomorphism that preserves unit. $X=\text{Spec}(A) , Y=\text{Spec}(B)$. Since $\phi^{-1}$ draws back prime ideals in $Y$ into prime ideals in $X$, $\phi$ introduces naturally a mapping $\phi^{*}: \text{Spec}(B) \to \text{Spec}(A)$ by $\mathsf{q} \mapsto \phi^{-1} (\mathsf{q})$

It is easy to verify that $(\phi^{*})^{-1}(X_{f}) =Y_{\phi(f)}$, which implies $\phi^{*}$ is continuous.
If $\phi$ is surjective, then $\phi^{*}$ is injective. By Fundamental theorem on homomorphisms, $A/\ker(\phi) \cong B$. Note in above we mention that $\phi$ is a homeomorphism mappping, thereby $\phi^{*}$ establish a homeomorphism between $Y=\text{Spec}(B)$ and $V\left(\ker(\phi)\right)$.
$\overline{\phi^{*}(V(\mathsf{b}))} = V(\phi^{*}(\mathsf{b}))$. Hence if $\phi$ is injective, then $\phi^{*}(Y)$ is dense in $X$. Let $\mathsf{q} \in V(\mathsf{b})$ and note that $^{*}() = ^{-1}() ^{-1}() $, so $\phi^{-1}(\mathsf{q}) \in V(\phi^{*}(\mathsf{b}))$ and $\mathsf{q} \in \phi(V(\phi^{*}(\mathsf{b})))$. Thus $V(\mathsf{b}) \subseteq \phi(V(\phi^{*}(\mathsf{b})))$. Then note that \[y \in \overline{\{x\}} \Leftrightarrow \overline{\{y\}} \subseteq \overline{\{x\}} \Leftrightarrow V(\mathsf{p}_{y}) \subseteq V(\mathsf{p}_{x}) \Leftrightarrow \mathsf{p}_{x} \subseteq \mathsf{p}_{y}\] which says $\overline{\{x\}} = V\left(\mathsf{p}(={x})\right)$; $V(\mathsf{b})$ is obviously the minimal closed set in $X$ which contains $\phi^{*}(V(\mathsf{b}))$ since $\phi^{*}(\mathsf{b}) \in \phi^{*}(V(\mathsf{b}))$.

\[\begin{align} \require{AMScd} \begin{CD} X @<<{\phi^*}< Y\\ @AA{\tiny \text{$\cup$}}A @AA{\tiny \text{$\cup$}}A\\ V_A(\phi^* (\mathsf{b})) @<<{\phi^*}< V_B(\mathsf{b})\\ @AA{\text{$*$}}A @AA{\text{$*$}}A\\ \text{Spec}(A/\phi^* (\mathsf{b})) @<<{\tilde\phi^*}< \text{Spec} (B/\mathsf{b}) \end{CD} \end{align}\]

Example: Let $A$ be an integral domain with just one non-zero prime ideal $\mathsf{p}$. and let $K$ be the field of fractions of $A$. Let $B=(A/\mathsf{p}) \times K$. Define $\phi: A \to B$ by $\phi(x) = (\bar{x}, x)$, where $\bar{x}$ is the image of $x$ in $A/\mathsf{p}$. $\phi^{*}$ is bijective but not a homeomorphism: Since $A/\mathsf{p}$ is also a field, hence $B = (A/\mathsf{p}) \times K$ has only two prime ideals \[\left\{(1)\times(0), (0) \times(1)\right\}\] Direct computation yields $\phi^{*}$ is a bijection. However, $\{(1) \times (0)\}$ is a closed set of $\text{Spec}(B)$ whereas $\{(0)\}$ is not closed in $\text{Spec}(A)$.

More on Separation Axiom [$T_{1}$, $T_{2}$]

We define \[\mathscr{V}^{\prime} = \left\{ \phi^{*}\left(\text{Spec}(B)\right): \text{ where $\phi:A \to B$ is ring homomorphism} \right\}\] It can be verified that $\mathscr{V}^{\prime}$ satisfies axiom of closed set. The topology in $\text{Spec}(A)$ is called constructable topology. which is finer than Zariski topology.

$A/\mathfrak{R}$ is absolutely flat ($\mathfrak{R}$ being the nilradical of $A$).
Every prime ideal of $A$ is maximal.
$\text{Spec} (A)$ is a $T_{1}$-space (i.e., every subset consisting of a single point is closed).
$\text{Spec} (A)$ is Hausdorff.
The Zariski topology and the constructible topology on $\text{Spec} (A)$ coincide

Proporsition 5 [Connectedness]

If a commutative ring $A$ with unit can be represented as direct product of non-zero ring, $A_{i}, i=1, 2, \cdots, n$, By a standard result in commutative ring, the ideals $A=\prod_{i=1}^{n}A_{i}$ has the form

\[\mathsf{a}_{1} \times \cdots \times \mathsf{a}_{i} \times \cdots \times \mathsf{a}_{n}, \text{ where $\mathsf{a}_{i}$ are ideals of $A_{i}$} \]
Especially, prime ideals in $A$ has the form

\[(1) \times \cdots \times \mathsf{p}_{i} \times \cdots \times (1), \text{ where $\mathsf{p}_{i}$ are prime ideals of $A_{i}$}\]

For ring $A=\prod_{i=1}^{n}A_{i}$, the spectrum $X = \text{Spec}(A)$ is the disjoint union of $X_{i}, i =1,2,\cdots,n$, where $X_{i}$ consists of prime ideals containing $(1) \times \cdots \times \underset{i}{(0)} \times \cdots \times (1)$, homeomorphic to $\text{Spec}(A_{i})$. Hence $X=\text{Spec}(A)$ is not connected. $\square$

Conversly, if the subsets $X_{1}, X_{2} \neq \emptyset, X_{1} \cap X_{2} = \emptyset$ of $X=\text{Spec}(A)$ satisfies $X_{1} \cup X_{2} = X$. According to the definition of spectrum, we have $X_{1} =V(\mathsf{a}_{1}), X_{2} =V(\mathsf{a}_{2}),$ where $\mathsf{a}_{1},\mathsf{a}_{2}$ are ideals of ring $A$, and \[ \begin{aligned} V(\mathsf{a}_{1}) \cap V(\mathsf{a}_{2}) = V(\mathsf{a}_{1} +\mathsf{a}_{2}) =\emptyset &\Rightarrow \mathsf{a}_{1} +\mathsf{a}_{2} = (1)\\ V(\mathsf{a}_{1}) \cup V(\mathsf{a}_{2}) = V(\mathsf{a}_{1} \cap \mathsf{a}_{2}) = X &\Rightarrow \mathsf{a}_{1} \cap \mathsf{a}_{2} \subseteq \mathfrak{R} \end{aligned} \] there exist $x_{1} \in \mathsf{a}_{1}, x_{2} \in \mathsf{a}_{2}$, and integer $n>0$ such that \[ \begin{aligned} x_{1} + x_{2} &=1\\ (x_{1} x_{2})^{n} &=0 \end{aligned} \] Due to elementary arithmetic it finds that \[r(x_{1}^{n}) + r(x_{2}^{n}) =1 \Rightarrow (x_{1}^{n}) + (x_{2}^{n}) =1 \]. Therefore we can find an idempotent $e\neq 0,1 \in (x_{1}^{n})$ such that $1 - e \in (x_{2}^{n})$, and $e(1 - e) = 0$, because the non-empty assumption of $X_{i}, i=1,2$. Thereby $A$ can be decomposed as the direct product of $eA$ and $(1-e) A$. $ $

We summarize the result below:

Let $A$ be any commutative ring with unit. The following statements are equivalent:

$X = \text{Spec} (A)$ is disconnected.
$A \cong A_{1} \times A_{2}$ where neither of the rings $A_{1}$ or $A_{2}$ is the zero ring.
$A$ contains an idempotent $\neq 0,1$.

Remark: Local ring, that is, ring with only one maximal ideal, is always connected.

Reference

[1] Michael Atiyah. Introduction to commutative algebra. CRC Press, 2018.

[2] Suzanne C Dieudonne. History Algebraic Geometry. CRC Press, 1985.

[3] Bartel L Van der Waerden. A history of algebra: From al-Khwārizmī to Emmy Noether. Springer Science & Business Media, 2013.

Zariski Topology