$$V(E) = \{\mathfrak{p}\in Spec(R)|E\subset \mathfrak{p}\}$$
as closed subset of the Zarisky topology on $Spec(R)$. Let $I=(E)$, the ideal generated by $E$, then one can see $V(E)=V(I)$, and hence, it is sufficient to define $V(I)$ where $I$ is an ideal of $R$. Or more precisely, one can look at radical ideal of $R$. Because $I\subset \mathfrak{p}$ iff $\sqrt{I}\mathfrak{p}$.
Before proving it is actually a topological space, we will look more close to this definition via example.
Example 1. Let $R=\mathbb{Z}$, then any ideal of $\mathbb{Z}$ is of the form $n\mathbb{Z}$, where $n$ is an integer. One can see $V(n\mathbb{Z})=\{p\mathbb{Z}|p\text{ is prime and } p|n\}$. By this, we can see $V(n\mathbb{Z})\cup V(m\mathbb{Z})=V(mn)$. Also, radical ideals of $\mathbb{Z}$ is of the form $p_1...p_k\mathbb{Z}$, where $p_i$ are distinct primes. By this $V(p_1...p_k\mathbb{Z})=\{p_i\mathbb{Z}|1\le i\le k\}$. Also, any non-zero prime ideal $p\mathbb{Z}$ of $\mathbb{Z}$ is maximal, and hence $V(p\mathbb{Z})=\{p\mathbb{Z}\}$ is closed in Zarisky topology defined above.
We are now ready to prove
Proposition 2. The collection of $V(I)$, where $I$ is an ideal of $R$ forms a topological space on $Spec(R)$.
Proof. It can be seen that $V(R) = \emptyset, V(0)=Spec(R)$. Furthermore, $V(I)\cup V(J) = \{\mathfrak{p}\in Spec(R)|I\subset \mathfrak{p} \text{ or } J\subset \mathfrak{p}\}=\{\mathfrak{p}\in Spec(R)|IJ\subset \mathfrak{p}\}$. The last equality occurs follows from the fact that $\mathfrak{p}$ is a prime ideal. Let $\cap_{\alpha}V(I_\alpha)$ is the intersection of the collection of ideals in $R$. By Zorn's lemma, there exists a smallest ideal $J$ that contains all $I_\alpha$. Then any prime ideal $\mathfrak{p}$ contains all $I_\alpha$ will also contain $J$, and vice versa. This implies $\cap_{\alpha}V(I_\alpha)=V(J)$. Via these steps, one can conclude the collection of $V(I)$ forms a topological space. (Q.E.D)
We now turn to the definition of dimension for a topological space. Our first purpose is to prove that the dimension of $Spec(R)$ via this definition is actually the Krull dimension of $R$.
Definition 3. Let $X$ be a topological space, then the dimension of $X$ is defined
$$\dim X=\sup_{n\in\mathbb{N}}\{\emptyset\ne Y_0 \subsetneq Y_1\subsetneq ...\subsetneq Y_n\subset X\}$$
where $Y_i$ are irreducible closed subsets of $X$.
Via this definition, we need to characterize the irreducible closed subset of $Spec(R)$.
Lemma 4. An irreducible closed subset of $Spec(R)$ is of the form $V(\mathfrak{p})$, where $\mathfrak{p}$ is prime ideal of $R$.
Proof. Assume that $V(I)$ is irreducible closed subset of $Spec(R)$ with $I$ is not prime ideal. Then there exists $a,b\notin I$ such that $ab\in I$. We can see $I\subsetneq I+aR,I\subsetneq I+bR$, and obviously, $V(I+aR)\cup V(I+bR) \subset V(I)$. Let $\mathfrak{p}\in V(I)$, we can see $\mathfrak{p}\in V((I+aR)(I+bR))=V(I+aR)\cup V(I+bR)$, or $V(I+aR)\cup V(I+bR)=V(I)$, a contradiction to the irreducibility of $V(I)$. Therefore, if $V(I)$ is irreducible, $I\in Spec(R)$. (Q.E.D)
In particular, if $\mathfrak{m}$ be a maximal ideal of $R$, then we can see $V(\mathfrak{m})=\{\mathfrak{m}\}$, or $\{\mathfrak{m}\}$ is a closed point in $Spec(R)$. And one can easily deduce that $\{\mathfrak{p}\}$ is closed in $Spec(R)$ iff $\mathfrak{p}$ is the maximal ideal of $R$. If there is a chain of irreducible closed subset of $Spec(R)$
$${\mathfrak{m}}\subsetneq V(E_1)\subsetneq V(E_2)\subsetneq ...\subsetneq V(E_n)\subset Spec(R) $$
Then by the previous lemma, there exists $\mathfrak{p_i}\in Spec(R)$ such that $V(E_i)=V(\mathfrak{p}_i)$, and actually $\mathfrak{m}$ is the maximal ideal of $R$. Or we obtain the following chain
$$\mathfrak{m}\subsetneq V(\mathfrak{p}_1)\subsetneq...\subsetneq V(\mathfrak{p}_n)\subset Spec(R)$$
We define the co-dimension of $\mathfrak{m}$ as
$$codim(\mathfrak{m})=\sup_{n\in \mathbb{N}}\{\mathfrak{m}\subsetneq V(\mathfrak{p}_1)\subsetneq...\subsetneq V(\mathfrak{p}_n)\subset Spec(R)\}$$
Now, we can see $\{\mathfrak{m}\}\subsetneq V(\mathfrak{p}_1)$ implies that $\mathfrak{p}_1\subsetneq \mathfrak{m}$. And $V(\mathfrak{p}_1)\subsetneq V(\mathfrak{p}_2)$ implies that $\mathfrak{p}_2\subsetneq \mathfrak{p}_1$, and so on. We finally obtain the corresponding chain of prime ideals in $R$
$$(0)\subset \mathfrak{p}_n\subsetneq ... \subsetneq\mathfrak{p}_2\subsetneq \mathfrak{p}_1\subsetneq \mathfrak{m}$$
From the fact that $ht(\mathfrak{m})=\sup_{n\in\mathbb{N}}\{(0)\subset \mathfrak{p}_n\subsetneq ... \subsetneq\mathfrak{p}_2\subsetneq \mathfrak{p}_1\subsetneq \mathfrak{m}\}$. We can see the co-dimension of $\mathfrak{m}$ is just the height of $\mathfrak{m}$. By the definition of Krull dimension, $\dim R = \sup_{\mathfrak{m}\in Max(R)}ht(\mathfrak{m})$, we have $\dim Spec(R)=\dim R$ iff $\dim Spec(R)=\sup_{\mathfrak{m}\in Max(R)}codim(\mathfrak{m})$. And the following lemma will finish our statement on the equality between the dimension of $Spec(R)$ and the Krull dimension of $R$.
Lemma 5. $\dim Spec(R)=\sup_{\mathfrak{m}\in Max(R)}codim(\mathfrak{m})$.
Proof. When $E\subset R$, and the ideal generated by $E$ strictly contained in $R$, we can see that any closed subset $V(E)$ always contain at least one maximal ideal $\mathfrak{m}$, for any proper ideal of $R$ is contained in a maximal ideal. Hence, for any chain of irreducible closed subset of $Spec(R)$ of the form $V(E_1)\subsetneq V(E_2)\subsetneq ...\subsetneq V(E_n)\subset Spec(R)$, there exists $\mathfrak{m}\in V(E_1)$, and obviously, $\mathfrak{m}$ is closed and irreducible, we obtain the chain $\mathfrak{m}\subsetneq V(E_2)\subsetneq...\subsetneq V(E_n)\subset Spec(R)$, this implies if $\dim Spec(R)\ge n$ then $codim(\mathfrak{m})\ge n$.
Conversely, if $codim(\mathfrak{m})\ge n$, then obviously $\mathfrak{m}\subsetneq V(e_2)\subsetneq...\subsetneq V(e_n)\subset Spec(R)$ is a chain of irreducible closed subset of $Spec(R)$. This implies $\dim Spec(R)\ge n$.
From this, $\dim Spec(R)=\sup_{\mathfrak{m}\in Spec(R)}codim(\mathfrak{m})$. (Q.E.D)
And hence, $\dim Spec(R) = \dim R$. An important characterization of the topology of $Spec(R)$ is the following
Proposition 6. $Spec(R)$ is compact.
Proof. Let $U$ be an open cover of $Spec(R)$. It is sufficient to consider the case $U$ is the collection of open subset of the form $D(I)=\{\mathfrak{p}\in Spec(R)|I\not\subset \mathfrak{p}\}$, where $I$ is an ideal of $R$. We then have $Spec(R) = \cup_{alpha}D(I_\alpha)$. The following lemma is necessary
Lemma 7. Let $\{I_\alpha\}$ be the collection of ideal of $R$, then $R=\sum_{\alpha}I_\alpha\Leftrightarrow Spec(R)=\cup_{\alpha}D(I_\alpha)$.
Proof. For any $\mathfrak{p}\in Spec(R)$, we can see $\mathfrak{p}\subsetneq \sum_\alpha I_\alpha$. This implies there exists $\beta$ such that $I_\beta \not \subset \mathfrak{p}$, or $\mathfrak{p}\in D(I_\beta)$, or $Spec(R) = \cup_{\alpha}D(I_\alpha)$. Conversely, let $I=\sum_{\alpha}I$, and assume that $I\subsetneq R$, then there exists a maximal ideal $\mathfrak{m}\in Spec(R)$ such that $I\subset \mathfrak{m}$, or $I_\alpha\subset \mathfrak{m}$. This implies $\mathfrak{m}\in \cap_{\alpha}V(I_\alpha)$. Taking the complement, we obtain $\cup_{\alpha}D(I_\alpha)\subsetneq Spec(R)$, a contradiction. This implies $I=Spec(R)$. (Q.E.D)
By the Lemma, we have $R=\sum_{\alpha}I_{\alpha}$, or $1$ can be expressed by a finite sum $1=i_1+...+i_n$ where $i_j\in I_j\in \{I_\alpha\}$. This implies $R=\sum_{j=1}^n I_j$. By the Lemma again, $Spec(R)=\cup_{j=1}^n I_j$, or there exists the finite sub-cover of $U$. And hence, $Spec(R)$ is compact. (Q.E.D)
Examples 1 ($\dim Spec(R)=\infty)$. Let us consider the ring of infinitely many variables $R=k[x_1,x_2,...]$. We now prove that $\dim Spec(R)=\infty$. By our previous argument, it suffices to prove $\dim R = \infty$. Let us consider the following chain of prime ideals in $R$
$$(x_1)\subset (x_1,x_2)\subset ... \subset (x_1,...,x_n)\subset ...$$
Assume that there exists a positive integer $n$ such that $(x_1,...,x_n)=(x_1,...,x_{n+1})$, then we there exists $h_1,...,h_n\in R$ such that $x_{n+1}=h_1x_1+...+h_nx_n$. Let us evaluate both sides at the points (0,0,...,0,1,0,...), where the unique $1$ lies at the $n+1^{\text{th}}$ position, and obtain $1=0$, a contradiction. Hence the chain of prime ideals above is strictly increasing. This implies $\dim R=\infty$.
Example 2 ($\dim Spec(R)=0$ and discrete). Let $R$ be a coordinate ring of an algebraic curves $f(x,y)=0$ over an algebraically closed $k$, and $\dim R=0$ (for example, $R=k[x]/(x^2)$), then it can be seen that $R$ is Noetherian. This implies $R$ is Artinian, and hence, $R$ has finitely many maximal ideals. By Hilbert's nullstellensatz, $R$ has finitely many points.
Example 3 ($\dim Spec(R)=0$, and not discrete). Let $R$ be a commutative ring, $R$ is called von Neumann ring if for any $r\in R$, there exists $s\in R$ such that $r=sr^2$. We can easily see any field is von Neumann ring, and arbitrary product of von Neumann rings is von Neumann ring. We now prove that any prime ideal of $R$ is maximal. Let $\mathfrak{p}$ be a prime ideal of $R$, we can see $R/\mathfrak{p}$ is also a von Neumann ring, for the identity $r=sr^2$ holds in $R$, we just need to transform them via the canonical projection from $R$ to $R/\mathfrak{p}$. Now $R/\mathfrak{p}$ is a domain, and for any $a\in R/\mathfrak{p}\setminus\{0\}$, there exists $s\in R/\mathfrak{p}$ such that $r=sr^2$. By the cancelation law, $1=rs$, or $s$ is the invert of $r$. This implies $R/\mathfrak{p}$ is a field, and hence $\mathfrak{p}$ is maximal. This implies $\dim R = 0$. Let us take an infinite product of field, then we will got a ring with infinitely many prime ideals. Because $Spec(R)$ is compact, it is discrete will implies it is finite. Hence, $Spec(R)$ is not discrete in this case.
Proposition 6. $Spec(R)$ is compact.
Proof. Let $U$ be an open cover of $Spec(R)$. It is sufficient to consider the case $U$ is the collection of open subset of the form $D(I)=\{\mathfrak{p}\in Spec(R)|I\not\subset \mathfrak{p}\}$, where $I$ is an ideal of $R$. We then have $Spec(R) = \cup_{alpha}D(I_\alpha)$. The following lemma is necessary
Lemma 7. Let $\{I_\alpha\}$ be the collection of ideal of $R$, then $R=\sum_{\alpha}I_\alpha\Leftrightarrow Spec(R)=\cup_{\alpha}D(I_\alpha)$.
Proof. For any $\mathfrak{p}\in Spec(R)$, we can see $\mathfrak{p}\subsetneq \sum_\alpha I_\alpha$. This implies there exists $\beta$ such that $I_\beta \not \subset \mathfrak{p}$, or $\mathfrak{p}\in D(I_\beta)$, or $Spec(R) = \cup_{\alpha}D(I_\alpha)$. Conversely, let $I=\sum_{\alpha}I$, and assume that $I\subsetneq R$, then there exists a maximal ideal $\mathfrak{m}\in Spec(R)$ such that $I\subset \mathfrak{m}$, or $I_\alpha\subset \mathfrak{m}$. This implies $\mathfrak{m}\in \cap_{\alpha}V(I_\alpha)$. Taking the complement, we obtain $\cup_{\alpha}D(I_\alpha)\subsetneq Spec(R)$, a contradiction. This implies $I=Spec(R)$. (Q.E.D)
By the Lemma, we have $R=\sum_{\alpha}I_{\alpha}$, or $1$ can be expressed by a finite sum $1=i_1+...+i_n$ where $i_j\in I_j\in \{I_\alpha\}$. This implies $R=\sum_{j=1}^n I_j$. By the Lemma again, $Spec(R)=\cup_{j=1}^n I_j$, or there exists the finite sub-cover of $U$. And hence, $Spec(R)$ is compact. (Q.E.D)
Examples 1 ($\dim Spec(R)=\infty)$. Let us consider the ring of infinitely many variables $R=k[x_1,x_2,...]$. We now prove that $\dim Spec(R)=\infty$. By our previous argument, it suffices to prove $\dim R = \infty$. Let us consider the following chain of prime ideals in $R$
$$(x_1)\subset (x_1,x_2)\subset ... \subset (x_1,...,x_n)\subset ...$$
Assume that there exists a positive integer $n$ such that $(x_1,...,x_n)=(x_1,...,x_{n+1})$, then we there exists $h_1,...,h_n\in R$ such that $x_{n+1}=h_1x_1+...+h_nx_n$. Let us evaluate both sides at the points (0,0,...,0,1,0,...), where the unique $1$ lies at the $n+1^{\text{th}}$ position, and obtain $1=0$, a contradiction. Hence the chain of prime ideals above is strictly increasing. This implies $\dim R=\infty$.
Example 2 ($\dim Spec(R)=0$ and discrete). Let $R$ be a coordinate ring of an algebraic curves $f(x,y)=0$ over an algebraically closed $k$, and $\dim R=0$ (for example, $R=k[x]/(x^2)$), then it can be seen that $R$ is Noetherian. This implies $R$ is Artinian, and hence, $R$ has finitely many maximal ideals. By Hilbert's nullstellensatz, $R$ has finitely many points.
Example 3 ($\dim Spec(R)=0$, and not discrete). Let $R$ be a commutative ring, $R$ is called von Neumann ring if for any $r\in R$, there exists $s\in R$ such that $r=sr^2$. We can easily see any field is von Neumann ring, and arbitrary product of von Neumann rings is von Neumann ring. We now prove that any prime ideal of $R$ is maximal. Let $\mathfrak{p}$ be a prime ideal of $R$, we can see $R/\mathfrak{p}$ is also a von Neumann ring, for the identity $r=sr^2$ holds in $R$, we just need to transform them via the canonical projection from $R$ to $R/\mathfrak{p}$. Now $R/\mathfrak{p}$ is a domain, and for any $a\in R/\mathfrak{p}\setminus\{0\}$, there exists $s\in R/\mathfrak{p}$ such that $r=sr^2$. By the cancelation law, $1=rs$, or $s$ is the invert of $r$. This implies $R/\mathfrak{p}$ is a field, and hence $\mathfrak{p}$ is maximal. This implies $\dim R = 0$. Let us take an infinite product of field, then we will got a ring with infinitely many prime ideals. Because $Spec(R)$ is compact, it is discrete will implies it is finite. Hence, $Spec(R)$ is not discrete in this case.
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