Commutative Algebra and Algebraic Geometry (4): Zariski topology on affine scheme

 

Zariski topology on Spec(R) (1) Galois connection between ideal and varietyCRing to TopThe connection between the spectrum of ring and spectrum of linear operatorCategory of Affine Scheme and Algebra-Geometry DualitySpec functor, redefined.Zariski Topology on Spec(R) (2)$\mathrm{Spec}(A)$ as $T_0$ spaceQuasi-compactIrreducible space and generic pointKrull Dimension: A first glanceFor commutative ringFor topological space

Zariski topology on Spec(R) (1)

Let us go back to calculus first. Consider the ring of continuous functions over $\mathbb{R}$$\mathbb{R}$, $C(\mathbb{R})$$C(\mathbb{R})$. Evaluating $f\in C(\mathbb{R})$$f\in C(\mathbb{R})$ at a point $p$$p$ induces a ring homomorphism $e{v}_p:f⟼f(p)$$ev_p:f\longmapsto f(p)$. The kernel of $e{v}_p$$ev_p$ is the maximal ideal ${\mathfrak{m}}_p:=\{g\in C(\mathbb{R})\mid g(p)=0\}$$\mathfrak{m}_p:=\{g\in C(\mathbb{R}) \mid g(p)=0\}$.

Hence evaluating a function at $p$$p$ corresponds to taking the quotient by ${\mathfrak{m}}_p$$\mathfrak{m}_p$ by the first isomorphism theorem.

Remark. The notation $x$$x$ and ${\mathfrak{p}}_x$$\mathfrak{p}_x$ refer the same thing. But when we use $x$$x$, we view it as a point, and when we use ${\mathfrak{p}}_x$$\mathfrak{p}_x$, we view it as a prime ideal.

Now let us consider a ring $R$$R$ and $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$. View $f\in R$$f\in R$ as a function on $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$ and for a point ${\mathfrak{p}}_x\in \mathrm{Spec}(R)$$\mathfrak{p}_x\in \mathrm{Spec}(R)$, define $f(x)={\pi }_{{\mathfrak{p}}_x}(f)\in R/{\mathfrak{p}}_x$$f(x)= \pi_{\mathfrak{p}_x}(f)\in R/\mathfrak{p}_x$. Hence the codomain of $f$$f$ is ${\displaystyle \coprod _{x\in \mathrm{Spec}(R)}R/{\mathfrak{p}}_x\subseteq \coprod _{x\in \mathrm{Spec}(\mathrm{R})}k(x)}$$\displaystyle\coprod_{ x\in \mathrm{Spec}(R)} R/\mathfrak{p}_x\subseteq\coprod_{x\in\mathrm{Spec(R)}}k(x)$.

Where $k(x)$$k(x)$ is the fraction fields of $R/{\mathfrak{p}}_x$$R/\mathfrak p_x$i.e.

$$\begin{array}{}\text{(1)}& f:\mathrm{Spec}(R)\to {\displaystyle \coprod _{x\in \mathrm{Spec}(R)}k(x)}\end{array}$$

Then whether $f(x)=0$$f(x)=0$ or not makes sense since that means $f\in {\mathfrak{p}}_x$$f\in \mathfrak{p}_x$​​.

Notice that $(fg)(x)=f(x)g(x)=0⟺f(x)=0\vee g(x)=0$$(fg)(x)=f(x)g(x)=0\iff f(x)=0\vee g(x)=0$ since ${\mathfrak{p}}_x$$\mathfrak p_x$ is prime ideal.

Hence we define the zero locus of $f$$f$ as

$$\begin{array}{}\text{(2)}& V(f)=\{x\in \mathrm{Spec}(R)\mid f\in {\mathfrak{p}}_x\}\end{array}$$

In addition, for a subset $I\subseteq R$$I\subseteq R$, define

$$\begin{array}{}\text{(3)}& V(I)=\{x\in \mathrm{Spec}(R)\mid I\subseteq {\mathfrak{p}}_x\}=\bigcap _{f\in I}V(f)\end{array}$$

to be the Zariski closed set on $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$.

Again, let $Y\subseteq \mathrm{Spec}(R)$$Y\subseteq\mathrm{Spec}(R)$, we get ${\displaystyle I(Y):=\bigcap _{{\mathfrak{p}}_x\in Y}{\mathfrak{p}}_x}$$\displaystyle I(Y):=\bigcap_{\mathfrak p_x\in Y}\mathfrak p_x$. i.e. the function that vanish on $Y$$Y$.

Hence we have $\sqrt{I(Y)}=I(Y)$$\sqrt {I(Y)}=I(Y)$. i.e. the codomain of $I(-)$$I(-)$ is the set of radical ideal.

Then we have the following bijections:

• Points of $X$$X$ (resp. $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$) $⟺$$\iff$ maximal ideals of $A$$A$ (resp. $A(X)$$A(X)$).

• Irreducible closed subsets of $X$$X$ (resp. $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$) $⟺$$\iff$ prime ideals of $A$$A$ (resp. $A(X)$$A(X)$).

Finally, it is worth mentioning that taking the zero locus reverses inclusions. $\mathfrak{a}\subseteq \mathfrak{b}⟺V(\mathfrak{a})\supseteq V(\mathfrak{b})$$\mathfrak{a}\subseteq \mathfrak{b} \iff V(\mathfrak{a})\supseteq V(\mathfrak{b})$​​​. This is in line with the fact that radical ideals correspond to affine varieties.

Proposition. ${\displaystyle \sqrt{(0)}=\bigcap _{{\mathfrak{p}}_x\in \mathrm{Spec}(R)}{\mathfrak{p}}_x}$$\displaystyle \sqrt{(0)}=\bigcap_{\mathfrak p_x\in\mathrm{Spec}(R)}\mathfrak p_x$.

Proof. Obviously ${\displaystyle \sqrt{(0)}\subseteq \bigcap _{{\mathfrak{p}}_x\in \mathrm{Spec}(R)}{\mathfrak{p}}_x}$$\displaystyle\sqrt{(0)}\subseteq \bigcap_{\mathfrak p_x\in\mathrm{Spec }(R)}\mathfrak p_x$. Since prime ideal is radical and each prime ideal contains $0$$0$. Hence if $a^n=0⟹a^n\in {\mathfrak{p}}_x⟹a\in {\mathfrak{p}}_x$$a^n=0\implies a^n\in\mathfrak p_x\implies a\in \mathfrak p_x$.

For the converse, see Nilradical of a ring - Wikipedia $◻$$\square$​

Hence ${\displaystyle V(\sqrt{(0)})=\bigcup _{{\mathfrak{p}}_x\in \mathrm{Spec}(R)}V({\mathfrak{p}}_x)=\mathrm{Spec}(R)}$$\displaystyle V(\sqrt{(0)})=\bigcup_{\mathfrak p_x\in\mathrm{Spec} (R)} V(\mathfrak p_x)=\mathrm{Spec}(R)$.

Remark.

If $f(x)=0\mathrm{\forall }{\mathfrak{p}}_x$$f(x)=0\forall\mathfrak p_x$, then ${\displaystyle f\in \bigcap _{{\mathfrak{p}}_x\in \mathrm{Spec}(R)}{\mathfrak{p}}_x=\sqrt{(0)}}$$\displaystyle f\in\bigcap_{\mathfrak p_x\in\mathrm{Spec}(R)}\mathfrak p_x=\sqrt{(0)}$. That is, the radical ideal of $(0)$$(0)$​ collect all the function that vanish at all the point.

Corollary. ${\displaystyle \sqrt{I}=\bigcap _{{\mathfrak{p}}_x\in \mathrm{Spec}(R),I\subseteq {\mathfrak{p}}_x}{\mathfrak{p}}_x}$$\displaystyle \sqrt{I}=\bigcap_{\mathfrak p_x\in\mathrm{Spec}(R),I\subseteq\mathfrak p_x}\mathfrak p_x$.

Proof. By the Galois connection of contraction and extension of ideals for $\pi :R\to R/I$$\pi:R\to R/I$ and ${\displaystyle \sqrt{(0)}=\bigcap _{{\mathfrak{p}}_x\in \mathrm{Spec}(R)}{\mathfrak{p}}_x◻}$$\displaystyle \sqrt{(0)}=\bigcap_{\mathfrak p_x\in\mathrm{Spec}(R)}\mathfrak p_x\quad\square$​

Galois connection between ideal and variety

Proposition. Galois connection.

Let $Y\subseteq \mathrm{Spec}(R)$$Y\subseteq\mathrm{Spec}(R)$ be a subset and $\mathfrak{a}\subseteq R$$\mathfrak a\subseteq R$ be an ideal. Then

$$\begin{array}{}\text{(4)}& IV(\mathfrak{a})=\sqrt{\mathfrak{a}},VI(Y)=\bar{Y}\end{array}$$

Proof. By definition, $V(\mathfrak{a})=\{x\in \mathrm{Spec}(R)|\mathfrak{a}\subseteq {\mathfrak{p}}_x\}$$V(\mathfrak a)=\{x\in\mathrm{Spec}(R)|\mathfrak a\subseteq\mathfrak p_x\}$, hence ${\displaystyle I(Y)=\bigcap _{{\mathfrak{p}}_x\in \mathrm{Spec}(R),\mathfrak{a}\subseteq {\mathfrak{p}}_x}{\mathfrak{p}}_x=\sqrt{\mathfrak{a}}\supseteq \mathfrak{a}}$$\displaystyle I(Y)=\bigcap_{\mathfrak p_x\in\mathrm{Spec}(R),\mathfrak a\subseteq\mathfrak p_x}\mathfrak p_x=\sqrt{\mathfrak a}\supseteq\mathfrak a$.

For $VI(Y)=\bar{Y}$$VI(Y)=\overline Y$, since ${\displaystyle I(Y)=\bigcap _{{\mathfrak{p}}_x\in Y}{\mathfrak{p}}_x}$$\displaystyle I(Y)=\bigcap_{\mathfrak p_x\in Y}\mathfrak p_x$, we have $VI(Y)=V(\bigcap _{{\mathfrak{p}}_x\in Y}{\mathfrak{p}}_x)=\{x\in \mathrm{Spec}(R)\mid \bigcap _{{\mathfrak{p}}_x\in Y}{\mathfrak{p}}_x\subseteq {\mathfrak{p}}_x\}\supseteq Y$$VI(Y)=V(\bigcap_{\mathfrak p_x\in Y}\mathfrak p_x)=\{x\in \mathrm{Spec}(R) \mid \bigcap_{\mathfrak p_x\in Y}\mathfrak p_x\subseteq \mathfrak{p}_x\}\supseteq Y$​.

Hence $I,V$$I,V$ form a pair of Galois connection, and both $IV,VI$$IV,VI$ are closure operator.

But to check that $VI$$VI$ is the Zariski closure operator, we need $(5)$$(5)$ in the next proposition. Our prove for $(5)$$(5)$ is neat. Use the property of Galois connection, we see that $VIV(\mathfrak{a})=V(\mathfrak{a})$$VIV(\mathfrak a)=V(\mathfrak a)$. But $V(IV)(\mathfrak{a})=V(\sqrt{\mathfrak{a}}).$$V(IV)(\mathfrak a)=V(\sqrt{\mathfrak a}).$ Hence we know that the image of $VI$$VI$ is all the closed set in $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$​.

Proof for $VI(Y)=\bar{Y}$$VI(Y)=\overline{Y}$. (1)

As we already see, $Y\subseteq VI(Y)$$Y\subseteq VI(Y)$, hence $\bar{Y}\subseteq \overline{VI(Y)}=VI(Y)$$\overline Y\subseteq \overline{VI(Y)}=VI(Y)$. Also, we know that $Y\subseteq \bar{Y}$$Y\subseteq \overline Y$, hence we have $VI(Y)\subseteq VI(\bar{Y})=\bar{Y}$$VI(Y)\subseteq VI(\overline Y)=\overline Y$. Hence we have $VI(Y)=\bar{Y}$$VI(Y)=\overline Y$.

Proof for $VI(Y)=\bar{Y}$$VI(Y)=\overline{Y}$. (2) Reader should review the closure operator and Galois connection

Since the image of $VI$$VI$ and $\overline{(-)}$$\overline{(-)}$ is $\mathrm{Clo}(\mathrm{Spec}(R))$$\mathrm{Clo}(\mathrm{Spec}(R))$,

Hence we have

$$\begin{array}{}\text{(5)}& VI(X)\subseteq Y⟺X\subseteq \iota (Y),\bar{X}\subseteq Y⟺X\subseteq \iota (Y)\end{array}$$

But the left adjoint of $\iota$$\iota$ is unique, hence $VI(X)=\bar{X}◻$$VI(X)=\overline X\quad\square$.

Proposition.

$\begin{array}{c}(1).V(0)=\mathrm{Spec}(R),V(1)=\mathrm{\varnothing }\\ (2)\mathfrak{a}\subseteq {\mathfrak{a}}^{\prime }⟹V(\mathfrak{a})\supseteq V({\mathfrak{a}}^{\prime })\\ (3)V\left(\sum _{i\in I}{\mathfrak{a}}_i\right)=\bigcap _{i\in I}V(\mathfrak{a})\\ (4)V(\mathfrak{a}{\mathfrak{a}}^{\prime })=V(\mathfrak{a})\cup V({\mathfrak{a}}^{\prime }),\mathfrak{a}{\mathfrak{a}}^{\prime }:=\{f{f}^{\prime }:f\in \mathfrak{a},f^{\prime }{\mathfrak{a}}^{\prime }\}\\ (5)V(\mathfrak{a})=V(\sqrt{\mathfrak{a}})\\ (6)V(\mathfrak{p})=\stackrel{―}{\{{\mathfrak{p}}_x\}}\cong \mathrm{Spec}(R/\mathfrak{p})\end{array}$$(1). V(0)=\mathrm{Spec}(R),V(1)=\empty\\(2)\mathfrak a\subseteq \mathfrak a'\implies V(\mathfrak a)\supseteq V(\mathfrak a')\\(3)V\left(\sum_{i\in I}\mathfrak a_i\right)=\bigcap_{i\in I}V(\mathfrak a)\\(4)V(\mathfrak a\mathfrak a')=V(\mathfrak a)\cup V(\mathfrak a'),{\mathfrak a \mathfrak a':=}\{ff':f\in \mathfrak a,f'\mathfrak a'\}\\(5)V(\mathfrak a)=V(\sqrt{\mathfrak a})\\(6) V(\mathfrak p)=\overline{\{{\mathfrak p_x}\}}\cong\mathrm{Spec}(R/\mathfrak p)$

Hence all the $V(\mathfrak{a})$$V(\mathfrak a)$ do form a topology, we call it Zariski topology on $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$.

Proof. $(1),(2)$$(1),(2)$ is obvious, $(3)$$(3)$ follows from the property of Galois connection directly. (map colimit to limit)

To prove $(4)$$(4)$, notice that $x\in \mathrm{Spec}(R)$$x\in\mathrm{Spec}(R)$ is prime ideal, hence $f{f}^{\prime }(x)=0⟺f(x)=0\vee f^{\prime }(x)=0$$ff'(x)=0\iff f(x)=0\vee f'(x)=0$​.

Our prove for $(5)$$(5)$ is neat. Use the property of Galois connection, we see that $VIV(\mathfrak{a})=V(\mathfrak{a})$$VIV(\mathfrak a)=V(\mathfrak a)$. But $V(IV)(\mathfrak{a})=V(\sqrt{\mathfrak{a}}).$$V(IV)(\mathfrak a)=V(\sqrt{\mathfrak a}).$​

For $(6)$$(6)$, let $\pi :R\to R/\mathfrak{p}$$\pi:R\to R/\mathfrak p$ be the quoient map. Then $V(f)\cap V(\mathfrak{p})$$V(f)\cap V(\mathfrak p)$ one-one corresponding to $V(\pi (f))\cap \mathrm{Spec}(R/\mathfrak{p})◻$$V(\pi(f))\cap\mathrm{Spec}(R/\mathfrak p)\quad\square$.

Remark. This induce the one-one corresponding between radical ideal and closed set.

By the property of Galois connection, we get ${\displaystyle I\left(\bigcup _{j\in J}{V}_j\right)=\bigcap _{j\in J}I(V_j)}$$\displaystyle I\left(\bigcup_{j\in J} V_j\right)=\bigcap_{j\in J} I(V_j)$. ${\displaystyle V\left(\sum _{i\in I}{\mathfrak{a}}_i\right)=\bigcap _{i\in I}V(\mathfrak{a})}$$\displaystyle V\left(\sum_{i\in I}\mathfrak a_i\right)=\bigcap_{i\in I}V(\mathfrak a)$.

CRing to Top

Let $h:R\to S$$h:R\to S$ be a ring homomorphism. Then we claim that $\mathrm{Spec}:\mathrm{CRing}\to \mathrm{Top}$$\mathrm{Spec}:\mathrm{CRing}\to\mathrm{Top}$ is a functor.

What we need to prove is $\mathrm{Spec}(\psi )={\psi }^{-1}$$\mathrm{Spec}(\psi)=\psi^{-1}$ is continuous.

Consider

$$\begin{array}{}\text{(6)}& {}^a\psi :=\mathrm{Spec}(\psi ):\mathrm{Spec}(S)\to \mathrm{Spec}(R)\end{array}$$

It induce a pull back

$$\begin{array}{}\text{(7)}& [f:\mathrm{Spec}(R)\to \coprod _{x\in \mathrm{Spec}(R)}R/{\mathfrak{p}}_x]⟼[{(}^a\psi {)}^{\ast }(f):\mathrm{Spec}(S)\to \coprod _{x\in \mathrm{Spec}(S)}S/{\mathfrak{p}}_x]]\end{array}$$

by

$$\begin{array}{}\text{(8)}& {(}^a\psi {)}^{\ast }(f)(x)=f{(}^a\psi (x))\end{array}$$

We need to check that the preimage of ${\psi }^{\ast }$$\psi^*$​​​ map closed set to closed set.

Lemma.

$$\begin{array}{}\text{(9)}& {(}^a\psi {)}^{-1}V(I)=V(\psi (I{)}^e)\end{array}$$

Proof.

$$\begin{array}{}\text{(10)}& {(}^a\psi {)}^{-1}(V(I))={(}^a\psi {)}^{-1}\{{\mathfrak{p}}_x\in \mathrm{Spec}(R)|f({\mathfrak{p}}_x)=0\mathrm{\forall }f\in I\}\end{array}$$

and

$$\begin{array}{}\text{(11)}& x\in {(}^a\psi {)}^{-1}\{{\mathfrak{p}}_x\in \mathrm{Spec}(R)|f({\mathfrak{p}}_x)=0\mathrm{\forall }f\in I\}⟺{}^a\psi (x)={\psi }^{-1}(x)\in \{{\mathfrak{p}}_x\in \mathrm{Spec}(R)|f({\mathfrak{p}}_x)=0\mathrm{\forall }f\in I\}\end{array}$$

Hence

$$\begin{array}{}\text{(12)}& \mathrm{\forall }f\in I,f({\psi }^{-1}(x))=0⟺f\in {\psi }^{-1}(x)⟺\psi (f)\in x⟺\psi (f)(x)=0\end{array}$$

i.e.

$$\begin{array}{}\text{(13)}& {(}^a\psi {)}^{-1}\{{\mathfrak{p}}_x\in \mathrm{Spec}(R)|f({\mathfrak{p}}_x)=0\mathrm{\forall }f\in I\}=\{{\mathfrak{p}}_x\in \mathrm{Spec}(S)|\psi (f)({\mathfrak{p}}_x)=0\mathrm{\forall }f\in I\}=V(\psi (I{)}^e)◻\end{array}$$

Hence ${}^a\psi$$^a\psi$ is continuous.

Proposition. $\overline{{}^a\psi (V(\mathfrak{a}))}=V({\psi }^{-1}(\mathfrak{a}))$$\overline{^a\psi(V(\mathfrak a))}=V(\psi^{-1}(\mathfrak a))$​

Proof. Since $VI=\overline{(-)},$$VI=\overline{(-)},$ we can write left hand side as $VI{(}^a\psi (V(\mathfrak{a})))$$VI(^a\psi(V(\mathfrak a)))$.

But

$$\begin{array}{}\text{(14)}& I{(}^a\psi (V(\mathfrak{a})))=\bigcap _{\mathfrak{q}\in V(\mathfrak{a})}{\psi }^{-1}(\mathfrak{q})=\sqrt{({\psi }^{-1}(\mathfrak{a}))}\end{array}$$

and apply $V(-)$$V(-)$ both side we complete the proof. $◻$$\square$​

Corollary. The nap ${}^a\psi$$^a\psi$ is dominant (i.e. the image is dense in $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$) iff $\mathrm{Ker}\psi \subseteq \sqrt{(0)}$$\mathrm{Ker}\psi\subseteq\sqrt{(0)}$.

Proof. $\overline{{}^a\psi (V(0))}=V({\psi }^{-1}(0))=V(\mathrm{Ker}\psi )\supseteq V(0)=\mathrm{Spec}(R).◻$$\overline{^a\psi(V(0))}=V(\psi^{-1}(0))=V(\mathrm{Ker}\psi)\supseteq V(0)=\mathrm{Spec}(R).\quad\square$

The connection between the spectrum of ring and spectrum of linear operator

Definition. Annhilator.

Let $M$$M$ be a $R-$$R-$module, the kernel of the ring homomorphism $f:R\to {\mathrm{End}}_{\mathrm{Ab}}(M)$$f:R\to\mathrm{End}_{\mathrm{Ab}}(M)$ is the annihilator of $M$$M$ respect to $R$$R$.

Let $K$$K$ be an algebraic closed field (e.g. $\mathbb{C}$$\mathbb C$), $V$$V$ be a finite dimension vector space over $K$$K$ and $T:V\to V$$T:V\to V$ be a linear map.

Then $f:K[X]\to K[T]$$f:K[X]\to K[T]$ makes $V$$V$ become a $K[X]$$K[X]$ module, and ${\mathrm{Ann}}_{K[X]}(V)$$\mathrm{Ann}_{K[X]}(V)$ is an ideal in $K[X]$$K[X]$. Since $K[X]$$K[X]$​ is PID,

$$\begin{array}{}\text{(15)}& {\mathrm{Ann}}_{K[X]}(V)=(m(X))\end{array}$$

Recall the spectrum of a linaer map refer to the set of eigen value of $T$$T$​, and spectrum of a ring refer to the set of prime ideal. A natural question is, what is the connection between two different spectrum?

Observe that $\lambda$$\lambda$ is a eigenvalue of $T$$T$ if and only if $T-\lambda$$T-\lambda$ is not invertible. Hence $(T-\lambda )$$(T-\lambda)$ is a proper ideal in $K[T]$$K[T]$ iff $\lambda$$\lambda$ is the eigenvalue of $T$$T$. The operator polynomial ring $K[T]\cong K[X]/(m(X))$$K[T]\cong K[X]/(m(X))$. Let $m(X)$$m(X)$ be a monic polynomial.

By $K$$K$ is algebraic closed field, $m(X)=(X-{\lambda }_1{)}^{e_1}...(X-{\lambda }_n{)}^{e_n}$$m(X)=(X-\lambda_1)^{e_1}...(X-\lambda_n)^{e_n}$.

Hence

$$\begin{array}{}\text{(16)}& \mathrm{Spec}(K[X]/(m(X))=\{(X-{\lambda }_1),...,(X-{\lambda }_n)\}\cong \{(T-{\lambda }_1),...,(T-{\lambda }_n)\}=\mathrm{Spec}(K[T])\end{array}$$

Then we see the corresponding between spectrum of ring and spectrum of linear operator.

Think about that, if we map a linear operator $T$$T$ on a finite dimension vector space to the algebraic curve defined by its characteristic polynomial, i.e.

$$\begin{array}{}\text{(17)}& y-\det (T-x\cdot I)\end{array}$$

What is the intersection multiplicity means?

What if we consider the minimal polynomial of $T$$T$​?

Category of Affine Scheme and Algebra-Geometry Duality

An affine scheme is a triple $(X,\alpha ,A)$$(X,\alpha,A)$, where $X$$X$ is a topological space, $A$$A$ a ring and ${\displaystyle \alpha :X\stackrel{\sim }{\to }\mathrm{Spec}A}$$\displaystyle\alpha:X \xrightarrow{\sim}\mathrm{Spec} A$ an isomorphism of topological spaces.

A morphism of affine scheme $(Y,\beta ,B)⟶(X,\alpha ,A)$$(Y,\beta,B)\longrightarrow(X,\alpha,A)$ is a pair $(f,\phi )$$(f,\varphi)$ is a pair, where $f:X\to Y$$f:X\to Y$ is a continuous function and $\phi :A\to B$$\varphi:A\to B$ is a ring homomorphism, such that the diagram commute.

$$\begin{array}{}\text{(18)}& \begin{array}{c}\begin{array}{ccc}Y& \stackrel{\beta }{\to }& \mathrm{Spec}(B)\\ f\downarrow & & \downarrow {}^a\phi & \\ X& \stackrel{\alpha }{\to }& \mathrm{Spec}(A)\end{array}\end{array}\end{array}$$

In deed, we define a functor from affine scheme to category of $\mathrm{Spec}$$\mathrm{Spec}$.

$$\begin{array}{}\text{(19)}& Y⟼\mathrm{Spec}(B),(f:Y\to X)⟼(\alpha \circ f\circ {\beta }^{-1}{=}^a\phi :\mathrm{Spec}(B)\to \mathrm{Spec}(A))\end{array}$$

Readers can see that actually the relation between category of affine scheme and category of specturum of comuutative ring is just like the relation between category of finite dimension vector space over $F$$F$ and category of $F^n$$F^n$.

Reader could get some geometry intuition from homotopy here (image a contraction from $X$$X$​ to Y).

Math Essays: ... Fundamental group and Homotopy as Natural Transformation (marco-yuze-zheng.blogspot.com)

Spec functor, redefined.

Let us review our point of view of $\mathrm{Spec}$$\mathrm{Spec}$. At the beginning, we observe that for a ring homomorphism $\phi :R\to S$$\varphi:R\to S$ and a prime ideal ${\mathfrak{p}}_x\in \mathrm{Spec}(S),$$\mathfrak p_x\in\mathrm{Spec}(S),$ ${\phi }^{-1}({\mathfrak{p}}_x)$$\varphi^{-1}(\mathfrak p_x)$ is a prime ideal as well, hence we get a functor as follows:

$$\begin{array}{}\text{(20)}& \mathrm{Spec}:{\mathrm{CRing}}^{\mathrm{op}}\to \mathrm{Set}\end{array}$$

Then we observe that we could view $f\in R$$f\in R$ as a function $f:{\displaystyle \mathrm{Spec}(R)\to \coprod _{x\in \mathrm{Spec}(R)}R/{\mathfrak{p}}_x}$$f:\displaystyle\mathrm{Spec}(R)\to\coprod_{x\in\mathrm{Spec}(R)} R/\mathfrak p_x$ and define vanishing set of ideals in $R$$R$ to be the Zariski closed sets, we assign a topology on $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$ and make it become a functor such that:

$$\begin{array}{}\text{(21)}& \mathrm{Spec}:{\mathrm{CRing}}^{\mathrm{op}}\to \mathrm{Top}\end{array}$$

But, actually, what we want is not a topological space, we want a ringed space. The object should record both Algebra and Geometry information. If we viewed $\mathrm{Spec}(-)$$\mathrm{Spec}(-)$ as a functor to $\mathrm{Top}$$\mathrm{Top}$, then every field $\mathbb{F}$$\mathbb F$ will be mped to $\{\ast \}$$\{*\}$, hence isomorphic to each other.

The proper point of view of $\mathrm{Spec}$$\mathrm{Spec}$ is a functor

$$\begin{array}{}\text{(22)}& \mathrm{Spec}:{\mathrm{CRing}}^{\mathrm{op}}\stackrel{\sim }{\to }\mathrm{Affine}\mathrm{Scheme}\end{array}$$

$$\begin{array}{}\text{(23)}& R⟼(\mathrm{Spec}(R),{\mathrm{Id}}_{\mathrm{Spec}}(R),R)\end{array}$$

Easy to see that this is essential full and faithful, hence it induce a anti categorical equivalence.

Corollary.

$$\begin{array}{}\text{(24)}& A\cong \prod _{i\in I}{A}_i⟺(\mathrm{Spec}(A),A)\cong (\coprod _{i\in I}\mathrm{Spec}(A_i),\prod _{i\in I}{A}_i)\end{array}$$

But actually I think we need add more details for that.

Proposition. The following conditions are equivalent:

$(i)$$(i)$ $\mathrm{Spec}(A)$$\mathrm{Spec} (A)$ is not connected.

$(ii)$$(ii)$ $A\cong A_1\times A_2,A_1,A_2\ne 0$$A\cong A_1\times A_2,A_1,A_2\ne 0$.

$(iii)$$(iii)$ There exists $a\in A,a\ne 1\wedge a\ne 0,a^2=a$$a\in A,a\ne 1\wedge a\ne 0,a^2=a$.

Proof.

$(ii)⟹(iii)$$(ii)\implies (iii)$ is easy since if $A\cong A_1\times A_2$$A\cong A_1\times A_2$, then we have $(1,0),(0,1)$$(1,0),(0,1)$.

$(iii)⟹(i)$$(iii)\implies (i)$ Let $f=f^2$$f=f^2$, then $f(f-1)=0$$f(f-1)=0$. Hence $V(0)=V(f(f-1))=V(f)\cup V(f-1)=\mathrm{Spec}(A)$$V(0)=V(f(f-1))=V(f)\cup V(f-1)=\mathrm{Spec}(A)$​.

Also $V(f)\cap V(f-1)=\mathrm{\varnothing }$$V(f)\cap V(f-1)=\empty$ since if $f(x)=0⟺(f-1)(x)=-1$$f(x)=0\iff (f-1)(x)=-1$.

$(i)⟹(ii)$$(i)\implies (ii)$​. We know that $\mathrm{Spec}(A)\cong \mathrm{Spec}(A/\sqrt{0})$$\mathrm{Spec}(A)\cong\mathrm{Spec}(A/\sqrt{0})$. Denote $A/\sqrt{(0)}$$A/\sqrt{(0)}$ as $A^{\prime }$$A'$ and the extension of ideals respect to the quotient map as $I^{\prime }$$I'$.

If $\mathrm{Spec}(A^{\prime })$$\mathrm{Spec}(A')$ is not connected, then $\mathrm{Spec}(A^{\prime })=V(I^{\prime })\cup V(J^{\prime })=V(I^{\prime }\cap J^{\prime })=V(0)$$\mathrm{Spec}(A')=V(I')\cup V(J')=V(I'\cap J')=V(0)$. Also, $V(I^{\prime })\cap V(J^{\prime })=\mathrm{\varnothing }$$V(I')\cap V(J')=\empty$.

Since $A^{\prime }=A/\sqrt{(0)}$$A'=A/\sqrt{(0)}$ is reduced, hence $I^{\prime }\cap J^{\prime }=0$$I'\cap J'=0$. Also, $V(I^{\prime })\cap V(J^{\prime })=V(I^{\prime }+J^{\prime })=\mathrm{\varnothing }=V(A^{\prime })$$V(I')\cap V(J')=V(I'+J')=\empty=V(A')$.

Hence $I^{\prime }+J^{\prime }=A^{\prime }$$I'+J'=A'$​. By Chinese Remainder Theorem, $A^{\prime }\cong A^{\prime }/I^{\prime }\times A^{\prime }/J^{\prime }$$A'\cong A'/I'\times A'/J'$​.

By the third isomorphism theorem. $A/I\cong A^{\prime }/I^{\prime },A/J\cong A^{\prime }/J^{\prime }$$A/I\cong A'/I',A/J\cong A'/J'$, hence $A\cong A/I\times A/J.$$A\cong A/I\times A/J.$ $◻$$\square$

Zariski Topology on Spec(R) (2)

Definition. We call $D(f):=D_A(f):=\mathrm{Spec}(A)-V(f)$$D(f):=D_A(f):=\mathrm{Spec}(A)-V(f)$ principal open sets of $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$.

It is clear that $D(0)=\mathrm{\varnothing },D(1)=\mathrm{Spec}(A)$$D(0)=\empty,D(1)=\mathrm{Spec}(A)$.

Notice that $D(f)$$D(f)$ is the points that $f(x)\ne 0$$f(x)\ne 0$. Hence it will correspoding to localization respect to $S=\{f,f^2,f^3,...\}$$S=\{f,f^2,f^3,...\}$. We will talk about that in next section.

Now, easy to see that $D(f)\cap D(g)=D(fg)$$D(f)\cap D(g)=D(fg)$ since $fg(x)\ne 0⟺f(x)\ne 0\wedge g(x)\ne 0$$fg(x)\ne0\iff f(x)\ne 0\wedge g(x)\ne0$​​​​.

In addition, $D(f+g)\subseteq D(f)\cup D(g)$$D(f+g)\subseteq D(f)\cup D(g)$ since $V(f+g)\supseteq V(f)\cap V(g)$$V(f+g)\supseteq V(f)\cap V(g)$.

Also, $D(f)=D(f^{\prime })⟺V(f)=V(f^{\prime })⟺V(\sqrt{(f)})=V(\sqrt{(f^{\prime })})⟺\sqrt{(f)}=\sqrt{(f^{\prime })}$$D(f)=D(f')\iff V(f)=V(f')\iff V(\sqrt{(f)})=V(\sqrt{(f')})\iff \sqrt{(f)}=\sqrt{(f')}$.

For example, $D(2)=D(4)=D(8)=D(16)=D(32)=D(64)\subseteq \mathrm{Spec}(\mathbb{Z})$$D(2)=D(4)=D(8)=D(16)=D(32)=D(64)\subseteq \mathrm{Spec}(\mathbb Z)$​.

$\mathrm{Spec}(A)$$\mathrm{Spec}(A)$ as $T_0$$T_0$ space

Proposition. Zariski topology on $X=\mathrm{Spec}(A)$$X=\mathrm{Spec}(A)$ is a Kolmogorov space($T_0$$T_0$ space), i.e. a topological space satisfying the following separation axiom:

Given two different points $x,y\in X$$x,y\in X$, there exists a open neighborhood of $x$$x$, denote as $U_x$$U_x$ such that $y\notin U_x$$y\notin U_x$, or an open neighborhood of $y$$y$ denote as $U_y$$U_y$ such that $x\notin U_y$$x\notin U_y$.

Proof. $x\ne y$$x\ne y$ in $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$ means ${\mathfrak{p}}_x\ne {\mathfrak{p}}_y⟺{\mathfrak{p}}_x⊈{\mathfrak{p}}_y\vee {\mathfrak{p}}_y⊈{\mathfrak{p}}_x$$\mathfrak p_x\ne \mathfrak p_y\iff \mathfrak p_x\nsubseteq \mathfrak p_y\vee \mathfrak p_y\nsubseteq \mathfrak p_x$. If ${\mathfrak{p}}_x⊈{\mathfrak{p}}_y$$\mathfrak p_x\nsubseteq \mathfrak p_y$, then $y\notin V({\mathfrak{p}}_x)=\overline{\{x\}}$$y\notin V(\mathfrak p_x)=\overline{\{x\}}$,

Hence $U_y=\mathrm{Spec}(A)-\overline{\{x\}}$$U_y=\mathrm{Spec}(A)-\overline{\{x\}}$ is an open neighborhood of $y$$y$ such that $x\notin U_y$$x\notin U_y$. $◻$$\square$

Quasi-compact

Definition. The word quasi-compact means compact but without Hausdorff separation axiom. Hence $U$$U$ is quasi-compact if every open covering of $U$$U$​ admit a finite subcover.

Lemma. ${\displaystyle D(g)\subseteq \bigcup _{i\in I}D(f_i)⟺(g)\subseteq \sqrt{\sum _{i\in I}(f_i)}}$$\displaystyle D(g)\subseteq\bigcup_{i\in I}D(f_i)\iff (g)\subseteq\sqrt{\sum_{i\in I}(f_i)}$ . In other word, ${\displaystyle \mathrm{\exists }n\in \mathbb{N},g^n\in \sum _{i\in I}(f_i)}$$\displaystyle\exists n\in\mathbb N,g^n\in\sum_{i\in I}(f_i)$.

Proof. ${\displaystyle D(g)\subseteq \bigcup _{i\in I}D(f_i)⟺V(g)\supseteq \bigcap _{i\in I}V(f_i)=V(\sum _{i\in I}(f_i))⟺IV(g)=\sqrt{(g)}\subseteq IV(\sum _{i\in I}(f_i))=\sqrt{\sum i\in I(f_i)}}$$\displaystyle D(g)\subseteq\bigcup_{i\in I}D(f_i)\iff V(g)\supseteq\bigcap_{i\in I} V(f_i)=V\left(\sum_{i\in I}(f_i)\right)\iff IV(g)=\sqrt{(g)}\subseteq IV\left(\sum_{i\in I}(f_i)\right)=\sqrt{\sum{i\in I}(f_i)}$

and

$$\begin{array}{}\text{(25)}& \sqrt{(g)}\subseteq \sqrt{\sum _{i\in I}(f_i)}⟺(g)\subseteq \sqrt{\sum _{i\in I}(f_i)}◻\end{array}$$

Let $g=1$$g=1$ we get that $D(f_i{)}_{i\in I}$$D(f_i)_{i\in I}$ is a covering of $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$ iff ${\displaystyle \sum _{i\in I}(f_i)=A}$$\displaystyle\sum_{i\in I}(f_i)=A$​​​.

Proposition.

Let $A$$A$ be a commutative ring. The principal open subsets $D(f)$$D(f)$ for $f\in A$$f\in A$ form a basis for the topology $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$. For all $f\in A$$f\in A$, the open sets $D(f)$$D(f)$ are quasi-compact. In particular, the space $\mathrm{Spec}(A)=D(1)$$\mathrm{Spec}(A)=D(1)$ is quasi-compact.

Proof. Since every closed set ${\displaystyle V(I)=\bigcap _{f\in I}V(f)}$$\displaystyle V(I)=\bigcap_{f\in I}V(f)$, hence every open set ${\displaystyle \mathrm{Spec}(A)-V(I)=\bigcup _{f\in I}D(f)}$$\displaystyle \mathrm{Spec}(A)-V(I)=\bigcup_{f\in I} D(f)$, we say that $D(f)$$D(f)$ form a basis. Let ${\displaystyle D(g)\subseteq \bigcup _{i\in I}D(f_i)}$$\displaystyle D(g)\subseteq \bigcup_{i\in I} D(f_i)$, then by the lemma, ${\displaystyle g^n=\sum _{i\in I}{a}_i{f}_i}$$\displaystyle g^n=\sum_{i\in I}a_i f_i$, where $a_i\in A$$a_i\in A$ and $a_i=0\mathrm{\forall }i\notin J,J\subseteq I$$a_i=0\forall i\notin J,J\subseteq I$ is a suitable finite subset of $I$$I$. Hence ${\displaystyle D(g)\subseteq \bigcup _{i\in J}D(f_i)}$$\displaystyle D(g)\subseteq\bigcup_{i\in J} D(f_i)$. $◻$$\square$​

Irreducible space and generic point

Definition. Let $X$$X$ be a topological space.

We say $X$$X$ is irreducible if $X$$X$ is not empty, and whenever $X=Z_1\cup Z_2$$X=Z_1\cup Z_2$ with $Z_1,Z_2$$Z_1,Z_2$ closed, we have $X=Z_1$$X=Z_1$ or $X=Z_2$$X=Z_2$. A subset $E\subseteq X$$E\subseteq X$ is irreducible when it is irreducible as a subspace of $X$$X$. Otherwise we say it is reducible.

Remark.

Let $X=\mathrm{Spec}(A)$$X=\mathrm{Spec}(A)$. Easy to see that $fg\in \sqrt{(0)}⟺V(fg)=V(f)\cup V(g)=V(0)=X$$fg\in\sqrt{(0)}\iff V(fg)=V(f)\cup V(g)=V(0)= X$.

Hence $X$$X$ is irreducible iff $V(f)=X=V(0)$$V(f)=X=V(0)$ or $V(g)=X=V(0)$$V(g)=X=V(0)$.

i.e. $f(x)=0\mathrm{\forall }x\in \mathrm{Spec}(A)$$f(x)=0\forall x\in\mathrm{Spec}(A)$ or $g(x)=0\mathrm{\forall }x\in \mathrm{Spec}(A)$$g(x)=0\forall x\in\mathrm{Spec}(A)$. i.e. $f\in \sqrt{(0)}\vee g\in \sqrt{(0)}$$f\in\sqrt{(0)}\vee g\in \sqrt{(0)}$. i.e. $\sqrt{(0)}$$\sqrt{(0)}$ is a prime ideal.

Radical of $(0)$$(0)$ is prime or not is a kind of generalization of $(0)$$(0)$ is prime or not, i.e. the ring $A$$A$ is integral domain of not.

Also, $fg\in \sqrt{(0)}$$fg\in \sqrt{(0)}$ or not could be thought as a generalization of zero divisor.

Remark.

An irreducible space is obviously connected.

Proposition. Let $f:X\to Y$$f:X\to Y$ be a continuous function, if $E\subseteq X$$E\subseteq X$ is irreducible subset, then $f(E)$$f(E)$ is irreducible as well.

Remark. Hence irreducible is a topological property, just like compact, path connected and connected, it preserved by homomorphism. In category of group, abelian property is also preserved by homomorphism.

Proof. Clearly we assume that $E=X$$E=X$ and $f(E)=Y$$f(E)=Y$. First, $X\ne \mathrm{\varnothing },Y\ne \mathrm{\varnothing }$$X\ne\empty,Y\ne \empty$. Now assume that $Y=Y_1\cup Y_2$$Y=Y_1\cup Y_2$ with $Y_1,Y_2$$Y_1,Y_2$ is closed. Then $X=f^{-1}(Y_1)\cup f^{-1}(Y_2)$$X=f^{-1}(Y_1)\cup f^{-1}(Y_2)$. By $X$$X$ is irreducible, $f^{-1}(Y_1)=X\vee f^{-1}(Y_2)=X$$f^{-1}(Y_1)=X\vee f^{-1}(Y_2)=X$. Hence $Y=Y_1\vee Y=Y_2$$Y=Y_1\vee Y=Y_2$. $◻$$\square$

Proposition. Let $X$$X$ be a non-empty topological space. The following assertions are equivalent.

$(i)$$(i)$ $X$$X$ is irreducible.

$(ii)$$(ii)$ Any two non-empty open subsets of $X$$X$ have a non-empty intersection.

$(iii)$$(iii)$ Every non-empty open subset is dense in $X$$X$​.

$(iv)$$(iv)$ Every non-empty open subset is connected.

$(v)$$(v)$​​ Every non-empty open subset is irreducible.

$(vi)$$(vi)$ For any set $|E|\ge 2$$|E|\ge 2$, the $E$$E$ valued constant function presheaf over $X$$X$ is a sheaf.

Proof. Taking complements then $(i),(ii)$$(i),(ii)$ is equivalent by definition of irreducible space.

To see $(ii)⟺(iii)$$(ii)\iff (iii)$, we need a lemma.

Lemma. Let $X$$X$ be a topological space and let $A\subseteq X$$A\subseteq X$ be a subset. $A$$A$ is dense if and only if for any non-empty open set $U\subseteq X$$U\subseteq X$, $A\cap U\ne \mathrm{\varnothing }$$A\cap U\ne \empty$. i.e. $A$$A$ is dense if and only if for any open set $U$$U$,$A\cap U=\mathrm{\varnothing }⟺U=\mathrm{\varnothing }$$A\cap U=\empty\iff U=\empty$​​.

Proof. $A\cap U=\mathrm{\varnothing }$$A\cap U=\empty$ iff $A\subseteq U^c$$A\subseteq U^c$. As $A$$A$ is dense, the only closed set contains $A$$A$ is $X$$X$. Hence $U^c=X$$U^c=X$, $U=\mathrm{\varnothing }$$U=\empty$. $◻$$\square$

Then $(ii)⟺(iii)$$(ii)\iff (iii)$​ is obviously.

Now we want to shows that $(ii)⟺(iv)$$(ii)\iff (i v)$, but this is obviously.

Then if we can prove $(i)⟺(v)$$(i)\iff(v)$, we finish. Obviously $(v)⟹(i)$$(v)\implies (i)$, now let us show that $(iii)⟹(v)$$(iii)\implies (v)$.

Let $U$$U$ be a non empty open set with subspace topology, we want to show that every open set in $U$$U$ is dense in $U$$U$, hence $U$$U$ is irreducible by $(iii)⟺(i)$$(iii)\iff (i)$. By $(iii)$$(iii)$, every non empty subset $V\subseteq X$$V\subseteq X$ is dense in $X$$X$ hence $V\cap U$$V\cap U$ is dense in $U$$U$. Hence $U$$U$​ is irreducible.

For $(vi)$$(vi)$, we prove that $(ii)⟺(vi)$$(ii)\iff (vi)$

If any two non-empty open subsets of $X$$X$ have a non-empty intersection, then let $U\subseteq X$$U\subseteq X$ be an open set. For every open covering of $U$$U$ $(U_i{)}_{i\in I}$$(U_i)_{i\in I}$ . If $s_i|U_i\cap U_j=s_j|U_i\cap U_j$$s_i| U_i\cap U_j= s_j| U_i\cap U_j$ for all $i,j\in I$$i,j\in I$, then there exist a uniqe constant function $s\in \mathcal{F}(U)$$s\in\mathcal F(U)$ such that $s{|}_{U_i}=s_i$$s|_{U_i}=s_i$. Since $U_i\cap U_j\ne \mathrm{\varnothing }$$U_i\cap U_j\ne \empty$ for any two non-empty open subsets. Conversely, if $X$$X$ is reducible, then there exists two non-empty open subsets $U,V\subseteq X$$U,V\subseteq X$ have an empty intersection. Let $Y=U\cup V$$Y=U\cup V$, and $U,V$$U,V$ is an open covering of $Y$$Y$. Consider $f\in \mathcal{F}(U),f(x)=a$$f\in\mathcal F(U),f(x)=a$ for all $x\in U$$x\in U$ and $g\in \mathcal{F}(V),g(x)=b$$g\in \mathcal F(V),g(x)=b$ for all $x\in V$$x\in V$. If $a\ne b$$a\ne b$, there exists not a constant function $h\in \mathcal{F}(Y)$$h\in \mathcal F(Y)$ such that $h_U=f,h_V=g$$h_U=f,h_V=g$. $◻$$\square$​​

Lemma. Let $X$$X$ be a topological space. Let $Y\subseteq X$$Y\subseteq X$ is irreducible iff $\bar{Y}$$\overline {Y}$​ is irreducible. Hence if $\bar{Y}={\bar{Y}}^{\prime }$$\overline Y=\overline Y'$ and $Y$$Y$ is irreducible then $Y^{\prime }$$Y'$ is irredducible as well.

Proof. By previous proposition $(ii)$$(ii)$, we can see that $Y$$Y$ is irreducible iff for any two open sets $U,V$$U,V$, we have $Y\cap U\ne \mathrm{\varnothing },Y\cap V\ne \mathrm{\varnothing }⟹Y\cap (U\cap V)\ne \mathrm{\varnothing }$$Y\cap U\ne\empty,Y\cap V\ne\empty\implies Y\cap(U\cap V)\ne\empty$. (Consider the subspace topology)

This implies the lemma because an open set $U\cap Y\ne \mathrm{\varnothing }⟺U\cap \bar{Y}\ne \mathrm{\varnothing }$$U\cap Y\ne\empty\iff U\cap\overline{Y}\ne\empty$​. $◻$$\square$

Remark. $U\cap \bar{Y}\ne \mathrm{\varnothing }⟹U\cap Y\ne \mathrm{\varnothing }$$U\cap\overline Y\ne\empty\implies U\cap Y\ne\empty$ since every points $x\in \bar{Y}$$x\in\overline Y$ is a limit point. As $x\in U$$x\in U$, $U$$U$ is a open neighborhood of $x$$x$, hence intercet with $Y$$Y$.

Proposition. Let $A$$A$ be a commutative ring. A subset $Y\subseteq \mathrm{Spec}(A)$$Y\subseteq\mathrm{Spec}(A)$ is irreducible iff $\mathfrak{p}:=I(Y)$$\mathfrak p:=I(Y)$ is a prime ideal. In this case, $\{\mathfrak{p}\}$$\{\mathfrak p\}$ is dense in $\bar{Y}$$\overline Y$ since $\overline{\{\mathfrak{p}\}}=VI(Y)=\bar{Y}$$\overline{\{\mathfrak p\}}=VI(Y)=\overline Y$.

Proof. Assume $Y$$Y$ is irreducible. Let $f,g\in A,fg\in \mathfrak{p}$$f,g\in A,fg\in\mathfrak p$. Then

$$\begin{array}{}\text{(26)}& Y\subseteq V(fg)=V(f)\cup V(g)\end{array}$$

Hence $Y=(Y\cap V(f))\cup (Y\cap V(g))$$Y=(Y\cap V(f))\cup (Y\cap V(g))$. As $Y$$Y$ is irreducible, $Y=Y\cap V(f)$$Y=Y\cap V(f)$ or $Y=Y\cap V(g)$$Y=Y\cap V(g)$. This implies that $V(f)\supseteq Y$$V(f)\supseteq Y$ or $V(g)\supseteq Y$$V(g)\supseteq Y$. Hence $IV(f)=\sqrt{(f)}\subseteq I(Y)=\mathfrak{p}$$IV(f)=\sqrt{(f)}\subseteq I(Y)=\mathfrak p$ or $IV(g)=\sqrt{(g)}\subseteq I(Y)=\mathfrak{p}$$IV(g)=\sqrt{(g)}\subseteq I(Y)=\mathfrak p$. Hence $f\in \mathfrak{p}$$f\in\mathfrak p$ or $g\in \mathfrak{p}$$g\in\mathfrak p$. Hence $\mathfrak{p}$$\mathfrak p$ is prime.

Conversely let $\mathfrak{p}$$\mathfrak p$ be a prime ideal, then $\bar{Y}=VI(Y)=V(\mathfrak{p})=\overline{\{\mathfrak{p}\}}$$\overline{Y}=VI(Y)=V(\mathfrak p)=\overline{\{\mathfrak p\}}$. Hence $Y$$Y$ is irreducible since $\{\mathfrak{p}\}$$\{\mathfrak p\}$ is irreducible by the previous lemma. $◻$$\square$​

Definition. We say $Z\subset X$$Z\subset X$ is an irreducible component of $X$$X$ if $Z$$Z$ is a maximal irreducible subset of $X$$X$.

Corollary. The map $\mathfrak{p}\mapsto V(\mathfrak{p})$$\mathfrak p\mapsto V(\mathfrak p)$ is a one-one corresponding from prime ideals to irreducible closed subset in $\mathrm{Spec}(A).$$\mathrm{Spec}(A).$ Via this bijection, the minimal prime ideals of $A$$A$ correspond to the irreducible component of $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$.

Definition. Let $X$$X$ be an arbitrary topological space.

$(i)$$(i)$ A point $x$$x$ is called closed if the set $\{x\}$$\{x\}$ is closed.

$(ii)$$(ii)$ We say that a point $\eta \in X$$\eta\in X$ is a generic point if $\overline{\{\eta \}}=X$$\overline{\{\eta\}}=X$.

$(iii)$$(iii)$ Let $x,y$$x,y$ be two points of $X$$X$. We say that $x$$x$ is a generization of $y$$y$ or that $y$$y$ is a specialization of $x$$x$ if $y\in \overline{\{x\}}$$y\in\overline{\{x\}}$.

$(iv)$$(iv)$ A point $x\in X$$x\in X$ is a maximal point if $\overline{\{x\}}$$\overline{\{x\}}$ is an irreducible component of $X$$X$.

Thus a point $\eta \in X$$\eta\in X$ is a generic point iff its a generization of every point of $X$$X$. As the closure of an irreducible set is again irreducible, the existence of a generic point implies that $X$$X$ is irreducible.

Let $X=\mathrm{Spec}(A)$$X=\mathrm{Spec}(A)$, then the notation introduced have the following algebraic meaning.

$(i)$$(i)$ A point $x\in X$$x\in X$ is closed iff ${\mathfrak{p}}_x$$\mathfrak p_x$ is maximal ideal.

$(ii)$$(ii)$ A point $\eta \in X$$\eta\in X$ is a generic point of $X$$X$ iff ${\mathfrak{p}}_{\eta }$$\mathfrak p_{\eta}$ is the unique minimal prime ideal. This exist iff the nillradical ideal is prime ideal. Hence we have $\mathrm{Spec}(A)$$\mathrm{Spec} (A)$ is irreducible $⟺\mathfrak{n}(A)=\sqrt{(0)}$$\iff\mathfrak n(A)=\sqrt{(0)}$ is prime $⟺A/\mathfrak{n}(A)$$\iff A/\mathfrak n(A)$ is integral domain.

$(iii)$$(iii)$ A point $x\in X$$x\in X$ is a generization of $y$$y$ iff ${\mathfrak{p}}_x\subseteq {\mathfrak{p}}_y$$\mathfrak p_x\subseteq\mathfrak p_y$.

$(iv)$$(iv)$ A point $x\in X$$x\in X$ is a maximal point iff ${\mathfrak{p}}_x$$\mathfrak p_x$ is a minimal prime ideal.

Krull Dimension: A first glance

For commutative ring

Definition. Let $R$$R$ be a commutative ring, we use the strictly accending chains of prime ideals in $R$$R$.

$$\begin{array}{}\text{(27)}& {\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset ...\subset {\mathfrak{p}}_n\end{array}$$

Where $n$$n$ is the length of this chain. We define the supremum length of those strictly accending chains of prime ideals in $R$$R$ to be the Krull dimension of $R$$R$. Notice that the it is $\sup \{0,1,2,3,...,n-1,n\}\subseteq \mathbb{N}\cup \{\pm \mathrm{\infty }\}$$\mathrm{sup}\{0,1,2,3,...,n-1,n\}\subseteq\mathbb N\cup\{\pm\infty\}$. For a Ring $R$$R$, we denote the Krull dimension of $R$$R$ as $\dim R$$\dim R$.

Example.

The Krull dimension of $0$$0$ is $-\mathrm{\infty }$$-\infty$. Since $sup$$\sup$ means least upper bound. But every elements is a upper bound of $\mathrm{\varnothing }$$\empty$. Hence $sup\mathrm{\varnothing }=-\mathrm{\infty }$$\sup\empty=-\infty$​​.

The Krull dimension of a field $\mathbb{F}$$\mathbb F$ is $0$$0$, $\dim (\mathbb{Z})=1$$\dim(\mathbb Z)=1$. The Krull dimension of $K[T_1,...,T_n]$$K[T_1,...,T_n]$ is $n$$n$. We only give a prove for

$\dim (K[T_1,...,T_n])\ge n$$\dim(K[T_1,...,T_n])\ge n$. Consider $(T_1)\subset (T_1,T_2)\subset ...\subset (T_1,...,T_n)$$(T_1)\subset (T_1,T_2)\subset...\subset(T_1,...,T_n)$.

For topological space

Let $X$$X$ be a topological space, we use the strictly accending chains of irreducible closed subset in $X$$X$

$$\begin{array}{}\text{(28)}& Z_n\supset Z_{n-1}\supset ...\supset Z_1\supset Z_0\end{array}$$

Where $n$$n$ is the length of this chain. We define the supremum length of those strictly accending chains of those strictly accending chains of irreducible closed subset in $X$$X$. For a topological space $X$$X$, we denote the Krull dimension of $X$$X$ as $\dim X$$\dim X$.

Proposition. Let $R$$R$ be a commutative ring, then $\dim R=\dim \mathrm{Spec}(R)$$\dim R=\dim\mathrm{Spec}(R)$.

Proof. It follows from $\mathfrak{p}\mapsto V(\mathfrak{p})$$\mathfrak p\mapsto V(\mathfrak p)$ is a bijection between prime ideals and irreducible closed sets directly.

The bijection, indeed, is an anti poset isomorphism between $(\mathfrak{p},\subseteq )$$(\mathfrak p,\subseteq)$ and $(Z,\subseteq )$$(Z,\subseteq)$. $◻$$\square$