From Calculus to Algebraic Geometry

Overview

Hello! Long time no see. I was dealing with the final and trying to figure out what I did this term.

The result is impressive: the ODE, an algebraic approach a king of D- Module, But it can also be related to Differential Galois theory. The functor ${\mathrm{Hom}}_{\mathrm{T}\mathrm{o}\mathrm{p}}(-,\mathbb{R}):{\mathrm{Top}}^{op}\to \mathrm{CRing}$$\mathrm{Hom}_{\rm Top}(-,\mathbb R):\mathrm{Top}^{op}\to\mathrm{CRing}$ related to Algebraic Geometry.

Math Essays: A contravariant functor from subcat of Top to Algebra, a kind of algebra-geometry duality (wuyulanliulongblog.blogspot.com)

Today, I aim to give a duality between the Category of Compact Hausdorff Space and $\mathbb{R}-\mathrm{A}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}\mathrm{r}\mathrm{a}$$\mathbb R-\rm {Algebra}$.

We will denote $C$$C$ as the Category of Compact Hausdorff Space, and $A$$A$ for the Category of $\mathbb{R}-\mathrm{A}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}\mathrm{r}\mathrm{a}$$\mathbb R-\rm {Algebra}$ in this Blog.

By the way, if you consider this functor, ${\mathrm{Hom}}_C(-,\mathbb{C})$$\mathrm{Hom}_{C}(-,\mathbb C)$, it will relate to Gelfand Duality. The Duality between $C^{\ast }$$C^*$ Algebra and Compact Hausdorff Space. If you consider ${\mathrm{Hom}}_C(-,\mathbb{H})$$\mathrm{Hom}_C(-,\mathbb H)$, $\mathbb{H}$$\mathbb H$ is Quaternion, it will relate to Noncommutative Geometry...

Spec as a Functor from A to C

The motivation I observe it comes from the trivial example $C[a,b]$$C[a,b]$ Again. I have heard that Zariski Topology is a way to define Topology over $\mathrm{Spec}(R).$$\mathrm{Spec}(R).$ And I know that the maximal ideal of $C[a,b]$$C[a,b]$ bijectively correspond to the point $x\in [a,b]$$x\in[a,b]$.

By the way, actually $x(f):=f(x)$$x(f):=f(x)$ is also a pullback. So, it has to be an Algebra Homomorphism.

Since we know that singleton set $\ast$$*$ is the final object in $\mathrm{T}\mathrm{o}\mathrm{p}$$\rm Top$, thus we can consider the ''point'' in Category Theory.

We can view each point $x$$x$ of a Topology Space as a continuous function $x:\ast \to X,\ast \mapsto x$$x:*\to X,*\mapsto x$.

Then $f(x)$$f(x)$ should be viewed as $f\circ x=x^{\ast }(f)$$f\circ x=x^*(f)$.

By the way, I feel that all the prime ideals in $C[a,b]$$C[a,b]$ are maximal ideal as well. Thus, I think I can reconstruct the Topology.

Then I find it is true; it has been proved. The Zariski Topology of ${\mathrm{Hom}}_C(X,\mathbb{R})$$\mathrm{Hom}_C(X,\mathbb R)$ is homeomorphic to $X$$X$ !

Let us see the idea on $C[a,b]$$C[a,b]$ first.

Proposition 1. In $C[a,b]$$C[a,b]$, the prime ideal is the maximal ideal.

Proof.

We know that $S\subset U⟼I(S)\supset I(U)$$S\subset U\longmapsto I(S)\supset I(U)$, Thus, each prime ideal at least corresponds to one point.

Suppose that $I(S)$$I(S)$ is a prime ideal, and $|S|>1$$|S|>1$, let $a,b\in S$$a,b\in S$, consider $f,g,f(a)=0\ne f(b),g(a)\ne 0=g(b)$$f,g,f(a)=0\ne f(b),g(a)\ne 0= g(b)$

Thus $f\notin I(S),g\notin I(S)$$f\notin I(S),g\notin I(S)$, but $fg\in I(S)$$f g\in I(S)$, contradiction. So In $C[a,b]$$C[a,b]$, every prime ideal is a maximal ideal.

Thus $\mathrm{Spec}(C[a,b])\cong [a,b]$$\mathrm{Spec}(C[a,b])\cong[a,b]$.

Then let me introduce the definition of Zariski Topology, and reconstruct the topology structure of $[a,b]$$[a,b]$.

Definition 1. $V(I)=\{p\in \mathrm{Spec}(\mathrm{R})|I\subseteq p\}$$V(I)=\{p\in \mathrm{Spec(R)}|I\subseteq p\}$ is the closed set of $\mathrm{Spec}(\mathrm{R})$$\mathrm{Spec(R)}$. ($I$$I$ could be a subset of $R$$R$)

Proposition 2. $\mathrm{Spec}(\mathrm{R})$$\mathrm{Spec(R)}$ with Zariski Topology is a Topology Space.

Proof.

$†.V(R)=\mathrm{\varnothing },V(\mathrm{\varnothing })=\mathrm{Spec}(\mathrm{R})$$\dagger.V(R)=\empty,V(\empty)=\mathrm{Spec(R)}$

$†.$$\dagger.$ $V(I_1)\cap V(I_2)=V(I_1+I_2)$$V(I_1)\cap V(I_2)=V(I_1+I_2)$

Since $I_1+I_2$$I_1+I_2$ is the least ideal contain $I_1$$I_1$ and $I_2$$I_2$. That is, the least upper bound.

The upper bound of $I_1,I_2$$I_1,I_2$ is just the upper bound of the least upper bound...

For any collection of $V$$V$, $\begin{array}{c}\bigcap _{j\in \alpha }V(I_j)=V\left(\sum _{j\in \alpha }{I}_j\right)\end{array}$$\bigcap_{j\in\alpha} V(I_j)=V\left(\sum_{j\in\alpha}I_j\right)\\$ as well.

$†.V(I)\cup V(J)=V(IJ)$$\dagger. V(I)\cup V(J)=V(IJ)$

Since $V(I)\cup V(J)$$V(I)\cup V(J)$ is the greatest lower bound of $V(I)$$V(I)$ and $V(J)$$V(J)$

If $I\subseteq p\vee J\subseteq p$$I\subseteq p\vee J\subseteq p$, then $IJ\subset p$$IJ\sub p$ , since $IJ\subseteq I\cap J\subseteq p$$IJ\subseteq I\cap J\subseteq p$.

If $IJ\subseteq p$$IJ\subseteq p$, then $I\subseteq p\vee J\subseteq p$$I\subseteq p\vee J\subseteq p$. Since if $ij\in p$$ij\in p$, then $i\in p\vee j\in p$$i\in p\vee j\in p$ by definition of prime ideal.

Proposition 3. $V(I[\alpha ,\beta ])=[\alpha ,\beta ]$$V(I[\alpha,\beta])=[\alpha,\beta]$

Proof. $V(I[\alpha ,\beta ])=\{p\in \mathrm{Spec}(\mathrm{R})|I\subseteq p\}$$V(I[\alpha,\beta])=\{p\in \mathrm{Spec(R)}|I\subseteq p\}$, that is , $\{I_x|x\in [\alpha ,\beta ]\}$$\{I_x|x\in[\alpha,\beta]\}$.

The homeomorphic $f:\mathrm{Spec}(C[a,b])\to [a,b]$$f:\mathrm{Spec}(C[a,b])\to [a,b]$ is $I_x\to x,f^{-1}(x)=I_x$$I_x\to x,f^{-1}(x)=I_x$

That is, $\mathrm{Spec}(C[a,b])\cong [a,b]$$\mathrm{Spec}(C[a,b])\cong [a,b]$, in $\mathrm{T}\mathrm{o}\mathrm{p}$$\rm Top$. You can generalise this result to any Compact Hausdorff Space.

So $\mathrm{Spec}$$\mathrm{Spec}$ should be a functor $F:\mathrm{R}\mathrm{i}\mathrm{n}{\mathrm{g}}^{\mathrm{o}\mathrm{p}}\to \mathrm{T}\mathrm{o}\mathrm{p}$$F:\rm Ring^{op}\to Top$, in particular, $F:A\to C$$F:A\to C$

If we denote $G:={\mathrm{Hom}}_C(-,\mathbb{R})$$G:=\mathrm{Hom}_C(-,\mathbb R)$, then $F\circ G={\mathrm{id}}_C$$F\circ G=\mathrm{id}_C$

$G\circ F=\mathrm{i}{\mathrm{d}}_{\mathrm{A}}$$G\circ F=\rm id_A$ should be true as well. After I learn point set topology, I will go back to prove it.

This result shows us that $A^{\mathrm{o}\mathrm{p}}\cong C$$A^{\rm op}\cong C$. Duality between Algebra and Geometry.

We still need to prove that $\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}$$\rm Spec$ is a functor from $\mathrm{R}\mathrm{i}\mathrm{n}{\mathrm{g}}^{\mathrm{o}\mathrm{p}}$$\rm Ring^{op}$ to $\mathrm{T}\mathrm{o}\mathrm{p}$$\rm Top$, that is, $f:R_A\to R_B,F(f)=\mathrm{Spec}(R_B)\to \mathrm{Spec}(R_A)$$f:R_A\to R_B, F(f)=\mathrm{Spec}(R_B)\to\mathrm{Spec}(R_A)$

And check that $F(\mathrm{i}{\mathrm{d}}_{\mathrm{R}})=\mathrm{i}{\mathrm{d}}_{\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}(\mathrm{R})}$$F(\rm id_R)=id_{Spec(R)}$, $F(f\circ g)=F(g)\circ F(f)$$F(f\circ g)=F(g)\circ F(f)$.

Theorem.

Let $f:R\to S$$f: R → S$ be a ring homomorphism for rings $R$$R$ and $S$$S$.

Then the map $f^{\star }:\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}(\mathrm{S})\to \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}(\mathrm{R})$$f^{\star} : \rm Spec(S) → Spec(R)$ defined by $f^{\star }(P)=f^{-1}(P)$$f^{\star}(P) = f^{−1} (P)$ is continuous.

Proof.

We need to prove that $f^{\star -1}(V(I))=V(f(I))$$f^{\star-1}(V(I))=V(f(I))$.

$f^{\star -1}(V(I))=\{p\in \mathrm{Spec}(S)|f^{\star }(p)\in V(I)\}=\{p\in \mathrm{Spec}(S)|I\subseteq f^{\star }(p)\}=\{p\in \mathrm{Spec}(S)|f(I)\subseteq p\}=V(f(I))$$f^{\star-1}(V(I))=\{p\in\mathrm{Spec}(S)| f^{\star}(p)\in V(I)\}=\{p\in \mathrm{Spec}(S)|I\subseteq f^{\star}(p)\}=\{p\in\mathrm{Spec}(S)| f(I)\subseteq p\}=V( f(I))$

$Q.E.D.$$Q.E.D.$

$F(\mathrm{i}{\mathrm{d}}_{\mathrm{R}})=\mathrm{i}{\mathrm{d}}_{\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}(\mathrm{R})}$$F(\rm id_R)=id_{Spec(R)}$ is obviously, to see $F(f\circ g)=F(g)\circ F(f)$$F(f\circ g)=F(g)\circ F(f)$, $(f\circ g{)}^{\star }=(f\circ g{)}^{-1}=g^{-1}\circ f^{-1}=g^{\star }\circ f^{\star }$$(f\circ g)^{\star}=(f\circ g)^{-1}=g^{-1}\circ f^{-1}=g^{\star}\circ f^{\star}$.

Presheaf

I remember when the first time I see the definition of topology, I do not understand it.

But I know that I can view preorder set as a Category, thus I get Open Set Category. $\mathrm{O}\mathrm{p}(\mathrm{T}\mathrm{o}\mathrm{p})$$\rm Op(Top)$

The object is open set and the morphism is $\subseteq$$\subseteq$, you can replace it by inclusion map.

In open set category, initial and final object exist, any coproduct exist, finitely many product exist.

In fact, this view is basic idea of presheaf.

Giving each open set subspace topology, since the subspace topology is the coarset topology makes inclusion map continuous.

Then consider the functor $\mathrm{H}\mathrm{o}{\mathrm{m}}_{\mathrm{O}\mathrm{p}(\mathrm{T}\mathrm{o}\mathrm{p})}(-,\mathbb{R}):\mathrm{O}\mathrm{p}((\mathrm{T}\mathrm{o}\mathrm{p}){)}^{\mathrm{o}\mathrm{p}}\to \mathrm{C}\mathrm{R}\mathrm{i}\mathrm{n}\mathrm{g}$$\rm Hom_{Op(Top)}(-,\mathbb R):Op((Top))^{op}\to CRing$ , pullback will give the ring homomorphism.

In general, presheaf is a functor ${\mathrm{C}}^{\mathrm{o}\mathrm{p}}\to \mathrm{S}\mathrm{e}\mathrm{t}$$\rm C^{op}\to Set$.

For example

$\mathrm{H}\mathrm{o}{\mathrm{m}}_{\mathrm{V}\mathrm{e}\mathrm{c}{\mathrm{t}}_{\mathbb{F}}}(-,\mathbb{F})$$\rm Hom_{Vect_\mathbb F}(-,\mathbb F)$, map each vector space to dual space, and linear map to dual map(dual map is just the pullback) is a presheaf.

An Algebraic approach to why some smooth functions not analytic.

We can consider a radical Ideal in $C[-1,1]$$C[-1,1]$, $\begin{array}{c}I=\{f\in C[-1,1]|\underset{x\to 0}{lim}\frac{f(x)}{x^k}=0,\mathrm{\forall }k\ge 0\}\end{array}$$I=\{f\in C[-1,1]|\lim_{x\to 0}\frac{f(x)}{x^k} =0,\forall k\ge 0\}\\$

To see it is radical, if $f(x{)}^n\in I$$f(x)^n\in I$, then $\underset{x\to 0}{lim}f(x{)}^n/x^{kn}=0,\mathrm{\forall }n,k\ge 0$$\lim_{x\to 0}f(x)^n/x^{kn}=0,\forall n,k\ge 0$, thus $\begin{array}{c}\underset{x\to 0}{lim}f(x)/x^k=\underset{x\to 0}{lim}\sqrt[n]{f(x{)}^n/x^{kn}}=0\end{array}$$\lim_{x\to0} f(x)/x^k= \lim_{x\to 0}\sqrt[n]{f(x)^n/x^{kn}}=0\\$

Lemma. $\mathrm{\forall }f\in I\cap C^{\mathrm{\infty }},f^{(n)}(0)=0\mathrm{\forall }n\ge 0$$\forall f\in I\cap C^{\infty},f^{(n)}(0)=0\forall n\ge 0$.

Proof. $\underset{x\to 0}{lim}f/x^n=f(0)/n!=0$$\lim_{x\to 0}f/x^n=f(0)/n!=0$

Proposition. $I$$I$ is prime ideal in $C^{\mathrm{\infty }}[-1,-1]$$C^{\infty}[-1,-1]$.

Proof.

Consider a ring homomorphism $\iota :C^{\mathrm{\infty }}\to \mathbb{R}[[X]]$$\iota:C^{\infty}\to\mathbb R[[X]]$ by $\begin{array}{c}f⟼\sum _{n=0}^{\mathrm{\infty }}{f}^{(n)}(0)\frac{x^n}{n!}\end{array}$$f\longmapsto\sum_{n=0}^{\infty}{f^{(n)}(0)}\frac{x^n}{n!}\\$, then $I\cap C^{\mathrm{\infty }}$$I\cap C^{\infty}$ is the kernel.

Since $\mathbb{R}[[X]]$$\mathbb R[[X]]$ is intergal domain, thus $I\cap C^{\mathrm{\infty }}$$I\cap C^{\infty}$ is prime ideal.

For example, $\begin{array}{c}f(x)=\{\begin{array}{l}0,x=0\\ e^{-1/x^2},x\ne 0\end{array}\end{array}$$f(x)=\begin{cases}0,x=0\\e^{-1/x^2},x\ne 0\end{cases}$ belongs to this ideal. The most famous example for $C^{\mathrm{\infty }}$$C^{\infty}$ but not analytic.

Since $\iota :C^{\mathrm{\infty }}\to \mathbb{R}[[X]]$$\iota :C^{\infty}\to\mathbb{R}[[X]]$ is not isomorphic. In fact, $C^{\mathrm{\infty }}$$C^{\infty}$ is not integral domain.

In general, we can define $\begin{array}{c}{I}_g=\{f\in C[-1,1]|\}\underset{x\to 0}{lim}\frac{f(x)}{g(x{)}^k}=0,\mathrm{\forall }k\ge 0\}\end{array}$$I_g=\{f\in C[-1,1]|\}\lim_{x\to 0}\frac{f(x)}{g(x)^k} =0,\forall k\ge 0\}\\$, $g(x)=0⟺x=0$$g(x)=0\iff x=0$ as the radical ideal as well.