Commutative Algebra and Algebriac Geometry (2) Zariski topology on affine space and basic algebraic curve

In [the last section], we define some basic concepts and their relation in AG.

Now let us delve deeper, and gain some geometric intuition from algebraic curves.

Zariski topology on affine spaceCoordinate ring functor is representableClassic Algebra-Geometry DualityIntroduction to Functorial Algebraic GeometryBasic Algebraic curves

Zariski topology on affine space

Definition: Zariski topology.

The set $V(\mathfrak{a})$$V(\mathfrak a)$, where $\mathfrak{a}$$\mathfrak a$ runs through the set of ideals of $K[T_1,...,T_n]$$K[T_1,...,T_n]$, are the closed sets of topology on ${\mathbb{A}}_K^n$$\mathbb A_K^n$, called Zariski topology on ${\mathbb{A}}_K^n$$\mathbb A_{K}^n$.

We still need to check that this definition satisfies the axioms of closed sets.

• $\mathrm{\varnothing }=V(1),{\mathbb{A}}_K^n=V(0)$$\emptyset=V(1),\mathbb A_K^n=V(0)$

• For every family $({\mathfrak{a}}_i{)}_{i\in I}$$(\mathfrak a_i)_{i\in I}$ of ideals ${\mathfrak{a}}_i\subseteq K[T_1,...,T_n]$$\mathfrak a_i\subseteq K[T_1,...,T_n]$, we have by the property of Galois connection.

$$\begin{array}{}\text{(1)}& \bigcap _{i\in I}V({\mathfrak{a}}_i)=V\left(\sum _{i\in I}{\mathfrak{a}}_i\right)\end{array}$$

• For any two ideals $\mathfrak{a},\mathfrak{b}$$\mathfrak a,\mathfrak b$, we have

$$\begin{array}{}\text{(2)}& V(\mathfrak{a})\cup V(\mathfrak{b})=V(\mathfrak{a}\cap \mathfrak{b})=V(\mathfrak{a}\mathfrak{b})\end{array}$$

To prove $(2)$$(2)$, it is clear that $\mathfrak{a}\mathfrak{b}\subseteq \mathfrak{a}\cap \mathfrak{b}\subseteq \mathfrak{a},\mathfrak{b}$$\mathfrak a\mathfrak b\subseteq \mathfrak a\cap\mathfrak b\subseteq \mathfrak a,\mathfrak b$, hence $V(\mathfrak{a}\mathfrak{b})\supseteq V(\mathfrak{a}\cap \mathfrak{b})\supseteq V(\mathfrak{a})\cup V(\mathfrak{b})$$V(\mathfrak a\mathfrak b)\supseteq V(\mathfrak a\cap\mathfrak b)\supseteq V(\mathfrak a)\cup V(\mathfrak b)$.

If $x\in V(\mathfrak{a}\mathfrak{b})$$x\in V(\mathfrak a\mathfrak b)$ and $x\notin V(\mathfrak{a})$$x\notin V(\mathfrak a)$, then $\mathrm{\exists }f\in \mathfrak{a},f(x)\ne 0$$\exists f\in\mathfrak a,f(x)\ne 0$. For all $g\in \mathfrak{b}$$g\in\mathfrak b$ we have $fg\in \mathfrak{ab}$$fg\in\mathfrak{ab}$, hence $fg(x)=f(x)g(x)=0\mathrm{\forall }g\in \mathfrak{b}$$fg(x)=f(x)g(x)=0 \forall g\in\mathfrak b$.

Therefore $g(x)=0\mathrm{\forall }g\in \mathfrak{b}$$g(x)=0 \forall g\in\mathfrak b$, hence $x\in V(\mathfrak{b})$$x\in V(\mathfrak b)$. Then $x\in V(\mathfrak{ab})⟹x\in V(\mathfrak{a})\cup V(\mathfrak{b})$$x\in V(\mathfrak {ab})\implies x\in V(\mathfrak a)\cup V(\mathfrak b)$. $◻$$\square$​

We will see that Zariski topology make the morphism between varieties become continuous.

Coordinate ring functor is representable

In the last section, we defined the coordinate ring of $X$$X$ as $A(X)=K[T_1,...,T_n]/I(X)$$A(X)=K[T_1,...,T_n]/I(X)$.

We assert that $A(X)\cong \mathrm{Hom}(X,{\mathbb{A}}_K^1)$$A(X)\cong \mathrm{Hom}(X,\mathbb A_K^1)$. Here, $\mathrm{Hom}(X,{\mathbb{A}}_K^1)$$\mathrm{Hom}(X,\mathbb A_K^1)$ denotes the set of all morphisms from the algebraic variety $X$$X$ to ${\mathbb{A}}_K^1$$\mathbb A_K^1$.

Addition and multiplication are defined pointwise, i.e., $(f+g)(x)=f(x)+g(x)$$(f+g)(x)=f(x)+g(x)$ and $(fg)(x)=f(x)g(x)$$(fg)(x)=f(x)g(x)$​​.

We will use the fact that $A(-)$$A(-)$ is representable to prove chinese remainder theorem.

Classic Algebra-Geometry Duality

Let $K$$K$ be an algebraic closed field. Let $V_K$$V_K$ refer to the category of affine $K$$K$ variety, $D_K$$D_K$ refer to the category of reduced finite generated $K-$$K-$ Algebra.

Then $\mathrm{Hom}(-,{\mathbb{A}}_K^1):V_K^{\mathrm{op}}\cong D_K$$\mathrm{Hom}(-,\mathbb A_K^1):V_K^{\mathrm{op}}\cong D_K$ is an anti category equvalience.

Proof. We want to show that the coordinate ring functor is fully faithful essential full.

Let us prove essential full first. Let $A$$A$ be an finite generated $K-$$K-$Algebra, then $A\cong K[T_1,...,T_n]/\mathfrak{I}$$A\cong K[T_1,...,T_n]/\mathfrak I$ for some $\mathfrak{I}$$\mathfrak I$ since $A$$A$ is finite generated. Also, $\mathfrak{I}$$\mathfrak I$ is a radical ideal since $A$$A$ is reduced. (See Math Essays: Commutative Algebra and Algebraic Geometry (3) ideals and their geometry (marco-yuze-zheng.blogspot.com)) Hence $A$$A$ isomorphic to a coordinate ring of an algebraic variety by Hilbert's Nullstellensatz.

Now, let us prove that $\mathrm{Hom}(-,{\mathbb{A}}_K^1)$$\mathrm{Hom}(-,\mathbb A_K^1)$ is fully faithful, i.e.

$$\begin{array}{}\text{(3)}& {\mathrm{Hom}}_{V_K}(X,Y)\cong {\mathrm{Hom}}_{D_K}(A(Y),A(X))\end{array}$$

Let $f:X\to Y$$f:X\to Y$ be a morphism, we can write $f$$f$ as $f=(f_1,...,f_m)$$f=(f_1,...,f_m)$.

Since we know that $A(Y)\cong K[T_1,...,T_m]/I(Y),A(X)\cong K[T_1,...,T_n]/I(X)$$A(Y)\cong K[T_1,...,T_m]/I(Y),A(X)\cong K[T_1,...,T_n]/I(X)$, then $f^{\ast }({\bar{y}}_i)={\bar{y}}_i\circ f=\overline{f_i}$$f^*(\overline y_i)=\overline y_i\circ f=\overline{f_i}$.

The class ${\bar{y}}_1\circ f,...,{\bar{y}}_m\circ f$$\overline y_1\circ f,...,\overline y_m\circ f$ determine $f$$f$. Hence the functor is faithful.

To see it is full, let $\alpha :A(Y)\to A(X)$$\alpha:A(Y)\to A(X)$ be a ring homomorphism, let $f_1,...,f_m\in K[T_1,...,T_m]$$f_1,...,f_m\in K[T_1,...,T_m]$ such that

$$\begin{array}{}\text{(4)}& \overline{f_1}=\alpha ({\bar{y}}_1),...,\overline{f_m}=\alpha ({\bar{y}}_m)\end{array}$$

It is clear that $f=(f_1,....,f_m):X\to {\mathbb{A}}_K^m$$f=(f_1,....,f_m):X\to\mathbb A_K^m$. To see its codomian is $Y$$Y$, let us consider aribitary $g\in I(Y)$$g\in I(Y)$ and consider

$$\begin{array}{}\text{(5)}& \alpha (g(y_1,...,y_m))=g(\alpha (y_1),...,\alpha (y_m))=g(\overline{f_1},...,\overline{f_m})=g\circ f=0\in I(X)\end{array}$$

Hence for all $x\in X$$x\in X$, $g\circ f(x)=0⟹f(x)\in Y$$g\circ f(x)=0\implies f(x)\in Y$. Hence $\alpha$$\alpha$ gives us a morphism $f:X\to Y$$f:X\to Y$ back and $f^{\ast }=\alpha$$f^*=\alpha$. $◻$$\square$

Introduction to Functorial Algebraic Geometry

Let $f_1,...,f_m\subseteq R[T_1,...,T_n]$$f_1,...,f_m\subseteq R[T_1,...,T_n]$ be a family of polynomials and $L$$L$ be a $R-$$R-$Algebra,

Let $X(L)=(t_j{)}_{j\in I}$$X(L)=(t_j)_{j\in I}$ be the solution systems of $f_1,...,f_m$$f_1,...,f_m$ in $L$$L$​. The elements in $X(L)$$X(L)$ is called $L-$$L-$​points.

For example, Let $R=\mathbb{Q},F=x^2+1$$R=\mathbb Q,F=x^2+1$, then $X(\mathbb{Q})=\mathrm{\varnothing }$$X(\mathbb Q)=\empty$ and $X(\mathbb{C})=\{\pm i\}$$X(\mathbb C)=\{\pm i\}$.

Then we claim that, $X(-):R-\mathrm{Alg}\to \mathrm{Set}$$X(-):R\mathrm{-Alg}\to\mathrm{Set}$ is a functor, and $X(L)\cong {\mathrm{Hom}}_{R-\mathrm{Alg}}(A(X),L)$$X(L)\cong\mathrm{Hom}_{R-\mathrm{Alg}}(A(X),L)$.

Where $A(X)=R[T_1,...,T_n]/(f_1,...,f_m)$$A(X)=R[T_1,...,T_n]/(f_1,...,f_m)$​​.

It is clear that $X(-)$$X(-)$ is a functor. For example, let $p(T)=a_n{T}^n+...+a_0$$p(T)=a_nT^n+...+a_0$ be a polynomial with $L-$$L-$​coefficient, and $f:L\to E$$f:L\to E$ be a $K-$$K-$algebra homomorphism.

If $p(t)=0$$p(t)=0$, then $f(p(t))=f(a_n)f(t{)}^n+...f(a_0)=0$$f(p(t))=f(a_n)f(t)^n+...f(a_0)=0$ as well. Hence if $t\in X(L)$$t\in X(L)$ then $f(t)\in X(E)$$f(t)\in X(E)$.

Now let us prove that $X(-)$$X(-)$ is representable. Let $(t_j{)}_{j\in J}\in X(L)$$(t_j)_{j\in J}\in X(L)$, then it could induce a $R-\mathrm{Alg}$$R-\mathrm{Alg}$ homomorphism by evaluation map. For example, consider $i\in X(\mathbb{C})$$i\in X(\mathbb C)$ and ${\mathrm{ev}}_i:\mathbb{Q}[T]/(T^2+1)\to \mathbb{C},p(T)⟼p(i)$$\mathrm{ev}_i:\mathbb Q[T]/(T^2+1)\to\mathbb C,p(T)\longmapsto p(i)$.

Conversely, let $h\in {\mathrm{Hom}}_{R-\mathrm{Alg}}(A(X),L)$$h\in\mathrm{Hom}_{R-\mathrm{Alg}}(A(X),L)$, and consider the quotient map $\pi :R[T_1,...,T_n]\to A(X)$$\pi:R[T_1,...,T_n]\to A(X)$​.

We get

$$\begin{array}{}\text{(6)}& h\circ \pi :K[T_1,...,T_n]\to L\end{array}$$

Then $h(\pi (T_1)),...,h(\pi (T_n))$$h(\pi (T_1)),...,h(\pi(T_n))$ is a solution of $f_1,...,f_m$$f_1,...,f_m$ in $L$$L$. For example, suppose that $F=T_1^2+T_2^2-1$$F=T_1^2+T_2^2-1$,

Then $h(\pi (T_1){)}^2+h(\pi (T_2){)}^2-1=h(\pi (T_1^2+T_2^2-1))=h(0)=0$$h(\pi(T_1))^2+h(\pi(T_2))^2-1=h(\pi(T_1^2+T_2^2-1))=h(0)=0$.

Basic Algebraic curves

Definition.

• An affine algebraic curve is a non-constant polynomial $F\in K[x,y]$$F\in K[x,y]$ modulo units. We call $V(F)=\{P\in {\mathbb{A}}_K^2|F(P)=0\}$$V(F)=\{P\in \mathbb A^2_K|F(P)=0\}$ the set of points of $F$$F$.

• The degree of a curve is its degree as a polynomial. Curves of degree 1, 2, 3,... are usually referred to as lines, quadrics, cubics, and so on.

• A curve is irreducible if it is as a polynomial, and reducible otherwise. Similarly, if $F=F_1^{a_1}\cdots F_n^{a_n}$$F=F_1^{a_1}\cdots F_n^{a_n}$ is the irreducible decomposition of the curve $F$$F$, then the curves $F_1,...,F_n$$F_1,...,F_n$ are called irreducible components of $F$$F$ and $a_1,...,a_n$$a_1,...,a_n$ their multiplicities. A curve is called reduced if $F=F_1\cdots F_n$$F=F_1\cdots F_n$, i.e., all its irreducible components have multiplicity 1. Readers may note that this resembles a square-free number.

Remark. An algebraically closed field has to be infinite, and $K[x,y]\cong K[x][y]$$K[x,y] \cong K[x][y]$.

If $K=\{c_1,...,c_n\}$$K=\{c_1,...,c_n\}$, then the polynomial $(x-c_1)\cdots (x-c_n)+1$$(x-c_1)\cdots(x-c_n)+1$ has no solution.

Lemma. Let $F$$F$ be an affine curve.

• If $K$$K$ is algebraically closed, then $V(F)$$V(F)$ is infinite.

• If $K$$K$ is infinite, then ${\mathbb{A}}_K^2\setminus V(F)$$\mathbb A_K^2 \setminus V(F)$ is infinite.

Proof. Suppose $F=a_n{x}^n+\cdots +a_{1}x+a_0$$F=a_nx^n+\cdots+a_1x+a_0$, where $a_i\in K[y]$$a_i \in K[y]$. Since for each $a_i\ne 0$$a_i\ne 0$, $a_i(y)=0$$a_i(y)=0$ only has finitely many zeros but $K$$K$ is infinite. Hence there exist infinitely many $y$$y$ such that $F(x,y)$$F(x,y)$ is a non-constant in $x$$x$.

• If $K$$K$ is algebraically closed, then for each $y$$y$, there exists $x$$x$ such that $F(x,y)=0$$F(x,y)=0$.

• If $K$$K$ is infinite, then for each $y$$y$, there exists $x$$x$ such that $F(x,y)\ne 0$$F(x,y)\ne 0$ since $F(-,y)$$F(-,y)$ only has finitely many solutions. $◻$$\square$

Proposition. If two curves $F$$F$ and $G$$G$ have no common component, then their intersection $V(F,G)$$V(F,G)$ is finite.

Proof. By assumption, $F$$F$ and $G$$G$ are coprime in $K[x,y]$$K[x,y]$. We claim that they are also coprime in $K(x)[y]$$K(x)[y]$.

In fact, if $G$$G$ and $F$$F$ had a common factor in $K(x)[y]$$K(x)[y]$, then after clearing denominators we would have $aF=H{F}^{\prime },aG=H{G}^{\prime }$$aF=HF',aG=HG'$ for some $H,F^{\prime },G^{\prime }\in K[x,y]$$H,F',G'\in K[x,y]$ of positive $y$$y$ degree and nonzero $a\in K[x]$$a\in K[x]$.

But $F,G\in K[x,y]$$F,G\in K[x,y]$ as well, i.e., $F=H{F}^{\prime }/a\in K[x,y]$$F=HF'/a\in K[x,y]$, $G=H{G}^{\prime }/a\in K[x,y]$$G=HG'/a\in K[x,y]$. Then every irreducible component of $a$$a$ must divide $H$$H$ or both $F^{\prime }$$F'$ and $G^{\prime }$$G'$ in $K[x,y]$$K[x,y]$. Hence $F=H^{\prime }{F}^{″},G=H^{\prime }{G}^{″}$$F=H'F'',G=H'G''$, contradicting the fact that $F$$F$ and $G$$G$ are coprime in $K[x,y]$$K[x,y]$.

Now, since $K(x)[y]$$K(x)[y]$ is a polynomial ring over the field $K(x)$$K(x)$, it is a PID. Hence we have $sF+tG=1$$sF + tG = 1$, where $s,t\in K(x)[y]$$s,t \in K(x)[y]$.

After clearing the denominators, we get $c=s^{\prime }F+t^{\prime }G$$c=s'F+t'G$ for $s^{\prime },t^{\prime }\in K[x,y]$$s',t'\in K[x,y]$ and nonzero $c\in K[x]$$c\in K[x]$.

But if $P\in V(F,G)$$P\in V(F,G)$, then $c(P)=s^{\prime }(P)F(P)+t^{\prime }(P)G(P)=0$$c(P)=s'(P)F(P)+t'(P)G(P)=0$ and $c\in K[x]$$c\in K[x]$ only has finitely many roots in $K$$K$.

i.e., $P_x$$P_x$ only has finitely many possibilities. By symmetry, we can see that $P_y$$P_y$ also has only finitely many possibilities. Hence $P=(P_x,P_y)$$P=(P_x,P_y)$ has only finitely many possibilities. i.e., $V(F)\cap V(G)=V(F,G)$$V(F)\cap V(G)=V(F,G)$ is finite. $◻$$\square$

Corollary. Let $F$$F$ be a curve over an algebraically closed field. Then for any irreducible curve $G$$G$, we have

$$\begin{array}{}\text{(7)}& G|F⟺V(G)\subseteq V(F)\end{array}$$

Proof.

Assume that $F=GH$$F=GH$ for some curve $H$$H$. If $P\in V(G)$$P\in V(G)$, then we also have $F(P)=G(P)H(P)=0$$F(P)=G(P)H(P)=0$. Hence $P\in V(F)$$P\in V(F)$.

Now assume $V(G)\subseteq V(F)$$V(G)\subseteq V(F)$. Then $V(F,G)=V(G)$$V(F,G)=V(G)$ is infinite since $K$$K$ is algebraically closed. Hence $F$$F$ and $G$$G$ have a common component; otherwise, $V(F,G)$$V(F,G)$ is finite. As $G$$G$ is irreducible, $G|F$$G | F$. $◻$$\square$