Commutative Algebra and Algebraic Geometry (1) Basic concepts of Algebraic Geometry

We will introduce some basic concepts in algebraic geometry, and see the algebra-geometry duality.

I higherly recommend readers read Math Essays: Galois Connection: The Initial One and Its Application to Hilbert's Nullstellensatz (marco-yuze-zheng.blogspot.com) and the internal link of that blog.

Definition. Let $K$$K$ be a field, and $n\in \mathbb{N}$$n\in\mathbb N$ , define the affine $n-$$n-$ spaces as:

$$\begin{array}{}\text{(1)}& {\mathbb{A}}_K^n:=\{(c_1,c_2,...,c_n):c_i\in K,1\le i\le n\}\end{array}$$

Remark. The difference between affine space and vector space.

As a set, there is no difference between ${\mathbb{A}}_K^n$$\mathbb A_K^n$ and $K^n$$K^n$. But usually we use $K^n$$K^n$ to refer the vector space. Let me explain the difference between affine space and vector space. The key point here is the elements of ${\mathbb{A}}_K^n$$\mathbb A_K^n$ is points, not vector!

Recall the definition of vector, they are elements of vector space. That means you should define $v+u$$v+u$ and $\lambda u$$\lambda u$ for $K^n$$K^n$ first, and say $v$$v$ is a vector in $(K^n,+,\cdot )$$(K^n,+,\cdot)$. If you define anorm $\parallel v{\parallel }_K$$\|v\|_{K}$ on $K^n$$K^n$, then you can talking about the length of each vector. Then each vector will be difference in the geometric sense. However, this is not what we want. In algebriac geometry, we study the zero of polynomials. For example, $3x^2+y^2-1=0$$3x^2+y^2-1=0$ for $3x^2+y^2-1\in \mathbb{R}[x,y]$$3x^2+y^2-1\in \mathbb R[x,y]$. What we want is points rather then vector. Each point should have same geometric property and $\{(x_1,...,x_n)\in {\mathbb{A}}_K^n|p(x_1,...,x_n)=0\}$$\{(x_1,...,x_n)\in\mathbb A_{K}^n|p(x_1,...,x_n)=0\}$ is a subset instead of subspace in general. That is the reason we do not work on the category of vector space over $K.$$K.$

Definition. Let $S$$S$ be a subset of $K[x_1,...,x_n]$$K[x_1,...,x_n]$. Then define the zero locus of $S$$S$ as

$$\begin{array}{}\text{(2)}& V(S):=\{x\in {\mathbb{A}}_K^n|f(x)=0\mathrm{\forall }f\in S\}\end{array}$$

Subset of ${\mathbb{A}}_K^n$$\mathbb A_K^n$ of this form are called affine variety. If $S=\{f_1,f_2,...,f_n\}$$S=\{f_1,f_2,...,f_n\}$, then we can write $V(S)$$V(S)$ as $V(f_1,...,f_n)$$V(f_1,...,f_n)$​.

Easy to see that $S\subseteq S^{\prime }⟹V(S)\supseteq V(S^{\prime })$$S\subseteq S'\implies V(S)\supseteq V(S')$. Notice that $\mathrm{\varnothing }$$\empty$ and ${\mathbb{A}}_K^n$$\mathbb A_K^n$ are algebraic variety since $V(1)=\mathrm{\varnothing },V(0)={\mathbb{A}}_K^n$$V(1)=\empty,V(0)=\mathbb A_K^n$.

Definition: The Adjoint of $V(-)$$V(-)$. Let $X\subseteq {\mathbb{A}}_K^n$$X\subseteq\mathbb A_{K}^n$, define the ideal associate to $X$$X$ as

$$\begin{array}{}\text{(3)}& I(X):=\{f\in K[x_1,...,x_n]|f(x)=0\mathrm{\forall }x\in X\}\end{array}$$

Propositin. $I(X)\supseteq S⟺X\subseteq V(S)$$I(X)\supseteq S\iff X\subseteq V(S)$. i.e. $I(-)$$I(-)$ is the left adjoint of $V(-)$$V(-)$.

Proof. According to the property of Galois Connection, we only need to prove that $X\subseteq VI(X),S\subseteq IV(S).$$X\subseteq VI(X),S\subseteq IV(S).$

Let $x\in X$$x\in X$, and for all $f\in I(X),f(x)=0$$f\in I(X), f(x)=0$. Hence $x\in VI(X)$$x\in VI(X)$. This claim that $X\subseteq VI(X)$$X\subseteq VI(X)$.

Let $g\in S$$g\in S$, and for all $x\in V(S)$$x\in V(S)$, $g(x)=0$$g(x)=0$. Hence $g\in IV(S)$$g\in IV(S)$. This claim that $S\subseteq IV(S)$$S\subseteq IV(S)$. $◻$$\square$​

Corollary. $I(X\cup Y)=I(X)\cap I(Y)$$I(X\cup Y)=I(X)\cap I(Y)$, $V(S\cup S^{\prime })=V(S)\cap V(S^{\prime })$$V(S\cup S')=V(S)\cap V(S')$. Since left adjoint preserve colimit and right adjoint preserve limit.

Remark. After we define more type of ideal, we will deduce much more corollary from this Galois connection.

Definition: the coordinate ring of $X$$X$.

Let $X\subseteq {\mathbb{A}}_K^n$$X\subseteq \mathbb A_{K}^n$, the coordinate ring of $X$$X$ is defined as

$$\begin{array}{}\text{(4)}& A(X):=K[x_1,...,x_n]/I(X)\end{array}$$

This definition is natural since sometimes we only care about the value of polynomial at $X$$X$.

Hence if $f(x)-g(x)=0\mathrm{\forall }x\in X$$f(x)-g(x)=0\forall x\in X$, then they should be same on $X$$X$.

In particular, $K[x_1,...,x_n]$$K[x_1,...,x_n]$ is the coordinate ring of ${\mathbb{A}}_K^n$$\mathbb A_K^n$.

Definition: Subvariety. Let $X\subseteq {\mathbb{A}}_K^n$$X\subseteq\mathbb A_K^n$ be an affine variety, $S\subseteq A(X)$$S\subseteq A(X)$. We could consider the zero locus defined as

$$\begin{array}{}\text{(5)}& V_X(S):=\{x\in X|f(x)=0\mathrm{\forall }f\in S\}\end{array}$$

We called $V_X(S)$$V_X(S)$ a subvariety of $X$$X$.

Similarly, for $Y\subseteq X$$Y\subseteq X$, we could define

$$\begin{array}{}\text{(6)}& I_X(Y):=\{f\in A(X)|f(y)=0\mathrm{\forall }y\in Y\}\end{array}$$

You can prove the Galois connection between $I_X(-)$$I_X(-)$ and $V_X(-)$$V_X(-)$ in the same way.

Definition : Morphism between variety.

Let $X\subseteq {\mathbb{A}}_K^n,Y\subseteq {\mathbb{A}}_K^m$$X\subseteq\mathbb A^n_K,Y\subseteq \mathbb A^m_K$ be two variety. A morphism between $X$$X$ and $Y$$Y$ is given by

$$\begin{array}{}\text{(7)}& \psi :=(f_1(x_1,...,x_n),...,f_m(x_1,...,x_n))\in Y\end{array}$$

Where each $f_i\in A(X)$$f_i\in A(X)$.

Now let us consider the category of affine algebraic variety $V_K$$V_K$, the coordinate ring $A(-)$$A(-)$ give us a contravarient functor

$$\begin{array}{}\text{(8)}& A(-):V_K^{\mathrm{op}}\to K-\mathrm{Alg},X⟼A(X),(\psi :X\to Y)⟼({\psi }^{\ast }:A(Y)\to A(X))\end{array}$$

Here ${\psi }^{\ast }$$\psi^*$ is the pullback. For $f:Y\to K\in A(Y),{\psi }^{\ast }(f)=f\circ \psi$$f:Y\to K\in A(Y),\psi^*(f)=f\circ \psi$.

To see it forms a $K-\mathrm{Alg}$$K-\mathrm{Alg}$ homomorphism, consider:

$$\begin{array}{}\text{(9)}& {\psi }^{\ast }(f+g)=(f+g)\circ \psi =f\circ \psi +g\circ \psi ={\psi }^{\ast }(f)+{\psi }^{\ast }(g)\end{array}$$

${\psi }^{\ast }(kf)=k{\psi }^{\ast }(f)$$\psi^*(kf)=k\psi^*(f)$ and ${\psi }^{\ast }(fg)={\psi }^{\ast }(f){\psi }^{\ast }(g)$$\psi^*(fg)=\psi^*(f)\psi^*(g)$ leaves to readers.

In the next essay, we will introduce more on the ideals of a ring $R$$R$, and see the Algebra-Geometery duality between Ideals and variety.