The Zariski Topology and The ring of Regular Function
We know generalized the notation of Algebraic Curve defined in [ALGE I]. Here $k$ is an algebraic closed field.
We say a subset of $\mathbb{A}^n$ is closed set if the elements of $X$ are the common roots of some (system) of polynomial equations in algebraic closed field $k$. That is $X$ is a common solutions of $F_1=F_2=\cdots=F_k=0$. We say $F_i$ is a defining equations of $X$.
Suppose that we are given $\{X_{\alpha}\}_{\alpha \in \mathbb{I}}$, the collection of closed sets, where for each $X_{alpha}$ is defined by system of equations $F_{\alpha_i} = 0$, then
\[\bigcap_{\alpha \in J} X_\alpha\]
is also a closed set, since it is defined by the whole systems (putting all together) $F_{\alpha_i}=0$ for all $\alpha \in I$ and for all $i$.
Also, for any two closed set $X_1$ dan $X_2$, with system $F_i=0$ for $X_1$ and system $G_i=0$ for $X_2$, the set $X_1 \cup X_2$ is defined by the system $F_i G_j=0$ for all $i,j$.
And finally, the empty set and $\mathbb{A}^n$ are closed set, each defined by equation $1=0$ dan $0=0$, respectively.
Therefore Closed Set $X$ form a topology, we called this topology Zariski Topology.
Example:
Consider the Zariski Topology in the affine line $A^1$, any closed set $X$ in this topology is defined by a polynomial of one variable $F(x)=0$. Indeed, for any closed set $X$, we can associate and ideal $\mathfrak{A}_X=\{f \in k[x]\, | f(x)=0 \, \forall x \in X\}$ of $k[x]$, thus in particular $F \in \mathfrak{A}_X$. Furthermore, since $k[x]$ is a PID, we should have $\mathfrak{A}_X =(F)$, for some polynomial $F$, and thus any closed set $X$ is defined by equation $F(x)=0$, and since $k$ is algebraic closed we have $F(x)=(x-a_1)(x-a_2)\cdots(x-a_k)$ so that $X=\{a_1,a_2,\cdots,a_k\}$, that is a closed set of this topology are those set with finitely many elements. Conversely for any finite set $X=\{a_1,a_2,\cdots,a_k\}$ we can form a polynomial $F(x)=(x-a_1)\cdots (x-a_k)$ as the equation defining $X$, and so $X$ is a closed set. We conclude that all closed set of the Zariski Topology in $A^1$ are those finite set of $k$.
The ideal $\mathfrak{A}_X$ introduced in the above example, is important on characterizing the closed set. Another way to look at it, is as follow:
Let $k[T_1,T_2,\cdots,T_n]$ be the ring of polynomial of several variables, for convenience we will denote it further as $k[T]$. For any closed set $X$, consider all function $f: X \rightarrow k$ such that there exists $F\in k[T]$ satisfy $F(x)=f(x)$ for all $x \in X$. Such a function $f$ is called regular function.
We can form the ring $k[X]$, consisting all regular function on $X$. Furthermore any $f \in k[T]$ we include $f|_X \in k[X]$ (the restriction function of $f$ in X). We can check that $k[X]$ form a ring with usual addition and multiplication of functions.
Consider the homomorphism $\psi : k[T] \rightarrow k[X]$, which send any $f \in k[T]$ to $f|X \in k[X]$, $\psi$ is easily checked to be a homomorphism and since $k[X]$ is regular $\psi$ is also onto. The kernel of $\psi$ is
\[\{F \in k[T] |\,F(P)=0 ,\forall \, P \in X \}\]
which is exactly $\mathfrak{A}_X$. Moreover by isomorphism theorem we have $k[T]/\mathfrak{A}_X \cong k[X]$.
This proves that $k[X]$ is determined by the ideal $\mathfrak{A}_X$.
Remark:By Hilbert Basis Theorem, every ideal in $R[T]$ where $R$ is a nonetherian ring is finitely generated. Here $R=k$ is a field (which has $(0)$ or $k$ only has ideals) therefore clearly nonetherian ring, thus $k[T]$ is finitely generated. Being the homomorphic image of $k[T]$, $k[X]$ must also finitely generated
More interesting from the remark is that $\mathfrak{A}_X$ being an ideal is also finitely generated, therefore any closed set $X$, can be described as the common roots of finitely many polynomial equations!
This could be noted more, that whenever $F_i$ are the defining equations of $X$, it is not always the case that $\mathfrak{A}_X=(F_1,\cdots,F_k)$. Clearly by definition $F_i \in \mathfrak{A}_X$. Since $\mathfrak{A}_X$ is finitely generated say by $G_i$ (which can be obtained as Grobner Bases), we can take $G_1=G_2=\cdots=G_l=0$ as the defining equation of $X$.
Furthermore, suppose that $f\in k[X]$ is vanish at all points for which $g_1,\cdots,g_k \in k[X]$ are also vanish, we can associate $F\in k[T]$ as an inverse image of $f$ under homomorphism $\psi$, and similarly $G_i\in k[T]$ for each $g_i$ mentioned above. By Hilbert Nullstellensatz, since $F$ vanish at the same the points as $G_i$, and also in $F_i$ (the defining equations of $X$) therefore we have $F^r \in (G_1,G_2,\cdots,G_k, F_1,F_2,\cdots,F_l)$, and we have $f^r \in (g_1,\cdots,g_k)$ (since all $F_i$ is associate to zero polynomials in $k[X]$ by $\psi$.)
Theorem
Let $X$ and $Y$ be closed sets. Then $Y \subset X$ if and only if $\mathfrak{A}_X \subset \mathfrak{A}_Y$
proof:
Suppose that $Y \subset X$, take $f\in \mathfrak{A}_X$, then $F(P)=0$ for all $P \in X$, since $Y \subset X$, therefore for every $Q\in Y$ we should have $F(Q)=0$, thus $f\in \mathfrak{A}_Y$.
Now suppose that $\mathfrak{A}_X \subset \mathfrak{A}_Y$ and $Y \not \subset X$, that is there exists $Q \in Y$ such that $Q \not \in X$. Suppose that $\mathfrak{A}_X=(G_1,G_2,\cdots,G_k)$ , then $G_i \in \mathfrak{A}_X \subset \mathfrak{A}_Y$, therefore we have $G_i(Q)=0$, and hence $F(Q)=0$ for all $F \in \mathfrak{A}_X$, thus in particular $Q$ is a common root of the defining equations of $X$, and hence $Q\in X$, a contradiction. Continue Reading
We say a subset of $\mathbb{A}^n$ is closed set if the elements of $X$ are the common roots of some (system) of polynomial equations in algebraic closed field $k$. That is $X$ is a common solutions of $F_1=F_2=\cdots=F_k=0$. We say $F_i$ is a defining equations of $X$.
Suppose that we are given $\{X_{\alpha}\}_{\alpha \in \mathbb{I}}$, the collection of closed sets, where for each $X_{alpha}$ is defined by system of equations $F_{\alpha_i} = 0$, then
\[\bigcap_{\alpha \in J} X_\alpha\]
is also a closed set, since it is defined by the whole systems (putting all together) $F_{\alpha_i}=0$ for all $\alpha \in I$ and for all $i$.
Also, for any two closed set $X_1$ dan $X_2$, with system $F_i=0$ for $X_1$ and system $G_i=0$ for $X_2$, the set $X_1 \cup X_2$ is defined by the system $F_i G_j=0$ for all $i,j$.
And finally, the empty set and $\mathbb{A}^n$ are closed set, each defined by equation $1=0$ dan $0=0$, respectively.
Therefore Closed Set $X$ form a topology, we called this topology Zariski Topology.
Example:
Consider the Zariski Topology in the affine line $A^1$, any closed set $X$ in this topology is defined by a polynomial of one variable $F(x)=0$. Indeed, for any closed set $X$, we can associate and ideal $\mathfrak{A}_X=\{f \in k[x]\, | f(x)=0 \, \forall x \in X\}$ of $k[x]$, thus in particular $F \in \mathfrak{A}_X$. Furthermore, since $k[x]$ is a PID, we should have $\mathfrak{A}_X =(F)$, for some polynomial $F$, and thus any closed set $X$ is defined by equation $F(x)=0$, and since $k$ is algebraic closed we have $F(x)=(x-a_1)(x-a_2)\cdots(x-a_k)$ so that $X=\{a_1,a_2,\cdots,a_k\}$, that is a closed set of this topology are those set with finitely many elements. Conversely for any finite set $X=\{a_1,a_2,\cdots,a_k\}$ we can form a polynomial $F(x)=(x-a_1)\cdots (x-a_k)$ as the equation defining $X$, and so $X$ is a closed set. We conclude that all closed set of the Zariski Topology in $A^1$ are those finite set of $k$.
The ideal $\mathfrak{A}_X$ introduced in the above example, is important on characterizing the closed set. Another way to look at it, is as follow:
Let $k[T_1,T_2,\cdots,T_n]$ be the ring of polynomial of several variables, for convenience we will denote it further as $k[T]$. For any closed set $X$, consider all function $f: X \rightarrow k$ such that there exists $F\in k[T]$ satisfy $F(x)=f(x)$ for all $x \in X$. Such a function $f$ is called regular function.
We can form the ring $k[X]$, consisting all regular function on $X$. Furthermore any $f \in k[T]$ we include $f|_X \in k[X]$ (the restriction function of $f$ in X). We can check that $k[X]$ form a ring with usual addition and multiplication of functions.
Consider the homomorphism $\psi : k[T] \rightarrow k[X]$, which send any $f \in k[T]$ to $f|X \in k[X]$, $\psi$ is easily checked to be a homomorphism and since $k[X]$ is regular $\psi$ is also onto. The kernel of $\psi$ is
\[\{F \in k[T] |\,F(P)=0 ,\forall \, P \in X \}\]
which is exactly $\mathfrak{A}_X$. Moreover by isomorphism theorem we have $k[T]/\mathfrak{A}_X \cong k[X]$.
This proves that $k[X]$ is determined by the ideal $\mathfrak{A}_X$.
Remark:By Hilbert Basis Theorem, every ideal in $R[T]$ where $R$ is a nonetherian ring is finitely generated. Here $R=k$ is a field (which has $(0)$ or $k$ only has ideals) therefore clearly nonetherian ring, thus $k[T]$ is finitely generated. Being the homomorphic image of $k[T]$, $k[X]$ must also finitely generated
More interesting from the remark is that $\mathfrak{A}_X$ being an ideal is also finitely generated, therefore any closed set $X$, can be described as the common roots of finitely many polynomial equations!
This could be noted more, that whenever $F_i$ are the defining equations of $X$, it is not always the case that $\mathfrak{A}_X=(F_1,\cdots,F_k)$. Clearly by definition $F_i \in \mathfrak{A}_X$. Since $\mathfrak{A}_X$ is finitely generated say by $G_i$ (which can be obtained as Grobner Bases), we can take $G_1=G_2=\cdots=G_l=0$ as the defining equation of $X$.
Furthermore, suppose that $f\in k[X]$ is vanish at all points for which $g_1,\cdots,g_k \in k[X]$ are also vanish, we can associate $F\in k[T]$ as an inverse image of $f$ under homomorphism $\psi$, and similarly $G_i\in k[T]$ for each $g_i$ mentioned above. By Hilbert Nullstellensatz, since $F$ vanish at the same the points as $G_i$, and also in $F_i$ (the defining equations of $X$) therefore we have $F^r \in (G_1,G_2,\cdots,G_k, F_1,F_2,\cdots,F_l)$, and we have $f^r \in (g_1,\cdots,g_k)$ (since all $F_i$ is associate to zero polynomials in $k[X]$ by $\psi$.)
Theorem
Let $X$ and $Y$ be closed sets. Then $Y \subset X$ if and only if $\mathfrak{A}_X \subset \mathfrak{A}_Y$
proof:
Suppose that $Y \subset X$, take $f\in \mathfrak{A}_X$, then $F(P)=0$ for all $P \in X$, since $Y \subset X$, therefore for every $Q\in Y$ we should have $F(Q)=0$, thus $f\in \mathfrak{A}_Y$.
Now suppose that $\mathfrak{A}_X \subset \mathfrak{A}_Y$ and $Y \not \subset X$, that is there exists $Q \in Y$ such that $Q \not \in X$. Suppose that $\mathfrak{A}_X=(G_1,G_2,\cdots,G_k)$ , then $G_i \in \mathfrak{A}_X \subset \mathfrak{A}_Y$, therefore we have $G_i(Q)=0$, and hence $F(Q)=0$ for all $F \in \mathfrak{A}_X$, thus in particular $Q$ is a common root of the defining equations of $X$, and hence $Q\in X$, a contradiction. Continue Reading