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.
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.
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