The Zariski Topology and The ring of Regular Function

The Zariski Topology and The ring of Regular Function

19:44:00   Algebraic Geometry,

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

[ALGAE IV] Rational Curve and Rational Function.

[ALGAE IV] Rational Curve and Rational Function.

22:41:00   Algebraic Geometry,

It has been months, and i actually has got quite far reading algebraic geometry, but i don't spare a time to share it here. Let's continue the previous posts (and to remind me more about what i have learned).

As we already learn that any irreducible rational curve $f(x,y)=0$ can be parameterized by $x=\varphi(t)$ and $y=\psi(t)$. Lets take a look this parameterization deeper,  Example in the [AlGAE II] is the place to look, the parameter $t$ help us to correspond a point in the curve to the point in the line $x=-1$, by a mapping (recall that $y=tx$)

\[(x,y) \mapsto (-1,-t) = \left(-1, -\frac{y}{x}\right)\]

For $x=0$ which also implies $y=0$, this map is undefined, and this is where the denominator of $\varphi(t)$ and $\psi(t)$ vanish. Ignoring these points, the map seems fine, and defines a bijection from the point in the curve to the line. In fact, there can be only finitely many points in the curve such that the parameter fails to represent (like our $(0,0)$ just know), those are actually the common roots of the denominator of $\varphi(t)$ and $\psi(t)$.

Let include the result obtained on [ALGAE III], where we can find the isomorphic map from $\mathbb{F}(X)$ to $\mathbb{F}(t)$, given that $X$ is a rational curve.
Let's see wether this result is vice versa, that is whenever $\mathbb{F}(X)\cong \mathbb{F}(t)$ then $X$ is rational.
Since $x, y \in \mathbb{F}(X)$,  they have a correspondence image in $\mathbb{F}(t)$, let say $x$ to $\varphi(t)$ and $y$ to $\psi(t)$. By isomorphism, the element $f(x,y)=0 \in \mathbb{F}(X)$ goes to the element $f(\varphi(t),\psi(t))=0$ (since $x$ and $y$ is related by an algebraic relation, note that this is not a mere substitution $x =\varphi(t)$ and $y =\psi(t)$). Therefore, such parametrization so that $f(\varphi(t),\psi(t))=0$ exists, and $X$ is rational.

Moreover, as for any $t_0 \in \mathbb{F}$ we should have $f(\varphi(t_0),\psi(t_0))=0$, but this can be failed for which $t_0$ is a common root of the denominator of $\varphi(t)$ and $\psi(t)$. Excluding these points  we can easily see that $x=\varphi(t)$ and $y=\psi(t)$ define a parameterization the point on a curve $X$.

Our goal is to represent any point $(x,y) \in X$ as $(x,y)=(\varphi(t),\psi(t))$ for some $t$, and conversely producing $(x,y) \in X$, for any given $t$ by using parametrization $x=\varphi(t)$ and $\psi(t)$. Unfortunately, this is so much a dream, as we can see our parametrization in [ALGAE II] failed to include/produce some points in $X$. We will resolve this as close as possible to our goal.

Let $\tau(x,y)$ is a function in $\mathbb{F}(x)$ which is mapped to the function $t \in \mathbb{F}(t)$ via the isomorphism $\mathbb{F}(X) \rightarrow \mathbb{F}(t)$.  Consider the map $s(t) \mapsto s(\tau(x,y))$, this map takes $t$ to $\tau(x,y)$,  also it is a isomorphism, thus the map defines an isomorphism $\mathbb{F}(t) \rightarrow \mathbb{F}(X)$.
Recall that $x\mapsto\varphi(t)$ and $y\mapsto\psi(t)$,  by using the field isomorphism properties we have $\tau(x,y) \mapsto \tau(\varphi(t),\psi(t))$, also by the same map $\tau(x,y) \mapsto t$, and we conclude that

\[t=\tau(\varphi(t),\psi(t))\]

Similarly, $t \mapsto \tau(x,y)$ implies $\varphi(t) \mapsto \varphi(\tau(x,y))$, and also by the same map $\varphi(t) \mapsto x$, thus we conclude that (with similar argument for $y \mapsto \psi(t)$ )

\[\varphi(\tau(x,y))=x \qquad \text{and} \qquad \psi(\tau(x,y))=y\]

for any points in the curve. But the thing is not so smooth here, suppose that $\tau(x,y)=\frac{p(x,y)}{q(x,y)}$, then for in any points $(x_0,y_0)$ such that $q(x_0,x_0)=0$, we can't find such $t_0$ we want.  Luckily, there is only finitely many points  $(x_0,y_0) \in X$, such that $q(x_0,y_0)=0$, since $q(x,y)$ is not divisible by $f(x,y)$. The value of $t$ for which the parametrization fails are all $t_0$ that is the common root of the denominator $\varphi(t)$ and $\psi(t)$ , thus when $t_0=\tau(x_0,y_0)$ is the common root  of the denominator of  $\varphi(t)$ and $\psi(t)$, we have $\frac{p(x_0,y_0)}{q(x_0,y_0)}=t_0$, we assert that there can be only finite points $(x_0,y_0)\in X$ such that this can happen. Suppose that $p(x,y)-t_0 q(x_y)=0$ for infinitely many points on $X$, thus $f(x,y) | p(x,y)-t_0 q(x,y)$, which implies $\tau(x,y)=t_0$ up to the equality on $\mathbb{F}(X)$, which is absurd.

Therefore we conclude that

Except for some finite points in $X$, any point $(x_0,y_0) \in X$ can be represented as parameterization $x_0 = \varphi(t_0)$ and $y_0=\psi(t_0)$ and also for any $t_0 \in\mathbb{F}$, but some finitely many points, we can produce the point  $(\varphi(t_0),\psi(t_0)) \in X$.


Actually we've got something more, from $t=\tau(\varphi(t),\psi(t))$ , we see that the point $t_0$ is uniquely determined by the point $(x_0,y_0) \in X$, therefore our representation is unique. Continue Reading

[ALGE III] The Field Of Rational Function of The Irreducible Variety

[ALGE III] The Field Of Rational Function of The Irreducible Variety

00:33:00   Algebraic Geometry,

Let $X$ be an irreducible curve given by $f(x,y)=0$ in the algebraic closed field $\mathbb{F}$, then we define the set
\[\mathbb{F}(X)=\left\{\left.\frac{p(x,y)}{q(x,y)} \right\lvert p,q \in \mathbb{F}[x,y] \mbox{ and } f(x,y)\not | q(x,y) \right\}\]
 Then $\mathbb{F}(X)$ contains the rational functions of two variables $x$ and $y$. We say two elements of $\mathbb{F}(X)$, $\frac{p_1(x,y)}{q_1(x,y)}$ and $\frac{p_2(x,y)}{q_2(x,y)}$ are equal in $X$ if and only if

\[f(x,y) | (p_1(x,y)q_2(x,y)-p_2(x,y)q_1(x,y))\]

It is easy to see that, along with this equality, $\mathbb{F}(X)$ is a field. We call $\mathbb{F}(X)$ is a Field of Rational Function on $X$ (or funcion field of algebraic variety $X$).
Before we proceed, we should aware that, the denominator of the rational function $\frac{p(x,y)}{q(x,y)} \in \mathbb{F}(X)$ can be zero only for finitely many points of $X$, that is $q(x,y)=0$ in a finite subset of $X$. This can be proved easily by using the Lemma in my first post [ALGE I]. It also can happen that, a rational function can be represented in two different expression $u=\frac{p(x,y)}{q(x,y)}$ and $\frac{p_1(x,y)}{q_1(x,y)}$ where $q(x,y)=0$ but $q_1(x,y) \neq 0$.  If $u$ has the exspresion $\frac{p(x,y)}{q(x,y)}$ where $q(P) \neq 0$, then we say $u$ is regular at $P$.

Now suppose that $X$ is rational curve, then by using the parametrization of $X$, we can substitue $x=\varphi(t)$ and $y=\psi(t)$ to the rational function $u=\frac{p}{q} \in \mathbb{F}(X)$. The substitution takes any element of $\mathbb{F}(X)$ to the rational function of one variable in $t$.  Notice that, the denominator $q(\varphi(t),\psi(t))$ can only equals zero in a finitely many times, by the same reason from the previous paragraph. Suppose that $u_1=\frac{p_1}{q_1}$ and $u_2=\frac{p_2}{q_2}$ are equals on $X$, then

\[p_1(x,y)q_2(x,y)- p_2(x,y) q_1(x,y) = \lambda(x,y) f(x,y)\]

substitute $x=\varphi(t)$ and $y=\psi(t)$, since $f(\varphi(t),\psi(t)) \equiv 0$ we have

\[\frac{p_1(\varphi(t),\psi(t))}{q_1(\varphi(t),\psi(t))}=\frac{p_2(\varphi(t),\psi(t))}{q_2(\varphi(t),\psi(t))}\]

thus, the substitution exhibit the map from $\mathbb{F}(X)$ to the $\mathbb{F}[t]$ (the field of rational function in $t$). This map is also an isomophism from $\mathbb{F}(X)$ to some subfield of $\mathbb{F}[t]$.
Now we can use Lüroth's Theorem to extend this property, that is it is indeed true that $\mathbb{F}(X)$ is isomorphic to $\mathbb{F}[t]$.

Continue Reading

[ALGE II] Rational Parametrization of Algebraic Curve

[ALGE II] Rational Parametrization of Algebraic Curve

07:52:00   Algebraic Geometry,

From the previous post, we know that any algebraic curve in the algebraic closed field can be decomposed into irreducible curves, which is a unique representation. So, it would be natural to look deeper on the irreducible curve. Recall that an algebraic curve $X$ defined by the equation $f(x,y)=0$, with irreducible polynomial $f(x,y) \in \mathbb{F}[x,y]$, where $\mathbb{F}$ is an algebraic closed field

We say an irreducible algebraic curve $X$ is  a rational curve, whenever there exist rational functions $\varphi$ and $\psi$ at least one is non-constant, such that \[f(\varphi(t), \psi(t)) \equiv 0.\]
$\varphi$ and $\psi$ are rational parametrization of $f$.

It is not hard to prove that every degree two irreducible curves $X$ (a conic) are rational curves, take a point $(x_0,y_0) \in X$ then substitute $y=t(x-x_0)+y_0$ to the equation $f(x,y)=0$, solving for $x$ we will find $x=\varphi(t)$ and substituting back to  $y=t(x-x_0)+y_0$ we will find $y=\psi(t)$.

Example: The Folium of Descartes is an algebraic curve $X$ defined by \[x^3+y^3-3axy=0\]
for the real number $a$.
Since $(0,0) \in X$, to get the rational parametrization we can try to substitute $y=tx$ yields
\[(t^3+1)x^3 - 3atx^2=0\]
 for the points other than $(0,0)$  we have from the above equation the parametrization
\[x=\varphi(t)=\frac{3at}{t^3+1}\qquad\text{and}\qquad y=tx=\psi(t)=\frac{3at^2}{t^3+1}\]
 
The geometric interpretation is also obvious. As we can see in the picture below, we can draw a line not contain the point $(0,0)$. and for any point $P \in X$ we project it to the line $x=-1$ by making the line joining $(0,0)$ and $P$. Hence, each point of the curve (other than $(0,0)$) produce a unique point on the line $x=-1$, and each point on the line $x=-1$ produce a unique point (correspondence with the origin) on the curve.

here the points on the line $x=-1$ act like $t$ in the parametrization, the parameter $t$ is the slope of the produced lines. So that the point $(-1,1)$ (which means $t=-1$) does not produce any point of the curve except $(0,0)$. This also can be seen in the parametrization $\varphi(t)$  and $\psi(t)$, where when $t=-1$, the denominator of both is zero.

We have seen that, the parametrization is important on determining the point on the curve. When the curve $X$ satisfy some criterion, for a suitable choice of $\varphi(t)$ and $\psi(t)$, we can get a one-to-one correspondence map $t \rightarrow (\varphi(t),\psi(t))$, by excluding some finite points of the curve. This is really interesting, one possible application is to the theory of indeterminte equation.  Suppose that $f(x,y)=0$ has rational coefficient, and suppose that the parametrization $\varphi(t)$ and $\psi(t)$ also have rational coefficient. If both $\varphi(t)$ and $\psi(t)$ are a bijection, we can list all the rational points of the curve, given we know only one point (for example, in the previous example we were given $(0,0)$ on the folium).

In the number theory, The Fermat Last Theorem, can be viewed as proving that a rational point of the of the curve $x^n+y^n-1=0$ does not exist.

Thus the problem whether a curve have a rational parametrization is essentially important.
Our next realm is to determine when a given curve is a rational curve?
 

Continue Reading

[ ALGE I ] Introduction to the Plane Algebraic Curve

[ ALGE I ] Introduction to the Plane Algebraic Curve

22:24:00   Algebraic Geometry,

Well i'm going to retell what i've  studied from the algebraic geometry books so far.. one of my favorite book is the Basic Algebraic Geometry: Varieties in Projective Space  By Shafarevich.

Let $\mathbb{F}$ be an algebraic closed field. An algebraic curve $X$ is the set of points $(x.y)$ defined by the equation \[f(x,y)=0\]
where $f(x,y) \in \mathbb{F}[x,y]$.
When  $f(x,y)$ is an irreducible polynomial, we say $X$ is an algebraic irreducible curve.
The degree of $X$ is equal to $\text{deg}f$. The condition that $\mathbb{F}$ must be an algebraic closed field is necessary for the uniqueness of the degree of $X$.
For example the curve $(0,0)$ in the field of real numbers is defined by infinitely many equations that is $x^{2n}+y^{2n}=0$ for any positive integers $n$. For even $n$ the curve is irreducible, while whenever $n$ is odd, the curve is reducible.This confusion doesn't happen when $\mathbb{F}$ is an algebraic closed field. As we can prove it by the following very useful lemma:

 Lemma:  Let $\mathbb{F}$ be a field, given an irreducible polynomial $f \in \mathbb{F}[x,y]$. Then for any polynomial $g \in \mathbb{F}[x,y]$ which is not divisible by $f$,  the system of equations $f(x,y)=g(x,y)=0$ has only finitely many solutions.


Proof:  We view $f$ and $g$ as the element of $u[y]$, where $u$ is the field of rational function of $x$ with coefficient in $\mathbb{F}$. By Gauss Lemma, it is easy to see that $f$ remains irreducible in $u[x]$  and also $f$ and $g$ are still relatively prime in $u[y]$. Thus there exists $\lambda, \mu \in u[y]$ such that
 \[f \lambda + g \mu =1\]
Multiplying by the common denominator $a(y) \in \mathbb{F}[y]$ we will get
\[f(x,y) \lambda(x,y) + g(x,y) \mu(x,y) = a(y)\]
If $(\alpha, \beta)$ is the solution of the system of equations. Then $a(\beta)=0$, and there are only finitely many possibility of such $\beta$. Thus $\alpha$ is the root of equation $f(x, \beta)=0$, and there are also a finitely many value of them. The result follows.

Applying the lemma, let $X$ be an irreducible algebraic curve in an algebraic closed field. If there are two such equations which define  $X$ say $f(x,y)=0$ and $g(x,y)=0$, since $f(x,y)=0$ has infinitely many solution in the algebraic closed field then the systems of equations $f(x,y)=g(x,y)=0$ also has infinitely many solutions. Hence the lemma says that $f(x,y)=c g(x,y)$. Thus the two equations are essentially the same.

The same is true for arbitrary curve $X$. Suppose that its equation is $h(x,y)=0$. Since $\mathbb{F}[x,y]$ is a Unique Factorization Domain,  therefore
\[h=f_1^{k_1} f_2^{k_2} \cdots f_n^{k_n}\]
for irreducible polynomial $f_i \in \mathbb{F}[x,y]$. The points such that $h(x,y)=0$ are also such points so that $f_i(x,y)=0$ for some $i$. Let $X_i$ be a curve defined by $f_i (x,y)=0$, then we have \[X= \bigcup_{i=1}^n X_i \]
where $X_i$ is an irreducible curves.
This decomposition is called the irreducible components decomposition. Since the equation of each $X_i$ is irreducible and defined by the equation which only may differ by constant multiple, so any other equation of $X$ is also differ from $h$ by the constant multiple. Continue Reading