[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 $\mathbb{F}[t]$.

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[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?
 

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[ 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