[ALGE II] Rational Parametrization of Algebraic Curve
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?
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.
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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