[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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