[ALGAE IV] Rational Curve and Rational Function.
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.
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.
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