Affine Maps « Elliptic Curves

Affine Maps

By ellipticcurves

What are acceptable maps between irreducible curves in ${\mathbb A}^2$ ? We start by defining rational maps, which are precisely what they sound like.

Definition. A rational map $\phi: C \to D$ between two irreducible curves is a map given on points of $C$ by $\phi(x, y) = (r(x,y), s(x,y))$ for some rational functions $r, s$ , such that

the denominators of $r, s$ do not identically vanish on $C,$
$\phi(P) \in D$ for all points $P \in C$ where $\phi$ is defined.

For now, we say that $\phi$ is defined over a field $K$ if we can choose $r, s \in K(x,y).$

The first requirement says that $\phi$ should be defined in at least one point of $C$ . We can say more.

Theorem. A rational map is defined on all but finitely many points of the curve $C$ .

Proof. Consider the (not necessarily) irreducible curves $X_1$ and $X_2$ defined by the denominators of $r$ and $s.$ The points of intersection of these curves with $C$ are precisely the points where $\phi$ is undefined. The first condition of the definition means that the defining polynomial $f(x, y)$ of $C$ does not divide the denominators of $r$ and $s,$ which implies that $C$ is not a component of $X_1$ and $X_2.$ By Bezout’s Theorem, each intersection $C \cap X_i$ is finite. So is their union, which is precisely the set at which $\phi$ is undefined. $\square$

Remark. The second condition can be restated as follows. If $D$ is given by $g(x,y)=0$ , then

$g(r(P), s(P))=0,$

for all points $P\in C$ where $\phi$ is defined.

Definition. Similarly, a rational map ${\mathbb A}^1 \to C$ is given by a pair of rational functions in one variable whose image lies in $C$ . A rational map $C \to {\mathbb A}^1$ is called a rational function on $C$ , because that is what it is. The only restriction is that the denominator of this function not identically vanish on $C$ .

We say that two rational maps $\phi_1, \phi_2: C \to D$ of irreducible curves are equivalent (write $\phi_1 = \phi_2$ ) if they agree on all but finitely many points of $C.$ From now on, we consider rational maps modulo this equivalence relation. In this way, domains of rational maps can sometimes be extended by switching to another pair of rational functions that define an equivalent function and are defined in the chosen point $P \in C.$ If this is possible, we say that $\phi$ is regular at $P.$

Example. For every curve $C,$ the identity map ${\mathrm id}_C: C \to C$ is a rational map.

Definition. Let $\phi: C \to D$ be a rational map. If there exists a rational map $\psi: D \to C$ such that $\phi\circ \psi = {\mathrm id}_D$ and $\psi\circ \phi = {\mathrm id}_C,$ we call $\phi$ a birational map, and say that curves $C$ and $D$ are birationally isomorphic, or just birational.

Definition. A rational map on a curve $C$ is said to be regular if it is regular at every point of $C.$ If a map $\phi: C\to D$ of curves is regular, with a regular inverse, we call it an isomorphism and say that curves $C$ and $D$ are isomorphic.

Example. Go back to the parameterization of points on the unit circle. Let $C: x^2 + y^2 = 1$ be the unit circle defined over ${\mathbb Q}$ , and consider maps

$\phi: {\mathbb A}^1 \to C$

given by

$\alpha \mapsto ((1-\alpha^2)/(1+\alpha^2), 2\alpha/(1-\alpha^2)),$

and

$\psi: C \to {\mathbb A}^1$

given by

$(x, y) \mapsto y / (x+1).$

Both maps are rational. The map $\phi$ is defined on the whole of ${\mathbb A}^1$ and therefore is regular. The map $\psi$ is undefined at $(-1, 0)$ and so it is not rational. These maps are inverses of each other. Therefore, in the affine plane, the circle is birationally isomorphic to an affine line, but is not isomorphic.

One of the important points about the previous example is that a map being a birational isomorphism and regular does not guarantee that the inverse is regular. In other words, a regular birational isomorphism is not necessarily an isomorphism.

Example. Generalizing the previous example, we showed using projections that any irreducible conic is birationally isomorphic to ${\mathbb A}^1.$

Example. Any line in ${\mathbb A}^2$ is isomorphic to ${\mathbb A}^1.$ Indeed, if $L:ax+by=c$ is the line in question with $b\neq 0$ , then rational maps $(x, y) \mapsto x$ and $t \mapsto (t, (c-at)/b)$ are regular and inverse to each other.

Tags: algebraic geometry, maps

This entry was posted on August 26, 2008 at 7:00 pm and is filed under algebraic geometry. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Elliptic Curves

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