Elliptic curves and Bezout's theorem

In the previous section we parameterized the rational points on the circle $x^2+y^2=1$ by choosing one rational point $P$, and then looking at the intersection of the circle with lines through $P$ having rational slope.

Rational points on elliptic curves cannot be parameterized in the same way. What goes wrong if we try to repeat the procedure that worked for the circle? To fix ideas, let us see what happens for the elliptic curve $E$ with equation

$$y^2 = x(x+5)(x-5).$$

(The polynomial $x(x+5)(x-5)$ has distinct roots, so this is an elliptic curve.) The point $S=(-4,6)$ is on the curve $E$. What happens if we intersect $E$ with the line $L$ of slope 1 through $S$?

The equation of $L$ is $y-6=1(x+4)$, i.e., $y=x+10$. Substituting this into the equation of $E$ yields

$$(x+10)^2 = x(x+5)(x-5)$$

$$= x^3 -x^2 -45x - 100$$ 0

$$= (x+4)(x^2-5x-25).$$ 0

The linear factor $x+4$ was expected; it reflects the fact that the point $S=(-4,6)$ is one of the intersection points. But this time the leftover factor is quadratic, not linear, so there is no reason to expect the other solutions to be rational. In fact, here they are not, because the discriminant of $x^2-5x-25$ is $(-5)^2-4(1)(-25)=125$, which is not the square of a rational number. Hence we do not obtain rational points on $E$ in this way.

Geometrically what has happened is that $L$ intersects $E$ in three points, one of which is $S$, and the best that can be said of the other two is that their coordinates will involve a single square root. It is not an accident that $L \cap E$ consisted of three points here, whereas the intersection of $L$ with a circle in the previous section had two points. These are special cases of the following general result:

Bezout's Theorem (almost): It is almost true that the intersection of a curve $f(x,y)=0$ of degree $m$ with a curve $g(x,y)=0$ of degree $n$ consists of exactly $mn$ points.

To make the theorem true, some care must be taken. For instance, the intersection of $xy=0$ and $(x^2+y^2)y=0$ has infinitely many points, not $2 \cdot 3 = 6$ as predicted, because both curves contain the curve $y=0$. Therefore one should insist that the two curves do not have a curve in common. Algebraically, this is equivalent to imposing the condition that $f(x,y)$ and $g(x,y)$ have no common factor.

With this assumption, it is now true that the curves intersect in at most $mn$ points. But to get exactly $mn$ points, three more modifications to the problem are required. As stated, the theorem gives the wrong answer for the number of real points in the intersection of $x-2=0$ with $x^2+y^2=1$. To get the correct number of intersection points ( $1 \cdot 2 = 2$), one should allow the intersection points $(2,\sqrt{-3})$ and $(2,-\sqrt{-3})$ with complex coordinates. Another problem is illustrated by the example in which one intersects $x-1=0$ with $x^2+y^2=1$. We again expect 2 intersection points, but there is only one, the point $(1,0)$ where the line is tangent to the circle. The fix this time is to count intersection points with multiplicity: points where two curves meet tangentially count extra! The third problem is that certain curves such as $y-1=0$ and $y-2=0$ do not meet as many times as they are supposed to. One finds the ``missing intersection points'' by adjoining a ``line of points at infinity'' to the plane, to form the projective plane ${\mathbb{P}}^2$. The lines $y-1=0$ and $y-2=0$ meet at one of the points on this line at infinity. (We will not discuss this in detail here.)

The precise version of Bezout's Theorem reads as follows:

Bezout's Theorem: Let $X$ and $Y$ be curves of degrees $m$ and $n$ in the projective plane over the complex numbers. If $X$ and $Y$ have no curves in common, then the number of intersection points in $X \cap Y$ with complex coordinates, counted with multiplicities, equals $mn$ exactly.