I am currently reading Introduction to Theory of Numbers by Niven, Zuckerman, Montgomery. It is a hard book to follow because the style is too terse and scattered. The author has introduced algebraic curves and intersection multiplicities. Some definitions as used by the author (I suspect they are pretty standard but still to clarify)
- Set of points $(x,y)$ that satisfy $f(x,y)=0$ constitute an algebraic curve, denoted by $\mathscr{C}_f$.
- Line $y=mx+r$ is denoted by $L$.
- The x-coordinates of intersection of $L$ with the curve are the roots of the polynomial $p(x)=f(x,mx+r)$
The author then assumes a non-singular conic $f(x,y)$ such that $f(x,y)$ contains rational points and the curve $\mathscr{C}_f(\mathbb R)$ contains a rational point $(x_0,y_0)$.
Let $m_0$ represent the slope of the tangent line to $\mathscr{C}_f$ at $(x_0,y_0)$. Hence, $m_0$ is rational.
For $m\ne m_0\in\mathbb{Q}$, line $L$ passing through $(x_0,y_0)$ having slope m has intersection multiplicity 1.
Now, here is the confusing part. The coefficient of $x^2$ in $p(x)$ defined above is $f(1,m)$.
$\color{red}{\text{This does not seem true}}$. What is the author trying to say?
Can anybody simplify this proof for me? What exactly is the author trying to prove in the end? Hence, there must exist rational values $m_1, m_2$ for which $f(1,m)=0$. Otherwise the conic would not be nonsingular.
Therefore, for such $m, p(x)$ is linear and $x=x_0$ is its only root. (Why? Because the line is a tangent?)
Then, the author proceeds to state that for all rational $m\ne m_0,m_1,m_2$, the polynomial $p(x)$ has rational coordinates, is of degree 2, and has a simple root at $x_0$.
Hence, the curve must have a second rational root $x_1$. (This is pretty obvious so no worries.)