Number of double points of a projectivity

Question

Let $K$ be a field and let $V$ be a vector space on $K$, with $\dim V = 3$. Let $r$ be a projective line in $P(V)$ and let $\{A , B ; C\}$ be a reference system in $r$, and we'll consider also $\{A' , B' ; C'\}$, being $A' = \sigma(A)$, $B' = \sigma(B)$ and $C' = \sigma(C)$, being $\sigma : r \to r$ a projectivity such that $\sigma \neq Id_r$. I have to solve to problems, but I have no idea: at first, why maximum number of double points of $\sigma$ is two? can I use fundamental theorem of the projective geometry to show that? and, supposing that $A = A'$, $B' \neq B \neq C'$ and $B' \neq C \neq C'$, can we determinate the number of double points of $\sigma$? Thank you very much.

Answer 1

A projectivity between two projective lines is determined by its action on three distinct points. Consequently, a projectivity with three distinct double points must to be the identity, thus a non-identical projectivity has at most two double points.

In the second question, you have a projectivity with at least a double point, $A$. To determine the number the double points, draw the axis of $\sigma$: take two points $S,S'$ out from $r$ such that $A\subset SS'$, and consider the line $a=(SB\cap S'B')(SC\cap S'C')$. Then $a$ contains $A$ if and only if $\sigma$ has one double point. Otherwise $\sigma$ has two double points, namely $A$ and $a\cap r$.