I sat an exam today on elementary algebraic geometry, which is now finished (it was a morning exam), and have been allowed to take the paper home (to be clear, the exam is now over). I've been wrestling with one of the questions all day but can't make any sense of it:
Suppose that $P,Q,R \in \mathbb{K}[x,z]$ are homogeneous, with $\deg P = d \gt 0, \; \deg R = d -1, \; \deg Q = d+1$, and that $P^2 + QR \neq 0$.
Define a rational map $\psi: \mathbb{P}^2 \dashrightarrow \mathbb{P}^2$ by
$\psi((x:y:z)) = (x(yR(x,z) - P(x,z)):yP(x,z) + Q(x,z):z(yR(x,z) - P(x,z)))$.
By considering $\psi \circ \psi$, or otherwise, show that $\psi$ is dominant.
Our definition of dominant: For $X$ and $Y$ projective algebraic sets, $\psi:X \dashrightarrow Y$ is dominant if there does not exist a projective algebraic set $W \subsetneq Y$ such that $\psi(p) \in W$ for every $p$ where $\psi$ is defined.
We have a lemma which is an alternative criterion for dominance, which says that $\psi$ is dominant if there exists a projective algebraic set $Z \subsetneq Y$ such that every $q \in Z^c$ can be written as $\psi(p)$ for some $p$ in $X$.
I can't see how to use either of these. I've tried composing $\psi$ with itself, and the expression I get is an awful mess. Does anyone have any clever ideas?
Let $U = \{ yR-P \ne 0\} \cap \{P^2+QR \ne 0\} \subseteq \mathbb{P}^2 $. Note that $\psi$ is a morphism on $U$. Let $p=[x:y:z] \in U$. Then $\psi(p) = (x: \frac{yP+Q}{yR-P}: z)$.
Observe that $\psi(p) \in U$, since if $\frac{yP+Q}{yR-P} R - P = 0$, then $QR + P^2 = 0$. So it follows that $\psi(U) \subseteq U$.
Now observe that we have $\psi \circ \psi (p) = \psi(x: \frac{yP+Q}{yR-P}: z) = (x: \frac{ \frac{yP+Q}{yR-P}P + Q}{\frac{yP+Q}{yR-P}R-P}: z) = (x:y:z) $. So it follows that $\psi: U \rightarrow U$ is an isomorphism.
This implies that the image of $\psi$ (as a rational map, where its domain may be larger than $U$) contains the open subset $U$. But this implies that the image of $\psi$ is dense (since $U$ is dense as it is an open subset of $\mathbb{P}^2$), hence dominant.