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This is problem $2.11$ page $133$ of Kenji Ueno's book.

Consider an irreducible hypersurface $V(F)$ where we assume that the homogeneous polynomial of degree $d$ satisfies the conditions $F(0,x_{1},..,x_{n}) \neq 0$ and $F(1,0,...0) \neq 0$.

Let $P=(1:0:...:0)$. For a point $Q \in V(F)$ let $R(Q)$ be the intersection of the line $PQ$ with the hyperplane $H_{\infty}: x_{0}=0$. Let $\phi_{P}: Q \in V(F) \rightarrow R(Q) \in H_{\infty}$.

Prove that for a point $R \in H_{\infty}$ there is a point $Q \in V(F)$ such that $R(Q)=R$.

I'm not sure how to write this. I tried looking at an example, say we are in $\mathbb{P}^{2}$ and $F(x,y,z)=xz+y^{2}$ then $\phi_{P}$ is the map that sends $(x:y:z)$ to $(0:y:z)$.

So suppose we have a point $(0:b:c) \in H_{\infty}$, we want to find a point $(u:v:w) \in V(xz+y^{2})$ such that $\phi_{P}(u:v:w)=(0:b:c)$. Now take $b=1,c=3$.

This implies there is some nonzero $\lambda$ such that $v=\lambda b$ and $w= \lambda c$. Since $b=1$ and $c=3$ then $v=\lambda$ and $w=3\lambda$ so we need to solve $uw+v^{2}=0$. Plugging these values yields:

$u(3\lambda)+\lambda^{2}=0$

So take $u=(-1/3)\lambda$. Thus setting $\lambda=-3$ gives $u=1$, $v=-3$, $w=-9$.

Then $(u:v:w) \in V(F)$ and the map sends this point to $(0:-3:-9)=(0:1:3)$ as desired.

How can we write this in a general way?

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Dear user, Unless your conventions are non-obvious, your chosen example is not a very good one, since $F(1,0,0) = 0$, contradicting the assumptions on $F$. Regards, –  Matt E Jun 19 '12 at 4:31
    
@Matt E: doh, you're right, missed that, thanks. –  user10 Jun 20 '12 at 1:08

1 Answer 1

up vote 3 down vote accepted

As Matt points out, your example doesn't work because $P\in V(F)$.
So I suggest modifying it slightly by taking $F(x,y,z)=x^2+yz$, while $H$ is still the line $x=0$.
The morphism we are interested in is $\phi:V(F)\to H:(x:y:z)\mapsto (0:y:z)$.

Now, if you fix $R=(0:b:c)\in H$, the points $Q\in V(F)$ such that $\phi(Q)=R$ are those $(x:b:c)$ with $x^2+bc=0$.
If the base field $k$ is algebraically closed the solutions are given by the formula $x=\pm \sqrt {-bc}$.
Supposing $char.k\neq 2$, we get :

a) Two distinct points $Q_i=(\pm \sqrt {-bc}:b:c)\in V(F) $ with $\phi(Q_i)=R$ in case $b,c\neq0$
b') A unique point point $Q'=(0:0:1)\in V(F)$ with $\phi(Q')=R'=(0:0:1)$.
And in this case the line $\overline {PQ'}$ is tangent to V(F) at $Q'=R'$.
b'') A unique point point $Q''=(0:1:0)\in V(F)$ with $\phi(Q'')=R''=(0:1:0)$.
And in this case the line $\overline {PQ''}$ is tangent to V(F) at $Q''=R''$.

A beautiful picture emerges from this: there are two tangents passing through $P$ to the conic $V(F)$ and the line joining the two contact points of these tangents with the conic is the line $x=0$ onto which you are projecting from $P$.
Do yourself a favor and draw that pleasant picture!

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