Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is a very elementary question but I can't find the error:

Denote $D$ the diagonal $\{[x_0:x_1],[x_0:x_1]\} \subset \mathbb P^1$x $\mathbb P^1$. Let $\phi: \mathbb P^1 \longrightarrow \mathbb P^2$ defined by $\phi([t_0:t_1])=(t_0^2:2t_0 t_1:t_1^2)$ and let P be its image. Similar to the Veronese map, this is an isomorphism and we have $P = \{[u_0: u_1: u_2] | u_0 u_2 = \frac {u_1^2} 4\}.$

Let $i$ be the natural isomorphism $\mathbb P^1 \longrightarrow D \subset\mathbb P^2$.

Let $\psi:\mathbb P^1$ x $\mathbb P^1 \longrightarrow \mathbb P^2$ be defined by $\psi([x_0:x_1],[y_0:y_1])= [x_0 y_0: x_0 y_1 + x_1 y_0 : x_1 y_1]$. One sees that $\psi(D) = P$.

We have $\psi \big |_{D}=\psi \big |_{im(i)}$ is an isomorphism to $P$, because $\phi$ is and $\psi \circ i = \phi$ (and it's easy to write down the inverse morphism explicitely).

But if I calculate it manually I get $\psi^{-1}(P) = V((x_0 y_1 - x_1 y_0)^2) \subset \mathbb P^1$x $\mathbb P^1$, which is 2 times the Diagonal $D$ (which must be true because we have a correspondence between biquadratic curves in $\mathbb P^1$x $\mathbb P^1$ and quadrics in $\mathbb P^2$). But this contradicts $\psi \big |_{D}$ being an isomorphism (and the latter in turn is used by an author to prove s.th., so I somehow hope it is not too far away from the truth / there is a way to fix his argument, but that's a different story).

share|improve this question
    
A nitpick: I would rather take $[t_0:t_1]$ as coordinates for the domain $\mathbb P^1$ of $\phi$ since $[x_0:x_1]$ already denotes the coordinates of the left-hand factor $\mathbb P^1$ of the product $ \mathbb P^1 \times \mathbb P^1 $. –  Georges Elencwajg May 3 '13 at 23:33
    
At first I took [s:t], but I wanted to make sure that $\psi(D)= P$ is really obvious. But I see your point. I think $[t_0:t_1]$ is a good compromise. –  Laugerizor May 4 '13 at 0:01

1 Answer 1

up vote 3 down vote accepted

Everything you write is correct and there is no contradiction!
An analogy may help:

Consider the morphism $ f: \mathbb C\to \mathbb C:z\mapsto z^2$.
It induces an isomorphism $\lbrace 0 \rbrace \to \lbrace 0 \rbrace$ but nevertheless the scheme-theoretic inverse image $f^{-1}(\lbrace 0 \rbrace )$ is twice ${\lbrace 0 \rbrace}$, since it has equation $z^2=0$.
Analogously, in your case the inverse image under $\psi$ of the cycle (actually a divisor) $P$ of $\mathbb P^2$ is the cycle (or divisor) $2D$ of $\mathbb P^1$x $\mathbb P^1$.
And this is perfectly compatible with $\psi|D:D\to P$ being an isomorphism : the point is to carefully distinguish the cycle $\psi^{-1}(P)$ of $\mathbb P^2$ from the image $(\psi|D)^{-1}(P)=D$ of the isomorphism of schemes $(\psi|D)^{-1}:P\to D$.

These considerations on cycles are basic in intersection theory: you might consult the first chapter of Fulton's Intersection Thory, the standard reference on the subject.

share|improve this answer
    
Thanks, I am somewhat familiar with basic intersection theory and in particular with a part of Fulton's book, but I seemed to be blind / confused, somehow. Your words were very clarifying. (There is a minor typo in your answer: there is no $\psi^{-1}(D)$). –  Laugerizor May 4 '13 at 0:18
    
Dear Laugerizor, I'm happy things are now clarified. Thanks for catching the typo: I have corrected it. –  Georges Elencwajg May 4 '13 at 6:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.