Existence of a morphism on given projective varitey (homework) In need help in my homework assignment: 
let $X =\left \{ (x_{0}:x_{1}:x_{2}:x_{3}:x_{4})\in \mathbb{P}^{4} : rk\begin{pmatrix}
x_{0} & x_{1} & x_{2}\\ 
x_{2} & x_{3} & x_{4}
\end{pmatrix}  < 2\right \}
$


*

*show that there is a morphism  $\varphi :X\rightarrow \mathbb{P}^2 $ which on the open subset of $X$ where $(x_{0},x_{1},x_{2}) \neq(0,0,0)$ is given by the projection $ (x_{0}:x_{1}:x_{2}:x_{3}:x_{4}) \mapsto (x_{0},x_{1},x_{2})$

*for each point $P \in \mathbb{P}^{2} $ determine the inverse image $\varphi ^{-1}(P)$
I thought at first to display X like V(I), but I got no intuition of what that could be, can you help?
 A: Here are some facts which you should prove and which solve your exercise:  


*

*Yes  $X=V(I)$ for  $I=(x_0x_3-x_1x_2,x_ox_4-x_2^2,x_1x_4-x_2x_3)$ .  

*The complete definition of  the morphism $\phi:X\to \mathbb P^2$ is:
$$\phi (x_0:x_1:x_2:x_3:x_4)=(x_0:x_1:x_2) \quad \text{if} \: (x_{0},x_{1},x_{2}) \neq(0,0,0)   $$ 
$$\phi (x_0:x_1:x_2:x_3:x_4)=(x_2:x_3:x_4) \quad \text{if} \: (x_{2},x_{3},x_{4}) \neq(0,0,0)$$  The rank $\lt2$ condition ensures that these definitions are compatible on $X$ and thus that $\phi$ is a well defined morphism  $X\to \mathbb P^2$.      

*Points in $\mathbb P^2$ have exactly one preimage except for $(0:1:0)$ which has as preimage the whole projective line $x_0=x_2=x_4=0$ [parametrically the points  $(0:x_1:0:x_3:0) $ ] included in $X$.  
Conclusion
The variety $X$ is isomorphic to $\mathbb P^2$ blown-up at $(0:1:0)$.
 Quite a pretty problem that!
A: This answer is a complement to the conclusion in Georges' answer.
First, construct a rational map $\sigma:\mathbb{P}^2\dashrightarrow\mathbb{P}^4$ defined by $$(x_0:x_1:x_2)\mapsto(x_0^2:x_0 x_1:x_0 x_2:x_1 x_2:x_2^2) $$
Obviously $\sigma$ is not defined only at $(0:1:0)$, and it is easy to check that the closure of the image of $\mathbb{P}^2$ under $\sigma$ is exactly $X$ (actually $\sigma=\varphi^{-1}$!).
Claim: $X$ is isomorphic to $\mathbb{P}^2$ blown-up at $(0:1:0)$, and $\sigma$ is exactly the blow-up map.
Consider the Veronese embedding $\bar\sigma:\mathbb{P}^2\rightarrow\mathbb{P}^5$ defined by $$
(x_0:x_1:x_2)\mapsto(x_0^2:x_0 x_1:x_0 x_2:x_1 x_2:x_2^2:x_1^2)
$$
The projection map $\pi:\mathbb{P}^5\dashrightarrow \mathbb{P}^4$ is defined by omitting the last homogeneous coordinate of $\mathbb{P}^5$, then obviously $\sigma=\pi\circ\bar\sigma$. Denote $\bar X=\bar\sigma(\mathbb{P}^2)$, thus we get a rational map $\pi:\bar X\dashrightarrow X$.
The center $p$ of projection $\pi$ is $p=(0:0:0:0:0:1)=\bar\sigma(0:1:0)\in \bar X$, hence $X$ is the projection of $\bar X$ with center $p\in\bar X$. Since $\bar X$ contains no lines (otherwise the preimage of this line must have degree $1/2$ in $\mathbb{P}^2$!) and every line through $p$ passes at most one other point in $\bar X$ (counting multiplicity, because $\bar X$ is the intersection of quadratic hypersurfaces), $\pi:\bar X\dashrightarrow X$ is exactly the blow-up map of $\bar X$ at $p$. Therefore, $\sigma=\pi\circ\bar\sigma$ is the blow-up map of $\mathbb{P^2}$ at $(0:1:0)$, and $\varphi$ is the blow-down map.
