Isomorphisms of blow-up

Question

Let $X = Bl_{C} \mathbb P^3$ and $Y = Bl_{pt} W$ where $C$ is a conic curve in $\mathbb P^3$ and $W$ is a smooth quadric in $\mathbb P^4$. I have to show that $X \cong Y$. I don't know what is the best way of doing this : if I should pick equation or try to reason geometrically.

Let's try to find equations : let's says the conic is the intersection of the quadric $xy - zw = 0$ with some hyperplane $l = 0$ : then $X$ is the subset of $\mathbb P^3 \times \mathbb P^1$ defined by the equations $u(xy - zw) = lv$ were $(u,v)$ are the standard coordinates on $\mathbb P^1$.

Now I take my smooth quadric to be the zeroes set of $Q(x_0, x_1, x_2, x_3, x_4) = \sum_{i=0}^3 x_i^2 + x_4(\sum_{i=0}^3 x_i)$. I agree this equation is a bit strange but I wanted my quadric to be smooth and $(0,0,0,0,1)$ be in my quadric so I can blow up simply in affine coordinates. The blow up of my quadric at a point is given by some affine equations : for example one of the equations (in the affine chart $x_4 = 1$ with coordinates $u_i = \frac{x_i}{x_3}$) is $\sum_{i=0}^2 x_3u_i(x_3u_i + 1) + x_3(x_3 + 1) = 0$.

They are lots of similar equations for $Bl_{pt}Q$ and I don't really see how to simply use them to get the isomorphism I want.

Any advices/help would be appreciated !

Answer 1

OK, let us do it with equations. First of all, your equation for the blowup of $P^3$ in a conic makes no sense, since its sides have different degrees in the coordinates on $P^3$. Do deal with that properly, you have to either replace $P^3 \times P^1$ by the projectivization of a nontrivial rank 2 bundle $O \oplus O(-1)$ on $P^3$ (and write the same equation), or, embedding $O \oplus O(-1) \hookrightarrow O \oplus O^4$ (the second summand is the tautological embedding), and thus embed the blowup into $P^3 \times P^4$. Let us use the second way. If $v,u_1,u_2,u_3,u_4$ are the coordinates on $P^4$ and $x_1,x_2,x_3,x_4$ are the coordinates on $P^3$, then the blowup is given by the equations $$ u_i(x_1x_2 - x_3x_4) = vl(x_1,x_2,x_3,x_4)x_i, \qquad u_ix_j = u_jx_i. $$ Now, if you forget the first equations, you will get precisely the equation of the blowup of $P^4$ in the point $P = (1,0,0,0,0)$. And if you take the strict transform of the quadric $$ u_1u_2 - u_3u_4 = vl(u_1,u_2,u_3,u_4) $$ (this is equal to the blowup of this quadric in $P$), and compare the equations in the standrard charts, you will see that they are the same.