TITLE: Is the quadric $3$-fold $v^2 + w^2 + x^2 + y^2 + z^2 = 0$ isomorphic to $P^3$?
QUESTION [7 upvotes]: The subset of projective $4$-space given by $5$-tuples $[v:w:x:y:z]$ with $v^2 + w^2 + x^2 + y^2 + z^2 = 0$ is birational to projective $3$-space.  I think it has the same cohomology as projective $3$-space.  Is it actually isomorphic to projective $3$-space$?$
In other words are there homogeneous polynomials $v(a,b,c,d), w(a,b,c,d), x(a,b,c,d), y(a,b,c,d), z(a,b,c,d)$ whose squares add to $0?$

REPLY [4 votes]: Let $Q$ be the quadric 3-fold in $\mathbb{P}^4$. By adjunction formula, $K_Q=\mathcal{O}_Q(-5+2)=\mathcal{O}_Q(-3)$, and $\text{deg}\mathcal{O}_Q(-3)=2\times(-3)=-6$. On the other hand, $K_{\mathbb{P}^3}=\mathcal{O}_{\mathbb{P}^3}(-4)$ and its degree is $-4$. So $Q$ and $\mathbb{P}^3$ are not isomorphic.