Is any two projective smooth quadric hypersurface (over an algebraically closed field with characteristic $\ne 2$) isomorphic? If so, what is the easiest way to prove this? I would probably go to the coordinate rings and show that we have an isomorphism of the two rings
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2No, $V(x^2+y^2)$ and $V(x^2-y^2)$ both inside $\Bbb P^1_{\Bbb R}$ aren't isomorphic because the first has no rational points while the second does. You will need more. – KReiser Sep 11 '19 at 06:15
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I am assuming an algebraically closed base field. – quantum Sep 12 '19 at 04:53
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1Then you should edit this in to your post. – KReiser Sep 12 '19 at 19:22
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Over an algebraically closed field of characteristic not 2, the result is true. In fact, more is true: every smooth quadric hypersurface is projectively equivalent, that is, there's a linear automorphism of $\Bbb P^n$ which sends any smooth quadric hypersurface to any other.
Any projective quadric hypersurface is given as the zero locus of a single homogeneous degree two polynomial. We can represent this quadratic form as a symmetric matrix, and as any symmetric matrix over an algebraically closed field of characteristic $\neq 2$ is diagonalizable, we can write the defining equation as $\sum_{i=0}^n c_ix_i^2$ after a linear change of coordinates. As our base field is algebraically closed, if $c_i\neq0$ we can replace $x_i$ by a scalar multiple so that $c_i=1$. Up to reordering, we our quadric is given by $\sum_{i=0}^d x_i^2$ where $1<d\leq n$. For such a quadric to be smooth, we must have $d=n$, as otherwise the Jacobian matrix $$\begin{pmatrix} 2x_0 & 2x_1 & \cdots & 2x_d & 0 & \cdots & 0\end{pmatrix}$$ is not full rank at $[0:\cdots:0:1]$ and thus our quadric is not smooth.
So up to a linear change of coordinates on $\Bbb P^n$, every smooth quadric hypersurface is the quadric given by $x_0^2+x_1^2+\cdots+x_n^2$. In particular, they're all isomorphic.
KReiser
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