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Show that any non-singular irreducible quadric in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$

I know that every non-singular and irreducible quadric in $\mathbb{P}^3$ can be written in the form $xy=zw$ after a suitable change of homogeneous coordinates.

So it is sufficient to prove that $Q=Z(xy-zw)$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$.

$\mathbb{P}^1 \times \mathbb{P}^1 \cong \sigma(\mathbb{P}^1 \times \mathbb{P}^1)$ where $\sigma$ is the Segre embedding.

So we now want to prove that $Q=\sigma(\mathbb{P}^1 \times \mathbb{P}^1)$.

I have some difficulties to prove it. And I don't know if I'm on the right way. Some help would be appreciated.

Thanks.

Leafar
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    Observe, in the Segre embedding, we have $(x_0:x_1)\times(y_0:y_1)\mapsto(x_0y_0:x_0y_1:x_1y_0:x_1y_1)\subset\mathbb{P}^3$. If we let the coordinates on $\mathbb{P}^3$ be $(x,z,w,y)$, we see that $xy=zw$ on this surface. – Michael Burr Jun 21 '15 at 19:25
  • @MichaelBurr Thanks a lot. Write an answer please so I can give you some points – Leafar Jun 21 '15 at 19:36
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    Dear Leafer, could you let me know the source of this problem (textbook or online notes)? I am asking because I am interested in doing problems like this to improve my knowledge about projective geometry. Thanks! – Prism Jun 22 '15 at 02:33
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    @Prism I'm using Andreas Gathmann notes (you can google it, there is a 2002 version that I'm using because it's a complete one and one of 2014 that is perfect for begginers). There are some examples and exercises, but not solved. Some of them require more knowledge than which is given in the notes, such as multilinear algebra. There are also many solved exercises that I found googleing. – Leafar Jun 22 '15 at 16:02
  • @Leafar: Thanks a lot! Much appreciated. – Prism Jun 22 '15 at 18:30

1 Answers1

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Observe, in the Segre embedding $\mathbb{P}\times\mathbb{P}\rightarrow\mathbb{P}^3$, we have $$ (x_0:x_1)\times(y_0:y_1)\mapsto(x_0y_0,x_0y_1,x_1y_0,x_1y_1). $$

If we choose the coordinates on $\mathbb{P}^3$ to be $(x,z,w,y)$, we see that $xy=zw$ on this surface.

Michael Burr
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