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I read the answer to the following question: Quadrics are birational to projective space

Here it is stated that: Over a field $k$ of characteristic ≠2 every irreducible quadric $Q \subset \mathbf P^n_k$ has equation $q(x)=x_0x_1+x^2_2+...+x^2_m=0$ in suitable coordinates .

Can anyone tell me why an irreducible quadric would look like this (and would not have the terms $x_1^2$ for example)?

Thank you in advance!

Georges Elencwajg
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TheBeiram
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    Dear Marie Beth, you can use LaTeX commands on this site in the usual way. I edited your answer to include them. –  Apr 23 '15 at 10:48

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The subtle point is the phrase "in suitable coordinates" in Georges's answer that you are referring to.

He is completely right that in suitable coordinates a quadric has that form; on the other hand, you are equally right that in other coordinates, there could be a term with $x_1^2$.

To be explicit, consider the following change of coordinates on projective space:

$$ \begin{align} y_0 &= \frac12 (x_0+x_1) \\ y_1&= \frac12 (x_0-x_1) \\y_i &= x_i \quad (i=2,\ldots,n) \end{align}$$

Exercise: check that after this change of coordinates, Georges's quadrics (in the variables $x_i$) turn into your quadrics (in the variables $y_i$).

  • Thank you for your answer Asal. I understand that by change of coordinates you can end up with a term $x_1^2$. But I still don't get how you get from a quadratic polynomial, which would look like $x_0+x_1+...+x_n+x_0^2+x_1^2+...+x_n^2+x_0x_1+x_0x_2+...+x_1x_2+..+C$ (so all the possible combinations of coordinates up to degree 2), to the coordinates of Georges. Is it also an explicit change of coordinates like above? Or does it have something to do with the irreducibility? – TheBeiram Apr 27 '15 at 11:10
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    Dear Marie Beth, the reason is that we are talking about a quadric in projective space. Such a thing is the zero set of a homogeneous quadratic polynomial. (In general, defining a subset of projective space by a polynomial equation only works if the equations is homogeneous.) –  Apr 27 '15 at 11:31
  • ah that was clearly very obvious! Thank you! – TheBeiram Apr 27 '15 at 19:27