Timeline for Quadratic forms over finite fields
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Mar 5, 2010 at 10:50 | comment | added | damiano | Note that if $k$ has characteristic two and $a_1,\ldots,a_n$ are elements of $k^\times$ that are independent in $k^\times/(k^\times)^2$, then the quadratic form $\sum a_i x_i^2$ is not of the form that you mention as soon as $n \geq 3$. | |
| Mar 5, 2010 at 10:27 | comment | added | damiano | It's ok, unless $k$ is imperfect of char 2. Let $R$ be the radical of $q$ and $W$ a complementary subspace. If $q|_R$ is non-zero, then ${\rm char}(k)=2$ and $q|_R=\sum a_ix_i^2$; if $k$ is perfect, this is the square of a linear form and hence $\dim(W)\geq o(q)-1$. Suppose that $k$ is a finite field and that $o(q)>2$. Then either $\dim(W)>2$, or $\dim(W)=2$ and $q$ is of the form $q'+ax^2$, with $a \in k^\times$ and $q'$ a form with $o(q')=2$. In either case there is a three dimensional subspace of $V$ where $q$ defines a smooth conic, and we conclude by the Chevalley-Warning Theorem. | |
| Mar 4, 2010 at 20:51 | comment | added | Wanderer | By the way, don't you need a statement in your last paragraph which is slightly (but only a very little bit) stronger to make your argument work? I mean the fact that every quadratic form of order at least three has a non-singular zero (and not just a non-trivial zero). | |
| Mar 4, 2010 at 20:24 | comment | added | damiano | Must've been me! Nice to meet you, d | |
| Mar 4, 2010 at 20:01 | comment | added | Wanderer | Thanks. I guess I've seen you giving a talk about "grosses surfaces rationnelles" earlier this year in Paris, by the way. | |
| Mar 4, 2010 at 19:58 | vote | accept | Wanderer | ||
| Mar 4, 2010 at 19:05 | history | edited | damiano | CC BY-SA 2.5 |
added 29 characters in body
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| Mar 4, 2010 at 18:53 | history | answered | damiano | CC BY-SA 2.5 |