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Let $F$ be a field of characteristic not equal to to $2$, $W(F)$ Witt ring of the quadratic forms. I've been trying to prove that $I^2(F)=0$ implies that every binary quadratic form over $F$ is universal.

EDIT: My original idea does not seem like the right one. After some toying around, it would seem like the following implication holds:

$\langle 1,a,b,ab\rangle$ isotropic for all $a,b\in F^\times$ $\Rightarrow \langle 1,a\rangle$ universal for all $a\in F^\times$.

Does anyone know if this is true or how to prove it?

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up vote 2 down vote accepted

To say that every binary form is universal is equivalent to saying that every form of dimension at least $3$ is isotropic, or in other words that every element of $W(F)$ has a representative of dimension at most $2$.

To show this, I suggest that you use Pfister's Theorem (Theorem 6 in these notes) that elements of $W(F)/I^2$ are classified by their mod $2$ dimension and signed discriminant. When $I^2 = 0$, this gives you a complete classification of quadratic forms up to Witt equivalence...

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As a comment, a few hours ago I answered another one of dst's Witt ring questions. Based on his/her commentary, I decided that s/he is rather on the ball with this material, so this time I decided to give less than complete detail. – Pete L. Clark Oct 25 '11 at 7:17
Got it. Thanks. – dst Oct 25 '11 at 7:53

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