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?


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...

  • $\begingroup$ 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. $\endgroup$ – Pete L. Clark Oct 25 '11 at 7:17

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