Pardon my ignorance, but I've been stuck with this silly problem for almost a whole day while trying to learn some quadratic forms theory. Let $F$ be a field of characteristic $\neq 2$. Szymiczek's book on the algebraic theory of quadratic forms contains an exercise where the equivalence of the following are supposed to be proven:

  1. $I^2(F)=0$.
  2. $\langle 1,a,b,ab\rangle=0\in W(F)$ for all $a,b\in F^\times$.
  3. $\langle 1,a,b,ab\rangle$ is hyperbolic.
  4. $\langle 1,a,b,ab\rangle$ is isotropic.
  5. $\langle 1,a\rangle$ is universal for all $a\in F^\times$.
  6. $\langle a,b\rangle$ is universal for all $a,b\in F^\times$.

For some reason I seem to have got stuck on showing $(4)\Rightarrow (5)\Rightarrow (6)$, everything else being trivial or easy.

EDIT: I actually have $(5)\Rightarrow (6)$ now, so I'm looking for $(4)\Rightarrow (5)$.


1 Answer 1


Here is one way to do it. Assume that (4) holds for all $a,b \in F^\times$, and now let $a \in F^\times$ be arbitrary. If $\langle 1,a \rangle$ is isotropic, then it is of course universal, so we may assume it is anisotropic, but then $\langle 1,a \rangle$ is the norm form $N$ of a separable quadratic extension $E/F$. So $\langle 1, a, b, ab \rangle = N \perp bN$ for all $b \in F^\times$.

Now let $t \in F^\times$ be arbitrary; then $N \perp (-t)N$ is isotropic by assumption, so there are elements $x,y \in E$, not both zero, such that $N(x) - tN(y) = 0$. But then $t = N(x y^{-1})$, so it is indeed represented by $\langle 1, a \rangle$. Since $t \in F^\times$ was arbitrary, this means that $\langle 1,a \rangle$ is universal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.