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I'm trying to figure out a proof in Lam's book on quadratic forms (he uses this to define a hyperbolic plane). He states that if $(V,q)$ is a 2-dimensional quadratic space over $F$, then the following are equivalent:

  1. $V$ is regular and isotropic.
  2. $V$ is regular and $\textrm{disc}(V)=-1\cdot F^{\times 2}$.
  3. $V$ is isometric to $\langle 1,-1\rangle$.
  4. $V$ corresponds to the equivalence class of the binary quadratic form $X_1X_2$.

I have trouble understanding his proof that $(2)\Rightarrow (3)$. He starts by stating that "We clearly have the diagonilization $V\simeq \langle a,-a\rangle$". I just don't see why this is clear assuming (2).

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Here is a trivial but surprisingly useful general fact: if you have a diagonalized quadratic form $q = \langle a_1,\ldots,a_n \rangle$ with discriminant $d$, then $a_n$ must lie in the same square class as $a_1 \cdots a_{n-1} d(q)$. This is true just because $d(q) = a_1 \cdots a_n$, and any $a \in K^{\times}$ lies in the same square class as $a^{-1}$.

In particular, for any binary quadratic form $q = \langle a,b \rangle$ of discriminant $d(q)$, we have $q \cong \langle a, a d(q) \rangle$. You are asking about the case where $d(q) \equiv -1 \pmod{K^{\times 2}}$.

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We know $q=\langle a,b\rangle$ with $a\neq 0\neq b$ by regularity. Now $ab\equiv -1\,(\textrm{mod }K^{\times 2})$, so $-1 = abu^2$ for some $u\in K^\times$ and depending on $K$ this could have tons of different solutions. I don't see what I'm missing. – eof Sep 26 '11 at 1:46
@eof: perhaps you are missing that for any diagonalized quadratic form $q = \langle a_1,\ldots, a_n \rangle$ and any $u \in K^{\times}$, $q \equiv \langle a_1,\ldots, u^2 a_n \rangle$, i.e., the isomorphism class of $q$ depends only on the square classes of its diagonal coefficients. – Pete L. Clark Sep 26 '11 at 2:24

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