Presentation of Grothendieck-Witt group $GW(\mathbb{F})$ in terms of generators and relations. Let $\mathbb{F}$ be a field, which for the sake of this discussion, is such that char $\mathbb{F} \neq 2$.
By Corollary 9.4 in Scharlau's Quadratic and Hermitian Forms, the Grothendieck-Witt group $GW(\mathbb{F})$ is generated by elements $\langle \alpha \rangle, \alpha \in \mathbb{F}^{\times}$, subject to the relations

*

*$\langle \alpha \rangle = \langle \alpha \beta^{2} \rangle$ for all $\alpha , \beta \in \mathbb{F}^{\times}$.

*$\langle \alpha \rangle + \langle \beta \rangle = \langle \alpha + \beta \rangle + \langle (\alpha + \beta)\alpha \beta \rangle$.

I understand the proof of this result as presented in Scharlau.
However, in Morel's $\mathbb{A}^{1}$ Algebraic topology over a field, lemma 2.9, he says that the second relation may be obtained from the first relation and the relation
$\langle \alpha \rangle + \langle -\alpha \rangle = 1 + \langle -1\rangle$.
I understand the motivation behind this relation (matrices of the form on the LHS are congruent to the hyperbolic plane when char $\mathbb{F} \neq 2$), but cannot formally derive the second relation using this and relation 1).
I am probably just missing a trick. Any help would be much appreciated!
 A: I think Morel's claim is not correct.
Consider the case $F=\mathbb{Q}_3$.
Let $M$ denote the free abelian group generated by $\langle\alpha\rangle$, $\alpha\in \mathbb{Q}_3^\times$.
We have $\mathbb{Q}_3^\times/(\mathbb{Q}_3^\times)^2=\{[1],[-1],[3],[-3]\}$, so the quotient of $M$ by the first relation is a free abelian group with basis $\langle 1 \rangle,\langle -1\rangle, \langle 3 \rangle,\langle -3\rangle$.
If Morel's claim is correct, then $\operatorname{GW}(F)$ must be the quotient of this group by the relation
$$
\langle 3\rangle + \langle -3\rangle = \langle 1\rangle + \langle -1\rangle.
$$
However, we have
$$
\langle 3\rangle + \langle 3\rangle\neq \langle 6\rangle + \langle 54\rangle \quad( = \langle -3\rangle + \langle -3\rangle)
$$
in this group, so the second relation does not hold.
On the other hand, it is true that the first and the second relation imply $\langle \alpha\rangle + \langle -\alpha\rangle=\langle 1\rangle + \langle -1\rangle$.
Let $x=\dfrac{\alpha+1}{2}$ and $y=\dfrac{\alpha-1}{2}$, so that $x^2-y^2=\alpha$.
By the second relation, we have
$$
\langle x^2\alpha\rangle + \langle -y^2\alpha\rangle = \langle (x^2-y^2)\alpha\rangle + \langle -x^2y^2(x^2-y^2)\alpha^3\rangle.
$$
Using the first relation and $x^2-y^2=\alpha$, we can rewrite this as
$$
\langle \alpha\rangle + \langle -\alpha\rangle = \langle 1\rangle + \langle -1\rangle.
$$
Perhaps this is what Morel wanted to note.
