Gille-Szamuely

Here are my exercise solutions and notes for Gille and Szamuely’s Central Simple Algebras and Galois Cohomology.

1.1. Since the map $\sigma$ is linear, for $q=x+yi+zj+wij$ we must have $\sigma(q)=r_0x+r_1yi+r_2zj+r_3wij.$ To have $\sigma(q)=1$ we must have $r_0=1.$ Computing the product $\sigma(q)q$ and taking $x=y=1$ , $z=w=0$ we find that $\sigma(q)q\in k$ only if $r_0=-r_1.$ Setting $z=w=1$ and $x=y=0$ we see that we must also have $r_2=r_3.$ Finally, setting $y=z=1$ and $x=w=0$ we see that $r_1=r_2.$ Thus $\sigma(q)=x-yi-zj-wij=\overline{q}.$

1.3. Let $K=k(\sqrt{a}).$ Recall that $Q=(a,b)$ for some $b\in K$ by Lemma $1.2.2.$ Therefore let us pick the basis $\{1,\sqrt{a}\}$ for $K$ and observe that $\{1,\sqrt{a},\sqrt{b},\sqrt{a}\sqrt{b}\}$ is a basis for $Q.$ Thus the quaternion norm agrees with the relative field norm $N_{K/k}$ for elements of $K\subset Q.$

1.5. Recall that $(-1,p)$ splits over $\mathbb{Q}$ if and only if the curve $C(-1,p)$ defined by $py^2=x^2+z^2$ has a $\mathbb{Q}$ -rational point. Observe that no $\mathbb{Q}$ -rational point on $C$ has $y$ -coordinate zero since such a point would have $x^2+y^2=0$ for both $x,y\ne 0$ , a contradiction. So all $\mathbb{Q}$ -rational points lie on the affine curve $p=x^2+y^2.$ By a classic result from the Gaussian integers, a prime $p$ may be written as the sum of two squares if and only if $p\equiv 1 \, (\text{mod } 4)$ (see the first section of the first chapter of Neukirch’s algebraic number theory book, for example).

Lemma. If $q\in (a,b)$ such that $q^2\in k^\times$ then $q+\overline{q}=0$ or $q\in k.$

Let $q=x+yi=zj+wij.$ Computing we find that $q^2=(x^2+y^2a+z^2b-w^2ab)+2xyi+2xzj+2xwij.$ Therefore $q^2=\in k^*$ if and only if $x=0$ or $y=z=w=0.$

1.6. Let $\varphi:(c,d)\to (a,b)$ be an isomorphism. Take $i,j\in (a,b)$ such that $i^2=a, j^2=b$ , and similarly $i&39;,j&39;\in (c,d)$ such that $(i&39;)^2=c,(j&39;)^2=d.$ Then let $I=\varphi(i&39;)$ and $J=\varphi(j&39;).$

Define $B_0=\{q\in (a,b) \, : \, q+\overline{q}=0\}.$ Observe that $B_0$ is in fact a subpsace of $(a,b)$ of dimension $3$ and is in fact the span of $\{i,j,ij\}$ (assuming that $\text{char}(k)\ne 2$ we see that $q=\overline{q}$ implies that $q=yi+zj+wij$ ).

Since $I^2,J^2\in k$ , by the lemma above we have $I,J\in B_0.$ Therefore we have $2$ -dimensional subspaces $\langle I,IJ\rangle$ and $\langle i,ij\rangle$ in $B_0.$ Thus there is a nonzero element $\epsilon$ in their intersection. Since $\epsilon\in\langle i,ij\rangle$ we see that $\epsilon j=-j\epsilon.$ Similarly, as $\epsilon\in\langle I,IJ\rangle$ we find $\epsilon J=-J\epsilon.$

Therefore $\{1,\epsilon,j,\epsilon j\}$ is an alternate basis for $(a,b)$ and $\{1,\varphi^{-1}(\epsilon),j&39;,\varphi^{-1}(\epsilon)j&39;\}$ is an alternate basis for $(c,d).$ Taking $e=\epsilon^2$ we therefore have $(a,b)\cong (e,b)$ and $(c,d)\cong (e,d).$