Posted on May 25, 2019

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).