Example of a central simple algebra

Question

Let $A$ be a finite dimensional central simple algebra over a field $F$ of characteristic $0$. So by Weddernburn's theorem, $A\cong M_n(D)$ for some division algebra $D$ over $F$. Let $\dim_F(D)=m^2$. Then $m$ is called the index of $A$.

Assume that $A$ is crossed product: there is a finite Galois extension $E$ of $F$, say of degree $n$, which is (isomorphically sitting) inside $A$ and $n$ invertible elements $u_1,\ldots, u_n\in A$ such that $Eu_1\oplus \cdots \oplus Eu_n=A$ (direct sum as $F$-vector space). The product of elements of $A$ from this decomposition are determined by an equivalence class of $2$-cocycle $f$ in $H^2({\rm Gal}(E/F), E^*)$ (together with action of Galois group on $E^*$).

Q. In general, it is known that the order of $[f]$ in $H^2({\rm Gal}(E/F), E^*)$ divides the index $m$ of $A$. Further, if $F$ is a number field, then equality holds. So, what is an example where order of $[f]$ properly divides the index $m$ of $A$?

The results mentioned in question are taken from Albert's book Structure of Algebras.

Answer 1

You can take: $F=\mathbb{C}(X_1,Y_1,\ldots,X_n,Y_n)$ and take the tensor product of quaternion algebras $$A=(X_1,Y_1)_F\otimes_F\cdots\otimes (X_n,Y_n)_F.$$ Here $A$ contains a subfield $E$ isomorphic to $F(\sqrt{Y_1},\ldots \sqrt{Y_n})$ for example.

Now $A$ is a division algebra, so it has index $2^n$, but it has order $2$ in $\mathrm{Br}(E/F)\simeq H^2(\mathrm{Gal}(E/F),E^\times)$.

PS. If you need details, I can post something this evening.

Edit. Here is a proof that $A$ is division. I will assume some knowledge on quadratic forms. Note that I replace $\mathbb{Q}$ by $\mathbb{C}$ to get rid of some ennoying coefficients (see below), but the result is the same.

If $A$ is a central simple algebra over a field $F$, let $T_A:A\to F$ the quadratic form defined by $T_A(a)=Trd_A(a^2)$.

Lemma. If $T_A$ is anisotropic, then $A$ is division.

Proof. Assume $A$ is not division, so $A\simeq M_r(D)$, where $D$ is division. The matrix $a_0$ filled with zero, except on the top right corner where the entry is $1_D$, satisfies $a_0^2=0$, hence $T_A(a_0)=0$, while $a_0\neq 0$. Hence, $T_A$ is isotropic.

Prop. Let $F$ be a field of characteristic different from $2$. Assume that $q_1,q_2$ are anistropic quadratic forms over $F$. Then the form $q_1+Xq_2$ is anistropic over $F(X)$.

Proof. Omitted. This is a standard result.

The rest of the proof is devoted to show that the trace form of our example is anisotropic.

Easy fact. We have $T_{A\otimes B}\simeq T_A\otimes T_B$ (by standard properties of the reduced trace)

Now if $Q=(a,b)_F$, the standard basis $1,i,j,ij$ is orthogonal wrt $T_Q$, and $$T_Q\simeq \langle 2,2a,2b,-2ab\rangle\simeq \langle 1,a,b,ab\rangle\simeq \langle 1,a\rangle\otimes \langle 1,b\rangle.$$ The second isomorphism comes from the fact that $2$ and $-2$ are squares in $\mathbb{C}$; this is why I changed.

Hence $T_A\simeq \langle 1,X_1\rangle\otimes \langle 1,Y_1\rangle\otimes\cdots\otimes \langle 1,X_n\rangle\otimes \langle 1,Y_n\rangle$.

Using the previous proposition and induction (starting form the fact that $\langle 1\rangle$ is anisotropic), we get the desired result.

Of course, you can do all of this for $n=2$...