# Example of a central simple algebra

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.

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 $$n=2$$...

• Thanks for the nice example; I want to see proof of the fact that $A$ is division algebra. If proof is length, can you suggest some elementary reference for proof? – Soluble Apr 23 '19 at 8:25