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2

The Brauer group of $\mathbb R$ is $Br(\mathbb R)\cong\mathbb Z/2\mathbb Z$; one way to see this is to note that there are exactly two (iso)classes of central simple f.d. real algebras, in view of Frobenius' theorem which classifies them, so the there is no choice for the group structure. It follows that the tensor square of every central simple real ...

6

It's a basic fact (here's a proof in the second proposition on page 157) that the tensor product of two central simple algebras is another central simple algebra. A proof should be available wherever central simple algebras are discussed. Another location in Jacobson's Basic Algebra II on page 218-219. Another location in Rowen's Ring theory Theorem ...

0

Suppose $k=4$ and $$\alpha=e_1'\wedge e_2'$$ in some basis $e_1', e_2', e_3', e_4'$. You are claiming that you can find another basis $e_1, e_2, e_3, e_4$ such that $$\alpha=e_1\wedge e_2+e_3\wedge e_4,$$ which implies $\alpha$ is not decomposable in the new basis (the basis vectors are linearly independent, so $\alpha\wedge\alpha\ne0$ in the new basis). But ...

1

If $B \wedge x = 0$, then $Bx^{-1} = B \cdot x^{-1}$ is some vector. $Bx^{-1} x = B$ on the one hand, but using a different associative grouping for the geometric product, you get $(B x^{-1}) \wedge x$ instead. Hence, $y = -Bx^{-1}$.

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