Crossed product algebra

Remarks

The cocycle condition is exactly what makes the multiplication associative. The algebra $(L/K, f)$ is a central simple algebra over $K$ of degree $n$, it contains $L$ as a maximal subfield, and $L$ splits it. Two factor sets give isomorphic algebras if and only if they differ by a coboundary, so the classes correspond to $H^2(G, L^\times)$ in Galois cohomology. This gives $\mathrm{Br}(L/K) \cong H^2(G, L^\times)$.

Example: take $L = \C$, $K = \R$, $G = \set{1, \sigma}$ with $\sigma$ complex conjugation, and $f(\sigma, \sigma) = -1$, $f = 1$ otherwise. With $j = u_\sigma$ one gets $j z = \bar{z} j$ and $j^2 = -1$, which are the relations of $\mathbb{H} = \C + \C j$. With $f(\sigma, \sigma) = 1$ instead one gets an algebra isomorphic to $M_2(\R)$.

Non-example: for $L = \C$, $K = \R$ as above, the map with $f(\sigma, \sigma) = i$ and $f = 1$ otherwise is not a factor set. The condition for $\sigma, \sigma, \sigma$ would give $i = \sigma(i) = -i$, and indeed $(u_\sigma u_\sigma) u_\sigma = i u_\sigma$ while $u_\sigma (u_\sigma u_\sigma) = \sigma(i) u_\sigma = -i u_\sigma$, so the product is not associative.

A central simple algebra is called a crossed product if it is isomorphic to some $(L/K, f)$; Amitsur showed that some central division algebras are not crossed products.