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For a Finite Galois extension $L/K$ and a finite-dimensional $L$-space $V$ with a semilinear action of its Galois group, the map $L{\otimes }_K{V}^G\to V$ is an isomorphism. The vectors ${\sum }_{\sigma }\sigma (a)\sigma (v)$ are invariant and span $V$ by the Artin independence theorem. Choosing an invariant $L$-basis then identifies the invariant space with its $K$-span. The proof does not divide by the extension degree.

Let $G=\mathrm{Gal}(L/K)$. A semilinear action on an $L$-vector space $V$ is an additive action with $\sigma (av)=\sigma (a)\sigma (v)$. The Galois descent of vector spaces theorem says that $V^G$ is a $K$-vector space and the natural mapis an isomorphism. In particular ${\dim }_K{V}^G={\dim }_{L}V$. Morphisms also descend: an $L$-linear equivariant map is the scalar extension of its restriction to invariant subspaces.

We prove that the invariant vectors span $V$ over $L$, without dividing by $|G|$. For $a\in L$ and $v\in V$, the vectoris invariant, because every $\tau \in G$ merely permutes the terms. Suppose their $L$-span were proper. There would be a nonzero $L$-linear functional $\lambda$ annihilating that span. For each fixed $v$, we would haveThe Artin independence theorem makes all coefficients zero, in particular $\lambda (v)=0$ for every $v$, a contradiction. To recall its elementary proof, take a nontrivial linear relation between distinct field automorphisms with the fewest nonzero coefficients. Replace the input $a$ by $ab$ and subtract one automorphism's value on $b$ times the original relation. Choose $b$ on which two participating automorphisms differ. The resulting nonzero relation has fewer terms, contradicting minimality.

Choose an $L$-basis $v_1,\dots ,v_m$ consisting of invariant vectors. If $v={\sum }_i{a}_i{v}_i$ is invariant, uniqueness of coordinates gives $\sigma (a_i)=a_i$ for every $\sigma$, so every $a_i$ lies in $K$. These vectors are therefore a $K$-basis of $V^G$, proving the displayed map is an isomorphism. Restriction and scalar extension are visibly inverse on equivariant morphisms, completing the descent theorem. The argument remains valid when the characteristic divides $|G|$.

Now put $B=M_n(L)$, and let ${\sigma }_0$ act on $B$ by applying $\sigma$ to each entry. The group ${\mathrm{Aut}}_L(B)$ is a $G$-group under $\sigma (h)={\sigma }_{0}h{\sigma }_0^{-1}$. Its nonabelian first cohomology consists of cocycles $c_{\sigma }$ satisfyingmodulo $c_{\sigma }^{\prime }=b{c}_{\sigma }\sigma (b{)}^{-1}$ for $b\in {\mathrm{Aut}}_L(B)$. This convention is equivalent to the usual one after replacing $b$ by its inverse. The matrix-automorphism calculation above identifies this coefficient group with ${\mathrm{PGL}}_n(L)$.

For a central simple algebra $A$ of degree $n$ split by $L$, choose $\varphi :A{\otimes }_{K}L\to B$. Transport the natural semilinear action to $T_{\sigma }=\varphi (1\otimes \sigma ){\varphi }^{-1}$ and setIt is $L$-linear and multiplicative, and $T_{\sigma }{T}_{\tau }=T_{\sigma \tau }$ gives the cocycle identity. Replacing $\varphi$ by $b\varphi$ changes $c$ by the displayed coboundary equivalence. Isomorphic $K$-algebras give the same class, so this defines the descent classification of central simple algebras map naturally.

Conversely, a cocycle defines a semilinear algebra action $T_{\sigma }=c_{\sigma }{\sigma }_0$. LetThe vector-space descent theorem gives $A_c{\otimes }_{K}L\cong B$, with the natural map also respecting multiplication and the identity. Hence ${\dim }_K{A}_c=n^2$. If $I$ is a nonzero two-sided ideal of $A_c$, then $I{\otimes }_{K}L$ is a nonzero ideal of $B$, so it is all of $B$. Dimensions force $I=A_c$, proving simplicity. If $z$ is central in $A_c$, it commutes with its $L$-span $B$, and thus is a scalar matrix with scalar $a\in L$. Twisted invariance of a scalar matrix means $\sigma (a)=a$, so $a\in K$. The center is exactly $K$. Thus $A_c$ is central simple, of degree $n$, and split by $L$.

Equivalent cocycles have conjugate semilinear actions: $T_{\sigma }^{\prime }=b{T}_{\sigma }{b}^{-1}$, so $b$ restricts to a $K$-algebra isomorphism between the fixed algebras. In the other direction, an isomorphism of fixed algebras extends to an $L$-algebra automorphism of $B$ and intertwines the two actions, giving exactly that equivalence relation. Starting from $A$ recovers the fixed algebra of $A\otimes L$, namely $A$; starting from a cocycle recovers its action. ThereforeThis is a bijection of pointed sets, with the split algebra corresponding to the trivial cocycle. Fixed-degree algebra classes do not carry the Brauer group's tensor-product group law, so a pointed-set interpretation is the appropriate meaning of the isomorphism here.

For an extension $L/K$ with Galois group $G$, a semilinear action on an $L$-space is an additive group action satisfying $\sigma (av)=\sigma (a)\sigma (v)$. Scalar coefficients are transformed along with vectors. Invariants form a $K$-space, and Galois descent of vector spaces reconstructs the original space from it.