Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 2 (Topics in Representation Theory)

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.

Because $A$ is hereditary, $J{P}_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its projective cover therefore supplies an epimorphismThis splits by projectivity of $J{P}_i$. Its kernel is a summand contained in the radical of the source; projection onto that summand and Nakayama lemma force the kernel to be zero. HenceEvery arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.

Choose lifts in $e_{j}J{e}_i$ of bases of the arrow spaces. Send the vertices and arrows of the path algebra to these idempotents and lifts, multiplying paths in the order $\beta \alpha$ for consecutive arrows $\alpha :i\to j$, $\beta :j\to l$. This defines an algebra homomorphism $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.

Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+{\sum }_j{a}_{ij}{N}_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. ThereforeThe surjective homomorphism is an isomorphism. This proves the basic hereditary algebra path-algebra theorem:

For each connected component $C$ of the underlying unoriented graph, $e_C={\sum }_{i\in C}{e}_i$ is a central idempotent. Paths never connect distinct components, so $kQ={⨁}_C{e}_{C}kQ$ as a direct sum of two-sided ideals. To prove each is a block, let $z$ be any central idempotent. Modulo the arrow ideal it has coordinates $c_i\in \{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent idempotent, hence zero; if they are one, apply that argument to $1-z$. Consequently

The finite representation type of a quiver condition means there are only finitely many isomorphism classes of indecomposable finite-dimensional $kQ$-modules. Equivalently, every finite-dimensional module is a direct sum of members of a finite list of indecomposables.

For the $E_6$ case, we give the reflection argument underlying the Gabriel theorem, rather than using its conclusion as the proof. Since the underlying graph is a tree, every orientation is acyclic. At a sink $i$, writefor the sum of the incoming arrow maps. If an indecomposable $V$ is not the simple $S_i$, then $h$ is surjective: a nonzero complementary subspace to its image at the sink would be a direct summand supported only at $i$. The Bernstein–Gelfand–Ponomarev reflection functor reverses the arrows and replaces $V_i$ by $\ker h$, with reversed maps the component projections. The source reflection replaces that kernel by the cokernel of its inclusion into $⨁V_j$, recovering $V_i$. The inclusion is injective, so the reflected representation has no direct summand $S_i$ at the new source. If it decomposed, applying the inverse would decompose $V$. Thus reflection preserves indecomposability except that it annihilates $S_i$.

On a dimension vector of a quiver representation, it acts by the integral simple reflectionLet $B=2I-\mathrm{Adj}(E_6)$. These reflections preserve the positive-definite quadratic form $d^{\mathsf{T}}Bd$. Positivity can be checked without a classification theorem: label the graph by the chain $1-2-3-4-5$ and attach vertex $6$ to $3$. The leading principal minors are $2,3,4,5,6,3$, all positive.

Choose a topological ordering so arrows point from smaller to larger indices. Reflect successively at sinks $6,5,\dots ,1$. This restores the orientation after one full cycle. The transformation of dimension vectors is the Coxeter element $C=s_1{s}_2\cdots s_6$. If $U$ has diagonal entries one and entry $-1$ above the diagonal at an arrow, direct multiplication of the reflection matrices givesThus $C$ has no fixed vector. It is an integral isometry of a positive definite form, so it has finite order $h$: the image of each lattice basis vector lies in a finite set of vectors of the same norm, giving only finitely many possible isometries. Since $C^h=I$ and $I-C$ is invertible,If an indecomposable survived all $h$ reflection cycles, its dimension vectors $d,Cd,\dots ,C^{h-1}d$ would all be nonzero and componentwise nonnegative, contradicting their zero sum. Therefore it reaches a simple at one of the finitely many reflection steps and is then annihilated. Applying the inverse reflections uniquely reconstructs it from that simple. There are at most $6h$ such stopping positions, provingThis supplies the requested finite-type proof sketch.

Take the quiver with one loop. Its path algebra has basis $1,a,a^2,\dots$ and multiplication $a^r{a}^s=a^{r+s}$, so it is the polynomial ring $k[a]$. HenceThe algebra generators need not form a spanning vector-space set: products produce the infinitely many independent paths.

Take the Kronecker quiver, with two arrows $\alpha ,\beta :1\to 2$. There are no paths of length two, so its path algebra has basis $e_1,e_2,\alpha ,\beta$ and dimension four.

For each $\lambda \in k$, take a quiver representation with one-dimensional spaces at both vertices and arrow maps $(1,\lambda )$. An endomorphism consists of scalars $a,b$ with $b=a$ from the first arrow; the second imposes no further condition. Its endomorphism ring is therefore $k$, which has no nontrivial idempotent. Thus the representation is indecomposable. An isomorphism to the representation $(1,\mu )$ again requires $b=a\ne 0$, and the second arrow then gives $\lambda =\mu$. Because an algebraically closed field is infinite, this gives infinitely many pairwise nonisomorphic indecomposables. Hence

For a left $\mathbb{Z}G$-module, group cohomology iswith $\mathbb{Z}$ the trivial module. Equivalently take a projective resolution $P_{\bullet }\to \mathbb{Z}$ and the cohomology of ${\mathrm{Hom}}_{\mathbb{Z}G}(P_{\bullet },M)$. This also describes the right-derived functors of the invariant-module functor $M\mapsto M^G$.

For an explicit model, the homogeneous bar free resolution has $P_n=\mathbb{Z}[G^{n+1}]$, with diagonal $G$-action and boundary the alternating sum of coordinate deletions. It is free over $\mathbb{Z}G$: each orbit has a unique representative with first coordinate $1$. The augmented complex is exact because inserting $1$ as a first coordinate supplies a contracting homotopy of underlying abelian groups; its lack of $G$-equivariance does not affect exactness. The associated inhomogeneous group cochains are functions $f:G^n\to M$, withFor group cochains of degree zero, $(dm)(g)=gm-m$. Adjacent deletions cancel in pairs, giving $d^2=0$ and hence $H^n=\ker d/\mathrm{im}d$.

In degree zero this givesIt is also ${\mathrm{Hom}}_{\mathbb{Z}G}(\mathbb{Z},M)$, since a map is determined by the invariant image of $1$.

In degree one the cocycle condition is $f(gh)=f(g)+gf(h)$, soThese are crossed homomorphisms modulo principal crossed homomorphisms. For trivial action they reduce to ordinary group homomorphisms, so $H^1(G,M)=\mathrm{Hom}(G_{\mathrm{ab}},M)$.

There are two further useful descriptions. If $I=\ker (\mathbb{Z}G\to \mathbb{Z})$ is the augmentation ideal, a crossed homomorphism gives an equivariant map $I\to M$ by $g-1\mapsto f(g)$, because $gh-1=(g-1)+g(h-1)$. Applying $\mathrm{Hom}(-,M)$ to $0\to I\to \mathbb{Z}G\to \mathbb{Z}\to 0$ therefore givesAlso a section of $M⋊G\to G$ has the form $g\mapsto (f(g),g)$ and is a group homomorphism exactly when $f$ is a crossed homomorphism. Conjugation of sections by an element of $M$ changes $f$ by a principal crossed homomorphism. Thus $H^1$ classifies the $M$-conjugacy classes of such sections, or equivalently of complements to $M$ in the semidirect product.

In degree two, a normalized two-cocycle $c:G\times G\to M$ satisfiesDefine multiplication on $M\times G$ byThe displayed cocycle identity is precisely associativity; normalization gives the identity and inverses exist. This produces a group extension $1\to M\to E_c\to G\to 1$ with the prescribed action on $M$.

Conversely a set-theoretic section $s:G\to E$, $s(1)=1$, gives $c(g,h)=s(g)s(h)s(gh{)}^{-1}$ in the additive kernel of a group homomorphism. Associativity gives the cocycle condition. Replacing $s(g)$ by $b(g)s(g)$ changes $c$ byThe resulting extensions are equivalent through a change of coordinates in $M\times G$. Conversely any extension equivalence preserving both kernel of a group homomorphism and quotient yields such a change of section. HenceThe zero class corresponds exactly to a split extension. With trivial action these are central group extensions; addition of classes corresponds to the Baer sum. This proves the second group cohomology classifies group extensions interpretation.

For $G={\mathbb{Z}}^2$ and trivial coefficients $\mathbb{Z}$, put $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The Koszul resolution for a rank-two free abelian group iswhere $d_1(a,b)=(x-1)a+(y-1)b$ and $d_2(c)=(-(y-1)c,(x-1)c)$. To check exactness at $R^2$, reduce $d_1(a,b)=0$ modulo $x-1$. Since $y-1$ is not a zero divisor in $\mathbb{Z}[y^{\pm 1}]$, write $b=(x-1)c$, and cancellation gives $a=-(y-1)c$. Injectivity of $d_2$ follows because $R$ is a domain. The augmentation kernel is generated by $x-1,y-1$, proving the other exactness statement.

Applying ${\mathrm{Hom}}_R(-,\mathbb{Z})$ makes both differentials zero, since $x,y$ act trivially. ThusDegree-one classes assign arbitrary integers to the two generators. Geometrically these are also the cohomology groups of the torus, a classifying space for ${\mathbb{Z}}^2$; its cohomology ring is the exterior algebra on the two degree-one classes.

The integer $n$ in degree two has an explicit extension representative. On triples $(a,b,m)\in {\mathbb{Z}}^3$, setThe bilinear term is a two-cocycle. With generators $x=(1,0,0)$, $y=(0,1,0)$ and central $z=(0,0,1)$,Every central extension of ${\mathbb{Z}}^2$ by $\mathbb{Z}$ has this form: choose lifts of the two generators and the fixed generator of the kernel of a group homomorphism, then their group commutator determines $n$ and every element has a unique form $z^m{y}^b{x}^a$. Changing the lifts by central elements leaves that commutator unchanged. Thus distinct integers give inequivalent extensions, and the cocycles $nad$ add with $n$. The case $n=0$ is ${\mathbb{Z}}^3$; $n=1$ is the integer Heisenberg group. This realizes the integral central extensions of a rank-two free abelian group concretely.

We first prove the simple Artinian case of the Artin–Wedderburn theorem. This part does not need finite dimension over $k$. Choose a minimal nonzero left ideal $L$, using the descending chain condition. Simplicity gives $LA=A$. Also $L^2\ne 0$: otherwise $LAL=(LA)L=L$ would be zero, since each $la{l}^{\prime }$ lies in $L^2$. Choose $x\in L$ for which right multiplication $r_x:L\to L$ is nonzero. By simplicity of $L$, it is an automorphism. Choose $e\in L$ with $ex=x$. Then $(e^2-e)x=0$, and injectivity of $r_x$ gives $e^2=e$. Thus $L=Ae$.

Right multiplication by $e$ splits the left regular module as $A=Ae\oplus A(1-e)$. Repeat within the remaining left ideal: if it contains a minimal left ideal $Af$ generated by an idempotent, the map $v\mapsto vf$ splits off $Af$. Each nonzero remaining complement is a strictly smaller left ideal, so the Artinian condition terminates the process. Therefore the regular module is a finite direct sum of minimal left ideals.

All these simple summands are isomorphic. For primitive idempotents $e,f$, simplicity gives $AeA=A$, so $fAe\ne 0$; otherwise $fAeA=fA$ would be zero. A nonzero element of $fAe$ defines a nonzero homomorphism $Af\to Ae$, which is an isomorphism between simple modules. Thus $A\cong L^{\oplus n}$ as a left module. Put $\Delta ={\mathrm{End}}_A(L)$, a division algebra by the Schur lemma. Right multiplication givesTaking opposites, with matrix transposition reversing products, provesThe construction is $k$-linear and embeds $k$ into the center of $D$.

A central simple algebra is a finite-dimensional simple $k$-algebra whose center is exactly $k$. The Brauer group consists of such algebras modulo matrix stabilization: $A,B$ represent the same class if $M_r(A)\cong M_s(B)$ for some positive integers $r,s$. Its operation and inverse areThe tensor product is central simple, and $A\otimes A^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ by $(a\otimes b)(x)=axb$, explaining the inverse. Both facts can be checked after a splitting field of a central simple algebra extension, where they are matrix-algebra identities, and descended by faithful scalar extension. Every class has a unique central division-algebra representative, by the regular-module description above.

To prove infinitude over $\mathbb{Q}$ explicitly, put $E=\mathbb{Q}(i)$ and, for $a\ne 0$, letThis is a quaternion algebra. It is central simple: the center is $\mathbb{Q}$, and after adjoining $i$ its generators have the faithful matrix modelThese generate $M_2(E)$; consequently a nonzero proper two-sided ideal over $\mathbb{Q}$ is impossible. The quaternion conjugation givesA nonzero element has an inverse exactly when this norm is nonzero. Thus $A_a$ is a division algebra when $a$ is not an $E/\mathbb{Q}$ norm; if $a$ is a norm it has a nonzero zero divisor and is $M_2(\mathbb{Q})$ by the classification just proved.

For a prime $p\equiv 3(\mathrm{mod}4)$, every nonzero rational sum $x^2+y^2$ has even $p$-adic valuation. Divide out the minimum valuation of $x,y$; cancellation modulo $p$ is impossible because $-1$ is not a square modulo $p$. Therefore $p$ is not an $E/\mathbb{Q}$ norm, and $A_p$ is a nontrivial division-algebra Brauer class. This is the quaternion norm obstruction at primes congruent to three modulo four.

We also prove the product rule needed to distinguish classes. In $A_a\otimes A_b$, write the generators as $i_1,j_1,i_2,j_2$. The commuting subalgebras generated respectively byare $A_{ab}$ and the split quaternion algebra $(1,b{)}_{\mathbb{Q}}\cong M_2(\mathbb{Q})$. The latter has the matrix model $\mathrm{diag}(1,-1)$ and $\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$. Together the two subalgebras generate the whole tensor product, since they recover $i_2$ and $j_1$. Dimensions are sixteen on both sides, givingQuaternion conjugation identifies $A_a$ with its opposite, so these classes have order at most two. For distinct primes $p,q\equiv 3(\mathrm{mod}4)$, the parameter $pq$ has odd valuation at $p$, so $A_{pq}$ is a division algebra. Thus $[A_p]\ne [A_q]$. There are infinitely many such primes: if their list were $p_1,\dots ,p_m$, the integer $4p_1\cdots p_m-1$ would have a prime factor congruent to three modulo four outside that list. We concludeIndeed the same valuation argument for any nonempty finite product gives independent quaternion classes over rational numbers.

For $\mathbb{R}$, we prove the required division-algebra classification. Let $D$ be finite-dimensional and central over $\mathbb{R}$. If $D\ne \mathbb{R}$, choose a nonreal element; its irreducible real minimal polynomial is quadratic, so translating and rescaling it gives $i\in D$ with $i^2=-1$. The centralizer $D_{+}$ of $i$ is a finite-dimensional division algebra over $\mathbb{C}=\mathbb{R}(i)$, with that field central. Every element has a split minimal polynomial over $\mathbb{C}$, hence belongs to $\mathbb{C}$, so $D_{+}=\mathbb{C}$.

Conjugation by $i$ is an involution, giving $D=D_{+}\oplus D_{-}$, where $D_{-}=\{x:ix=-xi\}$. If $D_{-}=0$, then $D=\mathbb{C}$, contrary to its center being $\mathbb{R}$. Choose $j\ne 0$ in $D_{-}$. For any $x\in D_{-}$, $x{j}^{-1}$ commutes with $i$, so $D_{-}=\mathbb{C}j$. Also $j^2\in \mathbb{C}$ and commutes with $j$, forcing $j^2\in \mathbb{R}$. It cannot be positive, since $(j-\sqrt{j^2})(j+\sqrt{j^2})=0$, or zero, since $D$ is a division algebra. Rescale to $j^2=-1$. ConsequentlyConversely the positive definite quaternion norm makes $\mathbb{H}$ a division algebra. The two possible central division representatives are therefore $\mathbb{R}$ and $\mathbb{H}$. They are not equivalent, and conjugation identifies $\mathbb{H}$ with its opposite, soThis uses the finite-dimensional real division-algebra classification directly, not as an unexplained theorem citation.

Finally let $k_s$ be a separable closure and $\Gamma =\mathrm{Gal}(k_s/k)$. The cohomological description isHere $k_s^{\times }$ has the discrete topology and its natural Galois action. For a finite Galois splitting field $L$, identify $A{\otimes }_{k}L$ with a matrix algebra. Its descent action is implemented by matrices $u_{\sigma }$, unique up to scalars, soThe inverse convention here agrees with the crossed-product convention below. The semilinear operators $u_{\sigma }\sigma$ commute with the descended algebra $A$; their crossed product has class $[A{]}^{-1}$ by the commuting-factor decomposition of the full endomorphism algebra. Inverting its scalar cocycle therefore gives the class $[A]$. Associativity gives the two-cocycle identity, and changing the representatives changes $c$ by a coboundary. Conversely a two-cocycle defines the crossed-product algebra of a Galois extension, with $u_{\sigma }a=\sigma (a)u_{\sigma }$ and $u_{\sigma }{u}_{\tau }=c(\sigma ,\tau )u_{\sigma \tau }$. It is central simple and split by $L$. Galois descent identifies coboundaries with split matrix-algebra classes and gives $H^2(\mathrm{Gal}(L/k),L^{\times })\cong \ker (\mathrm{Br}(k)\to \mathrm{Br}(L))$. Every central simple algebra has a finite separable, hence finite Galois, splitting field, yielding the displayed limit.