Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 21 (Galois Cohomology)

Write $d=\mathrm{deg}A$ for the degree of a central simple algebra, so ${\dim }_{K}A=d^2$. Choose a finite Galois splitting field of a central simple algebra $E/K$ for $A$ and an isomorphism $\varphi :A{\otimes }_{K}E\to M_d(E)$. The reduced norm isThe existence of a finite separable splitting field is a standard structural property of a central simple algebra; passing to its Galois closure supplies $E$.

Every $E$-algebra automorphism of $M_d(E)$ is inner. Here is the matrix-unit argument for the inner automorphisms of a matrix algebra result. For an automorphism $h$, choose $w\ne 0$ in the image of $h(E_{11})$, and set $w_i=h(E_{i1})w$. Then $h(E_{ij})w_k={\delta }_{jk}{w}_i$. The $w_i$ are nonzero and independent, and their number is $d$, so they form a basis. Relative to that basis $h$ acts on each matrix unit in the usual way. Thus $h(T)=PT{P}^{-1}$ for some $P\in {\mathrm{GL}}_d(E)$. Determinants are unchanged by conjugation, proving independence of $\varphi$.

For $\sigma \in \mathrm{Gal}(E/K)$, compare $\varphi$ with the isomorphism obtained by applying $\sigma$ to matrix entries and ${\sigma }^{-1}$ to the scalar factor of $A\otimes E$. Their difference is again inner. For $a\in A$, this shows $\sigma (\det \varphi (a))=\det \varphi (a)$. The determinant therefore belongs to $K$. In fact, for a $K$-basis $a_1,\dots ,a_{d^2}$, the same comparison shows that all coefficients ofare fixed by the Galois group, so the reduced norm is a homogeneous polynomial of degree $d$ over $K$. This coefficient argument also applies over finite fields, where equality merely as functions would not identify polynomials.

Finally, two finite splitting fields embed into a common finite splitting extension. Determinant commutes with scalar extension, and over that common extension the two matrix identifications differ by an inner automorphism. Hence the resulting polynomials agree over $K$. The reduced norm is independent of both the splitting field and the matrix identification. It is multiplicative, and $\mathrm{Nrd}(a)=0$ exactly when $a$ is not invertible: a matrix with nonzero determinant is invertible after scalar extension, and invertibility descends by the invertibility of the $K$-linear multiplication map. In a division algebra the only element of reduced norm zero is zero.

For a nonnegative integer $r$, a Cr field is a field over which every homogeneous polynomial of positive degree $d$ in $N>d^r$ variables has a nontrivial zero. Thus a C1 field uses the bound $N>d$, and a C2 field uses $N>d^2$.

The required field theorems are as follows. The Chevalley-Warning theorem implies that every finite field is $C_1$: the number of zeros of a polynomial with degree smaller than its number of variables is divisible by the characteristic, and for a homogeneous polynomial the origin is already a zero. The Lang-Nagata theorem for Ci fields states that a finitely generated extension of transcendence degree $s$ of a $C_r$ field is $C_{r+s}$. The Tsen theorem states that the function field of a curve over an algebraically closed field is $C_1$. In particular, finite fields, algebraically closed fields and fields such as $\mathbb{C}(t)$ are examples of $C_1$ fields. A function field of one variable over a finite field is $C_2$ by the Lang theorem. These are statements of the theorems; no theorem proof is needed here.

For the requested finite extension stability of C1 fields, let $m=[L:K]$ and choose a $K$-basis $u_1,\dots ,u_m$ of $L$. Given a homogeneous $f\in L[X_1,\dots ,X_N]$ of degree $d$ with $N>d$, substitute $X_i={\sum }_j{x}_{ij}{u}_j$ and take the field norm:The field norm is the determinant of multiplication on the $m$-dimensional $K$-space $L$, hence a homogeneous polynomial of degree $m$ in its coordinates. Consequently $F$ is homogeneous of degree $md$ in $mN>md$ variables over $K$. The $C_1$ property gives a nonzero coordinate vector $(x_{ij})$ with $F=0$. Its corresponding vector $(X_i)\in L^N$ is nonzero, since the $u_j$ are a basis. The norm of a field element vanishes only for the zero element, so $f(X_1,\dots ,X_N)=0$. Every finite extension of a $C_1$ field is $C_1$. This proof includes inseparable finite extensions, because the determinant definition of the field norm requires no separability.

Every Brauer class has a central division algebra representative $D$, and every central simple algebra is a matrix algebra over such a $D$. Put $d=\mathrm{deg}D$; its reduced norm is a homogeneous degree-$d$ polynomial on the $d^2$-dimensional vector space $D$.

If $d>1$, then $d^2>d$, so the $C_1$ property supplies a nonzero $x\in D$ with $\mathrm{Nrd}(x)=0$. But every nonzero element of a division algebra is invertible, and its reduced norm cannot be zero. Thus $d=1$, and a central division algebra of degree one is $K$ itself. Every central simple algebra over $K$ is therefore a matrix algebra over $K$, givingThis proves the trivial Brauer group of a C1 field. The essential ingredient is the anisotropy of a division algebra's reduced norm, not the norm polynomial of an arbitrary matrix algebra, which certainly can vanish on nonzero singular matrices.

Let $k$ be the full constant field of $K$. It is a finite extension of the given finite field, hence is itself finite. Since $k$ is perfect and algebraically closed in $K$, the one-variable extension $K/k$ is regular. In a common algebraic closure, put $E=K\overline{k}$. Then $E$ is the function field of a geometrically integral curve over the algebraically closed field $\overline{k}$.

The Tsen theorem makes $E$ a $C_1$ field, and the preceding argument gives $\mathrm{Br}(E)=0$. Thus $A{\otimes }_{K}E$ is a matrix algebra. A splitting isomorphism and its inverse involve only finitely many coefficients in $E$. All these coefficients lie in $K{k}^{\prime }$ for some finite extension $k^{\prime }/k$, since $E$ is the union of these finite constant extensions. The two isomorphism identities consequently already hold over $K{k}^{\prime }$, so $A$ splits there.

Finite extensions of a finite field are cyclic, and regularity gives $K\cap k^{\prime }=k$ and preserves their degree under this scalar extension. Thereforeis cyclic. A finite cyclic extension of constants splits $A$. This is cyclic splitting by extension of constants. The argument does not assert that the splitting extension has degree exactly the index of $A$; a larger cyclic constant extension is sufficient for the requested conclusion.

Put $d=\mathrm{deg}D$, choose a $K$-basis of $D$, and let $N(X_1,\dots ,X_{d^2})$ be its reduced norm polynomial. For $a\in K^{\times }$, considerThis has degree $d$ and $d^2+1>d^2$ variables. The $C_2$ property gives a nontrivial zero $(x,t)$. Necessarily $t\ne 0$: otherwise $N(x)=0$ forces $x=0$ in the division algebra, making the whole coordinate vector zero. Homogeneity now gives $N(x/t)=a$. The value zero is achieved by the zero element. HenceThis is the surjectivity of reduced norms over C2 fields, and the proof also gives the multiplicative surjection $D^{\times }\to K^{\times }$. The argument works for degree one as well.

For a complete discretely valued field with perfect residue field $k$, the Witt residue sequence is the split exact sequenceThe first map identifies the unramified Brauer classes. The residue map is canonical; a splitting is chosen by a uniformizer $\pi$ and sends an unramified character $\chi$ to the cyclic Brauer class $(\chi ,\pi )$. Changing $\pi$ may change this splitting by an unramified class. This is the version of Witt's theorem used here.

Now suppose $k={\mathbb{F}}_q$. Fix arithmetic Frobenius $F_k:a\mapsto a^q$. Define the local Brauer invariant byHere is an explicit cohomological construction. Set $E=K^{\mathrm{nr}}$ and $G=\mathrm{Gal}(E/K)\cong \hat{\mathbb{Z}}$. The permitted vanishing $\mathrm{Br}(E)=0$, together with Hilbert 90 and inflation, identifies $\mathrm{Br}(K)$ with $H^2(G,E^{\times })$. Normalize the valuation $v(\pi )=1$. Its induced map sends a multiplicative two-cocycle $c$ to the integer-valued two-cocycle $v(c)$, givingThe middle isomorphism is the connecting map for $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}\to 0$, since positive-degree continuous cohomology of the trivial module $\mathbb{Q}$ vanishes.

To see that the valuation map is an isomorphism, its module sequence splits using $\pi$, and the unit part has zero second cohomology. For each finite unramified cyclic extension $E_m/K$, cyclic cohomology identifies that unit contribution with ${\mathcal{O}}_K^{\times }/N_{E_m/K}{\mathcal{O}}_{E_m}^{\times }$. The norm on residue-field units is surjective. On the principal-unit quotient of level $r\ge 1$, the norm is the residue trace, through $N(1+{\pi }^{r}b)\equiv 1+{\pi }^r\mathrm{Tr}(\overline{b})(\mathrm{mod}{\pi }^{r+1})$. The trace of a finite separable field extension is surjective; successive corrections and completeness therefore make the norm surjective on all units. Passing to the direct limit gives the asserted vanishing. Finally, a continuous character of $\hat{\mathbb{Z}}$ is uniquely specified by its value on arithmetic Frobenius, and that value can be any element of $\mathbb{Q}/\mathbb{Z}$. The invariant map is an isomorphism.

In concrete terms, if $E_m/K$ is the unramified extension of degree $m$ and $\sigma$ lifts arithmetic Frobenius, the cyclic algebra with $ub=\sigma (b)u$ and $u^m=a$ hasIndeed its standard two-cocycle has value $a$ when the powers of $\sigma$ wrap past $m$ and value one otherwise. Valuation turns this into the connecting cocycle for the character with Frobenius value $v_K(a)/m$. This also fixes the sign of the normalization.

For restriction and corestriction of local Brauer invariants, let $l$ be the residue field of $L$, $e$ the ramification index and $f=[l:k]$. A finite extension of local fields has $[L:K]=ef$, $v_L{|}_K=e{v}_K$ and $v_K(N_{L/K}b)=f{v}_L(b)$. Every class of $\mathrm{Br}(K)$ is $(\chi ,{\pi }_K)$, with $\chi$ unramified. After restriction to $L$, its character has Frobenius value $f\chi (F_k)$ and its parameter has valuation $e$. ThusFor corestriction, write a class of $\mathrm{Br}(L)$ as $({\chi }_l,{\pi }_L)$. Choose a character ${\chi }_k$ of the finite-field absolute group such that ${\mathrm{Res}}_{l/k}{\chi }_k={\chi }_l$. This is possible because restriction is multiplication by $f$ on $\mathbb{Q}/\mathbb{Z}$, which is divisible. Inflate both characters to the local fields. The norm projection formula for cyclic Brauer pairings givesThis projection identity follows from the cohomological transfer and cup-product identity; on multiplicative degree-zero coefficients transfer is the field norm, and the identity in higher degree follows by the cochain definition or dimension shifting. The norm parameter has $K$-valuation $f$, so its invariant is $f{\chi }_k(F_k)={\chi }_l(F_l)$. ThereforeThis argument uses the character-and-norm description directly and does not cancel multiplication by $[L:K]$ in $\mathbb{Q}/\mathbb{Z}$, where such cancellation would be invalid. For purely inseparable steps in equal characteristic, absolute Galois groups identify, restriction uses inclusion of multiplicative coefficient modules, and Brauer corestriction uses the field norm on them. The same projection calculation applies. Factoring a general finite extension into separable and purely inseparable steps covers all finite extensions in the question.

Lastly, a class with invariant $a/b$ in lowest terms has Brauer period $b$. The unramified extension of degree $b$ splits it, so its division-algebra degree divides $b$. Conversely a maximal subfield of the division algebra splits it and has degree equal to that division-algebra degree; the restriction formula forces $b$ to divide that degree. These standard splitting-field properties are among the permitted inputs. Thus the index of a central simple algebra equals its period over a local field. The central division algebras of degree $n$ are exactly those with invariants $a/n$ and $\gcd (a,n)=1$. The Brauer class determines its division representative uniquely, so there areisomorphism classes, where $\phi$ is the Euler totient function. This includes the single degree-one algebra $K$.

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.

We use the standard characteristic-$p$ absolute Galois-group theorem, ${\mathrm{cd}}_p(K)\le 1$, and the cohomological identification $\mathrm{Br}(K)=H^2(K,K_s^{\times })$. Both are clearly stated inputs permitted for this question.

In characteristic $p$, the map $a\mapsto a^p$ on $K_s^{\times }$ is injective, although it need not be surjective: $p$th roots are purely inseparable. Thus it gives an exact sequence of discrete Galois modulesThe quotient $Q$ is killed by $p$. Since ${\mathrm{cd}}_p(K)\le 1$, $H^2(K,Q)=0$. The long exact cohomology sequence containsTherefore multiplication by $p$ on $\mathrm{Br}(K)$ is surjective. This is Brauer p-divisibility in characteristic p. The quotient-module sequence is the relevant one in characteristic $p$; the separable Kummer sequence with a nontrivial ${\mu }_p$ is not available there.

We first give the common reduction for the three conditions, then prove the cycle from (i) to (ii), from (ii) to (iii), and from (iii) back to (i). Assume here that $\mathrm{char}K\ne p$. The Galois module of roots of unity ${\mu }_p$ fits into the Kummer sequencewhich is surjective on the right because $p$th-root polynomials are separable. Restrict it to the absolute Galois group of any algebraic extension $L/K$. The long exact sequence gives the Kummer cohomology divisibility criterionContinuous cohomology of a profinite group in positive degree with discrete coefficients is torsion: a continuous normalized cocycle becomes zero on a sufficiently small open subgroup, and restriction followed by corestriction multiplies its class by the finite index. For an abelian group, p-primary torsion consists of elements killed by powers of $p$, while p-divisibility means that multiplication by $p$ is surjective. For a torsion group $A$, $A\{p\}=0$ if and only if $A[p]=0$, because a nonzero element of $p$-power order has a multiple of exact order $p$. Consequently the two conditions on the multiplicative cohomology groups are jointly equivalent toThe coefficient module $K_s^{\times }$ is used with its restricted Galois action as in the question. Purely inseparable algebraic extensions give equivalent absolute Galois groups; the ${\mu }_p$ reduction therefore also covers those extensions.

The cohomological dimension at a prime condition ${\mathrm{cd}}_p(K)\le n$ means that $H^q(G_K,M)=0$ for every discrete $p$-primary torsion Galois module $M$ and every $q>n$. Cohomological dimension does not increase on a closed subgroup. Since absolute Galois groups of algebraic extensions identify with closed subgroups, up to the purely inseparable identification just noted, condition (i) gives $H^{n+1}(L,{\mu }_p)=0$ for every algebraic $L/K$. The Kummer reduction proves (i) implies (ii).

Every finite extension of degree prime to $p$ is an algebraic extension. Thus the universal vanishing established for all algebraic extensions in (ii) holds in particular for the extensions appearing in (iii). Equivalently, the Kummer reduction gives $H^{n+1}(L,{\mu }_p)=0$ for each of these finite extensions. Hence (ii) implies (iii). No transfer or separability assumption is required for this implication.

Let $G=G_K$ and choose a profinite Sylow subgroup $P$ at the prime $p$. Its image in every finite quotient of $G$ is a Sylow $p$-subgroup. Put $F=K_s^P$. Every finite intermediate field $K\subseteq L\subseteq F$ has degree prime to $p$, because its corresponding open subgroup contains $P$. Condition (iii), together with the preceding Kummer reduction, gives $H^{n+1}(L,{\mu }_p)=0$ for each such field. The cohomology continuity at closed subgroups property saysfor a discrete $G$-module $M$. Applying it to $M={\mu }_p$ gives $H^{n+1}(P,{\mu }_p)=0$.

The action of the pro-p group $P$ on ${\mu }_p$ is trivial. Indeed it maps continuously to $\mathrm{Aut}({\mu }_p)\cong {\mathbb{F}}_p^{\times }$, whose order $p-1$ is prime to $p$; every finite image of $P$ is a $p$-group, so that image is trivial. Choosing a primitive root identifies this module with the trivial module ${\mathbb{F}}_p$. Thus $H^{n+1}(P,{\mathbb{F}}_p)=0$.

We spell out why this controls arbitrary $p$-primary coefficients. A finite discrete $p$-primary $P$-module has a composition series with trivial ${\mathbb{F}}_p$ factors: its action factors through a finite $p$-group, whose only simple module in characteristic $p$ is the trivial one. Induction through the long exact cohomology sequence therefore gives vanishing of $H^{n+1}$ for every such finite module. Any discrete $p$-primary $P$-module is the filtered union of finite $P$-stable submodules. The orbit of an element is finite by continuity, and its orbit generates a finite abelian $p$-group. Continuous cohomology commutes with these filtered unions, sofor every discrete $p$-primary module $M$. This is the mechanism behind trivial coefficients detect the cohomological dimension of a pro-p group.

Now let $M$ be any discrete $p$-primary $G$-module and $a\in H^{n+1}(G,M)$. Its restriction to $P$ is zero. Continuity at the closed subgroup $P$ makes its restriction zero in some open $U\supseteq P$. The index $m=[G:U]$ is prime to $p$, and the restriction-corestriction identity in group cohomology for the corestriction map in group cohomology givesBut $a$ has $p$-power order, so multiplication by $m$ is invertible on the cyclic group it generates. Hence $a=0$. This proves $H^{n+1}(G,M)=0$ for all discrete $p$-primary $M$. Dimension shifting through an acyclic coinduced module gives vanishing in every higher degree, which is exactly ${\mathrm{cd}}_p(K)\le n$.

Therefore (iii) implies (i), completing the equivalence of all three conditions. The key reason prime-to-$p$ extensions suffice is that they approximate a pro-$p$ Sylow subgroup, where the cyclotomic module becomes trivial; prime-to-$p$ transfer then detects every $p$-primary cohomology class.