Algebraic Number Theory

Consider the connecting homomorphisms $\delta$ and ${\delta }^{\vee }$ for the sequence

and its $\mathbb{Q}∕\mathbb{Z}$-dual

The image of $\overline{\sigma }$ in ${\widehat{H}}^{-1}(G,I_G)$ is the image of $\sigma -1$ in $I_G∕I_G^2$, and the inverse image of $\chi$ in ${\widehat{H}}^0(G,\mathrm{Hom}(I_G,\mathbb{Q}∕\mathbb{Z}))$ is class of the homomorphism $f$ that takes $\tau -1$ for $\tau \in G$ to $\chi (\tau )$. We then have

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Note that ${\alpha }_{L∕K}\cup {\rho }_{L∕K}(a)=\overline{a}$ by definition of ${\rho }_{L∕K}$, viewing its image as the group ${\widehat{H}}^{-2}(\mathrm{Gal}(L∕K),\mathbb{Z})$, where $\overline{a}$ denotes the image of $a$ in $A_K∕N_{L∕K}{A}_L$. By the associativity of cup products and property (iii) of their definition, we have

The composition of the canonical maps

is the isomorphism induced by multiplication by $[L:K]$ (since the norm for $\mathrm{Gal}(L∕K)$ acts as multiplication by $[L:K]$ on $\mathbb{Q}$). By definition of ${\alpha }_{L∕K}$ and the latter fact, we have

the final step from Lemma 8.1.9. □

Let $\chi :\mathrm{Gal}(L∕K)\to \mathbb{Q}∕\mathbb{Z}$ be a homomorphism, which we may view as a homomorphism on $\mathrm{Gal}(M∕K)$ (and its abelianization) as well. It suffices to show that for any such $\chi$, we have $\chi ({\rho }_{M∕K}(a))=\chi ({\rho }_{L∕K}(a))$. For this, it sufficient by Proposition 8.1.10 to show that

where abusing notation. Since ${\mathrm{inv}}_{L∕K}={\mathrm{inv}}_{M∕K}\circ \mathrm{Inf}$, where

is inflation, this is clear from the compatibility of cup products with inflation. □

Note that $G$ acts on the set of left $H$-cosets by left multiplication. If we replace a single $x_i$ by $x_{i}h$ for some $h\in H$, then $g{x}_{i}h=x_j{t}_i(g)h$, so $t_i(g)$ is replaced by $h^{-1}{t}_i(g)h$ if $g(i)=i$ and $t_i(g)h$ if $g(i)\ne i$. In the latter case, $g{x}_{g^{-1}(i)}=x_i{t}_{g^{-1}(i)}(g)$, and then $t_{g^{-1}(i)}(g)$ is replaced by $h^{-1}{t}_{g^{-1}(i)}(g)$. For all other $j$, the quantity $t_j(g)$ is unchanged. As the product $t_1(g)t_2(g)\cdots t_n(g)$ is taken in $H^{\mathrm{ab}}$, it is unchanged since the overall effect of the change is multiplication by $h\cdot h^{-1}$. Similarly, if the ordering of the $x_i$ is changed, then the order of the $t_i(g)$ is likewise changed, but this does not matter in $H^{\mathrm{ab}}$. Thus $V(g)$ is well-defined.

To see that $V$ is a homomorphism, we merely note that

and again that multiplication is commutative in $H^{\mathrm{ab}}$. □

Recall that the vertical isomorphism is given by the series of canonical isomorphisms

As restriction is a $\delta$-functor, we have a commutative diagram

That is, our restriction map factors through $I_G∕I_G{I}_H$. By the definition of restriction on $0$th cohomology groups, we have

where $\{x_1,x_2,\dots ,x_n\}$ is a set of left $H$-coset representatives in $G$. For $t_i(g)$ as in the definition of the transfer, this equals

Thus, the restriction map and the transfer agree.

The second statement follows easily from the commutative diagram

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Fix a Galois extension $M$ of $K$ containing $L$. The norm $N_{L∕K}:A_L\to A_K$ induces corestriction from $\mathrm{Gal}(M∕L)$ to $\mathrm{Gal}(M∕K)$ on zeroth Tate cohomology groups, and

coincides with corestriction on Tate cohomology groups of $\mathbb{Z}$ in degree $-2$ by Lemma 8.1.17. Part a then follows from Proposition 1.9.10c, which tells us that

for $\beta \in {\widehat{H}}^0(\mathrm{Gal}(M∕L),A_M)$, since $\mathrm{Res}({\alpha }_{M∕K})={\alpha }_{M∕L}$.

The injection $A_K\to A_L$ induces restriction on ${\widehat{H}}^0(\mathrm{Gal}(M∕K),A_L)$, and the transfer map $V_{L∕K}$ coincides with restriction on ${\widehat{H}}^{-2}(\mathrm{Gal}(M∕K),\mathbb{Z})$, again by Lemma 8.1.17. Part b then follows from Proposition 1.9.10a, which tells us that

for $\beta \in {\widehat{H}}^0(\mathrm{Gal}(M∕K),A_M)$.

We leave part c as an exercise for the reader. □

This is the following straightforward calculation using Proposition 1.3.6:

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It suffices to show that any $a\in {𝒩}_E$ is contained in ${𝒩}_L$. Let $M$ be a finite Galois extension of $K$ containing $L$. Set $G=\mathrm{Gal}(M∕K)$ and $H=\mathrm{Gal}(M∕L)$. By definition, we have ${\rho }_{E∕K}(a)=1$, so $\tau ={\rho }_{M∕K}(a)\in G^{\mathrm{ab}}$ maps trivially to $\mathrm{Gal}(E∕K)=G∕([G,G]H)$ by Corollary 8.1.11. In other words $\tau$ is the image of some element $\sigma$ in $H^{\mathrm{ab}}$. By the surjectivity of the reciprocity map ${\rho }_{M∕L}:A_L\to H^{\mathrm{ab}}$ and there exists $b\in A_L$ such that ${\rho }_{M∕L}(b)=\sigma$. By Proposition 8.1.18a, we then have

so $a-N_{L∕K}(b)=N_{M∕K}(c)$ for some $c\in A_M$. It follows that $a=N_{L∕K}(b-N_{M∕L}(c))$, as desired. □

a.

Let $a\in A_K$. We have $a\in {𝒩}_{\mathit{𝐿𝑀}}$ if and only if ${\rho }_K(a){|}_{\mathit{𝐿𝑀}}$ is trivial, which occurs if and only if ${\rho }_K(a){|}_L$ and ${\rho }_K(a){|}_M$ are both trivial, and so if and only if $a\in {𝒩}_L$ and $a\in {𝒩}_M$.

c.

By Lemma 8.2.3, if $L\subseteq M$, then ${𝒩}_M\subseteq {𝒩}_L$. On the other hand, if ${𝒩}_M\subseteq {𝒩}_L$, then by part (a) we have

$${𝒩}_{\mathit{𝐿𝑀}}={𝒩}_L\cap {𝒩}_M={𝒩}_M.$$

By Corollary 8.2.5, this implies that $[\mathit{𝐿𝑀}:K]=[M:K]$, so $\mathit{𝐿𝑀}=M$, and therefore $L\subseteq M$.

d.

Let $E=L^{{\rho }_{L∕K}(𝒜)}$. Then ${\rho }_{L∕K}$ induces an injective homomorphism

$$𝒜∕{𝒩}_L\to \mathrm{Gal}(L∕E)$$

that must be an isomorphism as ${\rho }_{L∕K}(𝒜)$ does not fix any larger subfield of $L$ than $E$. The kernel of ${\rho }_{E∕K}$, being that ${\rho }_{E∕K}$ is the composite of ${\rho }_{L∕K}$ with restriction, is then ${\rho }_{L∕K}^{-1}(\mathrm{Gal}(L∕E))=𝒜$. But we know from the reciprocity law that the kernel is ${𝒩}_E$.

b.

By part (d), the group $𝒜={𝒩}_L+{𝒩}_M$ is equal to ${𝒩}_E$ for some finite abelian $E∕K$ which by part (c) is contained in both $L$ and $M$. On the other hand, we clearly have that $𝒜$ is contained in ${𝒩}_{L\cap M}$, so again by (c), the field $E$ contains $L\cap M$ as well.

We have $a\in \ker {\rho }_L$ if and only if ${\rho }_{M∕L}(a)={\rho }_L(a){|}_M=1$ for all finite abelian $M$ over $L$, and $\ker {\rho }_{M∕L}=N_{M∕L}{A}_M$. □

Let $E_L$ be the set of finite abelian (or separable) extensions of $L$ in $K^{\mathrm{sep}}$ (and likewise with $K$). We have

the last step noting Lemma 8.2.3, and thus $N_{L∕K}{D}_L\subseteq D_K$.

Fix $a\in D_K$. For any finite separable $M∕L$, the set

is compact since $N_{M∕K}{A}_M$ is closed and $N_{L∕K}^{-1}(a)$ is compact by Definition 8.2.10(i). Since $a\in D_K$, there exists $b\in A_M$ with $N_{M∕K}(b)=a$, and then $N_{M∕L}(b)\in Y_M$. Thus $Y_M$ is nonempty. Note that if $M^{\prime }∕M$ is finite separable, then $Y_{M^{\prime }}\subseteq Y_M$, and so the $Y_M$ form a collection of subsets of the compact space $Y_L$ that satisfy the finite intersection property. We therefore have that the intersection of all $Y_M$ is nonempty, so contains an element $b$. Then $N_{L∕K}(b)=a$ , and $b\in D_L$ as it lies in every $N_{M∕L}{A}_M$. Thus, we have $N_{L∕K}(D_L)=D_K$. □

To see that $D_K$ is divisible, it suffices to show that $D_K=p{D}_K$ for all primes $p$. Fix $a\in D_K$. Let $L$ be a finite separable extension of $K$ containing $K_p$. Set

where ${\varphi }_p:A_L\to A_L$ is the multiplication-by-$p$ map. By (i) of Definition 8.2.10, the set ${𝒩}_L$ is closed, and by (ii), the set ${\varphi }_p^{-1}(a)$ is compact, so $X_L$ is compact. By Proposition 8.2.12, there exists $x\in D_L$ with $a=N_{L∕K}x$. By Definition 8.2.10(ii), there exists $y\in D_L$ with $\mathit{𝑝𝑦}=x$, and we set $b=N_{L∕K}y$. Then $b\in X_L$, so $X_L$ is nonempty. Again, if $M∕L$ is finite separable, then $X_M\subseteq X_L$. It follows as in the proof of Proposition 8.2.12 that the intersection of all $X_L$ as $L$ varies is nonempty, so contains some element $c$. By definition, we have $c\in D_K$ and $\mathit{𝑝𝑐}=b$. Thus $D_K=p{D}_K$.

Since $D_K\subset A_K$, we have

Suppose, on the other hand, that $a\in {\bigcap }_{n=1}^{\infty }n{A}_K$. Let $b\in A_K$ with $\mathit{𝑛𝑏}=a$ for $n\ge 1$. Take any finite separable extension $L∕K$, and set $n=[L:K]$ Then $N_{L∕K}(b)=\mathit{𝑛𝑏}=a$, so $a\in N_{L∕K}{A}_L$. Since $L$ was arbitrary, we have $a\in D_L$. □

If a subgroup is a norm group, then it is of finite index by the reciprocity law and is closed in $A_K$ by Definition 8.2.10(i).

Conversely, if $𝒩$ is a closed subgroup of $A_K$ of finite index, set $n=[A_K:𝒩]$. Then $n{A}_K\subseteq 𝒩$, so $D_K\subseteq 𝒩$ by Proposition 8.2.13. Let $U_K$ be as in Definition 8.2.10(iii). Then for any norm subgroup $M$ of $A_K$, the sets $M\cap U_K$ are compact. The intersection of the $M\cap U_K$ over all norm groups $M$ is equal to $D_K\cap U_K$, so is contained in the open set $𝒩$. Since the intersection of all $M\cap U_K$ with $A_K-𝒩$ would be nonempty if each intersection were, there exists a norm group $M$ with $M\cap U_K\subseteq 𝒩$.

Let $a\in M\cap (U_K+M\cap 𝒩)$, and write $a=u+v$ with $u\in U_K$ and $v\in M\cap 𝒩$. Then $u=a-v\in M$, so $u\in U_K\cap M$, so $u\in 𝒩$, and thus $a\in 𝒩$. Thus, we have

Since $M\cap 𝒩$ is closed of finite index, so is $U_K+M\cap 𝒩$, and thus it is a norm group by Definition 8.2.10(iii). Then $M\cap (U_K+M\cap 𝒩)$ is a norm group by Proposition 9.2.8a, and $𝒩$ is a norm group by Proposition 9.2.8d. □

To see that ${\rho }_K$ is continuous, we need only note that the inverse image ${𝒩}_L$ of the open neighborhood $\mathrm{Gal}(K^{\mathrm{ab}}∕L)$ of $1$, for $L∕K$ finite abelian, is open by property (i) of Definition 8.2.10.

The closure of the image of ${\rho }_K$ is obviously a closed subgroup of $G_K^{\mathrm{ab}}\cong {\underleftarrow{\lim }}_L\mathrm{Gal}(L∕K)$, so equal to $\mathrm{Gal}(K^{\mathrm{ab}}∕M)$ for some $M∕K$ abelian. As it also surjects onto each of the finite quotients $\mathrm{Gal}(L∕K)$ since each ${\rho }_{L∕K}$ is surjective, we must have $M=K$. Thus, ${\rho }_K$ has dense image. □

For positive $n$, we have

and the latter group is zero since the image of any continuous homomorphism of a compact Hausdorff group with values in a discrete group is finite, and the only finite subgroup of $\mathbb{Z}$ is trivial.

Consider the exact sequence

For $i\ge 1$, we have

where the direct limit is taken over multiples of $n$, since $H^i(\mathrm{Gal}({𝔽}_{q^m}∕{𝔽}_{q^n}),\mathbb{Q})$ has exponent dividing $\frac{m}{n}$ but is also a $\mathbb{Q}$-vector space. Thus, we have isomorphisms

the latter map being given by evaluation at $1$, and through these we obtain a map

For $n\mid m$, we have a commutative diagram

To see the commutativity of the lower square, note that the homomorphism that sends the Frobenius element ${\phi }_n$ in $G_{{𝔽}_{q^n}}$ to $1$ restricts to a homomorphism sending the Frobenius ${\phi }_m={\phi }_n^{m∕n}$ to $\frac{m}{n}$, and the invariant map for $n$ (resp., $m$) sends the homomorphism that takes ${\phi }_n$ (resp., ${\phi }_m$) to $1$ to the element $1\in \mathbb{Q}∕\mathbb{Z}$. Thus, $(\mathbb{Z},\mathrm{inv})$ is a class formation for ${𝔽}_q$. □

Let $n\ge 1$, and consider the homomorphism $\chi :\mathrm{Gal}({𝔽}_{q^n}∕{𝔽}_q)\to \mathbb{Q}∕\mathbb{Z}$ that takes (the restriction of) $\phi$ to $\frac{1}{n}$. Let $\delta$ denote the connecting homomorphism arising from the sequence $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}∕\mathbb{Z}\to 0$. By Proposition 8.1.10, as required, we have

the last equality following from the construction of the invariant map in (8.3.1). □