Kummer Theory - Elliptic Curves - Cambridge III notes

11 Kummer Theory

Let $K$ be a field and $char K ∤ n$ . Assume $μ_{n} \subset K$ .

Lemma 11.1. Assuming that:

$Δ \subset K^{*} ∕ {(K^{*})}^{n}$ a finite subgroup
$L = K (\sqrt[n]{Δ})$

Then

L ∕ K

is Galois and

Gal (L ∕ K) ≅ Hom (Δ, μ_{n})

Proof. $μ_{n} \subset K$ gives $L ∕ K$ normal, and $char K ∤ n$ gives that $L ∕ K$ is separable. So $L ∕ K$ is Galois.

Define the Kummer pairing

\begin{array}{l} ⟨, ⟩ : Gal (L ∕ K) \times Δ & \to μ_{n} \\ (σ, x) & \mapsto \frac{σ (\sqrt[n]{x})}{\sqrt[n]{x}} \end{array}

Well-defined: Suppose $α, β \in L$ with $α^{n} = β^{n} = x$ . Then ${(\frac{α}{β})}^{n} = 1$ , so $\frac{α}{β} \in μ_{n} \subset K$ . Then $σ (\frac{α}{β}) = \frac{α}{β}$ for all $σ \in Gal (L ∕ K)$ , hence $\frac{σ (α)}{α} = \frac{σ (β)}{β}$ for all $σ \in Gal (L ∕ K)$ .

Bilinear:

\begin{array}{l} ⟨ σ τ, x ⟩ & = \frac{σ (τ \sqrt[n]{x})}{τ \sqrt[n]{x}} \frac{τ \sqrt[n]{x}}{\sqrt[n]{x}} & = ⟨ σ, x ⟩ ⟨ τ, x ⟩ \\ ⟨ σ, x y ⟩ & = \frac{σ \sqrt[n]{x y}}{\sqrt[n]{x y}} \\ = \frac{σ \sqrt[n]{x}}{\sqrt[n]{x}} \frac{σ \sqrt[n]{y}}{\sqrt[n]{y}} \\ = ⟨ σ, x ⟩ ⟨ σ, y ⟩ \end{array}

Non-degenerate: Let $σ \in Gal (L ∕ K)$ . If $⟨ σ, x ⟩ = 1$ for all $x \in Δ$ , then

σ \sqrt[n]{x} = \sqrt[n]{x} \forall x \in Δ

adn hence $σ$ fixes $L$ pointwise, i.e. $σ = 1$ .

Let $x {(K^{*})}^{n} \in Δ$ . If $⟨ σ, x ⟩ = 1$ for all $σ \in Gal (L ∕ K)$ , then

σ \sqrt[n]{x} = \sqrt[n]{x} \forall σ \in Gal (L ∕ K) .

So $\sqrt[n]{x} \in K$ , so $x \in {(K^{*})}^{n}$ , i.e. $x {(K^{*})}^{n} \in Δ$ is the identity element.

We get injective group homomorphisms

(i) $Gal (L ∕ K) ↪ Hom (Δ, μ_{n})$ .
(ii) $Δ ↪ Hom (Gal (L ∕ K), μ_{n})$ .

(i) implies $Gal (L ∕ K)$ is abelian and of exponent dividing $n$ .

Fact: If $G$ is a finite abelian group of exponent dividing $n$ then $Hom (G, μ_{n}) ≅ G$ (non canonically).

| Gal (L ∕ K) | \overset{(i)}{\leq} | Δ | \overset{(ii)}{\leq} | Gal (L ∕ K) | .

Therefore (i) and (ii) are isomorphisms. □

Example. $Gal (ℚ (\sqrt{2}, \sqrt{3}, \sqrt{5}) ∕ ℚ) ≅ {(ℤ ∕ 2 ℤ)}^{3}$ .

Definition (Abelian extension). We say $L ∕ K$ is abelian if it is Galois, and has abelian Galois group.

Similarly for other group terminology (e.g. we can say that $L ∕ K$ has exponent dividing $n$ to mean that it is Galois, with Galois group having exponent dividing $n$ ).

Theorem 11.2. There is a bijection

\begin{array}{l} {\begin{array}{c} finite subgroups \\ Δ \subset K^{*} ∕ {(K^{*})}^{n} \end{array}} & \leftrightarrow {\begin{array}{c} finite abelian extension L ∕ K \\ of exponent dividing n \end{array}} \\ Δ & \mapsto K (\sqrt[n]{Δ}) \\ \frac{{(L^{*})}^{n} \cap K^{*}}{{(K^{*})}^{n}} & \leftarrow ↤ L \end{array}

Proof.

(i) Let $Δ \subset K^{*} ∕ {(K^{*})}^{n}$ be a finite subgroup. Let $L = K (\sqrt[n]{Δ})$ and $Δ^{'} = \frac{{(L^{*})}^{n} \cap K^{*}}{{(K^{*})}^{n}}$ .
We must show $Δ = Δ^{'}$ .

Clearly $Δ \subset Δ^{'}$ . So
$L = K (\sqrt[n]{Δ}) \subset K (\sqrt[n]{Δ^{'}}) \subset L .$

So $K (\sqrt[n]{Δ}) = K (\sqrt[n]{Δ^{'}})$ . So Lemma 11.1 gives $| Δ | = | Δ^{'} |$ .

Since $Δ \subset Δ^{'}$ , it follows that $Δ = Δ^{'}$ .
(ii) Let $L ∕ K$ be a finite abelian extension of exponent dividing $n$ . Let $Δ = \frac{{(L^{*})}^{n} \cap K^{*}}{{(K^{*})}^{n}}$ . Then $K (\sqrt[n]{Δ}) \subset L$ and we aim to prove this inclusion is an equality.
Let $G = Gal (L ∕ K)$ . Thu Kummer pairing gives an injection $Δ ↪ Hom (G, μ_{n})$ .

Claim: This map is surjective.

Granted the claim,
$\begin{array}{l} [K (\sqrt[n]{Δ}) : K] & \overset{Lemma 11.1}{=} | Δ | \\ \overset{by claim}{=} | G | \\ = [L : K] \end{array}$
Since $K (\sqrt[n]{Δ}) \subset L$ it follows that $L = K (\sqrt[n]{Δ})$ .

Proof of claim: Let $χ : G \to μ_{n}$ be a group homomorphism. Distinct automorphisms are linearly independent. So there exists $a \in L$ such that
$\underset{= y}{\underset{⏟}{\sum_{τ \in G} χ {(τ)}^{- 1} τ (a)}} \neq 0 .$

Let $σ \in G$ . Then
$\begin{array}{l} σ (y) & = \sum_{τ \in G} χ {(τ)}^{- 1} σ τ (a) \\ = \sum_{τ \in G} χ {(σ^{- 1} τ)}^{- 1} τ (a) \\ = χ (σ) \sum_{τ \in G} χ {(τ)}^{- 1} τ (a) \\ = χ (σ) \cdot y & (*) \end{array}$
Therefore $σ (y^{n}) = y^{n}$ for all $σ \in G$ . Hence $y^{n} \in K$ .

Let $x = y^{n}$ . Then $x \in K^{*} \cap {(L^{*})}^{n}$ . Then $x {(K^{*})}^{n} \in Δ$ .

Also, by ( $*$ ), $χ : σ \mapsto \frac{σ (y)}{y} = \frac{σ \sqrt[n]{x}}{\sqrt[n]{x}}$ . So the map $Δ ↪ Hom (G, μ_{n})$ sends $x \mapsto χ$ . This proves the claim. □

Proposition 11.3. Assuming: - $K$ a number field - $μ_{n} \subset K$ - $S$ a finite set of promes of $K$ Then there are only finitely many extensions $L ∕ K$ such that

(i) $L ∕ K$ is a finite abelian extension of exponent dividing $n$
(ii) $L ∕ K$ is unramified at all $𝔭 \notin S$

Proof. Theorem 11.2 gives $L = K (\sqrt[n]{Δ})$ for some $Δ \subset K^{*} ∕ {(K^{*})}^{n}$ a finite subgroup.

Let $𝔭$ be a prime of $K$ .

𝔭 O_{L} = P_{1}^{e_{1}} \dots P_{r}^{e_{r}}

for $P_{1}, \dots, P_{r}$ some distinct primes in $O_{L}$ .

If $x \in K^{*}$ represents an element of $Δ$ then

n v_{P_{i}} (\sqrt[n]{x}) = v_{P_{i}} (x) = e_{i} v_{𝔭} (x) .

If $𝔭 \notin S$ then all $e_{i} = 1$ , so $v_{𝔭} (x) \equiv 0 (m o d n)$ . Therefore $Δ \subset K (S, n)$ where

K (S, n) = {x \in K^{*} ∕ {(K^{*})}^{n} | v_{𝔭} (x) \equiv 0 (m o d n) \forall 𝔭 \notin S} .

The proof is completed by the next lemma. □

Lemma 11.4. $K (S, n)$ is finite.

Proof. The map

\begin{array}{l} K (S, n) & \to {(ℤ ∕ n ℤ)}^{| S |} \\ x & \mapsto {(v_{𝔭} (x) (m o d n))}_{𝔭 \in S} \end{array}

is a group homomorphism with kernel $K (\emptyset, n)$ .

Since $| S | < \infty$ , it suffices to prove the lemma with $S = \emptyset$ .

If $x \in K^{*}$ represents an element of $K (\emptyset, n)$ then $(x) = 𝔞^{n}$ for some fractional ideal $𝔞$ . There is a short exact sequence

\begin{array}{l} 0 \to O_{K}^{*} ∕ {(O_{K}^{*})}^{n} \to K (\emptyset, n) & \to {Cl}_{K} \to 0 \\ x {(K^{*})}^{n} & \mapsto [𝔞] \end{array}

$| {Cl}_{K} | < \infty$ and $O_{K}^{*}$ being a finitely generated abelian group (Dirichlet’s unit theorem) gives us that $K (ϕ, n)$ is finite. □