Different and discriminant - Local Fields

12 Different and discriminant

Notation. Let $x_{1}, \dots, x_{n} \in L$ . Set

\begin{array}{l} Δ (x_{1}, \dots, x_{n}) & = \det ({Tr}_{L ∕ K} (x_{i} x_{j})) \in K \\ = \det (\sum_{k = 1}^{n} σ_{k} (x_{i}) σ_{k} (x_{j})) \\ = \det (B B^{⊤}) \end{array}

where $σ_{k} : L \to \bar{K}$ are distinct embeddings and $B = (σ_{i} (x_{j}))$ .

Note:

If $y_{i} = \sum_{j = 1}^{n} a_{i j} x_{j}$ , $a_{i j} \in K$ , then
$Δ (y_{1}, \dots, y_{n}) = \det {(A)}^{2} Δ (x_{1}, \dots, x_{n})$

where $A = (a_{i j})$ .
If $x_{1}, \dots, x_{n} \in O_{L}$ , then $Δ (x_{1}, \dots, x_{n}) \in O_{K}$ .

Lemma 12.1. Assuming that:

$k$ a perfect field
$R$ a $k$ -algebra which is finite dimensional as a $k$ -vector space

Then the Trace form

\begin{array}{l} (∙, ∙) : R \times R & \to R \\ (x, y) & \mapsto {Tr}_{R ∕ k} (x y) (: = {Tr}_{k} (mult (x y))) \end{array}

is non-degenerate if and only if $R = k_{1} \times \dots \times k_{r}$ where $k_{i} ∕ k$ is a finite separable extension of $k$ .

Proof. Example Sheet 3. □

Theorem 12.2. Assuming that:

$0 \neq 𝔭 \subseteq O_{K}$ prime ideal

Then

(i) If $𝔭$ ramifies in $L$ , then for every $x_{1}, \dots, x_{n} \in O_{L}$ , we have $Δ (x_{1}, \dots, x_{n}) \equiv 0 mod 𝔭$ .
(ii) If $𝔭$ is unramified in $L$ , then there exists $x_{1}, \dots, x_{n}$ such that $𝔭 ∤ (Δ (x_{1}, \dots, x_{n}))$ .

Proof.

(i) Let $𝔭 O_{L} = P_{1}^{e_{1}} \dots P_{r}^{e_{r}}$ , $0 \neq P_{i} \subseteq O_{L}$ distinct prime ideals, $e_{i} > 0$ . Define
$R : = O_{L} ∕ 𝔭 O_{L} \overset{CRT}{=} \prod_{i = 1}^{r} O_{L} ∕ P_{i}^{e_{i}} .$

If $𝔭$ ramifies, then $O_{L} ∕ 𝔭 O_{L}$ has nilpotents. Hence
$Δ ({\bar{x}}_{1}, \dots, {\bar{x}}_{n}) = 0 \forall {\bar{x}}_{i} \in O_{L} ∕ 𝔭 O_{L} .$

Then using the fact that
commutes, we get that
$Δ (x_{1}, \dots, x_{n}) \equiv 0 mod 𝔭 \forall x_{i} \in O_{L} ∕ 𝔭 O_{L} .$
(ii) $𝔭$ unramified implies $R = O_{L} ∕ 𝔭 O_{L}$ is a product of finite extensions of $k$ . By Lemma 12.1, we get that the Trace form is non-degenerate, hence for ${\bar{x}}_{1}, \dots, {\bar{x}}_{n}$ a basis of $O_{L} ∕ 𝔭 O_{L}$ as a $k$ vector space, we have $Δ ({\bar{x}}_{1}, \dots, {\bar{x}}_{n}) \neq 0$ . So thee exist $x_{1}, \dots, x_{n} \in O_{L}$ such that
$Δ (x_{1}, \dots, x_{n}) ⁄ \equiv 0 mod 𝔭 . □$

Definition 12.3 (Discriminant). The discriminant is the ideal $d_{L ∕ K} \subseteq O_{K}$ generated by $Δ (x_{1}, \dots, x_{n})$ for all choices of $x_{1}, \dots, x_{n} \in O_{L}$ .

Corollary 12.4. $𝔭$ ramifies $L$ if and only if $𝔭 | d_{L ∕ K}$ . In particular, only finitely many primes ramify in $L$ .

Definition 12.5 (Inverse different). The inverse different is

D_{L ∕ K}^{- 1} = {y \in L : {Tr}_{L ∕ K} (x y) \in O_{K} \forall x \in O_{L}},

an $O_{L}$ submodule of $L$ .

Lemma 12.6. $D_{L ∕ K}^{- 1}$ is a fractional ideal in $L$ .

Proof. Let $x_{1}, \dots, x_{n} \in O_{L}$ a $K$ -basis for $L ∕ K$ . Set

d : = Δ (x_{1}, \dots, x_{n}) = \det ({Tr}_{L ∕ K} (x_{i} x_{j})),

which is non-zero since separable.

For $x \in D_{L ∕ K}^{- 1}$ write $x = \sum_{j = 1}^{r} λ_{j} x_{j}$ with $λ_{j} \in K$ . We show $λ_{j} \in \frac{1}{d} O_{K}$ . We have

{Tr}_{L ∕ K} (x x_{i}) = \sum_{j = 1}^{n} λ_{j} {Tr}_{L ∕ K} (x_{i} x_{j}) \in O_{K} .

Set $A_{i j} = {Tr}_{L ∕ K} (x_{i} x_{j})$ . Multiplying by $Adj (A) \in M_{n} (O_{K})$ , we get

d (\begin{matrix} λ_{1} \\ ⋮ \\ λ_{n} \end{matrix}) = Adj (A) (\begin{matrix} {Tr}_{L ∕ K} (x x_{1}) \\ ⋮ \\ {Tr}_{L ∕ K} (x x_{n}) \end{matrix})

Since $λ_{i} \in \frac{1}{d} O_{K}$ , we have $x \in \frac{1}{d} O_{L}$ . Thus $D_{L ∕ K}^{- 1} \leq \frac{1}{d O_{K}}$ , so $D_{L ∕ K}^{- 1}$ is a fractional ideal. □

The inverse $D_{L ∕ K}$ of $D_{L ∕ K}^{- 1}$ is the different ideal.

Remark. $D_{L ∕ K} \leq O_{L}$ since $O_{L} \subseteq D_{L ∕ K}^{- 1}$ .

Let $I_{L}$ , $I_{K}$ be the groups of fractional ideals.

Theorem 9.7 gives that

I_{L} ≅ \underset{\begin{array}{c} 0 \neq P \\ prime ideals in O_{L} \end{array}}{\otimes} ℤ, I_{K} ≅ \underset{\begin{array}{c} 0 \neq P \\ prime ideals in O_{K} \end{array}}{\otimes} .

Define $N_{L ∕ K} : I_{L} \to I_{K}$ induced by $P \mapsto 𝔭^{f}$ for $𝔭 = P \cap O_{K}$ and $f = f (P ∕ 𝔭)$ .

Fact:

(Use Corollary 10.10 and

v_{𝔭} (N_{L_{P} ∕ K_{𝔭}} (x)) = f_{P ∕ 𝔭} v_{(} x)

for

x \in msub \times

where

v_{𝔭}

and

v_{P}

are the normalised valuations for

L_{P}

K_{𝔭}

Theorem 12.7. $N_{L ∕ K} (D_{L ∕ K}) = d_{L ∕ K}$ .

Proof. First assume $O_{K}$ , $O_{L}$ are PIDs. Let $x_{1}, \dots, x_{n}$ be an $O_{K}$ -basis for $O_{L}$ and $y_{1}, \dots, y_{n}$ be the dual basis with respect to trace form. Then $y_{1}, \dots, y_{n}$ is a basis for $D_{L ∕ K}^{- 1}$ . Let $σ_{1}, \dots, σ_{n} : L \to \bar{K}$ be the distinct embeddings. Have

\sum_{i = 1}^{n} σ_{i} (x_{j}) σ_{i} (y_{k}) = Tr (x_{j} y_{k}) = δ_{j k} .

But

Δ (x_{1}, \dots, x_{n}) = \det {(σ_{i} (x_{j}))}^{2} .

Thus

Δ (x_{1}, \dots, x_{n}) Δ (y_{1}, \dots, y_{n}) = 1 .

Write $D_{L ∕ K}^{- 1} = β O_{L}$ since $β \in L$ . Then

\begin{array}{l} d_{L ∕ K}^{- 1} & = (Δ {(x_{1}, \dots, x_{n})}^{- 1}) \\ = (Δ (y_{1}, \dots, y_{n})) \\ = (Δ (β x_{1}, \dots, β x_{n})) & change of basis matrix is invertible in O_{K} \\ = N_{L ∕ K} (β^{2}) Δ (x_{1}, \dots, x_{n}) & change of basis matrix is [mult (β)] \end{array}

Thus

d_{L ∕ K}^{- 1} = N_{L ∕ K} {(D_{L ∕ K}^{- 1})}^{2} d_{L ∕ K}

N_{L ∕ K} (D_{L ∕ K}) = d_{L ∕ K} .

In general, localise at $S = O_{K} ∖ 𝔭$ and use $S^{- 1} D_{L ∕ K} = D_{S^{- 1}} O_{L} ∕ S^{- 1} O_{K}$ . Then $S^{- 1} d_{L ∕ K} = d_{S^{- 1} O_{L} ∕ S^{- 1} O_{K}}$ . Details omitted. □

Theorem 12.8. Assuming that:

$O_{L} = O_{K} [α]$
$α$ has monic minimal polynomial $g (X) \in O_{K} [X]$

Then

D_{L ∕ K} = (g^{'} (α))

Proof. Let $α = α_{1}, \dots, α_{n}$ be the roots of $g$ . Write

\frac{g (X)}{X - α} = β_{n - 1} X^{n - 1} + \dots + β_{1} X + β_{0}

with $β_{i} \in O_{L}$ and $β_{n - 1} = 1$ . We claim

\sum_{i = 1}^{n} \frac{g (X)}{X - α_{i}} \frac{α_{i}^{n}}{g^{'} (α_{i})} = X^{r}

for $0 \leq r \leq n - 1$ .

Indeed the difference is a palynomial of degree $< n$ , which vanishes for $X = α_{1}, \dots, α_{n}$ . Equate coefficients of $X^{s}$ , which gives

{Tr}_{L ∕ K} (\frac{α^{r} β_{s}}{g^{'} (α)}) = δ_{r s} .

Since $1, α, \dots, α^{n - 1}$ is an $O_{K}$ basis for $O_{L}$ , $D_{L ∕ K}^{- 1}$ has an $O_{K}$ basis

\frac{β_{0}}{g^{'} (α)}, \frac{β_{1}}{g^{'} (α)}, \dots, \underset{\frac{1}{g^{'} (α)}}{\underset{⏟}{\frac{β_{n - 1}}{g^{'} (α)}}} .

Note all of these are $O_{L}$ multiples of the last term, since the $β_{i}$ are in $O_{L}$ . So $D_{L ∕ K}^{- 1} = \frac{1}{(g^{'} (α))}$ , hence $D_{L ∕ K} = (g^{'} (α))$ . □

$P$ a prime ideal of $O_{L}$ , $𝔭 = O_{K} \cap P$ . $D_{L_{P}} ∕ K_{𝔭}$ using $O_{K_{𝔭}}$ , $O_{L_{P}}$ . We identify $D_{L_{P}} ∕ K_{𝔭}$ with a power $P$ .

Theorem 12.9. $D_{L ∕ K} = \prod_{P} D_{L_{P}} ∕ K_{𝔭}$ (finite product, see later).

Proof. Let $x \in L$ , $𝔭 \subseteq O_{K}$ . Then

Tr L ∕ K (x) = \sum_{P | 𝔭} {Tr}_{L_{P} ∕ K_{𝔭}} (x) (∗)

(of Corollary 10.10).

Let $r (P) = v_{P} (D_{L ∕ K})$ , $s (P) = v_{P} (D_{L_{P}} ∕ K_{𝔭})$ .

$\subseteq$ (i.e. $r (P) \geq s (P)$ ). Let $x \in L$ with $v_{P} (x) \geq - s (P)$ for all $P$ . Then ${Tr}_{L_{P} ∕ K_{𝔭}} (x y) \in O_{K_{𝔭}}$ , for all $y \in L$ and for all $P$ . Using ( $*$ ) we get
${Tr}_{L ∕ K} (x y) \in O_{K_{𝔭}} \forall y \in O_{L}, \forall P .$

Thus
${Tr}_{L ∕ K} (x y) \in O_{K} \forall y \in O_{L}$

so $D_{L ∕ K} \subseteq \prod_{P} D_{L_{P}} ∕ K_{𝔭}$ .
$\supseteq$ (i.e. $r (P) \leq s (P)$ ). Fix $P$ and let $x \in P^{- r (P)} ∖ P^{- r (P) + 1}$ . Then $v_{P} (x) = - r (P)$ , $v_{P^{'}} (x) \geq 0$ for all $P^{'} \neq P$ . By ( $*$ ) , we have
${Tr}_{L_{P} ∕ K_{𝔭}} (x y) = {Tr}_{L ∕ K} (x y) - \sum_{\begin{array}{c} P^{'} | 𝔭 \\ P^{'} \neq P \end{array}} {Tr}_{L_{P} ∕ K_{𝔭}} (x y) \forall y \in O_{L}$

hence
${Tr}_{L_{P} ∕ K_{𝔭}} (x y) \in O_{K_{𝔭}} \forall y \in O_{L_{P}} .$

Hence $x \in msubsup ∕ K_{𝔭} - 1$ , i.e. $- v_{P} (x) = r (P) \leq s (P)$ . So $D_{L ∕ K} \supseteq \prod_{P} D_{L_{P}} ∕ K_{𝔭}$ . □

Corollary 12.10. $d_{L ∕ K} = \prod_{P | 𝔭} d_{L_{P} ∕ K_{𝔭}}$ .

Proof. Apply $N_{L ∕ K}$ to $D_{L ∕ K} = \prod_{P | 𝔭} D_{L_{P}} ∕ K_{𝔭}$ . □