Title: Daniel's Flashcards / Year 4 / Term 1 / Local Fields

html: Definition 1.1 (Absolute value). Let K be a field. An absolute value on K is a function | ∙ | : K → ℝ ≥ 0 such that
- (i) | x | = 0 if and only if x = 0.
- (ii) | x y | = | x | | y | for all x , y ∈ K.
- (iii) | x + y | ≤ | x | + | y | ∀ ⁡ x , y ∈ K (triangly inequality).
We say ( K , | ∙ | ) is a valued field.
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html: Definition 1.2 (p-adic absolute value). Let K = ℚ, and p be a prime. For 0 ≠ x ∈ ℚ, write x = p n a b, where ( a , p ) = 1, ( b , p ) = 1. The p-adic absolute value is defined to be
| x | p = { 0 x = 0 p − n x = p n a b
nothing: True

html: Definition 1.3 (Place). Let | ∙ |, | ∙ | ′ be absolute values on a field K. We say | ∙ | and | ∙ | ′ are equivalent if they induce the same topology. An equivalence class of absolute values is called a place.
nothing: True

html: Proposition 1.4. Assuming that:
- | ∙ |, | ∙ | ′ are (non-trivial) absolute values on K.
Then the following are equivalent:
- (i) | ∙ | and | ∙ | ′ are equivalent.
- (ii) | x | < 1 ⟺ | ∙ | ′ < 1 for all x ∈ K.
- (iii) There exists c ∈ ℝ > 0 such that | x | c = | ∙ | ′ for all x ∈ K.
proof: Proof.
- (i) ⟹ (ii) | x | < 1 ⟺ x n → 0 w.r.t | ∙ | ⟺ x n → 0 w.r.t | ∙ | ′ ⟺ | x | ′ < 1
- (ii) ⟹ (iii) Note: | x | c = | x | ′ ⟺ c log ⁡ | x | = log ⁡ | x | ′. Let a ∈ K × such that | a | > 1 (exists since | ∙ | is non-trivial). We need that ∀ ⁡ x ∈ K ×,
log ⁡ | x | log ⁡ | a | = log ⁡ | x | ′ log ⁡ | a | ′ .
Assume that
log ⁡ | x | log ⁡ | a | < log ⁡ | x | ′ log ⁡ | a | ′ .
Choose m , n ∈ ℤ (with n > 0) such that
log ⁡ | x | log ⁡ | a | < m n < log ⁡ | x | ′ log ⁡ | a | ′ .
Then we have
n log ⁡ | x | < m log ⁡ | a | n log ⁡ | x | ′ > m log ⁡ | a | ′
Hence | x n a m | < 1 and | x n a m | > 1, contradiction. Similarly for the case where
log ⁡ | x | log ⁡ | a | > log ⁡ | x | ′ log ⁡ | a | ′ .
- (iii) ⟹ (i) Clear.
□
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html: Definition 1.5 (Non-archimedean). An absolute value | ∙ | on K is said to be non-archimedean if it satisfies the ultrametric inequality:
| x + y | ≤ max ⁡ ( | x | , | y | ) .
If | ∙ | is not non-archimedean, then it is archimedean.
nothing: True

html: Lemma 1.6. Assuming that:
- ( K , | ∙ | ) is non-archimedean
- x , y ∈ K
- | x | < | y |
Then | x − y | = | y |.
proof: Proof.
| x − y | ≤ max ⁡ ( | x | , | y | ) = | y |
and
| y | ≤ max ⁡ ( | x | , | x − y | ) ≤ | x − y | .
□
cloze1: True

html: Proposition 1.7. Assuming that:
- ( K , | ∙ | ) is non-archimedean
- ( x n ) n = 1 ∞ a sequence in K
- | x n − x n + 1 | → 0
Then ( x n ) n = 1 ∞ is Cauchy. In particular, if K is in addition complete, then ( x n ) n = 1 ∞ converges.
proof: Proof. For 𝜀 > 0, choose N such that | x n − x n + 1 | < 𝜀 for n > N. Then N < n < m,
| x n − x m | = | ( x n − x n + 1 ) + ⋯ + ( x n − 1 ) − x m ) | < 𝜀 .
The “In particular” is clear. □
cloze1: True

html: Lemma 1.9. Assuming that:
- ( K , | ∙ | ) is a non-archimedean valued field
Then
- (i) If z ∈ B ( x , r ), then B ( z , r ) = B ( x , r ) – so open balls don’t have a centre.
- (ii) If z ∈ B ¯ ( x , r ) then B ¯ ( x , r ) = B ¯ ( z , r ).
- (iii) B ( x , r ) is closed.
- (iv) B ¯ ( x , r ) is open.
proof: Proof.
- (i) Let y ∈ B ( x , r ). Then | x − y | < r hence | z − y | = | ( z − x ) + ( x − y ) | ≤ max ⁡ ( | z − x | , | x − y | ) < r
Thus B ( x , r ) ⊆ B ( z , r ). ⊇ follows by symmetry.
- (ii) Same as (i).
- (iii) Let y ∉ B ( x , r ). If z ∈ B ( x , r ) ∩ B ( y , r ) then B ( x , r ) = B ( z , r ) = B ( y , r ) Hence y ∈ B ( x , r ). Hence B ( x , r ) ∩ B ( y , r ) = ∅.
- (iv) If z ∈ B ¯ ( x , r ), then B ( z , r ) ⊆ B ¯ ( z , r ) = B ¯ ( x , r ). □
cloze1: True

html: Definition 2.1 (Valuation). Let K be a field. A valuation on K is a function v : K × → ℝ such that
- (i) v ( x y ) = v ( x ) + v ( y )
- (ii) v ( x + y ) ≥ min ⁡ ( v ( x ) , v ( y ) )
nothing: True

html: Lemma 2.6. Assuming that:
- v is a valuation on K
Then the following are equivalent:
- (i) v is discrete
- (ii) O K is a PID
- (iii) O K is Noetherian
- (iv) m is principal
proof: Proof.
- (i) ⟹ (ii) O K is an integral domain since it is a subset of K, which is an integral domain.
Let I ⊆ O K be a non-zero ideal. Let x ∈ I such that v ( x ) = min ⁡ { v ( a ) | a ∈ I }, which exists since v is discrete. Then we claim
x O K = { a ∈ O K | v ( a ) ≥ v ( x ) }
is equal to I.
  - ⊆ (I is an ideal)
  - ⊇ Let y ∈ I. Then v ( x − 1 y ) ≥ 0. Hence y = x ( x − 1 y ) ∈ x O K.
- (ii) ⟹ (iii) Clear.
- (iii) ⟹ (iv) Write m = x 1 O K + ⋯ + x n O K. Without loss of generality,
v ( x 1 ) ≤ v ( x 2 ) ≤ ⋯ ≤ v ( x n ) .
Then x 2 , … , x n ∈ x 1 O K. Hence m = x 1 O K.
- (iv) ⟹ (i) Let m = π O K for some π ∈ O K and let c = v ( π ). Then if v ( x ) > 0, x ∈ m hence v ( x ) ≥ c. Thus v ( K × ) ∩ ( 0 , c ) = ∅. Since v ( K × ) is a subgroup of ( ℝ , + ), we deduce v ( K × ) = ℤ. □
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html: Definition 2.7 (Discrete valuation ring). A ring R is called a discrete valuation ring (DVR) if it is a PID with exactly one non-zero prime ideal (necessarily maximal).
nothing: True

html: Proposition 3.2 (Universal property of inverse limits). Assuming that:
- B is a set / group / ring
- ψ n are homomorphisms ψ n : B → A n such that
commutes for all n
Then there exists a unique homomorphism ψ : B → lim ⁡ n [ ] ← A n such that 𝜃 n ∘ ψ = ψ n.
proof: Proof. Define
ψ : B → ∏ n = 1 ∞ A n b ↦ ∏ n = 1 ∞ ψ n ( b )
Then ψ n = φ n ∘ ψ n + 1 implies that ψ ( b ) ∈ lim ⁡ n [ ] ← A n. The map is clearly unique (determined by ψ n = 𝜃 n ∘ ψ) and is a homomorphism of sets / groups / rings. □
nothing: True

html: Definition 3.3 (I-adic completion). Let I ⊆ R be an ideal (R a ring). The I-adic completion of R is the
R ^ : = lim ⁡ R [ ] ← ∕ I n
where R ∕ I n + 1 → R ∕ I n is the natural projection.
nothing: True

html: Proposition 3.4. Assuming that:
- K is complete with respect to | ∙ |
Then
- (i) Then O K ≅ lim ⁡ n [ ] ← O K ∕ π n O K (O K is π-adically complete)
- (ii) Every x ∈ O K can be written uniquely as x = ∑ ⁡ i = 0 n a i π i, a i ∈ A, where A ⊆ O K is a set of coset representatives for O K ∕ π O K.
proof: Proof.
- (i) K is complete and O K is closed, so O K is complete.
x ∈ ⋂ ⁡ n = 1 ∞ π n O K impies v ( x ) ≥ n v ( π ) for all n, and hence x = 0. Hence O K → lim ⁡ O [ ] ← K ∕ π n O K is injective.
Let ( x n ) n = 1 ∞ ∈ lim ⁡ O [ ] ← K ∕ π n O K and for each n, let y n ∈ O K be a lifting of x n ∈ O K ∕ π n O K. Then y n − y n + 1 ∈ π n O K so that v ( y n − y n + 1 ) ≥ n v ( π ).
Thus ( y n ) n = 1 ∞ is a Cauchy sequence in O K. Let y n → y ∈ O K. Then y maps to ( x n ) n = 1 ∞ in the lim ⁡ n [ ] ← O K ∕ π n O K. Thus O K → lim ⁡ n [ ] ← O K ∕ π n O K is surjective.
- (ii) Exercise on Example Sheet 1. □
cloze1: True

html: Theorem 4.1 (Hensel’s Lemma version 1). Assuming that:
- ( K , | ∙ | ) is a complete discretely valued field
- f ( X ) ∈ O K [ X ]
- assume ∃ ⁡ a ∈ O K such that | f ( a ) | < | f ′ ( a ) | 2
Then there exists a unique x ∈ O K such that f ( x ) = 0 and | x − a | < | f ′ ( a ) |.
proof: Proof. Let π ∈ O K be a uniformiser and let r = v ( f ′ ( a ) ), with v the normalised valuation (v ( π ) = 1). We construct a sequence ( x n ) n = 1 ∞ in O K such that:
- (i) f ( x n ) ≡ 0 ( m o d π n + 2 r )
- (ii) x n + 1 ≡ x n ( m o d π n + r )
Take x 1 = a: then f ( x 1 ) ≡ 0 ( m o d π 1 + 2 r ).
Now we suppose we have constructed x 1 , … , x n satisfying (i) and (ii). Define
x n + 1 = x n − f ( x n ) f ′ ( x n ) .
Since x n ≡ x 1 ( m o d π r + 1 ), we have
v ( f ′ ( x n ) ) = v ( f ′ ( x i ) ) = r ,
and hence
f ( x n ) f ′ ( x n ) ≡ 0 ( m o d π n + r )
by (i).
It follows that x n + 1 ≡ x n ( m o d π n + r ), so (ii) holds. Note that letting X , Y be indeterminates, we have
f ( X + Y ) = f 0 ( X ) + f 1 ( X ) Y + f 2 ( X ) Y 2 + ⋯ ,
where f i ( X ) ∈ O K [ X ] and f 0 ( X ) = f ( X ), f 1 ( X ) = f ′ ( X ). Thus
f ( x n + 1 ) = f ( x n ) + c f ′ ( x n ) + c 2 f 2 ( x N ) + c 2 f 2 ( x n ) + ⋯ ⏟ ∈ π n + 2 r + 1
where c = − f ( x n ) f ′ ( x n ).
Since c ≡ 0 ( m o d π n + r ) and v ( f i ( x n ) ) ≥ 0 we have
f ( x n + 1 ) ≡ f ( x n ) + f ′ ( x n ) c ≡ 0 ( m o d π n + 2 r + 1 ) ,
so (i) holds.
Property (ii) implies that ( x n ) n = 1 ∞ is Cauchy, so let x ∈ O K such that x n → x. Then f ( x ) = lim ⁡ n → ∞ f ( x n ) = 0 by (i).
Moreover, (ii) impies that
a = x 1 ≡ x n ( m o d π r + 1 ) ∀ ⁡ n ⟹ a ≡ x ( m o d π r + 1 ) ⟹ | x − a | < | f ′ ( a ) |
This proves existence.
Uniqueness: suppose x ′ also satisfies f ′ ( x ) = 0, | x ′ − a | < | f ′ ( a ) |. Set δ = x ′ − x ≠ 0. Then
| x ′ − a | < | f ′ ( a ) | | x − a ′ | < | f ′ ( a ) | ,
and the ultrametric inequality implies
| δ | = | x − x ′ | < | f ′ ( a ) | = | f ′ ( x ) | .
But
0 = f ( x ′ ) = f ( x + δ ) = f ( x ) ⏟ = 0 + f ′ ( x ) δ + ⋯ ⏟ | ∙ | ≤ | δ | 2 .
Hence | f ′ ( x ) δ | ≤ | δ | 2, so | f ′ ( x ) | < | δ |, a contradiction. □
nothing: True

html: Theorem 4.4 (Hensel’s Lemma version 2). Assuming that:
- ( K , | ∙ | ) is a complete discretely valued field
- f ( X ) ∈ O K [ X ]
- f ¯ ( X ) : = f ( X ) ( m o d m ) ∈ k [ X ] factorises as f ¯ ( X ) = g ¯ ( X ) h ¯ ( X ) in k [ X ]
- g ¯ ( X ) and h ¯ ( X ) coprime.
Then there is a factorisation
f ( X ) = g ( X ) h ( X )
in O K [ X ], with g ¯ ( X ) ≡ g ( X ) ( m o d m ), h ¯ ( X ) ≡ h ( X ) ( m o d m ) and deg ⁡ g ¯ = deg ⁡ g.
proof: Proof. Example Sheet 1. □
nothing: True

html: Definition 5.1 (Perfect). A ring R of characteristic p > 0 (prime) is a perfect ring if the Frobenius x ↦ x p is a bijection. A field of characteristic p is a perfect field if it is perfect as a ring.
nothing: True

html: Theorem 5.2. Assuming that:
- ( K , | ∙ | ) is a complete discretely valued field
- such that k : = O K ∕ m is a perfect field of characterist p
Then there exists a unique map [ ∙ ] : k → O K such that
- (i) a ≡ [ a ] mod m for all a ∈ k
- (ii) [ a b ] = [ a ] [ b ] for all a , b ∈ k
Moreover if characteristic ⁡ O K = p, then [ ∙ ] is a ring homomorphism.
proof: Proof. Let x = y + u π k with u ∈ O K. Then
x p = ∑ i = 0 p p i y p − i ( u π k ) i = y p + ∑ i = 1 p p i y p − 1 ( u π k ) i
Since O K ∕ π O K has characteristic p, we have p ∈ π O K. Thus
p i ( u π k ) i y p − i ∈ π k + 1 O K ∀ ⁡ i ≥ 1 ,
hence x p ≡ y p mod π k + 1. □
cloze1: True

html: Lemma 5.4. Assuming that:
- ( K , | ∙ | ) is a complete discretely valued field
- such that k : = O K ∕ m is a perfect field of characterist p
- π ∈ O K a fixed uniformiser
- x , y ∈ O K such that x ≡ y mod π k (k ≥ 1)
Then x p ≡ y p mod π k + 1.
proof: Proof. Let x = y + u π k with u ∈ O K. Then
x p = ∑ i = 0 p p i y p − i ( u π k ) i = y p + ∑ i = 1 p p i y p − 1 ( u π k ) i
Since O K ∕ π O K has characteristic p, we have p ∈ π O K. Thus
p i ( u π k ) i y p − i ∈ π k + 1 O K ∀ ⁡ i ≥ 1 ,
hence x p ≡ y p mod π k + 1. □
cloze1: True

html: Lemma 5.5. Assuming that:
- ( K , | ∙ | ) complete discretely valued field
- k = O K ∕ m ⊆ 𝔽 p ¯
- a ∈ k ×
Then [ a ] is a root of unity.
proof: Proof.
a ∈ k × ⟹ a ∈ 𝔽 p n × for some n ⟹ [ a ] p n − 1 = [ a p n − 1 ] = [ 1 ] = 1 □
cloze1: True

html: Theorem 5.6. Assuming that:
- ( K , | ∙ | ) complete discretely valued field
- characteristic ⁡ ( K ) = p > 0
- k is perfect
Then K = k ( ( t ) ) (k = O K ∕ m).
proof: Proof. Since K = Frac ⁡ ( O K ), it suffices to show O K ≅ k [ [ t ]. Fix π ∈ O K a uniformiser, and let [ ∙ ] : k → O K be the Teichmüller map and define
φ : k [ [ t ] ] → O K φ ( ∑ i = 0 ∞ a i t i ) = ∑ i = 0 ∞ [ a i ] π i
Then φ is a ring homomorphism since [ ∙ ] is, and it is a bijection by Proposition 3.4(ii). □
cloze1: True

html: Theorem 6.1. Assuming that:
- ( K , | ∙ | ) is a complete discretely valued field
- L ∕ K a finite extension of degree n
Then
- (i) | ∙ | extends uniquely to an absolute value | ∙ | L on L defined by
| y | L = | N L ∕ L ( y ) | 1 n ∀ ⁡ y ∈ L .
- (ii) L is complete with respect to | ∙ | L.
proof: Proof. Let ( v i ) i = 1 ∞ be a Cauchy sequence in V, and let e 1 , … , e n be a basis for V.
Write v i = ∑ ⁡ j = 1 n x j i e j. Then ( x j i ) i = 1 ∞ is a Cauchy sequence in K. Let x j i → x j ∈ K, then v i → v : = ∑ ⁡ j = 1 n x j e j. □
cloze1: True

html: Definition 6.2 (Norm). Let ( K , | ∙ | ) be a non-archimedean valued field, V a vector space over K. A normon V is a function ∥ ∙ ∥ : V → ℝ ≥ 0 satisfying:
- (i) ∥ x ∥ = 0 ⟺ x = 0.
- (ii) ∥ λ x ∥ = ∥ λ ∥ ∥ x ∥ for all λ ∈ K, x ∈ V.
- (iii) ∥ x + y ∥ ≤ max ⁡ ( ∥ x ∥ , ∥ y ∥ ) for all x , y ∈ V.
nothing: True

html: Definition 6.3 (Equivalent norms). Two norms ∥ ∙ ∥ 1 and ∥ ∙ ∥ 2 on V are equivalent if there exists C , D ∈ ℝ > 0 such that
C ∥ x ∥ 1 ≤ ∥ x ∥ 2 ≤ D ∥ x ∥ 1 ∀ ⁡ x ∈ V .
nothing: True

html: Proposition 6.4. Assuming that:
- ( K , | ∙ | ) is a complete non-archimedean valued field
- V a finite dimensional vector space over K
Then V is complete with respect to ∥ ∙ ∥ sup ⁡.
proof: Proof. Let ( v i ) i = 1 ∞ be a Cauchy sequence in V, and let e 1 , … , e n be a basis for V.
Write v i = ∑ ⁡ j = 1 n x j i e j. Then ( x j i ) i = 1 ∞ is a Cauchy sequence in K. Let x j i → x j ∈ K, then v i → v : = ∑ ⁡ j = 1 n x j e j. □
cloze1: True

html: Theorem 6.5. Assuming that:
- ( K , | ∙ | ) is a complete non-archimedean valued field
- V a finite dimensional vector space over K
Then any two norms on K are equivalent. In particular, V is complete with respect to any norm (using Proposition 6.4).
proof: Proof. Since equivalence defines an equivalence relation on the set of norms, it suffices to show that any norm ∥ ∙ ∥ is equivalent to ∥ ∙ ∥ sup ⁡.
Let e 1 , … , e n be a basis for V, and set D : = max ⁡ i ∥ e i ∥ > 0. Then for x = ∑ ⁡ i = 1 n x i e i, we have
∥ x ∥ ≤ max ⁡ i ∥ x i e i ∥ = max ⁡ i | x i | ∥ e i ∥ ≤ D max ⁡ i | x i | = D ∥ x ∥ sup ⁡ .
To find C such that C ∥ ∙ ∥ sup ⁡ ≤ ∥ ∙ ∥, we induct on n = dim ⁡ V.
For n = 1: ∥ x ∥ = ∥ x 1 e 1 ∥ = | x 1 | ∥ e 1 ∥, so take C = ∥ e 1 ∥.
For n > 1: set V i = span ⁡ ⟨ e 1 , … ⁡ , e i − 1 , e i + 1 , … ⁡ , e n ⟩. By induction, V i is complete with respect to ∥ ∙ ∥, hence closed.
Then e i + V i is closed for all i, and hence
S : = ⋃ i = 1 n e i + V i
is a closed subset not containing 0. Thus there exists c > 0 such that B ( 0 , C ) ∩ S = ∅ where B ( 0 , C ) = { x ∈ V | ∥ x ∥ < C }.
Let 0 ≠ x = ∑ ⁡ i = 1 n x i e i and suppose | x j | = max ⁡ i | x i |. Then ∥ x ∥ sup ⁡ = | x j |, and 1 x j ∈ S. Thus ∥ x i x j ∥ ≥ C, and hence
∥ x ∥ ≥ C | x j | = C ∥ x ∥ sup ⁡ .
V is complete since it is complete with respect to ∥ ∙ ∥ sup ⁡ (see Proposition 6.4). □
cloze1: True

html: Definition 6.6 (Integral closure). Let R be a subring of S. We say s ∈ S is integral over R if there exists a monic polynomial f ( X ) ∈ R [ X ] such that f ( s ) = 0.
The integral closure R i n t ( S ) of R inside S is defined to be
R i n t ( S ) = { s ∈ S | s integral over R } .
We say R is integrally closed in S if R i n t ( S ) = R.
nothing: True

html: Lemma 6.8. Assuming that:
- ( K , | ∙ | ) is non-archimedean valued field
Then O K is integrally closed in K.
proof: Proof. Let x ∈ K be integral over O K. Without loss of generality, x ≠ 0. Let f ( X ) = X n + a n − 1 X n − 1 + ⋯ + a 0 ∈ O K [ X ] such that f ( x ) = 0. Then
x = − a n − 1 1 x − ⋯ − a 0 1 x n − 1 .
If | x | > 1, we have | − a n − 1 1 x − ⋯ − a 0 1 x n − 1 | < 1. Thus | x | ≤ 1 ⟹ x ∈ O K. □
cloze1: True

html: Proposition 6.12. Assuming that:
- L ∕ K finite extension of complete discretely valued fields.
- (i): O K is compact.
- (ii): The extension of residue fields k L ∕ k is finite and separable.
Then there exists α ∈ O L such that O L = O K [ α ].
proof: Proof. We’ll choose α ∈ O L such that:
- there exists β ∈ O L [ α ] a uniformiser for O L
- O K [ α ] → k L surjective
k L ∕ k separable tells us that there exists α ¯ ∈ k L such that k L = k ( α ¯ ).
Let α ∈ O L a lift of α ¯, and g ( X ) ∈ O K [ X ] a monic lift of the minimal polynomial of α ¯.
Fix π L ∈ O L a uniformiser. Then g ¯ ( X ) ∈ k [ X ] irreducible and separable, hence g ( α ) ≡ 0 mod π L and g ′ ( α ) ⁄ ≡ 0 mod π L.
If g ( α ) ≡ 0 mod π L 2, then
g ( α + π L ) ≡ g ( α ) + π L g ′ ( α ) mod π L 2 .
Thus
v L ( g ( α + π L ) ) = v L ( π L g ′ ( α ) ) = v L ( π L ) = 1 .
(v L normalised valuation on L).
Thus either v L ( g ( α ) ) = 1 or v L ( g ( α + π L ) ) = 1. Upon possibly replacing α by α + π L, we may assume v L ( g ( α ) ) = 1.
Set β = g ( α ) ∈ O K [ α ] a uniformiser. Then O K [ α ] ⊆ L is the image of a continuous map:
O K n → L ( x 0 , … , x n − 1 ) ↦ ∑ i = 0 n x i a i
where n = [ K ( α ) : K ]. Since O K is compact, O K [ α ] ⊆ L is compact, hence closed. Since k L = k ( α ¯ ), O K [ α ] contians a set of coset representatives for k L = O L ∕ β O L.
Let y ∈ O L. Then Proposition 3.4 gives us
y = ∑ i = 0 ∞ λ i β i , λ i ∈ O K [ α ]
Then y m = ∑ ⁡ i = 0 m λ i β i ∈ O K [ α ]. Hence y ∈ O K [ α ], since O k [ α ] is closed. □
cloze1: True

html: Definition 7.1 (Local field). Let ( K , | ∙ | ) be a valued field. Then K is a local field if it is complete and locally compact.
nothing: True

html: Proposition 7.2. Assuming that:
- ( K , | ∙ | ) is a non-archimedean complete valued field
Then the following are equivalent:
- (i) K is locally compact
- (ii) O K is compact
- (iii) v is discrete and k = O K ∕ m is finite.
proof: Proof.
- (i) ⟹ (ii) Let U ∋ 0 be a compact neighbourhood of 0 (0 ∈ U ⊆ Z with U open, Z compact). Then there exists x ∈ O K such that x O K ⊆ U. Since x O K is closed, x O K is compact. Hence O K is compact (x O K → ⋅ x − 1 O K is a homeomorphism).
- (ii) ⟹ (i) O K compact implies a + O K is compact for all a ∈ K. So K is locally compact.
- (ii) ⟹ (iii) Let x ∈ m, and A x ⊆ O K be a set of coset representatives for O K ∕ x O K. Then O K = ⋃ ⁡ y ∈ A y + x O K is a disjoint open cover. So A x is finite by compactness of O K. So O K ∕ x O K is finite, hence O K ∕ m O K is finite.o
Suppose v is not discrete. Then let x 1 , x 2 , … such that
v ( x 1 ) > v ( x 2 ) > ⋯ > 0 .
Then x O K ⊊ x 2 O K ⊊ x 3 O K ⊊ ⋯ ⊊ O K. But O K ∕ x O K is finite so can only have finitely many subgroups, contradiction.
- (iii) ⟹ (ii) Since O K is a metric space, it suffices to prove O K is sequentially compact.
Let ( x n ) n = 1 ∞ be a sequence in O K, and fix π ∈ O K a uniformiser. Since π i O K ∕ π i + 1 O K ≅ k, O K ∕ π i O K is finite for all i (O K ⊇ π O K ⊇ ⋯ ⊇ π i O K). Since O K ∕ π O K is finite, there exists a 1 ∈ O K ∕ π O K and a subsequence ( x n ) n = 1 ∞ such that x 1 n ≡ a mod π for all n.
Since O K ∕ π 2 O K is finite, there exists a 2 ∈ O K ∕ π 2 O K and a subsequence ( x 2 n ) n = 1 ∞ of ( x 1 n ) n = 1 ∞ such that x 2 n ≡ a 2 ( m o d π ) 2 O K. Continuing, this, we obtain sequences ( x i n ) n = 1 ∞ for i = 1 , 2 , … such that
  - (1) ( x ( i + 1 ) n ) n = 1 ∞ is a subsequence of ( x i n ) n = 1 ∞
  - (2) For any i, there exists a i ∈ O K ∕ π i O K such that x i n ≡ a i mod π i for all n.
Then necessarily a i ≡ a i + 1 mod π i for all i.
Now choose y i = x i i. This defines a subsequence of ( x n ) n = 1 ∞. Moreover, y i ≡ a i ≡ a i + 1 ≡ y i + 1 mod π i. Thus y i is Cauchy, hence converges by completeness. □
cloze1: True

html: Definition 7.3 (Profinite topology). Assume A n is finite. The profinite topology on A : = lim ⁡ n [ ] ← A n is the weakest topology on A such that 𝜃 n : A → A n is continuous for all n, where A n is equipped with the discrete topology.
nothing: True

html: Proposition 7.4. Assuming that:
- K is a non-archimedean local field
Then under the isomorphism O K ≅ lim ⁡ n [ ] ← O K ∕ π n O K (π ∈ O K a uniformiser), the topology on O K coincides with the profinite topology.
proof: Proof. One checks that the sets
B : = { a + π n O K | n ∈ ℕ ≥ 1 , a ∈ O K }
is a basis of open sets in both topologies.
For | ∙ |: clear.
For profinite topology: O K ∕ O K ∕ π n O K is continuous if and only if a + π n O K is open for all a ∈ O K. □
cloze1: True

html: Lemma 7.5. Assuming that:
- K is a non-archimedean local field
- L ∕ K a finite extension
Then L is a local field.
proof: Proof. Theorem 6.1 implies that L is complete and discretely valued. Suffices to show k L : = O L ∕ m L is finite. Let α 1 , … , α n be a basis for L as a K vector space.
∥ ∙ ∥ sup ⁡ (sup norm) equivalent to | ∙ | L implies that there exists r > 0 such that
O L ⊆ { x ∈ L : ∥ x ∥ sup ⁡ ≤ r } .
Take a ∈ K such that | a | ≥ r, then
O L ⊆ ⊕ i = 1 n a α i O K ≤ L .
Then O L is finitely generated as a module over O K, hence k L is finitely generated over k. □
cloze1: True

html: Definition 7.6 (Equal characteristic). A non-archimedean valued field ( K , | ∙ | ) has equal characteristic if characteristic ⁡ ( K ) = characteristic ⁡ ( k ). Otherwise it has mixed characteristic.
nothing: True

html: Theorem 7.7. Assuming that:
- K is a non-archimedean local field of equal characteristic p > 0
Then K ≅ 𝔽 p n ( ( t ) ) for some n ≥ 1.
proof: Proof. K complete discretely valued, characteristic ⁡ K > 0. Moreover, k ≅ 𝔽 p n is finite, hence perfect.
By Theorem 5.6, K ≅ 𝔽 p n ( ( t ) ). □
cloze1: True

html: Lemma 7.8. Assuming that:
- K a field
Then an absolute value | ∙ | is non-archimedean if and only if | n | is bounded for all n ∈ ℤ.
proof: Proof.
- ⇒ Since | − 1 | = 1, | − n | = | n |, it suffices to show that | n | bounded for n ≥ 1. Then note that
| n | = | 1 + 1 + ⋯ + 1 | ≤ 1 .
- ⇐ Suppose | n | ≤ B for all n ∈ ℤ. Let x , y ∈ K with | x | ≤ | y |. Then we have | x + y | m = | ∑ i = 0 m m i x i y m − i | ≤ ∑ i = 0 m | m i x i y m − i | ≤ | y | m B ( m + 1 )
Taking m-th roots gives
| x + y | ≤ | y | [ B ( m + 1 ) ] 1 m .
The right hand side tends to | y | as m → ∞, hence
| x + y | ≤ | y | = max ⁡ ( | x | , | y | ) □ .
cloze1: True

html: Theorem 7.9 (Ostrowski’s Theorem). Assuming that:
- | ∙ | is a non-trivial absolute value on ℚ
Then | ∙ | is equivalent to either the usual absolute value | ∙ | ∞ or the p-adic absolute value | ∙ | p for some prime p.
proof: Proof. Case: | ∙ | is archimedean. We fix b > 1 an integer such that | b | > 1 (exists by Lemma 7.8). Let a > 1 be an integer and write b n in base a:
b n = c m a m + c m − 1 a m − 1 + ⋯ + c 0
with 0 ≤ c i < a, c m ≠ 0. Let B = max ⁡ 0 ≤ c < a − 1 ( | c | ), and then we have
| b n | ≤ ( m + 1 ) B max ⁡ ( | a | m , 1 ) ⟹ | b | ≤ [ n ( log ⁡ ⁡ a b + 1 ) B ] 1 ∕ n ⏟ → 1 max ⁡ ( | a | log ⁡ ⁡ a b , 1 ) m ≤ log ⁡ a b n ⟹ | b | ≤ max ⁡ ( | a | log ⁡ ⁡ a b , 1 )
Then | a | > 1 and
| b | ≤ | a | log ⁡ a b . (∗)
Switching roles of a and b, we also obtain
| a | ≤ | b | log ⁡ b a . (∗∗)
Then (∗) and (∗ ∗) gives (using log ⁡ a b = log ⁡ b log ⁡ a):
log ⁡ | a | log ⁡ a = log ⁡ | b | log ⁡ b = λ ∈ ℝ > 0 .
Hence | a | = a λ for all a ∈ ℤ > 1, hence | x | = | x | ∞ λ for all x ∈ ℚ.
Case 2: | ∙ | is non-archimedean. As in Lemma 7.8, we have | n | ≤ 1 for all n ∈ ℤ. Since | ∙ | is non-trivial, there exists n ∈ ℤ > 1 such that | n | < 1. Write n = p 1 e 1 ⋯ p r e r decomposition into prime factors. Then | p | < 1, for some p ∈ { p 1 , … , p r }. Suppose | q | < 1 for some prime q, q ≠ p. Write 1 = r p + s q with r , s ∈ ℤ. Then
1 = | r p + s q | ≤ max ⁡ ( | r p | , | s q | ) < 1
contradiction. Thus | p | = α < 1 and | q | = 1 for all primes q ≠ p. Hence | ∙ | is equivalent to | ∙ | p. □
nothing: True

html: Theorem 7.10. Assuming that:
- ( K , | ∙ | ) is a non-archimedean local field of mixed characteristic
Then K is a finite extension of ℚ p.
proof: Proof. K mixed characteristic implies that characteristic ⁡ K = 0, hence ℚ ⊆ K. K non-archimedean implies that | ∙ | | ℚ = | ∙ | p for some prime p. Since K is complete, ℚ p ⊆ K. Suffices to show that O K is finite as a ℤ p-module.
Let π ∈ O K be a uniformiser, v a normalised valuation and set v ( p ) = e. Then O K ∕ p O K ≅ O K ∕ π e O K is finite since π i O K ∕ π i + 1 O K ≅ O K ∕ π O K is finite. Since 𝔽 p ≅ ℤ ∕ p ℤ ↪ O K ∕ p O K we have O K ∕ p O K a finite dimensional vector space over 𝔽 p.
Let x 1 , … , x n ∈ O K be coset representatives for 𝔽 p-basis of O K ∕ p O K. Then
{ ∑ i = 1 n a i x i | a i ∈ { 0 , … , r − 1 } }
is a set of coset representatives for O K ∕ p O K. Let y ∈ O K. Proposition 3.4(ii) tells us that
y = ∑ i = 0 ∞ ( ∑ j = 1 n a i j x j ) p i ( a i j ∈ { 0 , … , p − 1 } ) = ∑ j = 1 n ( ∑ i = 0 ∞ a i j p i ) x j ∈ ℤ p
Hence O K is finite over ℤ p. □
cloze1: True

html: Definition 8.1 (Global field). A global field is a field which is either:
- (i) An algebraic number field
- (ii) A global function field, i.e. a finite extension of 𝔽 p ( t ).
nothing: True

html: Lemma 8.2. Assuming that:
- ( K , | ∙ | ) is a complete discretely valued field
- L ∕ K a finite Galois extension with absolute value | ∙ | L extending | ∙ |.
Then for x ∈ L and σ ∈ Gal ⁡ ( L ∕ K ), we have | σ ( x ) | L = | x | L.
proof: Proof. Since x ↦ | σ ( x ) | L is another absolute value on L extending | ∙ | on K, the result follows from uniqueness of | ∙ | L. □
cloze1: True

html: Lemma 8.3 (Kummer’s Lemma). Assuming that:
- ( K , | ∙ | ) a complete discretely valued field
- f ( X ) ∈ K [ X ] a separable irreducible polynomial with roots α 1 , … , α n ∈ K sep (K sep is the separable closure of K)
- β ∈ K sep with
| β − α 1 | < | β − α i |
for i = 2 , … , n.
Then α 1 ∈ K ( β ).
proof: Proof. Let L = K ( β ), L ′ = L ( α 1 , … , α n ). Then L ′ ∕ L is a Galois extension. Let σ ∈ Gal ⁡ ( L ′ ∕ L ). We have
| β − σ ( α 1 ) | = | σ ( β − α 1 ) | = | β − α 1 |
using Lemma 8.2. Hence σ ( α 1 ) = α 1, so α 1 ∈ K ( β ). □
nothing: True

html: Proposition 8.4. Assuming that:
- ( F , | ∙ | ) is a complete discretely valued field
- f ( X ) = ∑ ⁡ i = 0 n a i X i ∈ O K [ X ] a separable irreducible monic polynomial
- α ∈ K sep a root of f
Then there exists 𝜀 > 0 such that for any g ( X ) = ∑ ⁡ i = 0 n b i X i ∈ O K [ X ] monic with | a i − b i | < 𝜀 for all i, there exists a root β of g ( X ) such that K ( α ) = K ( β ).
proof: Proof. Let α 1 , … , α n ∈ K sep be the roots of f which are necessarily distinct. Then f ′ ( α 1 ) ≠ 0. We choose 𝜀 sufficiently small such that | g ( α 1 ) | < | f ′ ( α 1 ) | 2 and | f ′ ( α 1 ) − g ′ ( α 1 ) | < | f ′ ( α 1 ) |. Then we have | g ′ ( α 1 ) | < | f ′ ( α 1 ) | 2 = | g ′ ( α 1 ) | 2 (the equality is by Lemma 1.6).
By Hensel’s Lemma version 1 applied to the field K ( α 1 ) there exists β ∈ K ( α 1 ) such that g ( β ) = 0 and | ∙ − α 1 | < | g ′ ( α 1 ) |. Then
| g ′ ( α 1 ) | = | f ′ ( α 1 ) | = ∏ j = 1 n | α 1 − α j | ≤ | α 1 − α i |
for i = 2 , … , n. (Use | α 1 − α i | ≤ 1 since α i integral). Since | β − α 1 | < | α 1 − α i | = | β − α i | using Lemma 1.6, we have that Kummer’s Lemma gives that α 1 ∈ K ( β ) and hence K ( α 1 ) = K ( β ). □
cloze1: True

html: Theorem 8.5. Assuming that:
- K is a local field
Then K is the completion of a global field.
proof: Proof. Case 1: | ∙ | is archimedean. Then ℝ is the completion of ℚ, and ℂ is the completion of ℚ ( i ) (with respect to | ∙ | ∞).
Case 2: | ∙ | non-archimedean, equal characteristic. Then K ≅ 𝔽 q ( ( t ) ) is the completion of 𝔽 q ( t ) with respect to the t-adic valuation.
Case 3: | ∙ | non-archimedean mixed characteristic. Then K = ℚ p ( α ), with α a root of a monic irreducible polynomial f ( X ) ∈ ℤ p [ X ]. Since ℤ is dense in ℤ p, we choose g ( X ) ∈ ℤ [ X ] as in Proposition 8.4. Then K = ℚ ( β ) with β a root of g ( X ). Since ℚ ( β ) dense in ℚ p ( β ) = K, and K is complete, we must have that K is the completion of ℚ ( β ). □
cloze1: True

html: Definition 9.1 (Dedekind domain). A Dedekind domain is a ring R such that
- (i) R is a Noetherian integral domain.
- (ii) R is integrally closed in Frac ⁡ ( R ).
- (iii) Every non-zero prime ideal is maximal.
nothing: True

html: Lemma 9.3. Assuming that:
- R is a Noetherian ring
- I ⊆ R a non-zero ideal
Then there exists non-zero prime ideals p 1 , … , p r such that p 1 , … , p r ⊆ I.
proof: Proof. Suppose not. Since R is Noetherian, we may choose I maximal with this property. Then I is not prime, so there exists x , y ∈ R ∖ I such that x , y ∈ I.
Let I 1 + ( x ), I 2 = I + ( y ). Then by maximality of I, there exist p 1 , … , p r and q 1 , … , q s such that p 1 ⋯ p r ⊆ I 1 and q 1 ⋯ q s ⊆ I 2. Then p 1 ⋯ p r q 1 ⋯ q s ⊆ I 1 I 2 ⊂ I. □
cloze1: True

html: Lemma 9.4. Assuming that:
- R is an integral domain
- R is integrally closed in K = Frac ⁡ ( R )
- 0 ≠ I ⊆ R a finitely generated ideal
- x ∈ K
Then if x I ⊆ I, we have x ∈ R.
proof: Proof. Let I = ( c 1 , … , c n ). We write
x c i = ∑ j = 1 n a i j c j
for some a i j ∈ R. Let A be the matrix A = ( a i j ) 1 ≤ i , j ≤ n and set B = x id ⁡ n − A ∈ M n × n ( K ).
Then in K n
B ( c 1 ⋮ c n ) = 0 .
Multiply by adj ⁡ ( B ), the adjugate matrix for B. We have
det ⁡ ( B ) id ⁡ n ( c 1 ⋮ ⁡ c n ) = 0 .
Hence det ⁡ ( B ) = 0. But det ⁡ B is a monic polynomial with coefficients in R. Then x is integral over R, hence x ∈ R. □
cloze1: True

html: Definition 9.6 (Valuation on a Dedekind domain). If R is a Dedekind domain, and p ⊆ R a non-zero prime ideal, we write v p for the normalised valuation on Frac ⁡ ( R ) = Frac ⁡ ( R ( p ) ) corresponding to the discrete valuation ring R ( p ).
nothing: True

html: Theorem 9.7. Assuming that:
- R is a Dedekind domain
- I ⊆ R a non-zero ideal
Then Ican be written uniquely as aproduct of prime ideals:
I = p 1 e 1 ⋯ p r e r
(with p i distinct).
proof: Proof (Sketch). We quote the following properties of localisation:
- (i) I = J ⟺ I R ( p ) = J R ( p ) for all prime ideals p.
- (ii) If R a Dedekind domain, p 1 , p 2 non-zero ideals, then
p 1 R ( p 2 ) = { p 2 R ( p 2 ) p 1 = p 2 R ( p 2 ) p 1 ≠ p 2
Let I ⊆ R be a non-zero ideal. By Lemma 9.3, there are distinct prime ideals p 1 , … , p r such that p 1 β 1 ⋯ p r β r ⊆ I, where β i > 0.
Let 0 ≠ p be a prime ideal, p ∉ { p 1 , … , p r }. Then property (ii) gives that p i R ( p ) = R ( p ), and hence I R ( p ) = R ( p ).
Corollary 9.5 gives I R ( p ) = ( p i R ( p i ) ) α i = p i α i R ( p i ) for some 0 ≤ α i ≤ β i. Thus I = p 1 α 1 ⋯ p r α r by property (i).
For uniqueness, if I = p 1 α 1 ⋯ p r α r = p 1 γ 1 ⋯ p r γ r then p i α i R ( p i ) = p i γ i R ( p i ) hence a i = γ i by unique factorisation in discrete valuation rings. □
cloze1: True

html: Lemma 10.1. Assuming that:
- L ∕ K a finite separable extension of fields
Then the symmetric bilinear pairing ( ∙ , ∙ ) → K ( x , y ) ↦ Tr ⁡ L ∕ K ( x y )
is non-degenerate.
proof: Proof. L ∕ K separable tells us that L = K ( α ) for some α ∈ L. Consider the matrix A for ( ∙ , ∙ ) in the K-basis for L given by 1 , α , … α n − 1.
Then A i j = Tr ⁡ L ∕ K ( α i + j ) = [ B B ⊤ ⁡ ] i j where
B = ( 1 1 ⋯ 1 σ 1 ( α ) σ 2 ( α ) ⋯ σ n ( α ) ⋮ ⋮ ⋱ ⋮ σ 1 ( α n − 1 ) σ 2 ( α n − 1 ) ⋯ σ n ( α n − 1 ) )
So
det ⁡ A = det ⁡ ( B ) 2 = [ ∏ 1 ≤ i < j ≤ n ( σ i ( α ) − σ j ( α ) ) ] 2
(Vandermonde determinant), which is non-zero since σ i ( α ) ≠ σ j ( α ) for i ≠ j by separability. □
cloze1: True

html: Theorem 10.2. Assuming that:
- O K a Dedekind domain
- L a finite separable extension of K = Frac ⁡ ( O K )
Then the integral closure O L of O K in L is a Dedekind domain.
proof: Proof. O L a subring of L, hence O L is an integral domain.
Need to show:
- (i) O L is Noetherian.
- (ii) O L is integrally closed in L.
- (iii) Every ≠ 0 prime ideal P in O L is maximal.
Proofs:
- (i) Let e 1 , … , e n ∈ L be a K-basis for L. Upon scaling by K, we may assume e i ∈ O L for all i.
Let f i ∈ L be the dual basis with respect to the trace form ( ∙ , ∙ ). Let x ∈ O L, and write x = ∑ ⁡ i = 1 n λ i f i , λ i ∈ K. Then λ i = Tr ⁡ L ∕ K ( x e i ) ∈ O K.
(For any z ∈ O L, Tr ⁡ L ∕ K ( z ) is a sum of elements in K ¯ which are integral over O K. Hence Tr ⁡ L ∕ K ( z ) ∈ K is integral over O K, hence Tr ⁡ L ∕ K ( z ) ∈ O K.)
Thus O L ⊆ O K f 1 + ⋯ + O K f n ⊆ L. Since O K is Noetherian, O L is finitely generated as an O K-module, hence O L is Noetherian.
- (ii) Example Sheet 2.
- (iii) Let P be a non-zero prime ideal of O L, and p : = P ∩ O K be a prime ideal of O K. Let 0 ≠ x ∈ P. Then x satisfies an equation
x n + a n − 1 x n − 1 + ⋯ + a 0 = 0 , a i ∈ O K ,
with a 0 ≠ 0. Then a 0 ∈ P ∩ O K is a non-zero element of p, hence p is non-zero, hence p is maximal.
We have O K ∕ p ↪ O L ∕ P, and O L ∕ P is a finite dimensional vector space over O K ∕ p. Since O L ∕ P is an integral domain and finite, it is a field. □
cloze1: True

html: Lemma 10.4. Assuming that:
- 0 ≠ x ∈ O ) K
Then
( x ) = ∏ p ≠ 0 prime ideal p v p ( x ) .
proof: Proof. x O K , ( p ) = ( p O K , ( p ) ) v p ( x ) by definition of v p ( x ).
Lemma follows from property of localisation
I = J ⟺ I O K , ( p ) = J O K , ( p )
for all prime ideals p. □
cloze1: True

html: Theorem 10.5. Assuming that:
- O K, O L, K, L as usual
- for p a non-zero prime ideal of O K, we write p O L P 1 e 1 ⋯ P r e r
Then the absolute values on L extending | ∙ | p (up to equivalence) are precisely | ∙ | P 1 , … , | ∙ | P L.
proof: Proof. By Lemma 10.4 for any 0 ≠ x ∈ O K and i = 1 , … , r we have v P i ( x ) = e i v p ( x ). Hence, up to equivalence, | ∙ | P i extends | ∙ | p.
Now suppose | ∙ | is an absolute value on L extending | ∙ | p. Then | ∙ | is bounded on ℤ, hence is non-archimedean. Let R = { x ∈ L | | x | ≤ 1 } ≤ L be the valuation ring for L with respect to | ∙ |. Then O K ⊆ R, and since R is integrally closed in L (Lemma 6.8), we have O L ⊆ R. Set
P : = { x ∈ O L | | x | < 1 } = m R ∩ O L
(where m R is the maximal ideal of R).
Hence P a prime ideal in O L. It is non-zero since p ⊆ P. Then O L , ( p ) ⊆ R, since s ∈ O L ∖ P ⟹ | s | = 1.
But O L , ( p ) is a discrete valuation ring, hence a maximal subring of L, so O L , ( p ) = R. Hence | ∙ | is equivalent to | ∙ | p. Since | ∙ | extends | ∙ | p, P ∩ O K = p so P 1 e 1 ⋯ P r e r ⊆ P, so P = P i for some i. □
cloze1: True

html: Lemma 10.8 (Chinese remainder theorem). Assuming that:
- R a ring
- I 1 , … , I n ⊆ R ideals
- I i + I j = R for all i ≠ j
Then
- (i) ⋂ ⁡ i = 1 n = ∏ i = 1 n I i (= I say).
- (ii) R ∕ I ≅ ∏ ⁡ I = 1 n R ∕ I i.
proof: Proof. Example Sheet 2. □
nothing: True

html: Definition 11.1 (Ramification). Let 0 ≠ 𝔭 be a prime ideal of O K, and
𝔭 O L = P 1 e 1 ⋯ P r e r
with P i distinct prime ideals in O L, and e i > 0.
- (i) e i is the ramification index of P i over 𝔭.
- (ii) We say 𝔭 ramifies in L if some e i > 1.
nothing: True

html: Definition 11.2 (Residue class degree). f i : = [ O L ∕ P i : O K ∕ 𝔭 ] is the residue class degree of P i over 𝔭.
nothing: True

html: Definition 11.7 (Decomposition). Let L ∕ K be a finite Galois extension. The decomposition at a prime P of O L is the subgroup of Gal ⁡ ( L ∕ K ) defined by
G P = { σ ∈ Gal ⁡ ( L ∕ K ) | σ ( P ) = P } .
nothing: True

html: Proposition 11.8. Assuming that:
- O K a Dedekind domain
- L ∕ K a finite Galois extension
- 0 ≠ P ⊆ O L a prime ideal
- P | 𝔭 ⊆ O K
Then
- (i) L P ∕ K 𝔭 is Galois.
- (ii) There is a natural map
res ⁡ : Gal ⁡ ( L P ∕ K 𝔭 ) → Gal ⁡ ( L ∕ K )
which is injective and has image G P.
proof: Proof.
- (i) L ∕ K Galois implies that L is a splitting field of a separable polynomial f ( X ) ∈ K [ X ]. Hence L P is the splitting field of f ( X ) ∈ K [ X ], hence L P ∕ K 𝔭 is Galois.
- (ii) Let σ ∈ Gal ⁡ ( L P ∕ K 𝔭 ), then σ ( L ) = L since L ∕ K is normal, hence we have a map res ⁡ : Gal ⁡ ( L P ∕ K 𝔭 ) → Gal ⁡ ( L ∕ K ), σ ↦ σ | L. Since L is dense in L P, res ⁡ is injective. By Lemma 8.2, we have
| σ ( x ) | P = | x | P
for all σ ∈ Gal ⁡ ( L P ∕ K 𝔭 ) and x ∈ L P. Hence σ ( P ) = P for all σ ∈ Gal ⁡ ( L P ∕ K 𝔭 ) and hence res ⁡ ( σ ) ∈ G P for all σ ∈ Gal ⁡ ( L P ∕ K 𝔭 ).
To show surjectivity, it suffices to show that
| G P | = e f = [ L P : K 𝔭 ] .
Write 𝔭 O L = P 1 e 1 ⋯ P r e r, f = [ O L ∕ P : O K ∕ 𝔭 ]. Then
  - | G P | = | Gal ⁡ ( L ∕ K ) | r = e f r r = e f (using Corollary 11.5).
  - [ L P : K 𝔭 ] = e f. Apply Corollary 11.6 to L P ∕ K 𝔭, noting that e , f don’t change when we take completions. □
cloze1: True

html: 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 ( ∙ , ∙ ) : R × R → R ( x , y ) ↦ Tr ⁡ R ∕ k ( x y ) ( : = Tr ⁡ k ( mult ⁡ ( x y ) ) )
is non-degenerate if and only if R = k 1 × ⋯ × k r where k i ∕ k is a finite separable extension of k.
proof: Proof. Example Sheet 3. □
cloze1: True

html: Theorem 12.2. Assuming that:
- 0 ≠ 𝔭 ⊆ O K prime ideal
Then
- (i) If 𝔭 ramifies in L, then for every x 1 , … , x n ∈ O L, we have Δ ( x 1 , … , x n ) ≡ 0 mod 𝔭.
- (ii) If 𝔭 is unramified in L, then there exists x 1 , … , x n such that 𝔭 ∤ ( Δ ( x 1 , … , x n ) ).
proof: Proof.
- (i) Let 𝔭 O L = P 1 e 1 ⋯ P r e r, 0 ≠ P i ⊆ O L distinct prime ideals, e i > 0. Define
R : = O L ∕ 𝔭 O L = CRT ∏ i = 1 r O L ∕ P i e i .
If 𝔭 ramifies, then O L ∕ 𝔭 O L has nilpotents. Hence
Δ ( x ¯ 1 , … , x ¯ n ) = 0 ∀ ⁡ x ¯ i ∈ O L ∕ 𝔭 O L .
Then using the fact that
commutes, we get that
Δ ( x 1 , … , x n ) ≡ 0 mod 𝔭 ∀ ⁡ x i ∈ 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 x ¯ 1 , … , x ¯ n a basis of O L ∕ 𝔭 O L as a k vector space, we have Δ ( x ¯ 1 , … , x ¯ n ) ≠ 0. So thee exist x 1 , … , x n ∈ O L such that
Δ ( x 1 , … , x n ) ⁄ ≡ 0 mod 𝔭 . □
cloze1: True

html: Definition 12.3 (Discriminant). The discriminant is the ideal d L ∕ K ⊆ O K generated by Δ ( x 1 , … , x n ) for all choices of x 1 , … , x n ∈ O L.
nothing: True

html: Definition 12.5 (Inverse different). The inverse different is
D L ∕ K − 1 = { y ∈ L : Tr ⁡ L ∕ K ( x y ) ∈ O K ∀ ⁡ x ∈ O L } ,
an O L submodule of L.
nothing: True

html: Theorem 12.8. Assuming that:
- O L = O K [ α ]
- α has monic minimal polynomial g ( X ) ∈ O K [ X ]
Then D L ∕ K = ( g ′ ( α ) ).
proof: Proof. Let α = α 1 , … , α n be the roots of g. Write
g ( X ) X − α = β n − 1 X n − 1 + ⋯ + β 1 X + β 0
with β i ∈ O L and β n − 1 = 1. We claim
∑ i = 1 n g ( X ) X − α i α i n g ′ ( α i ) = X r
for 0 ≤ r ≤ n − 1.
Indeed the difference is a palynomial of degree < n, which vanishes for X = α 1 , … , α n. Equate coefficients of X s, which gives
Tr ⁡ L ∕ K ( α r β s g ′ ( α ) ) = δ r s .
Since 1 , α , … , α n − 1 is an O K basis for O L, D L ∕ K − 1 has an O K basis
β 0 g ′ ( α ) , β 1 g ′ ( α ) , … , β n − 1 g ′ ( α ) ⏟ 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 = 1 ( g ′ ( α ) ), hence D L ∕ K = ( g ′ ( α ) ). □
cloze1: True

html: Lemma 13.1. Assuming that:
- M ∕ L ∕ K finite separable extensions of local fields
Then
- (i) f M ∕ K = f L ∕ K f M ∕ L
- (ii) e M ∕ K = e L ∕ K f M ∕ L
proof: Proof.
- (i) f M ∕ K = [ k M : k ] = [ k M : k L ] [ k L : k ] = f M ∕ L f L ∕ K.
- (ii) (i) and (∗). □
cloze1: True

html: Definition 13.2 (Unramified / ramified / totally ramified). The extension L ∕ K is said to be:
- unramified if e L ∕ K = 1 (equivalently f L ∕ K = [ L : K ]).
- ramified if e L ∕ K > 1 (equivalently f L ∕ K < [ L : K ]).
- totally ramified if e L ∕ K = [ L : K ] (equivalently f L ∕ K = 1).
nothing: True

html: Theorem 13.3. Assuming that:
- L ∕ K a finite separable extension of non-archimedean local fields
Then there exists a field K 0, K ⊆ K 0 ⊆ L and such that
- (i) K 0 is unramified
- (ii) L ∕ K 0 is totally ramified
Moreover [ L : K 0 ] = e L ∕ K, [ K 0 : K ] = f L ∕ K and K 0 ∕ K is Galois.
proof: Proof. Let k = 𝔽 q, so that k L = 𝔽 q f, f L ∕ K = f. Set m = q f − 1, [ ∙ ] : 𝔽 q f → L the Teichmüller map for L.
Let ζ m : = [ α ] for α a generator of 𝔽 q f ×. ζ m a primitive m-th root of unity. Set K 0 = K ( ζ m ) ⊆ L, then K 0 ∕ K is Galois and has residue field k 0 = 𝔽 q ( α ) = k L. Hence f L ∕ K 0 = 1, i.e. L ∕ K 0 is totally ramified.
Let res ⁡ : Gal ⁡ ( K 0 ∕ K ) → Gal ⁡ ( k 0 ∕ k ) be the natural map. For σ ∈ Gal ⁡ ( K 0 ∕ K ). We have σ ( ζ m ) = ζ m if σ ( ζ m ) ≡ ζ m mod m (since μ m ( K 0 ) ≅ μ m ( k 0 ) by Hensel’s Lemma version 1). Hence res ⁡ is injective. Thus | Gal ⁡ ( K 0 ∕ K ) | ≤ | Gal ⁡ ( k 0 ∕ k ) | = f K 0 ∕ K, so [ K 0 : K ] = f K 0 ∕ K.
Hence res ⁡ is an isomorphism, and K 0 ∕ K is unramified. □
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html: Theorem 13.4. Assuming that:
- k = 𝔽 q
- n ≥ 1
Then there exists a unique unramified L ∕ K of degree n. Moreover, L ∕ K is Galois and the natural Gal ⁡ ( L ∕ K ) → Gal ⁡ ( k L ∕ k ) is an isomorphism. In particular, Gal ⁡ ( L ∕ K ) ≅ ⁡ ⟨ Frob ⁡ L ∕ K ⟩ is cyclic, where Frob ⁡ L ∕ K ( x ) = x q mod m L for all x ∈ O L.
proof: Proof. For n ≥ 1, take L = K ( ζ m ) where m = q n − 1.
As in Theorem 13.3:
Gal ⁡ ( L ∕ K ) → ∼ Gal ⁡ ( k L ∕ K ) ≅ ⁡ Gal ⁡ ( 𝔽 q n ∕ 𝔽 q ) .
Hence Gal ⁡ ( L ∕ K ) is cyclic, generated by a lift of x ↦ x q.
Uniqueness: L ∕ K of degree n unramified. Then Teichmüller gives ζ m ∈ L, so L = K ( ζ m ). □
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html: Definition 13.6 (Inertial subgroup). The inertial subgroup is
I L ∕ K = ker ⁡ ( Gal ⁡ ( L ∕ K ) ↠ Gal ⁡ ( k L ∕ k ) ) .
nothing: True

html: Definition 13.7 (Eisenstein polynomial). f ( x ) = x n + a n − 1 x n − 1 + ⋯ + a 0 ∈ O K [ x ] is Eisenstein if v K ( a i ) ≥ 1 for all i, and v K ( a 0 ) = 1.
nothing: True

html: Proposition 13.9. Assuming that:
- r > e p − 1
Then exp ⁡ ( x ) = ∑ ⁡ n = 0 ∞ x n n ! converges on π r O K and induces an isomorphism
( π r O K , + ) → ∼ ( 1 + π r O K , × ) .
proof: Proof.
v K ( n ! ) = e v p ( n ! ) = e ( n − s p ( n ) ) p − 1 Example Sheet 1 ≤ e ( n − 1 p − 1 )
For x ∈ π r O K and n ≥ 1,
v K ( x n n ! ) ≥ n r − e ( n − 1 ) p − 1 = r − ( n − 1 ) ( r − e p − 1 ) ⏟ > 0
Hence v K ( x n n ! ) → ∞ as n → ∞. Thus exp ⁡ ( x ) converges.
Since v K ( x n n ! ) ≥ r for all n ≥ 1, exp ⁡ ( x ) ∈ 1 + π r O K.
Consider log ⁡: 1 + π r O K → π r O K.
log ⁡ ( 1 + x ) = ∑ n = 1 ∞ ( − 1 ) n − 1 n x n
which converges as before.
Recall identities in ℚ [ [ X , Y ] ]:
exp ⁡ ( X + Y ) = exp ⁡ ( X ) exp ⁡ ( Y ) exp ⁡ ( log ⁡ ⁡ ( 1 + X ) ) = 1 + X log ⁡ ( exp ⁡ ( X ) ) = X
Thus exp ⁡ ; ( π r O K , + ) → ∼ ( 1 + π r O K , × ) is an isomorphism. □
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html: Definition 13.10 (s-th unit group). For s ∈ ℤ, the s-th unit group U K ( s ) is defined by
U K ( s ) = ( 1 + π s O K , × ) .
Set U K ( 0 ) = U K. Then we have
⋯ ⊆ U K ( s ) ⊆ U K ( s − 1 ) ⊆ ⋯ ⊆ U K ( 0 ) = U K .
nothing: True

html: Definition 14.1 (s-th ramification group). Let v L be a normalised valuation in O L. For s ∈ ℝ ≥ − 1, the s-th ramification group is
G s ( L ∕ K ) = { σ ∈ Gal ⁡ ( K ) | v L ( σ ( x ) − x ) ≥ s + 1 ∀ ⁡ x ∈ O L } .
nothing: True

html: Theorem 14.5. Assuming that:
- [ K : ℚ p ] < ∞
- L ∕ K finite
- D L ∕ K = ( π δ ( L ∕ K ) )
Then δ ( L ∕ K ) ≥ e L ∕ K − 1, with equality if and only if tamely ramified. In particular, L ∕ K unramified if and only if D L ∕ K = O L.
proof: Proof. Example Sheet 3 shows D L ∕ K = D L ∕ K 0 ⋅ D K 0 ∕ K. Suffices to check 2 cases:
- (i) L ∕ K unramified. Then ?? gives that O L = O K [ α ], for some α ∈ O L with k L = k ( a ¯ ).
Let g ( X ) ∈ O K [ X ] be the minimal polynomial of α. Since [ L : K ] = [ k L : k ], we have that g ¯ ( X ) ∈ k [ X ] is the minimal polynomial of a ¯. g ¯ ( X ) separable and hence g ′ ( α ) ⁄ ≡ 0 ( m o d π ) L. Theorem 12.8 implies D L ∕ K = ( g ( α ) ) = O L.
- (ii) L ∕ K totally ramified. Say [ L : K ] = e, O L = O K [ π L ], π L a root of
g ( X ) = X e + ∑ i = 0 e − 1 a i X i ∈ O K [ X ]
is Eisenstein. Then
g ′ ( π L ) = e π L e − 1 ⏟ ≥ e − 1 + ∑ i = 1 e − 1 i a i π L i − 1 ⏟ v L ≥ e .
Thus v L ( y ′ ( π L ) ) ≥ e − 1. Equality if and only if p ∤ e. □
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html: Definition 15.1 (Infinite Galois definitions).
- L ∕ K is separable if ∀ ⁡ α ∈ L, the minimal polynomial f α ( X ) ∈ K [ X ] for α is separable.
- L ∕ K is normal if f α ( X ) splits in L for all α ∈ L.
- L ∕ K is Galois if it is separable and normal. Write Gal ⁡ ( L ∕ K ) : = Aut ⁡ K ( L ) in this case. If L ∕ K is a finite Galois extension, then we have a Galois correspondence:
{ subextensions K ⊆ K ′ ⊆ L } ↔ { subgroups of Gal ⁡ ( L ∕ K ) } K ′ ↦ Gal ⁡ ( K ∕ K ′ )
nothing: True

html: Proposition 15.3. Assuming that:
- L ∕ K Galois
Then
- (i) The set I = { F ∕ K finite | F ⊆ L , F Galois } is a directed set under ⊆.
- (ii) For F , F ′ ∈ I, F ⊆ F ′ there is a restriction map res ⁡ F , F ′ : Gal ⁡ ( F ′ ∕ K ) ↠ Gal ⁡ ( F ∕ K ) and the natural map
Gal ⁡ ( L ∕ K ) → lim ⁡ F [ ∈ I ] ← Gal ⁡ ( F ∕ K )
is an isomorphism.
proof: Proof. Example Sheet 4. □
cloze1: True