Local Fields

Tim Browning Notes by Florian Bouyer

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Contents

1 Foundations 2 1.1 Completion ...... 6

2 The p-adic 8

3 Non-archimedean Local Fields 10 3.1 Hensel's Lemma ...... 12

4 Extensions of local elds 14 4.1 Normed vector spaces ...... 14 4.2 Extension of Absolute Values ...... 15 4.3 Ramication ...... 15

5 Algebraic Closure 20

6 Algebraic Number Fields 22

7 Diaphantine Equations 24 7.1 Quadratic forms ...... 24 7.2 Cubic forms ...... 24

1 1 Foundations

Absolute Values

Let K be a eld.

Denition 1.1. An absolute value on K is a map | · | : K → R>0 1. |x| = 0 ⇐⇒ x = 0

2. |xy| = |x| · |y|∀x, y ∈ K

3. |x + y| ≤ |x| + |y| (the 4 inequality)

Denition. An absolute value on K is called non-archimedean if also

1. |x + y| ≤ max{|x|, |y|} (the ultrametric inequality) Otherwise we say the absolute value is archimedean

Example.

1. K = Q and | · | to be the usual absolute value given by inclusion Q ,→ R. This is an archimedean absolute value. ( 1 x 6= 0 2. Take |x| = . This is non-archimedean absolute value. The trivial absolute value 0 x = 0

3. and a prime. For ∗ the -adic valuation is if r u for K = Q p x ∈ Q p νp(x) = r x = p v u, v ∈ Z, r ∈ Z and p - uv. We extend to all of Q by setting νp(0) = +∞

Check: νp(xy) = νp(x) + νp(y) and νp(x + y) ≥ min{νp(x), νp(y)} (∗) ( pνp(x) x 6= 0 Dene the p-adic absolute value on Q to be | · |p : Q → R≥0 by |x|p = . This satises 0 x = 0 the axioms of being a non-archimedean absolute value (using (∗)) n −n n Note. |p |p = p so p → 0 as n → ∞. n o 4. Let be any eld. Put P (T ) . Dene the valuation K F = K(T ) = Q(T ) : P,Q ∈ K[T ],Q 6= 0 ( P (T ) deg Q − deg P P 6= 0 ν = Q ∞ Q(T ) P +∞ Q = 0 Check this satises (∗). If c > 1, then we get a non-archimedean absolute value on F given by −ν∞(f(T )) |F (T )|∞ := c .

Note. If K = Fq then convenient to take c = q. Lemma 1.2. Let | · | be an absolute value on a eld K. Then

1. |1| = 1

2. x ∈ K such that xn = 1, then |x| = 1.

2 3. x ∈ K, then | − x| = |x|

4. K is a nite eld then the absolute value has to be the trivial absolute value Proof.

1. Note that x 6= 0 ⇒ x > 0. We have |1| = |12| = |1| · |1|. So 1. holds.

2. Note that 1 = |1| = |xn| = |x|n ⇒ |x| = 1

3. Note that −x = −1 · x

4. Follows from 2. since any non-zero element x of a nite eld satises xn = 1 for some n.

The following result gives a criterion for checking whether an absolute value is non-archimedean.

Lemma 1.3. Let | · | be an absolute value on a eld K. Then | · | is non-archimedean if and only if |e| ≤ 1 for all e in the additive ring generated by 1. Proof. ⇒ Since |n| = | − n| we may as well assume that n ≥ 1. Then |n| = |1 + ··· + 1 | ≤ |1| = 1 | {z } n times ⇐ Suppose |e| ≤ 1 for all elements e in the additive ring generated by 1. Let x, y ∈ K, then

m m X m j m−j |x + y| = x y j j=0 m X m j m−j ≤ |x| |y| j j=0 m X j m−j m ≤ |x| |y| by assumption ≤ 1 j j=0 ≤ max({|x|, |y|}m

Take mth root and let m → ∞ (since (m + 1)1/m → 1 as m → ∞)

Corollary 1.4. If char(K) 6= 0 then all absolute values are non-archimedean Proof. The ring in Lemma 1.3 is a nite eld. Then apply Lemma 1.2 part 4.

Corollary 1.5. Suppose F ⊂ K is a subeld of K and | · | is an absolute value on K. Then | · | is non-archimedean on K if and only if | · | is non-archimedean on F

Topology

Let K be a eld with absolute value | · | on K. Then we get a metric on K induced by | · |. Call it d : K × K → R≥0 dened by d(x, y) 7→ |x − y|. Exercise. Check this is a metric. The notion of distance on elds with non-archimedean values is weird.

3 Lemma 1.6. Let K be a eld with non-archimedean absolute value. If x, y ∈ K with |x|= 6 |y|, then |x + y| = max{|x|, |y|}

Proof. Without loss of generality assume |x| > |y|. Then |x+y| ≤ max{|x|, |y|} = |x| and |x| = |x+y−y| ≤ max{|x + y|, |y|}. Hence |x| ≤ |x + y| ≤ |x|.

Denition 1.7. Let K be a eld with absolute value | · |. Let a ∈ K and r ∈ R≥0. The open ball of radius r and centre a is B(a, r) = {x ∈ K : |x − a| < r}. The closed ball of radius r and centre a is B(a, r) = {x ∈ K : |x − a| ≤ r}. A set U ⊂ K is open if and only if ∀x ∈ U there exists an open ball around x contained in U. A set is closed if and only if its complement in K is open

Lemma 1.8. Let K be a eld with non-archimedean absolute value | · |. Then

1. b ∈ B(a, r) ⇒ B(a, r) = B(b, r)

2. b ∈ B(a, r) ⇒ B(a, r) = B(b, r)

3. B(a, r) ∩ B(a0, r0) 6= 0 ⇐⇒ B(a, r) ⊂ B(a0, r0) or B(a, r) ⊃ B(a0, r0)

4. B(a, r) ∩ B(a0, r0) 6= 0 ⇐⇒ B(a, r) ⊂ B(a0, r0) or B(a, r) ⊃ B(a0, r0)

5. B(a, r) is both open and closed

6. B(a, r) is both open and closed. Proof. We prove 1. 3. 5. only. 1. b ∈ B(a, r) and c ∈ B(b, r). |c − a| ≤ max{|c − b|, |b − a|} < r, i.e., B(b, r) ⊂ B(a, r). Reverse inclusion follows from symmetry since a ∈ B(b, r). 3. Follows form 1. 5. b ∈ B(a, r) implies B(b, r) ⊂ B(a, r), so any open ball is open. To show that it is closed, note that b∈ / B(a, r) ⇒ a∈ / B(b, r). So neither ball is contained in the other and they are disjoint. Hence B(b, r) ⊂ K \ B(a, r) and the complement of B(a, r) in K is open.

Remark. Recall that a set S is said to be disconnected if there exists open sets U, V such that

• U ∩ V = ∅,

• S ⊂ U ∪ V

• S ∩ U 6= ∅ and S ∩ V 6= ∅

Otherwise S is connected. If x ∈ K then the connected component of x is the union of all connected sets containing it.

Example. K = R with usual absolute value, then connected component of any x ∈ R is R. Exercise. If | · | is a non-archimedean absolute value on a led K, then the connected component of any x ∈ K is {x}, i.e., K is totally disconnected topological space.

4 Equivalence

Denition 1.9. Two absolute values | · |1 and | · |2 on a eld K are equivalent if they induce the same topology on K. (i.e., every set which is open with respect to | · |1 is open with respect to | · |2) Given an absolute value |·| on a eld K, a sequence {an}n in K converges to a in the induced topology if and only if ∀ > 0∃N ∈ N such that for n > N, |an − a| < . Equivalently, for all open sets U containing a, there exists N such that an ∈ U for n > N Thus the notion of convergence depends on the topology induced by the absolute value.

Lemma 1.10. Let | · |1, | · |2 be absolute values on eld K, with | · |1 non-trivial. Then the following are equivalent

1. | · |1 , | · |2 are equivalent

2. ∀x ∈ K, |x|1 < 1 ⇐⇒ |x|2 < 1 3. such that , α. ∃α > 0 ∀x ∈ K |x|1 = |x|2 Proof.

α 3. ⇒ 1. Then |x − a|2 < r ⇐⇒ |x − a|1 < r . So any open ball with respect to | · |2 is an open ball with respect to | · |1. Hence the topology must be the same and the absolute value are equivalent.

n 1. n 1. ⇒ 2. |x|1 < 1 ⇐⇒ x → 0 as n → ∞ with respect to | · |1 ⇐⇒ x → 0 as n → ∞ with respect | · |2 ⇐⇒ |x|2 < 1

−1 −1 2. ⇒ 3. Now |x|1 > 1 ⇐⇒ |x | < 1 ⇐⇒ |x |2 < 1 ⇐⇒ |x|2 > 1. Also |x|1 = 1 ⇐⇒ |x|2 = 1. ∗ Now pick (and x) a ∈ K such that |a|1 < 1 (which is possible since | · |1 is non-trivial). Then log |a|1 ∗ also |a|2 < 1. Let α = > 0. Choose b ∈ k log |a|2 1. then and α |b|1 = 1 |b|2 = 1 1 = 1

log|a|i 2. |b| < 1 by assumption |b| < 1. Dene βi = for o = 1, 2. We show that β1 = β2 1 2 log|b|i log|b| log|a| which implies 1 = 1 = α. log|b|2 log|a|2 m m n −m Suppose that β1 > β2, then ∃ ∈ Q such that β2 ≤ < β1. Set x = a b ∈ k, then n n m , hence we have a contradiction with log |x|i = n log |a|i − m log |b|i = n log |b|i βi − | {z } n <0 | {z } > 0 i = 1

< 0 i = 2 and . Similarly if . Hence |x|1 < 1 |x|2 > 1 β2 > β1 β1 = β2 3. If , , replace by −1 and get −1 and −1 |b|1 > 1 |b|2 > 1 b b b 1 < 1 b 2 < 1

How independent inequivalent absolute value are

Lemma 1.11. Let be non trivial inequivalent absolute values on . Then there exists | |1 ,..., | |J K x ∈ K such that and for . |x|1 > 1 |x|j < 1 2 ≤ j ≤ J

5 Proof. By induction on J. Since are non-trivial and non-equivalent, by the previous lemma there exists such J = 2 | |1 , | | y ∈ K that and , and such that and . Let −1, then |y|1 < 1 |y|2 ≥ 1 z ∈ K |z|1 ≥ 1 |z|2 < 1 x = zy −1 and −1 |x|1 = |z|1 |y|1 > 1 |x|2 = |z|2 |y|2 < 1 By induction, there exists such that , for and , J > 2 y, z ∈ K |y|1 > 1 |y|j < 1 2 ≤ j < J |z|1 < 1 for . Consider and we have dierent cases: |z|j > 1 2 ≤ j < J |y|J 1. so take |y|J < 1 x = y 2. so take n for large enough |y|J = 1 x = y z n ( n 1 j = 2,...,J n 3. , then y 1 . So Let y for large |y|J > 1 n = −n → x = n z 1+y j 1+y j n→∞ 0 else 1+y enough n

Theorem 1.12 (Weak Approximation). Let be non trivial inequivalent absolute values on . | |1 ,..., | |J K Let for and let . Then there exists such that for all bj ∈ K j = 1,...,J > 0 x ∈ K |x − bj|j < j = 1,...,J. Proof. By Lemma 1.11, there exists such that but for . Consider xj ∈ K |xj|j > 1 |xj|i < 1 i 6= j ( xn 1 i = j xn j . Take PJ j , so take for large enough. 1+xn → wn = j=1 bj 1+xn → bj x = wn n j j n→∞ 0 else j n→∞

Remark. This is clearly related to the Chinese Remainder Theorem. Let p1, . . . , pj be distinct primes and . Then there exists such that mj . Using the Theorem above, mj ∈ N, bj ∈ Z x ∈ Z x ≡ bj mod pj −mj with -adic absolute value |x − bj|j < pj pj

1.1 Completion Denition 1.13.

1. A sequence {xn} is a eld K is called Cauchy if ∀ > 0, ∃N > 0 such that ∀m, n > N, |xm − xn| < 2. (K, |·|) is complete if every Cauchy sequence is convergent

3. A subset S ⊂ K is dense if ∀x ∈ K, ∀ > 0, B(x, )∩S 6= 0. That is, ∀x ∈ K, there exists a sequence {xn} ∈ S such that {xn} → x. 4. A eld K,b || || is a completion of (K, | |) if

(a) There exists an embedding ι : K → Kb which respect absolute values (b) im(K) is dense in Kb (c) K,b || || is complete

6 Theorem 1.14. Let (K, | |) be a eld. Then there exists a completion (K,b || ||) of K and it is unique as any two completions are canonically isomorphic. That is if (Kbj, || ||j) for j = 1, 2 then there exists a ∼ unique isomorphism of Kb1 = Kbj which is the identity of K and preserves || ||1 = || ||2 Proof.

Existence of Completion Let K be the set of all Cauchy Sequences in K. This is a ring as {an} + {bn} = {an + bn}, {an} × {bn} = {anbn} and id = {1}. Dene || || : K → R>0 by {an} → limn→∞ |an| (R is complete). Let N ⊂ K be the subset of all null sequences (||an|| = 0). Then N is a maximal ideal (Exercise). Hence K/N is a eld Kb. We have || || (not an absolute value since ||an|| = 0 for non zero elements) only depends on K/N . We get a well dened functions || || : Kb → R>0. This is an absolute value. Dene ι : K → Kb by a 7→ {a} mod N . Then ι(K) is dense and (K,b || ||) is complete.

Uniqueness Suppose (Kc0, || ||0) is complete and is a completion, ι0 : K → Kb 0 satisfy the embedding properties above.

Claim. ι0 extends uniquely to an embedding λ : Kb → Kb 0 such that

ι0 K / K0 bO ι λ Kb

Let x ∈ Kb and {xn} is a sequence in K such that {ι(xn)} converges to x (dense). Dene 0 0 0 λ(x) = limn→∞{ι (xn)}. Construct λ : Kb → Kb in the same way

Corollary 1.15. Let K be a eld and | |j (j ≤ J) be non-trivial and inequivalent absolute values on K. Let be the respective completions, let Q dened by . Then is dense, Kbj ∆ : K,→ j Kbj x 7→ (ιj(x)) ∆(K) i.e., its closure is Q . ∆(K) j Kj

Remark. We have Q ,→ R but Q ,→ R × R is not dense.

7 2 The p-adic

Theorem 2.1 (Ostrowski ). Every non trivial absolute value on Q is equivalent to | |v where v = p a prime or v = ∞.

Proof. Let | | be an absolute value on Q and a > 1, b > 0 be integers. Let t = max{|0| , |1|,..., |a − 1|}, m with , and log b . Then Pm j b = bma + ··· + b1a + b0 bi ∈ {0, . . . , a − 1} bm 6= 0 m ≤ log a |b| ≤ j=0 bja ≤ (m + 1)t max{1, |a|m} ≤ (log b/ log a + 1)t max{1, |a|m}. Replace b by bn and take nth root,

log b 1/n |b| ≤ n + 1 t1/n max{1, |a|}log b/ log a log a | {z } → 1 n→∞

Take the limit as n → ∞, then |b| ≤ max{1, |a|}log b/ log a (∗). We have two cases 1. | | is archimedean, then there exists |b| > 1 for some b by Lemma 1.3. So apply (∗), then |a| > 1 for all a > 1, so |b| ≤ |a|log b/ log b. Reversing a and b we get |a| ≤ |b|log a/ log b . Hence |a|1/ log a = |b|log b ,so log|a| log|b| , and it is independent of and . Hence α α for all . But log a = log b = α > 0 a b |a| = a = |a|∞ a ∈ N , hence α for all . Let a , hence true for all | ± 1| = 1 |a| = |a|∞ a ∈ Z q = b q ∈ Q 2. | | is non-archimedean. Then there exists a ∈ N such that |a| < 1. Let b be the such least integer. Claim. b = p a prime number We prove this by contradiction. Suppose b is not a prime, b = uv. Now |uv| < 1, but as b is the least such number, we have |u| = |v| = 1, hence |b| = 1 a contradiction.

So b is a prime, let b = p. Claim. p|a if and only if |a| < 1.

⇒: Let a = up, then |a| = |u||p|, hence |u| < 1 and |p| < 1. ⇐: Suppose that if p - a then a = up + r where r < p. By minimality of p, |r| = 1, |up| < 1, hence |a| = max{|up|, |r|} = 1

So let log|p| , α. For all we have r 0 where 0, hence 0 0 . α == log p |p| = |p|p a ∈ Z a = p a p - a |a | = |a |p = 1 Therefore, r 0 rα α. And α for all |a| = |p a | = |p|p = |a|p |q| = |q|p q ∈ Q

Denition 2.2.

1. The eld of p-adic numbers Qp is the completion of Q with respect to | |p. (Qp, | |) is a non archimedean complete eld.

2. The ring of p-adic integers Zp is Zp = {x ∈ Qp||x|p ≤ 1} = B(0, 1) (check it is a ring, by using non archimedean properties)

Lemma 2.3. Z is dense in Zp

8 Proof. is dense in , is open in so is dense in . Now a . Let Q Qp Zp Qp Q ∩ Zp Zp Q ∩ Zp = b ∈ Q|p - b a be such that for pick such that n ( is a unit in ). Then b ∈ Q p - b n ≥ 1 yn ∈ Z byn ≡ 1 mod p b Zp byn → 1 as n → ∞. Hence Z is dense in Q ∩ Zp, hence dense in Zp

What do elements of Qp look like? Let , let then by density there exists a such that a −n. But x ∈ Zp n ∈ N, q = b ∈ Q x − b p ≤ p then a a . Hence and there exists 0 such that 0 n. But b p ≤ max{|x|p , x − b p} p - b b ∈ Z bb ≡ 1 mod p n o then a 0 a 0 −n. Hence 0 a a 0 −n. Now let b − ab p = b (1 − bb ) p ≤ p |x − ab |p ≤ max x − b p , b − ab p ≤ p α ∈ {0, . . . , pn − 1} be the unique integer such that ab0 ≡ α mod pn. n n Conclusion: ∀x ∈ Zp, ∀n ∈ N, ∃α ∈ {0, . . . , p − 1} such that x ≡ α mod p

n p φn n Lemma 2.4. For all n ∈ N there exists an exact sequence of rings 0 → Zp → Zp → Z/(p Z) → 0 n n Proof. Note that ker p = {z ∈ Zp|p z = 0} = {0} (take absolute value on both side). We have that φn n is surjective since {0, . . . , p − 1} ⊂ Zp. n n n We show that im(p ) = ker(φn). Suppose that x ∈ im(p ), then x = p y for some y ∈ Zp, then n −n. Thence and . |p y − 0|p ≤ p φn(x) = 0 x ∈ ker φn −n −n n −n n Conversely, let x ∈ ker(φn). Then |x|p = |x−0|p ≤ p , hence |p x| ≤ 1 so x = p p x ∈ im(p ) | {z } ∈Zp n ∼ n Hence Zp/(p Zp) = Z/(p Z). We will see in a more general context that elements of Qp can be uniquely written as a Laurent series expansion in p. Later we will consider the extensions of Qp. In a Global setting: [k : Q] < ∞, OK is the integral closure of Z in k. In a local setting: [k : Qp] < ∞, Ok is the integral closure of Zp in k. But in the global setting Ok is not necessarily a Unique Factorisation Domain while in a local setting it always is.

9 3 Non-archimedean Local Fields

We will examine a general theory of elds which are complete with respect to a non archimedean absolute value.

Theorem 3.1 (Ostrowski ). Let K be a eld complete with respect to a archimedean absolute value. Then ∼ ∼ K = R or K = C and the absolute value are equivalent to the usual absolute value. Proof. See Chapter 3 of Cassels Local Fields

Basics Let K be a eld with a non-trivial non-archimedean absolute value | |.

•OK = {x ∈ K| |x| ≤ 1}, the ring of integers of K

•PK = {x ∈ K| |x| < 1}

Check that OK is an integral domain and that PK is maximal. (If J ) PK is an ideal of OK then there exists x ∈ J such that |x| = 1. Then |x−1| = 1 and |1 = xx−1| ∈ J)

•UK = {x ∈ K| |x| = 1}, the group of units in K

• kK = OK /PK , the residue eld of K

The characteristic of kK is the residual characteristic.

Note. In general charkK 6= charK.

∗ • ΓK = {|x| |x ∈ K }, the value group of | | on K.

This is a multiplicative subgroup of R>0. ∼ Denition 3.2. A non-archimedean absolute value is discrete if ΓK is discrete. (i.e., ΓK = Z) Lemma 3.3. A non-archimedean absolute value is discrete if and only if the maximal ideal is principal.

Proof. By problem A.6 ΓK is discrete if and only if ΓK is cyclic.

⇐: Suppose that PK is principal, say hπi. Let γ = |π| < 1. Hence for all x ∈ PK , we have |x| ≤ γ (since x = πy with y ∈ OK ). So for all x ∈ K, there exists n ∈ Z such that n n−1 n−1 1−n 1−n γ ≤ |x| < γ . Dividing through by γ we get that γ ≤ xπ < 1, whence xπ ∈ PK . 1−n 1−n n So γ ≤ xπ ≤ γ, thus xπ = γ. So |x| = γ , hence ΓK is cyclic generated by γ.

⇒: Suppose that ΓK is cyclic with generator γ < 1 say. Let π ∈ K be such that |π| = γ. Clearly n hπi ⊂ PK . Conversely, for x ∈ PK , then |x| = γ for some n ≥ 1 since ΓK is cyclic. So −1 n−1 −1 |xπ | = γ ≤ 1, i.e. xπ ∈ OK and x ∈ hπi

From now on | | is a discrete non-archimedean absolute value on a eld K. So by the previous lemma ∗ PK = hπi .We call π the uniformiser for the absolute value. Any x ∈ K can be written as x = πn with n ∈ Z and ∈ UK . We write VK (x) = n ∈ Z for the order of x. This gives a valuation VK : K → Z ∪ {∞} by setting VK (0) = ∞.

10 Lemma 3.4. Let be an integral ideal. Then n for some . 0 6= I ⊂ OK I = PK := {x1 . . . xn|xi ∈ PK } n ∈ N n Proof. The subset {|x| |x ∈ I} ⊂ ΓK is bounded and so it attains its maximal at x0 = π ,say (ΓK is discrete). Then n I = hx0i = PK

This implies that PK is the unique non-zero prime ideal in OK . Furthermore, OK is a PID and a local ring (with a unique maximal ideal) Let be the completion of with respect to the absolute value . Let be the ring of integers K K | | OK , PK and the maximal ideal of respectively. then and . There is an inclusion K OK = OK ∩ K PK = PK ∩ K map and so a map dened by . The kernel of this map is OK ,→ OK OK → kK := OK /PK x 7→ x + PK , so it induces a natural map ∼ . PK kK → kK Claim. This map is an isomorphism. It suces to show that it is surjective

Proof. Let x ∈ O . By density of K in K, there exists x ∈ K such that |x − x| < 1. Then x − x ∈ P ( K ) K and . Thus |x| ≤ max |x − x|, |x| ≤ 1 x ∈ K ∩ OK = OK <1 ≤1 Denition 3.5. A non-archimedean local eld is a eld which is complete with respect to a non-trivial discrete non-archimedean absolute value such that the residue class kK is nite.

K Qp Fq ((T )) Completion of Q Fq(T ) Example. OK Zp Fq [[T ]] PK pZp (T ) ∼ kK = Fp Fq From now on, K is a non-archimedean local eld. Say that an innite sum P∞ , , converges to if PN n=0 xn xn ∈ K s s = limN→∞ n=0 xn Lemma 3.6. P∞ converges if and only if as n=0 xn xn → ∞ n → ∞ Proof. Exercise

Lemma 3.7. Let π be a uniformiser of K and let A ⊂ OK be set of representative of OK /PK . Then P∞ n OK = { n=0 xnπ : xn ∈ A} Proof. By Lemma 3.6 we have P∞ converges and lies in . Conversely, if , then there exists a n=0 OK x ∈ OK unique x0 ∈ A such that |x − x0| < 1. Hence x = x0 + πy1 for some y1 ∈ OK . Continue inductively with y1 etc Suppose ∗, Then −N for some . Apply Lemma 3.7 to get ∗ P∞ n x ∈ K π x ∈ OK N ∈ Z K = { n=N xnπ : xn ∈ A,N ∈ Z, xN 6= 0} Let us return topology. A subset V ⊂ K is said to be compact if whenever we have a family Uλ (λ ∈ Λ) of open sets of such that , then there exists a nite subset such that . K V ⊂ ∪λ∈ΛUλ Λ0 ⊂ Λ V ⊂ ∪λ∈Λ0 Uλ We say that K locally compact if every point of K has a compact neighbourhood. (i.e., ∀x ∈ K there exists Vx ⊂ K which is compact and contains B(x, r) for some r > 0)

Lemma 3.8. Let K be a non-archimedean local eld. Then OK is compact, and hence K is locally compact.

11 Proof. First we prove that OK is compact. Let Uλ (λ ∈ Λ) be open sets covering OK . Suppose that there does not exists a nite subcovering. Now OK = ∪x∈A(x + πOK ) where A is set of representation for (nite eld) OK /PK . Then there exists x0 ∈ A such that x0 + πOK is not covered by nitely many 2 Uλ. Similarly there exists x1 ∈ A such that x0 + x1π + π OK is not nitely covered and so on. Let 2 . There exists such that . Since is open it follows that x = x0 + x1π + x2π + · · · ∈ OK λ0 ∈ Λ x ∈ Uλ0 Uλ0 n for some , which is a contradiction. x + π OK ∈ Uλ0 N Next we prove that K is locally compact. Put Vx = B(x, 1) = x + OK . Remark. In fact: F locally compact with respect non-archimedean absolute value ⇐⇒ F non-archimedean local eld.

3.1 Hensel's Lemma

Theorem 3.9. Let K be a non-archimedean local eld and f ∈ OK [X]. Suppose x0 ∈ OK satises 0 2 |f(x0)| |f(x )| < |f (x )| . Then there exists a unique x ∈ O such that f(x) = 0, |x − x | ≤ 0 . 0 0 K 0 |f (x0)|

Proof. Dene fj ∈ OK [X] via

2 f(X + Y ) = f(X) + f1(X)Y + f2(X)Y + ... (3.1)

0 0 In particular f1(X) + f (X). Dened y0 ∈ OK by f(x0) + y0f (x0) = 0. Then

j |f(x0 + y0)| ≤ max fj(x0)y0 By (3.1) j≥2

j ≤ max y0 j≥2 2 ≤ |y0| 2 f(x0) = 0 f (x0) < |f(x0)|

Similarly |f1(x0 + y0) − f1(x0)| ≤ |y0| < |f1(x0)|. Then |f1(x0 + y0)| = |f1(x0)|. Put x1 = x0 + y0. Then 2 |f(x0)| , and |f(x0)| . So repeat the process and obtain a sequence |f(x1)| ≤ 2 |f1(x1)| = |f1(x0)| |x1 − x0| = 0 |f1(x0)| |f (x0)| 2 2 of such that and |f(xn)| |f(xn)| . So as xn+1 = xn + yn |f1(xn)| = |f1(x0)| |f(xn+1)| ≤ 2 = 2 f(xn) → 0 |f1(xn)| |f1(x0)| . Finally |f(xn)| as . So is Cauchy and it has a limit as n → ∞ |xn+1 − xn| = |yn| = |f (x )| → 0 n → ∞ {xn} required. 1 n Now suppose that we have another solution x + z with z 6= 0 and |z| ≤ |f(x0)| < |f (x )| = |f (x)|. |f1(x0)| 1 0 1 2 Then, putting X = x and Y = z in equation (3.1), we get 0 = f(x + z) − f(x) = xf1(x) + z f2(x) + ... . j j But |zf1(x)| > z ≥ z fj(x) for all j ≥ 2. Which gives a contradiction. Example.

1. Squares in Qp.

. Suppose that ∗. If there exists such that 2 then there exists • p 6= 2 y ∈ Zp x0 ∈ Zp x0 − y < 1 2 2 0 x ∈ Zp such that x = y . (Take f(X) = X − y, so |f(x0)| < 1 but |f (x0)| = |2x0| = 1).

12 Theorem. Any with , is a square in z z ∈ Z p - z Zp ⇐⇒ p = +1 Claim. ∗ ∗ 2 has elements represented by where is a quadratic Qp/(Qp) 4 1, c, p, cp c ∈ {1, . . . , p−1} non-residue.

Corollary. It follows that Qp has exactly 3 quadratic extensions. Proof of the claim. Suppose ∗. We may assume or for (on multiplying x ∈ Qp x = u pu u ∈ Zp 2 ∗2 x by a power of p ∈ Q ). Let α ∈ {1, . . . , p − 1} be such that u ≡ α mod pZp (i.e. u = α + v −1 −1 for some v ∈ pZp). Then u = α(1 + α v) and 1 + α v ≡ 1 mod pZp which is a square. Thus we may assume . But u ∗2, otherwise is in ∗2. u = α p = 1 ⇒ u ∈ Qp uc Qp

• p = 2. See exercise B.1

2. Since residue eld is nite, it follows that ∗ is cyclic group of order where r for kK kK q − 1 q = p some prime p. Now show there exists an alternative set of representative for kK = OK /PK , besides {0, . . . , q − 1}. Note and so ∗ . For each ∗ , let ∗ such that and p · 1 ∈ OK q − 1 ∈ OK α ∈ kK x0 ∈ OK x0 ≡ α mod p q−1 0 q−2 consider f(x) = x − 1. Then |f(x0)| < 1, |f (x0)| = |q − 1| · |x0| = 1. Hence by Theorem (3.9), there exists a unique Teichmuller representative ∗ of such that and αb ∈ OK α f(αb) = 0 αb ≡ α . We can take ∗ as a set of representative for . mod p {0} ∪ {αb : α ∈ kK } kK Dene principal congruence subgroup n ∗ n n . Then and UK = {u ∈ UK = OK : u − 1 ∈ PK } = 1 + PK UK n are open and closed and compact in ∗ (with induced topology). UK K We have isomorphism of topological groups:

∗ • K /UK → Z dened by xUK 7→ VK (x). 1 ∗ dened by v 1 v where is a primitive th root of unity in and •UK /UK → kK ξ UK 7→ g ξ (q − 1) K g is a generator for ∗ . kK Hence any ∗ can be uniquely written as u v for 1 , i.e., ∗ ∼ 1 . x ∈ K π ξ ∈ UK K = Z × Z/(q − 1)Z × UK

13 4 Extensions of local elds

We consider eld extensions of non-archimedean local elds. We would like to show that these extension are non-archimedean local elds.

4.1 Normed vector spaces

Let K be a non-archimedean local eld

Denition 4.1. Let V be a vector space over K. A function k k : V → R≥0 is a norm if 1. kxk = 0 if and only if x = 0

2. kx + yk ≤ kxk + kyk

3. kλxk = |λ| · kxk for all λ ∈ K

Note. The norm induces a metric d(x, y) = kx − yk on V , which gives a topology.

Denition 4.2. Two norms k k1 and k k2 one a vector V are equivalent if ∃c1,c2 > 0 such that c1kxk2 ≤ kxk1 ≤ c2kxk2 ∀x ∈ V

Exercise. Show that k k1, k k2 are equivalent if and only if they induce the same topology on V . Lemma 4.3. Let V be a nite dimensional vector space over K. Then any 2 norms on V are equivalent. Moreover, V is complete with respect to the induced metric.

Proof. We proof by induction on n = dimK V n = 1 Trivial n > 1 Let e1, . . . , en be a basis for V over K. Put a = a1e1 + ··· + anen (aj ∈ K) and dene kak0 := maxj |aj|. Check that k k0 is a norm and that V is complete with respect to it. It will suce to show any norm on is equivalent to . Firstly P k k V k k0 kak ≤ j |aj| · kejk ≤ c2kak0 P with c2 = kejk.

We now need to show ∃c > 0 such that kak0 ≤ ckak for all a ∈ V (∗). If not, ∀ > 0, there exists b = b ∈ V such that kbk ≤ kbk0. Assume without loss of generality that kbk0 = |bn|. Replacing by −1 we have where . b bn b b = c + en c ∈ he1, . . . , en−1iK Summary: false, implies we can nd a sequence (m) such that (∗) c ∈ W = he1, . . . , en−1iK (m) (m) (l) kc + enk → 0 as m → ∞. But then kc − c k → 0. So now use induction hypothesis. Since dim W = n−1, it is complete under k k. Thus there exists c∗ ∈ W such that kc(m)−ck = 0. ∗ (m) ∗ Hence kc + enk = limm→∞ kc + enk = 0. Therefore c + en = 0, which is impossible. Hence (∗) hold and so k k and k k0 are equivalent.

Corollary 4.4. V nite dimensional normed vector space over K. Then V is locally compact. (i.e., v ∈ V has a compact neighbourhood)

Proof. By Lemma 4.3 we can assume k k is k k0, with respect to some xed basis e1, . . . , en. Now imitate the proof of Lemma 3.8 to show that {v ∈ V : kvk0 ≤ 1} is compact.

14 4.2 Extension of Absolute Values

Let K be a non-archimedean local eld and L ⊃ K an extension. We say that an absolute value k k L extends to the absolute value | | on K if kλk = |λ| for all λ ∈ K Theorem 4.5. Let L ⊃ K be a nite extension. Then there exists a unique extension k k of | | to L. Moreover, L, k k is a non-archimedean local eld. Proof.

Uniqueness: Suppose k k1, k k2 extend | | to L. Then, regarding L as a nite dimensional vector space over K, Lemma 4.3implies k k1 and k k2 are equivalent and some dene the same topology on L. But then they are equivalent as absolute values and so by Lemma 1.10, there exists α such that α . But the two absolute values are equal on , so that . kxk1 = kxk2 ∀x ∈ L K α = 1 Second_part Apply 4.4 and converse of Lemma 3.8

1/n Existence We will show that the extension of k k of | | to L is given by kxk = |NL/K (x)| for x ∈ L, where n = [L : K]. Here NL/K : L → K is the norm map. (Recall: Thinking of L as a vector space over K, multiplication by α ∈ L gives a linear map mα : K → L, with matrix n 1/n Aα ∈ Mn(K). Put NL/K (α) := det Aα). For x ∈ K, kxk = |x | = |x|. So k k does extend | |. ∗ For x ∈ L , the linear map mx : L → L is invertible with inverse mx−1 . Thus the matrix Ax is invertible, and det Ax 6= 0. Hence kxk= 6 0. Multiplicativity follows from the multiplicativity of the norm map. Remains to prove the ultrametric inequality. Suces to show kxk ≤ 1, then k1+xk ≤ 1. (Then, assuming then x ). Suppose . Let be the kxk ≤ kyk kx + yk = kyk · k y + 1k ≤ kyk kxk ≤ 1 χ(x) r r−1 characteristic polynomial of the linear map mx : L → L. Let f(X) = X + fr−1X + ··· + f0 be the minimal polynomial of x. Here r is the degree of x over K. Then χ(X) = f(X)n/r (where n/r is the degree of L over K(x), a proof of this can be found in Cassels book Lemma B.3). Then power hence . Since is irreducible and monic it follows |f0 | = |NL/K (x)| ≤ 1, |f0| ≤ 1 f from consideration of Newton polygon associated to it that |fi| ≤ 1 (See Cassels chapter 4). n Hence f ∈ OK [X] and also χ ∈ OK [X]. Now NL/K (1 + x) = det(In + Ax) = (−1) χ(−1). So k1 + xk = |χ(−1)|1/n ≤ 1. This completes the proof

Since the absolute value on L is unique, we will usually write it as | | instead of k k.

Corollary 4.6. | |p on Qp extends uniquely to an absolute value on algebraic closure Qp.

4.3 Ramication

Suppose L/K is a nite extension of non-archimedean local elds of degree n = [L : K].

Lemma 4.7. There exists a natural injection kK → kL such that kL is an extension of kK of degree f = f(L/K) := [kL : kK ] ≤ n.

15 Proof. There is certainly an inclusion OK ,→ OL. But PK = OK ∩PL, this induces the injection kK → kL. Let α1, . . . , αn+1 ∈ kL. Show that there are linearly dependent over kK . Then we will have shown that f = dim k ≤ n. Let α ,..., α ∈ O such that α = α + P for 1 ≤ i ≤ n + 1. Since dim L = n, kK L c1 [n+1 L i bi L K there are linearly dependent over , i.e., there exists not all zeroes such that Pn+1 . K λi ∈ K i=1 λiαbi = 0 Without loss of generality, we assume that . Let be the reduction of −1 modulo . λn+1 6= 0 µi ∈ kK λiλn+1 PL Then Pn , as required. i=1 µiαi + αn+1 = 0 Denition 4.8. If f = f(L/K) = n, we say L/K is unramied. If f = f(L/K) = 1, we say L/K is totally ramied. If f = f(L/K) < n, we say L/K is ramied.

Remark. If K ⊂ L ⊂ E is a tower of extensions then f(E/K) = f(E/L) · f(L/K). We shall see that unramied extensions are easy to characterise.

Theorem 4.9. Let . Then there exists such that and α ∈ kL = OL/PL αb ∈ OL αb + PL = α [K(αb): K] = [kK (α): kK ]. Furthermore, the eld depends on . K(αb) α Remark. The extension is unramied. K(αb)/K

Proof. Let φ ∈ kx[x] be the minimal polynomial of α. Let Φ ∈ K[X] be any lift of φ (i.e., deg φ = deg Φ and , meaning coecients of are reduced modulo ). Let be an element of the φ = Φ Φ PK αc0 ∈ OL residue class of . Then and 0 0 (since is a nite eld so it is α Φ(αc0) = φ(α) = 0 Φ (αb0) = φ (α) 6= 0 kK perfect). Thus and 0 . Hence by Hensel's lemma, with as the ground eld, |Φ(αc0)| < 1 |Φ (αc0)| = 1 K(αb0) implies there exists such that , . Hence in residue class of and αb ∈ K(αb0) ⊂ L Φ(αb) = 0 |αb − αc0| < 1 αb α since is irreducible. [K(αb): K] = [kK (α): kK ] Φ Now suppose that 0 is also in the residue class of and satises 0 . Then the αb α [K(αb ): K] = [kK (α): kK ] above argument implies 0 and so 0 . But we must have equality since the degrees αb ∈ K(αb ) K(αb) = K(αb ) are the same.

Corollary 4.10. There exists a bijection between intermediate elds E (with K ⊂ E ⊂ L) which are unramied and the elds k with kK ⊂ k ⊂ kL, given by E → kE = E ∩ OL/E ∩ PL Proof. The previous theorem gives one direction.

Let k be an intermediate eld kK ⊂ k ⊂ kL. Let q = #k. Then k = kK (α) for some (q − 1)th root of unity α. Then apply Theorem 4.9

Corollary 4.11. Let K be a non-archimedean local eld. For all n ∈ N there exists a unique (up to isomorphism) unramied extension of degree . It is the splitting eld over of q where n , n K X − X q = qK with qK = #kK Proof. Let be unramied extension of degree . Then has n elements. Then contains a L/K n kL q = qK L full set of (q − 1)th roots of unity (By example 2 after Hensel's lemma). In particular Xq − X splits in L and so L contains its splitting eld, say F . However qL = qF = q and so by Corollary 4.10 we must have F = L

Corollary 4.12. Let f ∈ OK [X] be monic of degree n and reduction f mod PK is irreducible. Then 1. if L = K(α) and α has minimal polynomial f, then L/K is unramied

16 2. The splitting eld of f over K is unramied and has degree n. Proof.

1. Note that kL ⊃ kK (α), where α is the reduction of α mod PL. Moreover, kK (α) has degree n over kK . Hence f(L/K) ≥ n. But we also have f(L/K) ≤ n = [L : K] by Lemma 4.7. 2. Let L be the splitting eld of f over K and let α, β be roots of f in L. Then part 1. implies that K(α) and K(β) are both unramied extensions of degree n. Then Corollary 4.10 implies they are equal, therefore L = K(α).

Summary. Unramied extensions of K are obtained by adjoining a root of unity of order coprime to residual characteristic of K. Now let us look at ramied extensions. Suppose L/K is a nite extension of non-archimedean local elds. Consider the relationship between ∗ value groups ΓL = {|x| : x ∈ L } is a discrete (cyclic) subgroup of R>0.

Denition 4.13. The ramication index of L/K is e = e(L/K) = [ΓL :ΓK ].

If πL, πK are uniformisers for L and K respectively. Recall |πK | < 1 is a generator for ΓK and similarly e πL for ΓL. Then |πK | = |πL| . This implies e(E/K) = e(E/L)e(L/K) for any tower K ⊂ L ⊂ E. Theorem 4.14. L/K be a nite extensions of non-archimedean local elds of degree n. Then n = ef.

It follows from this that L/K is unramied if and only if e(L/K) = 1 L/K is totally ramied if and only if e(L/K) = n It is ramied if and only if e(L/K) > 1

Proof. Let πL be a uniformiser of L and let α1,..., αf be any lift to OL of a basis for kL/kK . (As in n b b o Theorem 4.9) We will prove that j is a basis for . (In fact we B = αbiπL : 1 ≤ i ≤ f, 0 ≤ j ≤ e − 1 L/K will prove that it is an OK basis for OL.) Firs suppose that B is no