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P-ADIC

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PAUL JAMES HOLLOWAY

A Thesis Submitted in Partial Fulfilment of the Requirements for the Degree of

BACHELOR OF SCIENCE in the Undergraduate Academic Unit of and Statistics

Supervisor: Dr. Colin Ingalls

THE UNIVERSITY OF NEW BRUNSWICK

May 2014

c Paul James Holloway, 2014 Abstract

This paper focuses on the p-adic numbers and some of the properties of them. We start by defining a and the p-adic and prove that it is a norm on the ra- tional numbers. In doing do we also show that the p-adic norm is a non-Archimedean norm which has strange properties compared to the ordinary (Archimedean) norm. Then we define the p-adic numbers and show that it is a field that is complete with respect to the p-adic norm. Then we introduce Hensel’s lemma which we can use for showing that p-adic solutions exist to and then find them. Then we expand Hensel’s lemma to apply to a system of multivariable , and show that Fermat’s last theorem has p-adic solutions. Finally we extend the p-adic norm to finite field extensions of the p-adic numbers.

ii Table of Contents

Abstract ii

Table of Contents iii

Chapter 1. p-adic Numbers 1 1.1 Norms ...... 1 1.2 p-adic Norm ...... 2 1.3 Review of building up Complex numbers ...... 5 1.4 The field of p-adic numbers ...... 7 1.5 Arithmetic in Qp ...... 12 Chapter 2. Hensel’s Lemma 17 2.1 Hensel’s Lemma ...... 17 2.2 Expanding Hensel’s Lemma ...... 19 2.3 Hensel’s Lifting ...... 24

Chapter 3. extensions of Qp 29 3.1 Field extensions of Qp ...... 29 Bibliography 37

iii Chapter 1 p-adic Numbers

Before we can use the p-adic numbers we first have to define them, but to do that we will first need to define a few other basic terms.

1.1 Norms

Definition 1.1.1 Let X be a non-empty . A on X is a d taking a pairs of elements (x, y) of X to the non-negative real numbers such that:

(1) d(x, y) = 0 iff x = y

(2) d(x, y) = d(y, x)

(3) d(x, y) ≤ d(x, z) + d(z, y), ∀z ∈ X ( ).

A set with a metric, such as d in the definition above, is called a metric .

Definition 1.1.2 A norm, denoted k k, on a field F is a map from F to the non- negative real numbers such that:

(1) kxk = 0 iff x = 0

(2) kx · yk = kxk · kyk

1 (3) kx + yk ≤ kxk + kyk.

The metric d we are dealing with is d(x, y) = kx−yk. The normal absolute value, |x|, is an example of a norm on the rational numbers and the metric d(x, y) = |x − y| is the usual concept of distance on the on the .

1.2 p-adic Norm

Now that we have a norm defined, we can work towards defining the p-adic norm.

Definition 1.2.1 Let p be any . For any non-zero a, let ordp a be the higest power p which divides a, i.e., the greatest m such that a ≡ 0 (mod pm). Recall that a ≡ b (mod p) ⇒ c | (a − b).

Example 1.2.2 ord5 125 = 3, ord5 50 = 2, ord2 32 = 5, ord2 96 = 5, ord7 98 = 2,

ord7 99 = 0 (we agree that if a = 0 then ordp a = ∞).

Note: These behave in a similiar fashion to :

ordp(a · b) = ordp a + ordp b

ordp(a/b) = ordp a − ordp b

Also ordp(ab + cd) ≥ min(ordp ab + ordp cd), because if n is the minimum order then pn|ab, pn|cd, and pn|(ab + cd).

Definition 1.2.3 Now we define the map | |p on Q as:   ord x 1/p p , if x 6= 0 |x|p =  0, if x = 0 We call this map the p-adic absolute value.

2 1 1 Example 1.2.4 |25|5 = 52 = 25 1 |25|3 = 50 = 1 18 33 27 | 27 |3 = 32 = 9 = 3 1 1 |1 − 50|7 = |−49|7 = 72 = 49 (7-adic “distance” between 1 and 50) 1 1 25 50 1 1 1 | 9 + 16 |5 = | 144 |5 = 52 = 25 (5-adic “distance” between 9 and − 16 ).

Proposition 1.2.5 The p-adic absolute value | |p is a norm on Q.

Proof: We need to check the three properties of a norm.

Property (1) says |x|p = 0 iff x = 0. Let us assume x = 0, then by the definition of

the p-adic norm |x|p = 0. Now we assume |x|p = 0, then:

ordp x |x| = 1/p = 0 ⇒ ordp x = ∞ ⇒ x = 0.

This proves property (1).

The second property of a norm states |x · y|p = |x|p · |y|p, then: 1 1 1 1 1 |x · y|p = = = = · = |x|p · |y|p pordp(x·y) pordp x+ordp y pordp x · pordp y pordp x pordp y this proves property (2).

Property (3) states |x+y|p ≤ |x|p +|y|p. If x = 0, y = 0, or x+y = 0 then property

a c (3) becomes trivial, so let us assume they are all non-zero. Let x = b and y = d ,

then x + y = (ad + bc)/bd. This implies that ordp(x + y) = ordp(ad + bc) − ordp bd. Now the higest power of p dividing x + y is greater than or equal to the minimum of the higest power dividing x and the higest power dividing y:

ordp(x + y) ≥ min(ordp ad, ordp bc) − ordp bd

= min(ordp a + ordp d, ordp b + ordp c) − ordp b − ordp d

= min(ordp a − ordp b, ordp c − ordp d)

= min(ordp x, ordp y).

3 ordp(x+y) ordp x ordp y So, |x + y|p = 1/p ≤ max(1/p , 1/p ) = max(|x|p, |y|p) ≤ |x|p + |y|p.

This proves property (3) and thus finishes the proof. 

So we have now defined | |p to be the p-adic norm. But you might notice, from the proof, that this norm has a stronger property than an ordinary norm.

Definition 1.2.6 If kx + yk ≤ max(kxk, kyk) for a norm always holds, then that norm is called non-Archimedean. If d(x, y) ≤ max(d(x, z), d(z, y)), then the metric defined by d is non-Archimedean. More particularly, a non-Archimedean metric is a metric that is induced by a non-Archimedean norm: d(x, y) = |x − y| = |(x − z) + (z − y)| ≤ max(|x − z|, |z − y|) = max(d(x, z), d(z, y)).

This is the stronger property that is mentioned above. In the proof above we showed that |x + y|p ≤ max(|x|p, |y|p). Thus, | |p is a non-Archimedean norm on Q.

A norm or metric is called Archimedean simply if said norm or metric is not non-

Archimedean. An example of an Archimedean norm on Q is the ordinary absolute value.

In any X, a Cauchy {a0, a1, a2,...} of elements of X is defined that for any ε > 0, there exists an N such that d(am, an) < ε, when n, m > N.

If we have a metric d1 on X and there is a sequence S that is Cauchy with respect to d1, then a metric d2 on X is equivalent to d1 if and only if S is also Cauchy with respect to d2. We say two norms are equivalent if they induce equivalent metrics.

The way we usually think about distance is obviously based on the Archimedean norm k k. The non-Archimedean metric has properties that seem unusual or strange, here is an example:

4 For any field the (kx + yk ≤ kxk + kyk) says that the sum of two sides is greater than or equal to the third side. But what about when we have a non-Archimedean norm on a field? The triangle inequality for a non-Archimedean norm is kx + yk ≤ max(kxk, kyk). Now suppose that kxk < kyk, then kx + yk ≤ kyk. But kyk = k(x+y)−xk ≤ max(kx+yk, kxk) and we have kyk > kxk so kyk ≤ kx+yk. This implies that kyk = kx + yk. This means that at least two sides of a “triangle” in a non-Archimedean field are equal, which means that every “triangle” is isosceles.

In the case of | |p on Q, this says that if two rational numbers are divisible by different powers of p, then the sum of the two numbers is divisible by the lower power of p.

The “isosceles triangle principle” is what we call the non-Archimedean principle that kx ± yk ≤ max(kxk, kyk), with equality holding when kxk= 6 kyk.

1.3 Review of building up Complex numbers

Now we have a new concept of distance between two rational numbers: if two rational numbers have a difference divisible by a large power of p, they are considered to be close. For us to work with the “p-adic metric” we have to expand Q, the in a way similar to how the real numbers, R, and then the , C, are constructed in the usual Archimedean metric | |.

Let us now review how this is done:

Let us start with the natural numbers, N = {1, 2, 3,...}. Evey step in going from N to C can be analyzed in terms of a desire to do two things:

(1) Solve polynomial equations

(2) Find the limit of Cauchy (i.e., “complete” the number system such

5 that every Cauchy sequence has a limit in the new number system).

From N we can introduce the , Z, as solutions to equations of the form:

a + x = b a, b ∈ N.

Now the rationals can be introduced as solutions to equations of the form:

ax = b a, b ∈ Z.

One way to build up the real numbers is to consider the set S of Cauchy sequences of rational numbers. Let s1 = {aj} ∈ S and s2 = {bj} ∈ S be two Cauchy sequences that are equivalent, and write s1 ∼ s2 if |aj − bj| → 0 as j → ∞. We now define R to be the set of equivalence classes of Cauchy sequences of rational numbers.

Showing that R is a field by defining , , and the additive and multiplicative inverses of the equivalence classes of the Cauchy sequences is not hard.

√ Next, we wanted numbers that could solve x2 + 1 = 0. Then when i = −1 was introduced and the field of complex numbers of the form:

a + bi a, b ∈ R. was defined, it turned out that C was algebraically closed (this means that every polynomial with coefficients in C have solutions in C. This is the fun- damental theorem of ) and C is complete with respect to a unique norm √ (|a + bi| = a2 + b2) that extends the usual norm | | on R.

Since C is complete the process of finding field extensions stops here. The comple- tion of C was done with only a quadratic extension of R, which means it was obtained by adjoining solutions to the x2 + 1 = 0. So C is algebraically

6 closed with respect to the Archimedean metric, but it is not algebraically closed with

respect to | |p.

1.4 The field of p-adic numbers

Let {ai} be a sequence of rational numbers such that, if given an ε > 0, there exists

an N such that |ai − aj|p < ε if both i, j > N. For two Cauchy sequences {ai} and

{bi}, if |ai − bi|p → 0 as i → ∞, then the two Cauchy sequences are equivalent. The

set of equivalence classes of Cauchy sequences we call Qp, the p-adic numbers.

Definition 1.4.1 The p-adic norm of an equivalence class a is limi→∞ |ai|p, where

{ai} is any representative of a.

If we have two equivalence classes a and b of Cauchy sequences with representa-

tives {ai} and {bi} respectively, we can define a · b to be the equivalence class that

0 0 is represented by the Cauchy sequence {aibi}. If we also have {ai} and {bi} then we would have

0 0 0 0 0 |aibi − aibi|p = |ai(bi − bi) + bi(ai − ai)|p

0 0 0 ≤ max(|ai(bi − bi)|p, |bi(ai − ai)|p).

0 0 The expressions approach |a|p · lim |bi − bi|p = 0 and |b|p · lim |ai − ai|p = 0 as i → ∞. 0 0 This implies that {aibi} ∼ {aibi}.

The {1/ai} will be Cauchy unless |ai|p → 0. And if {ai} ∼

0 0 0 0 {ai}, then ai = ai for all i, then 1/ai = 1/ai, which implies {1/ai} ∼ {1/ai} (if all ai 0 and ai are non-zero).

7 Now we define a + b to be equivalence class represented by the Cauchy sequence

{ai + bi}. Once again if we have another two Cauchy sequences then we would have:

0 0 0 0 |(ai + bi) − (ai + bi)|p = |ai − ai + bi − bi|p

0 0 ≤ max(|ai − ai|p, |bi − bi|p)

0 0 As i → ∞ the two expressions go to lim(|ai − ai|p) = 0 and lim(|bi − bi|p) = 0 0 0 respectively. Hence {ai + bi} ∼ {ai + bi}. The is defined very easily.

Now with addition, multiplication,and inverses as defined above we show that the set of equivalence classes of Cauchy sequences, Qp, is a field.

Let the Cauchy sequences {ai}, {bi}, {ci} be representatives of a, b, c ∈ Qp.

Now let us check for the distributive law; a(b + c) is the equivalence class of:

{ai(bi + ci)} = {aibi + aici}.

So ab + ac is also an equivalence class. Hence a(b + c) = ab + bc, so the distributive law holds.

Now the associative law; a + (b + c) is the equivalence class of:

{ai + (bi + ci)} = {ai + bi + ci} = {(ai + bi) + ci}.

So (a + b) + c is also an equivalence class. Hence a + (b + c) = (a + b) + c, so the associative law holds.

Now commutative law; a + b is the equivalence class of:

{ai + bi} = {bi + ai}.

So b + a is also an equivalence class. Hence a + b = b + a, so the commutative law holds. This proves that Qp is a field.

8 Proposition 1.4.2 The field of p-adic numbers, Qp, is complete with respect to | |p.

Proof: Suppose we have a Cauchy sequence {xn} in Qp. Then given an ε > 0, there k exists an N such that |xi − xj| < ε when i, j > N. Now let ε = 1/p , then since {xi}

k is Cauchy, then |xi − xj|p < 1/p which implies:

ordp(xi−xj ) k ordp(xi−xj ) k 1/p < 1/p =⇒ p > p =⇒ ordp(xi − xj) > k.

But this implies that the first k terms of xi and xj are the same, and this is the same as making the sequence terms arbitrarily close to some p-adic number. So any sequence in Qp that is Cauchy with respect to | |p will converge. Thus the field Qp is Complete. 

Lemma 1.4.3 If x ∈ Q and |x|p ≤ 1, then for any i there exists an integer α such i i that |α − x|p ≤ p . We can choose α the set {0, 1, 2, . . . , p − 1}.

a Proof: Let x = b be in lowest terms. Now p does not divide b since |x|p ≤ 1, so b and are relatively prime. So now we can find integers n, m such that mb − npi = 1. Now let α = am and we have: a a a |α − x|p = am − = (mb − 1) = |mb − 1|p b p b p b p

≤ |mb − 1|p

i i = |np |p = |n|p/p

= 1/pi.

Now, we can add a multiple of pi to α to get a number in the set {0, 1, 2, . . . , pi − 1}

1 for which |α − x|p ≤ pi still holds. This proves the lemma. 

Theorem 1.4.4 Evey equivalence class a ∈ Qp for which |a|p ≤ 1 has exactly one

representative Cauchy sequence of the form {ai} for which:

9 i (1) 0 ≤ ai < p , for i = 1, 2, 3,...

i (2) ai ≡ ai+1 (mod p ), for i = 1, 2, 3,...

0 Proof: First we have to prove uniqueness. Assume {ai} 6= {ai} is also a sequence

0 0 i0 that satisfies properties (1) and (2). If ai0 6= ai0 then ai0 6≡ ai0 (mod p ), since 0 i both ai0 , ai0 are in the set {0, 1, 2, . . . , p − 1}. Then, for all i ≥ i0, we will have

0 0 i0 0 ai ≡ ai0 6≡ ai ≡ ai0 (mod p ), which implies ai 6≡ ai. So

0 1 |ai − ai|p ≤ , for all i ≥ i0 pi0

0 Hence, {ai} 6∼ {ai}.

Now suppose that we have a Cauchy sequence {bi}, we want to find a similar

sequence {ai} that satisfy (1) and (2). For every j = 1, 2, 3,..., let N(j) be a natural

j 0 0 number such that |bi − bi0 |p ≤ 1/p , when i, i ≥ N(j). For all i ≥ N(1), if i ≥ N(1) then

|bi|p ≤ max(|bi0 |p, |bi − bi0 |p) 1 ≤ max(|b 0 | , ) i p p

0 which implies that |bi|p ≤ 1 and as i → ∞, |bi0 |p → |a|p ≤ 1.

Using the lemma above we can find a sequence of integers aj ∈ {0, 1, 2, . . . , pj},

j such that |aj − bN(j)|p ≤ 1/p . For {aj} to be such a sequence we need aj+1 ≡ aj

j (mod p ) and {bi} ∼ {aj}.

|aj+1 − aj|p = |aj+1 − bN(j+1) + bN(j+1) − bN(j) − (aj − bN(j))|p

≤ max(|aj+1 − bN(j+1)|p, |bN(j+1) − bN(j)|p, |aj − bN(j)|p)

≤ max(1/pj+1, 1/pj, 1/pj)

= 1/pj.

10 j This implies that aj+1 ≡ aj (mod p ).

|ai − bi|p = |ai − aj + aj − bN(j) − (bi − bN(j))|p

≤ max(|ai − aj|p, |aj − bN(j)|p, |bi − bN(j)|p)

≤ max(1/pj, 1/pj, 1/pj)

= 1/pj.

As i → ∞, |ai − bi|p → 0 which implies that {ai} ∼ {bi}. This proves the theorem. 

If a ∈ Qp does not satisfy |a|p ≤ 1, then we can multiply a by the power of th 0 m p that equals |a|p, say the m power of p, to get an a = ap ∈ Qp that does 0 0 0 satisfy |a |p ≤ 1. Then, as in the theorem, a is represented by the sequence {ai} and 0 m a = a /p is represented by the sequence {ai}.

0 0 We can now write all ai in the sequence {ai} to the p with coefficients b0, b1, . . . , bi−1 ∈ {0, 1, 2, . . . , p − 1}:

0 2 i−1 ai = b0 + b1p + b2p + ... + bi−1p .

0 0 i The condition ai ≡ ai+1 (mod p ) means that:

0 2 i−1 i ai+1 = b0 + b1p + b2p + ... + bi−1p + bip .

0 So ai can be thought of as a number to the base p.

Then the original p-adic number a can be thought of as decimal number to the base p that has finitely many digits corresponding to negative powers of p, but infinitely many digits corresponding to positive powers of p:

b b b b a = 0 + 1 + 2 + ... + m−1 + b + b p + b p2 + ... pm pm−1 pm−2 p m m+1 m+2

We call this equality the “p-adic expansion” of of a.

11 We define the set of p-adic integers to be:

Zp = {x ∈ Qp |x|p ≤ 1}.

ordp a Let a ∈ Qp then, |a|p = 1/p ≤ 1 =⇒ ordp a ≥ 0 =⇒ all powers of p are positive. So if the p-adic expansion of a numbers in Qp has no negative powers of p, that number is a p-adic integer. Addition and multiplication in Zp is closed so Zp is a of Qp.

Now we define the p-adic units to be:

∗ Zp = {x ∈ Zp 1/x ∈ Zp} = {x ∈ Zp |x|p = 1} = {x ∈ Zp x 6≡ 0 (mod p)}.

Theorem 1.4.5 A converges in Qp iff its terms approach zero.

Proof: Suppose that {ai} is a sequence in Qp such that |ai|p → 0 as i → ∞. Then the partial sums Sn = a0 + a1 + ... + an are Cauchy since for n > m:

|Sn − Sm|p = |am+1 + ... + an|p ≤ max(|am+1|p,..., |an|p) which goes to zero and both n, m → ∞. Since Qp is complete this implies convergence.

Now suppose we have a series that converges in Qp, then it is Cauchy. Given an

ε > 0 there exists an N such that for n, m > N, |Sn − Sm|p < ε. In particular |Sn+1 −

Sn|p = |an+1|p < ε. Thus as n → ∞, the terms approach zero. 

1.5 Arithmetic in Qp

Addition, , multiplication, and of the p-adic numbers works a lot like the same operations on decimals, but with p-adic numbers you work from left to right instead of right to left.

12 Example 1.5.1 Some examples in Q7:

6 + 3 · 7 + 2 · 72 + . . . × 4 + 1 · 7 + 5 · 72 + . . . 3 + 1 · 7 + 3 · 72 + . . . (a) 6 · 7 + 3 · 72 + . . . + 2 · 72 + . . . 3 + 0 · 7 + 2 · 72 + . . .

3 · 7−1 + 2 · 70 + 2 · 71 + . . . (b) − 4 · 7−1 + 4 · 70 + 5 · 71 + . . . 6 · 7−1 + 4 · 70 + 3 · 71 + . . .

5 + 1 · 7 + 6 · 72 + . . .  3 + 5 · 7 + 1 · 72 + ... 1 + 2 · 7 + 4 · 72 + . . .

1 + 6 · 7 + 1 · 72 + . . . (c) 3 · 7 + 2 · 72 + . . . 3 · 7 + 5 · 72 + . . . 4 · 72 + . . . 4 · 72 + . . .

P∞ i If a = i=k aip is the p-adic expansion of a, where k < 0, then another way to represent the p-adic expansion is . . . a4a3a2a1a0.a−1a−2 . . . ak. An example is if b = 2p−2 + 6p−1 + 1 + 2p + 3p2 + 5p3 + ... , then b can also be represented as b = ... 5321.62.

13 The example 1.5.1(a) above can also be represented like this:

. . . 236 × . . . 514 . . . 313 . . . 360 + . . . 200 . . . 346 Which as you can see makes it easier to represent more digits in a p-adic number.

Here are a few interesting things about p-adic numbers, 7-adics particularly in this case: . . . 3333334 . . . 6666666 + . . . 3333334 + . . . 0000001 . . . 0000001 . . . 0000000 Since a 7-adic integer added to itself equals 1, this implies that “one-half” is a 7-adic integer. Also ... 666666 is −1 in the 7-adics since we added 1 to it and got 0.

√ Now let us try to see if n is in Qp, where n ∈ Z. In other words we want to find 2 2 a0, a1, a2,... ; 0 ≤ ai ≤ (p − 1), such that (a0 + a1 · p + a2 · p + ...) = n

√ Example 1.5.2 Let us try with n = 2 and p = 7 (i.e., is 2 in Q7?).

2 2 2 (a0 + a1 · 7 + a2 · 7 + ...) = 2 + 0 · 7 + 0 · 7 + ...

2 2 (a0 + a1 · 7 + a2 · 7 + ...) = 2

2 ⇒ a0 ≡ 2 (mod 7)

2 a0 ≡ 9 (mod 7)

a0 ≡ 3 (mod 7)

So set a0 = 3

14 2 2 So now we have: (3 + a1 · 7) ≡ 2 (mod 7 )

2 9 + 6a1 · 7 ≡ 2 (mod 7 )

2 6a1 · 7 ≡ −7 (mod 7 )

6a1 ≡ −1 (mod 7)

6a1 ≡ 6 (mod 7)

a1 ≡ 1 (mod 7)

So set a1 = 1

2 2 3 Now we have: (3 + 1 · 7 + a2 · 7 ) ≡ 2 (mod 7 )

2 3 (10 + a2 · 7 ) ≡ 2 (mod 7 )

2 3 100 + 20a2 · 7 ≡ 2 (mod 7 )

2 3 20a2 · 7 ≡ −98 (mod 7 )

20a2 ≡ −2 (mod 7)

6a2 ≡ 12 (mod 7)

a2 ≡ 2 (mod 7)

So set a2 = 2

By Hensel’s lemma, introduced in the next chapter, we can continue on this pattern and we get: √ 2 (3 + 1 · 7 + 2 · 7 + ...) = 2 ∈ Q7.

Example 1.5.3 What about n = 3 and p = 7?

2 2 2 (a0 + a1 · 7 + a2 · 7 + ...) = 3 + 0 · 7 + 0 · 7 + ...

2 2 (a0 + a1 · 7 + a2 · 7 + ...) = 3

2 a0 ≡ 3 (mod 7)

15 12 = 1, 22 = 4, 32 = 9 ≡ 2, 42 = 16 ≡ 2, 52 = 25 ≡ 4, 62 = 36 ≡ 1 (mod 7).

2 As you can see no squares are equivalent to 3 (mod 7), so a0 6≡ 3 (mod 7), which √ implies 3 ∈/ Q7.

16 Chapter 2 Hensel’s Lemma

2.1 Hensel’s Lemma

We now want to find p-adic solutions to polynomial equations, to do this we use Hensel’s lemma.

Theorem 2.1.1 (Hensel’s Lemma):

2 n Let F (x) = c0 + c1x + c2x + ... + cnx be a polynomial whose coefficients are

0 2 n−1 integers, with formal F (x) = c1 + 2c2x + 3c3x + ... + ncnx . Let a0 be

0 in {0, 1, ··· p − 1} such that F (a0) ≡ 0 (mod p) and F (a0) 6≡ 0 (mod p). Then there

exists a unique p-adic integer a such that F (a) = 0 and a ≡ a0 (mod p).

Proof: We need to find a p-adic integer a such that F (a) ≡ 0 (mod pn) for all n

which implies F (a) = 0. We need to construct a sequence of integers a1, a2, a3,..., such that the following 3 properties all hold when n ≥ 1:

n+1 (1) F (an) ≡ 0 (mod p )

n (2) an ≡ an−1 (mod p )

n+1 (3) 0 ≤ an < p .

We use induction on n to prove that such an an exists.

17 Let us begin with n = 1, then there is a unique integera ˜0 ∈ {0, 1, . . . , p − 1} such thata ˜0 ≡ a0 (mod p). We need an a1 that satisfies (2) and (3), so we need

2 2 a1 ≡ a0 (mod p ) where a1 ∈ {0, 1, . . . , p − 1}, this implies a1 =a ˜0 + b1p, where

b1 ∈ {0, 1, . . . , p − 1}. Now we check property (1); since we are looking for solutions mod p2, then any where the power of p is 2 or more can be ignored:

n X i F (a1) = F (˜a0 + b1p) = ci(˜a0 + b1p) i=0 n X i i−1 2 = (cia˜0 + icia˜0 b1p + terms divisible by p ) i=0 n n X i X i−1 2 ≡ cia˜0 + ( icia˜0 )b1p (mod p ) i=0 i=0

0 = F (˜a0) + F (˜a0)b1p.

Since a0 ≡ a˜0 (mod p), then F (a0) ≡ F (˜a0) ≡ 0 (mod p), then we can write F (˜a0) ≡

2 2 αp (mod p ) for some α ∈ {0, 1, . . . , p − 1}. So in order to get F (a1) ≡ 0 (mod p )

0 2 0 we must get αp + F (˜a0)b1p ≡ 0 (mod p ) which is equivalent to α + F (˜a0)b1 ≡ 0

0 0 (mod p). But, since F (˜a0) ≡ F (a0) 6≡ 0 (mod p), then we can solve for c1. By using

−α lemma 1.4.3 then we let b1 ∈ {0, 1, . . . , p − 1} be a unique integer so that b1 ≡ 0 F (˜a0)

(mod p). This b1 for a1 =a ˜0 + b1p satisfies properties (1), (2), and (3).

Continuing with the induction, suppose we already have already found a1, a2, . . . , an−1

that satisfy (1), (2), and (3). We want to find an that satisfy (2) and (3), So an =

n n an−1 + bnp with bn ∈ {0, 1, . . . , p − 1}. We expand F (an−1 + bnp ) like we did in the first step of induction, but now terms that divisible by pn+1 are ignored:

n F (an) = F (an−1 + bnp )

0 n n+1 ≡ F (an−1) + F (an−1)bnp (mod p ).

18 n We know that F (an−1) ≡ 0 (mod p ) so we can write, by the induction assumption,

n n+1 n+1 F (an−1) ≡ βp (mod p ), now F (an) ≡ 0 (mod p ) becomes:

n 0 n n+1 βp + F (an−1)bnp ≡ 0 (mod p )

0 β + F (an−1)bn ≡ 0 (mod p).

0 0 Since an−1 ≡ a0 (mod p), then like before F (an−1) ≡ F (a0) 6≡ 0 (mod p), now we

an find bn ∈ {0, 1, . . . , p − 1}. Proceeding like we did when looking for b1, we solve

−β n bn ≡ 0 (mod p). This an = an−1 + bnp satisfies (1), (2), and (3) and proves F (an−1) the claim. Now, let

2 a =a ˜0 + b1p + b2p + ...

n+1 n+1 Notice that a ≡ an (mod p ), so F (a) ≡ F (an) ≡ 0 (mod p ), for all n, it follows

2 that the p-adic number F (a) = 0. Also a =a ˜0 + b1p + b2p + ... gives a unique

sequence of an as in the claim, and that uniqueness implies the uniqueness of a. This

proves Hensel’s lemma.  √ Now to apply Hensel’s lemma. Finding 2 ∈ Q7, like in example 1.5.2, is the same as finding p-adic solutions to the polynomial x2 − 2 = 0.

Example 2.1.2 Let f(x) = x3 +2x2 +2x+4, let us look for 5-adic solutions. You will

0 2 notice that x0 = 3 is a solution to f(x0) ≡ 0 (mod 5). Also f (x0) = 3x0 + 4x0 + 2 = 27 + 12 + 2 = 41 6≡ 0 (mod 5). So f(x) satisfies Hensel’s lemma, this implies that there exists a unique p-adic number, x, that solves f(x) = 0.

2.2 Expanding Hensel’s Lemma

Now we will expand Hensel’s lemma to apply to a system of multivariable polynomials, as long as there are as many polynomials as variables.

19 Theorem 2.2.1 Let f(x, y) and g(x, y) be multivariable polynomials with integer coefficients and a Jacobian of:   ∂f ∂f J(x, y) =  ∂x ∂y   ∂g ∂g  ∂x ∂y

Let x0, y0 ∈ {0, 1, . . . , p − 1} such that f(x0, y0) ≡ g(x0, y0) ≡ 0 (mod p) and  det J(x0, y0) 6≡ 0 (mod p). Then there exists a unique x and y belonging to the

p-adic integers such that f(x, y) = g(x, y) = 0 and x ≡ x0 and y ≡ y0 (mod p).

Proof: First we construct two series x1, x2, x3,... and y1, y2, y3,... all in {0, 1, . . . , p− 1} such that these three properties all hold for when n ≥ 1:

n+1 (1) f(xn, yn) ≡ g(xn, yn) ≡ 0 (mod p )

n (2) xn ≡ xn−1, yn ≡ yn−1 (mod p )

n+1 (3) 0 ≤ xn, yn < p .

Let n = 1 to start this off. Choose x0, y0 ∈ {0, 1, . . . , p − 1} that satisfy f(x0, y0) ≡

g(x0, y0) ≡ 0 (mod p) and det(J(x0, y0)) 6≡ 0 (mod p). Now we need x1, y1 that

satisfy properties (2) and (3) above, so x1 = x0 + t1p and y1 = y0 + s1p, where

t1, s1 ∈ {0, 1, . . . , p − 1}. Then using multivariable we get

f(x1, y1) = f(x0 + t1p, y0 + s1p) ∂f(x , y ) ∂f(x , y ) ≡ f(x , y ) + 0 0 [(x + t p) − x ] + 0 0 [(y + s p) − y ] (mod p2) 0 0 ∂x 0 1 0 ∂y 0 1 0 ∂f(x , y ) ∂f(x , y ) ≡ f(x , y ) + 0 0 t p + 0 0 s p (mod p2). 0 0 ∂x 1 ∂y 1

We ignore the later terms of the multivariable Taylor series because all of the later terms are divisible by p2. The exact same process for g.

20 Since f(x0, y0) ≡ g(x0, y0) ≡ 0 (mod p) this implies that f(x0, y0) ≡ α1p (mod

2 2 p ) and g(x0, y0) ≡ β1p (mod p ), for some α1, β1 ∈ {0, 1, . . . , p − 1}. Since we are

2 looking for x1, y1 to satisfy property (1), we are looking for f(x1, y1) ≡ 0 (mod p ). This implies that:

f(x1, y1) = f(x0 + t1p, y0 + s1p) ∂f(x , y ) ∂f(x , y ) ≡ α p + 0 0 t p + 0 0 s p ≡ 0 (mod p2) 1 ∂x 1 ∂y 1 ∂f(x , y ) ∂f(x , y ) ≡ α + 0 0 t + 0 0 s ≡ 0 (mod p) 1 ∂x 1 ∂y 1 ∂f(x , y ) ∂f(x , y ) ≡ 0 0 t + 0 0 s ≡ −α (mod p). ∂x 1 ∂y 1 1

Do the same for g and get:

∂g(x , y ) ∂g(x , y ) g(x , y ) ≡ 0 0 t + 0 0 s ≡ −β (mod p). 1 1 ∂x 1 ∂y 1 1

Now we have:       ∂f(x0,y0) ∂f(x0,y0) t1 −α1  ∂x ∂y    ≡    ∂g(x0,y0) ∂g(x0,y0)      ∂x ∂y s1 −β1

and by using Cramer’s rule we can uniquely solve for t1 and s1. Since t1 and s1 are unique it implies that x1 = x0 + t1p and y1 = y0 + s1p are also unique. These x1, y1 satisfy (1), so this proves this step of induction.

Now assume we found that the three properties hold for when n = 2, 3, . . . , i − 1,

we will now see if it holds for n = i. We want xi, yi that satisfy properties (2) and (3)

i i above, so xi = xi−1 + tip and yi = yi−1 + sip , where ti, si ∈ {0, 1, . . . , p − 1}. Then

i i f(xi, yi) = f(xi−1 + tip , yi−1 + sip ) ∂f(x , y ) ∂f(x , y ) ≡ f(x , y ) + i−1 i−1 t pi + i−1 i−1 s pi (mod pi+1). i−1 i−1 ∂x i ∂y i

The exact same process for g.

21 i Since f(xi−1, yi−1) ≡ g(xi−1, yi−1) ≡ 0 (mod p ) this implies that f(xi−1, yi−1) ≡

i i+1 i i+1 αip (mod p ) and g(xi−1, yi−1) ≡ βip (mod p ), where αi, βi ∈ {0, 1, . . . , p − 1}.

i+1 We are looking for f(xi, yi) ≡ 0 (mod p ). This implies that:

∂f(x , y ) ∂f(x , y ) f(x , y ) ≡ α p + i−1 i−1 t pi + i−1 i−1 s pi ≡ 0 (mod pi+1) i i i ∂x i ∂y i ∂f(x , y ) ∂f(x , y ) ≡ α + i−1 i−1 t + i−1 i−1 s ≡ 0 (mod p) i ∂x i ∂y i ∂f(x , y ) ∂f(x , y ) ≡ i−1 i−1 t + i−1 i−1 s ≡ −α (mod p). ∂x i ∂y i i

Do the same for g and get:

∂g(x , y ) ∂g(x , y ) g(x , y ) ≡ i−1 i−1 t + i−1 i−1 s ≡ −β (mod p). i i ∂x i ∂y i i

Then like before we can use Cramer’s rule to uniquely solve for ti and si. Since

ti and si are unique it implies that xi and yi are also unique. These xi, yi satisfy (1), so this proves this step of induction and therefore the induction. So the two

sequences {xi} and {yi} exist for all i and are unique, so x and y exist and are

unique. 

We can expand this version of Hensel’s lemma to apply to n equations with n variables, instead of just two equations with two variables. The theorem and proof are the same as above but with more equations, variables, and notation.

Theorem 2.2.2 Let f1(x1, . . . , xn), . . . , fn(x1, . . . , xn) be multivariable polynomials with integer coefficients and a Jacobian matrix of:   ∂f1 ∂f1 ··· ∂f1  ∂x1 ∂x2 ∂xn     ∂f2 ∂f2 ··· ∂f2  J(x , . . . , x ) =  ∂x1 ∂x2 ∂xn  1 n  . . . .   . . .. .      ∂fn ∂fn ··· ∂fn ∂x1 ∂x2 ∂xn

22 Let x10 , . . . , xn0 ∈ {0, 1, . . . , p − 1} such that:

f1(x10 , . . . , xn0 ) ≡ ... ≡ fn(x10 , . . . , xn0 ) ≡ 0 (mod p)

and det(J(x10 , . . . , xn0 )) 6≡ 0 (mod p). Then there exists a unique x1, . . . , xn belonging to the p-adic integers such that f1(x1, . . . , xn) = ... = fn(x1, . . . , xn) = 0 and x1 ≡ x10

(mod p) , . . . , xn ≡ xn0 (mod p).

The proof the this is similar to the proof for Theorem 2.2.1.

Now lets apply the multivariable Hensel’s lemma.

Example 2.2.3 Let look for 11-adic solutions to:   f(x, y, z) = x3 + y3 − z3   g(x, y, z) = y + z    2 h(x, y, z) = z + 2

You will notice that x0 = 1 , y0 = 3, z0 = 8 are solutions to f(x0, y0, z0) ≡

g(x0, y0, z0) ≡ h(x0, y0, z0) ≡ 0 (mod 11). Also:     ∂f(x0,y0,z0) ∂f(x0,y0,z0) ∂f(x0,y0,z0) 3x2 3y2 3z2  ∂x ∂y ∂z   0 0 0   ∂g(x0,y0,z0) ∂g(x0,y0,z0) ∂g(x0,y0,z0)    J(x0, y0, z0) =   =  0 1 1   ∂x ∂y ∂z     ∂h(x0,y0,z0) ∂h(x0,y0,z0) ∂h(x0,y0,z0)    ∂x ∂y ∂z 0 0 2z0

If you expand along the first column you get:

1 1 2 2 2 2 det(J(x0, y0, z0)) = 3x0 = 3x0[2z0−0] = 6x0z0 = 6(1) (8) = 48 6≡ 0 (mod 11). 0 2z

Which implies that there exists unique 11-adic numbers, x, y, z, that solve f(x, y, z) = g(x, y, z) = h(x, y, z) = 0.

23 In the example above you will notice that f(x, y, z) = 0 is a form of Fermat’s last theorem when n = 3. This implies that Fermat’s last theorem has 11-adic solutions.

2.3 Hensel’s Lifting

We can also use Hensel’s lemma to actually find solutions, not just show that they exist. To do this we find solutions modulo p and then “lift” them to solutions modulo p2 and further “lift” those solutions to solutions modulo p3. Lifting all the way modulo pi, for any i belonging to the positive integers, if need be.

Example 2.3.1 Lets try to solve x2 +y2 = 3 and xy = 4. This has no solutions in R, but what about in Qp? Let us try to find solutions in Q7. First we need our equations to be homogeneous, so set f(x, y) = x2 + y2 − 3 = 0 and g(x, y) = xy − 4 = 0. You will notice that x = 1, y = 4 are solutions to these two equations modulo 7, because

2 2 (1) + (4) − 3 = 14 ≡ 0 and (1)(4) − 4 = 0, so set x0 = 1 and y0 = 4. Now we can “lift” it to solutions modulo 72 and then solutions modulo 73 and so on. To do this we need:

2 f(x1, y1) ≡ 0 (mod 7 )

2 f(x0 + 7t1, y0 + 7t2) ≡ 0 (mod 7 ) ∂f(x , y ) ∂f(x , y ) f(x , y ) + 7t 0 0 + 7t 0 0 ≡ 0 (mod 72) 0 0 1 ∂x 2 ∂y

2 14 + 7t1(2x0) + 7t2(2y0) ≡ 0 (mod 7 )

2 + t1(2x0) + t2(2y0) ≡ 0 (mod 7)

t1(2) + t2(8) ≡ −2 (mod 7)

2t1 + 1t2 ≡ 5 (mod 7).

24 Now we need to do the same to g:

2 g(x1, y1) ≡ 0 (mod 7 )

2 g(x0 + 7t1, y0 + 7t2) ≡ 0 (mod 7 ) ∂g(x , y ) ∂g(x , y ) g(x , y ) + 7t 0 0 + 7t 0 0 ≡ 0 (mod 72) 0 0 1 ∂x 2 ∂y

2 0 + 7t1(y0) + 7t2(x0) ≡ 0 (mod 7 )

0 + t1(x0) + t2(y0) ≡ 0 (mod 7)

t1(4) + t2(1) ≡ 0 (mod 7)

4t1 + t2 ≡ 0 (mod 7).   2t1 + 1t2 ≡ 5 We now have two equations and two variables:  4t1 + t2 ≡ 0

We can now solve for t1 and t2 using matrices and Cramer’s rule:       2 1 t 5    1        ≡   4 1 t2 0

We label these matrices J · t ≡ d respectively.     5 1 2 5 Now J ≡   and J ≡   and t1   t1   0 1 4 0

det(J) = (2)(1) − (1)(4) = 2 − 4 = −2

det(Jt1 ) = (5)(1) − (1)(0) = 5 − 0 = 5

det(Jt2 ) = (2)(0) − (5)(4) = 0 − 20 = −20.

By Cramer’s rule: det(J ) 5 5 t = t1 = ≡ ≡ 1 and 1 det(J) −2 5 det(J ) −20 1 15 t = t2 = ≡ ≡ ≡ 3. 2 det(J) −2 5 5

25 Now that we have t1 and t2 we can solve for x1 and y1.

x1 = x0 + 7t1 = 1 + 7(1) = 8, y1 = y0 + 7t2 = 4 + 7(3) = 25.

Let us now check if x1 = 8 and y1 = 25 do indeed give solutions to f(x, y) and g(x, y) modulo 72.

2 2 2 f(x1, y1) = f(8, 25) = (8) + (25) − 3 = 64 + 625 − 3 = 686 ≡ 0 (mod 7 )

2 g(x1, y1) = g(8, 25) = (8)(25) − 4 = 200 − 4 = 196 ≡ 0 (mod 7 ).

2 So x1 = 8 and y1 = 25 do give solutions modulo 7 . Now we can “lift” again to solutions modulo 73. To do this wee need:

3 f(x2, y2) ≡ 0 (mod 7 )

2 2 3 f(x1 + 7 t3, y1 + 7 t4) ≡ 0 (mod 7 )

2 ∂f 2 ∂f 3 f(x1, y1) + 7 t3 + 7 t4 ≡ 0 (mod 7 ) ∂x1 ∂y1

2 2 3 686 + 7 t3(2x1) + 7 t4(2y1) ≡ 0 (mod 7 )

2 2 3 686 + 7 t3(16) + 7 t4(50) ≡ 0 (mod 7 )

14 + 16t3 + 50t4 ≡ 0 (mod 7)

2t3 + 1t4 ≡ 0 (mod 7).

26 and

3 g(x2, y2) ≡ 0 (mod 7 )

2 2 3 g(x1 + 7 t3, y1 + 7 t4) ≡ 0 (mod 7 )

2 ∂g 2 ∂g 3 g(x1, y1) + 7 t3 + 7 t4 ≡ 0 (mod 7 ) ∂x1 ∂y1

2 2 3 196 + 7 t3(y1) + 7 t4(x1) ≡ 0 (mod 7 )

2 2 3 196 + 7 t3(25) + 7 t4(8) ≡ 0 (mod 7 )

4 + 25t3 + 8t4 ≡ 0 (mod 7)

4t3 + 8t4 ≡ −4 (mod 7)

4t3 + 1t4 ≡ 3 (mod 7).

Once again we have two equations and two variables so we can solve using matrices:       2 1 t 0    3        ≡   4 1 t4 3

We lable these matrices J · t0 ≡ d0 respectively.     0 1 2 0 Now J ≡   and J ≡   and t3   t4   3 1 4 3

det(J) = (2)(1) − (1)(4) = 2 − 4 = −2

det(Jt3 ) = (0)(1) − (1)(3) = 0 − 3 = −3

det(Jt4 ) = (2)(3) − (0)(4) = 6 − 0 = 6.

By Cramer’s rule: det(J ) −3 4 25 t = t3 = ≡ ≡ ≡ 5 3 det(J) −2 5 5 det(J ) 6 6 20 t = t4 = ≡ ≡ ≡ 4. 4 det(J) −2 5 5

27 Now that we have t3 and t4 we can solve for x2 and y2.

2 2 2 2 x2 = x1+7 t3 = 8+7 (5) = 8+245 = 253, y2 = y1+7 t4 = 25+7 (4) = 25+196 = 221

Let us now check if x2 = 253 and y2 = 221 do indeed give solutions to f(x2, y2) and

3 g(x2, y2) modulo 7 .

2 2 3 f(x2, y2) = f(253, 221) = (253) +(221) −3 = 64009+48841−3 = 112847 ≡ 0 (mod 7 )

3 g(x2, y2) = g(253, 221) = (253)(221) − 4 = 55913 − 4 = 55909 ≡ 0 (mod 7 )

3 So x2 = 253 and y2 = 221 do give solutions to f and g modulo 7 . We can also “lift” this to get solutions modulo higher powers of 7 by doing exactly what we did above, but this quickly becomes time consuming.

The ti and si are the coefficients to the p-adic expansions of x and y.

28 Chapter 3 Field extensions of Qp

3.1 Field extensions of Qp

For this section we let F be a locally compact field with a non-Archimedean norm k k.

Definition 3.1.1 Let V be a finite dimensional over F . Then a norm on V , called a field norm, is a map k kV , from V to the non-negative real numbers, satisfying:

(1) kxkV = 0 iff x = 0

(2) kaxkV = kakkxkV , for all a ∈ F and x ∈ V

(3) kx + ykV ≤ kxkV + kykV .

Theorem 3.1.2 If V is a finite dimensional vector space over a locally compact field F , then all norms on V are equivalent.

Corollary 3.1.3 Let V = K be a field. Then there is at most one norm of K, denoted k kK , as a field that extends k k of F . i.e., such that kakK = kak for a ∈ F . proof for this theorem and corollary is available in [1 pg 58-59].

29 Before we get into extensions of fields we must first define a different type of “norm” and then extend the p-adic norm to these fields.

Definition 3.1.4 Let K = F (α) be a finite extension of a field F generated by an

n n−1 element α which satisfies the monic irred(α) = x + a1x +

... + an−1x + an = 0 for ai ∈ F . Then the norm of α from K to F is:

NK/F (α) = det(Aα)

where the matrix for the F - σ : K → K given by σ(x) = αx is Aα.

Proposition 3.1.5 The norm of α form K to F , NK/F (α) = det(Aα) is equivalent the following:

n (1) NK/F (α) = (−1) an, where an is the constant term of irred(α) and n is its degree.

Qn (2) NK/F (α) = i=1 αi, where αi are the conjugates of α = α1 over F .

Proof: Let us choose {1, α, . . . , αn−1} as the for K over F . With this basis then:   0 0 0 ··· 0 −a  n     1 0 0 ··· 0 −a   n−1       0 1 0 ··· 0 −an−2  Aα =      0 0 1 ··· 0 −an−3     ......   ......      0 0 0 ··· 1 −a1 which, if you expand along the first row it is easy to see that:

n+1 n+2 n det(Aα) = (−1) (−an) = (−1) an = (−1) an.

30 This proves (1). Factoring the monic irreducible polynomial gets: n n n−1 Y irred(α) = x + a1x + ... + an−1x + an = (x − αi) i=1 and it is easy to see that: n Y n αi = (−1) an i=1 this proves (2). 

Notice that proposition 3.1.5 above implies that NK/F (α) belongs to the p-adic

numbers, Qp. Also for any α ∈ K, NK/F (α) is defined as the of the matrix of multiplication by α in K, it follows that:

NK/F (αβ) = NK/F (α)NK/F (β).

n n−1 Now suppose α has degree n, i.e., irred(α) = x + a1x + ... + an−1x + an Let

K be the field obtained by adjoining α and its conjugates to Qp, this is a finite Galois

extension of Qp. Suppose we have an extension k k of | |p to K. Then by corollary 3.1.3, such a field norm k k is unique. Let σ be an that takes α to a

conjugate α0. The map k k0 : F → R given by kxk0 = kσ(x)k is clearly a field norm 0 0 0 which extends | |p on K. This means k k = k k, so kαk = kαk = kσ(α)k = kα k. This implies that all conjugates have the same norm. Then the norm of |NQp(α)/Qp (α)|p ,

which is in Qp is:

|NQp(α)/Qp (α)|p = kNQp(α)/Qp (α)k n Y = αi i=1 n Y = kαik i=1 = kαkn.

31 1/n Which implies that kαk = |NQp(α)/Qp (α)|p .

If K is any field containing α, then:

[K:Qp(α)] NK/Qp (α) = (NQp(α)/Qp (α)) , and [K : Qp] n = [Qp(α) : Qp] = , [K : Qp(α)] so we can also define kαk to be

1/[K:Qp(α)] |NK/Qp (α)|p .

From now on we use | |p to denote the extension of the p-adic norm, instead of k k.

Theorem 3.1.6 Let K be a finite extension of Qp. Then there exists a field norm on K which extends the norm | |p on Qp.

Proof for this theorem is available in [1 pg 61].

Both theorem 3.1.6 and corollary 3.1.3 imply that there can be only one finite

field extension of the p-adic norm to any finite extension field K on Qp.

Definition 3.1.7 Let K be a finite extension field of Qp, and let A be the set of all x ∈ K such that x is a solution to polynomials of the form:

n n−1 x + a1x + ... + an−1x + an = 0 with ai ∈ Zp. Then we call A the integral of Zp in K.

32 Proposition 3.1.8 Let k be a finite extension of Qp of degree n, and let:

A = {x ∈ K |x|p ≤ 1}

M = {x ∈ K |x|p < 1}.

Then A is a , which is the integral closure of Zp in K. M it its unique maximal , and A/M is a finite extension of Fp of at most degree n.

Proof: First we need to show that A is a ring. A has the structure of K, so we just need to check if it is closed under addition and multiplication. If x, y ∈ A, then

|xy|p ≤ 1, so xy ∈ A. We also have |x + y|p ≤ max(|x|p, |y|p) ≤ 1, so x + y ∈ A. And A is closed under addition and multiplication so A is a ring. For a ∈ A and m ∈ M we have |am|p = |a|p|m|p < 1 which implies am ∈ M. M is an ideal because it is an additive subgroup of A.

Now to show that A is the integral closure os Zp in K. Suppose α ∈ A, then |α|p ≤ 1.

Since α ∈ K, then α is algebraic over Qp with: m m m−1 Y irred(α) = x + am−1x + ... + a0 = (x − αi) i=1 where αi are conjugates of α. All the conjugates of α have the same norm so |αi|p ≤ 1 and αi ∈ A. Also the coefficients of irred(α) are products and sums of the αi, which implies |ai|p ≤ 1. Which implies ai ∈ Zp, and irred(α) ∈ Zp[x].

Also if α ∈ K is the root of a polynomial in Zp[x], it also clear that irred(α) ∈ Zp[x]. m m−1 Now we let the degree of irred(α) be m so that α +am−1α +...+a0 = 0. Suppose

|α|p > 1, then:

m m m−1 |α|p = |α |p = | − am−1α − ... − a0|p

m−1 ≤ max(|am−1α |p,..., |a0|p |p)

m−1 ≤ max(|α |p,..., 1), since |ai|p ≤ 1

m−1 = |α |p

33 but since |α|p > 1 this is a contradiction. So then |α|p ≤ 1 and α ∈ A. Then A is

then integral closure of Zp in K. Now we try to see if M is a . Suppose N is an ideal such that M ⊂ N ⊂

A. Then there exists an α ∈ N but α∈ / M. Which implies |α|p = 1 so |1/α|p = 1

1 and 1/α ∈ A. But then α α = 1 ∈ N, but this is a contradiction since N 6= A. So M is a maximal ideal in A. A is a ring and M is maximal, so A/M is a field. From the definition of M we

have M ∩ Zp = pZp. If a ∈ Zp, then |a|p ≤ 1 so a ∈ A. If a, b ∈ Zp, then

a − b ∈ M ∩ Zp = pZp iff a + M and b + M represent the same . But this means

that for a + pZp ∈ Zp/pZp = Fp, there exists a corresponding a + M ∈ A/M, i.e., Fp

lies in A/M. This means that A/M is an extension field of Fp.

Now we need to show that [A/M : Fp] ≤ n, to do this we need to show that any linear

combination of n + 1 elements over Fp is linearly dependent. So take:

a1 + M, a1 + M, . . . , an+1 + M ∈ A/M, for ai ∈ A.

Since each ai ∈ K then the ai are linearly dependent over Qp. So:

b1a1 + ... + bn+1an+1 = 0, for bi ∈ Qp.

If we let m = min(ordp b1,..., ordp bn+1), the we can multiply the above equation by

−m p and obtain a similar equation where all coefficients are in Zp and at least one is |m| not in pZp (if m < 0 then we multiply by p ). Thus if be map the equation above onto A/M, then we get an equation of the form:

b1a1 + ... + bn+1an+1 + M = 0, for bi ∈ Fp.

Any bi ∈ pZp is mapped to 0+M in A/M, but this is not the case for all the bi. Thus,

the {ai} are linearly dependent. This proves the claim. 

34 Definition 3.1.9 The field A/M as described in theorem 3.1.8 is called the residue field of K.

Putting theorem 3.1.2 and 3.1.6 together, we can conclude that | |p has a unique

extension to any finite field extension of Qp, that we also denoted | |p. Since the

of Qp, denoted Qp, is the union of such extensions, then | |p extends n n−1 uniquely to Qp. This means that if α ∈ Qp has irred(α) = x + a1x + ... + an, 1/n then |α|p = |an|p .

Let K be an extension of Qp of degree n. For α ∈ K we define: 1 ord α = − log |α| = − log | (α)|1/n = − log | (α)| . p p p p NK/Qp p n p NK/Qp p

For any α ∈ Qp, this agrees with our earlier definition of ordp, and clearly it still has the same multiplicative property.

The image of K under the ordp map is contained in:

(1/n)Z = {x ∈ Q nx ∈ Z}

and the image is an additive subgroup of (1/n)Z, so for some positive integer e dividing n, it is of the form (1/e)Z.

Definition 3.1.10 The integer e mentioned above is called the index of ramifica-

tion of K over Qp. If e = 1, then K is called an unramified extension of Qp. If e = n, then K is called totally ramified.

Our upcoming theorem will involve an “Eisenstein equation”. It is well known

that such equations are irreducible over Q, but the same happens to be true over Qp. We will not prove it, but it is a useful fact for the proof of the upcoming theorem.

35 Theorem 3.1.11 If K is totally ramified and α ∈ K has the property ordp α = (1/e), then α satisfies an “Eisenstein equation”:

e e−1 x + ae−1x + ... + a0 = 0, ai ∈ Zp

2 where ai ≡ 0 (mod p) for all i, and a0 6≡ 0 (mod p ). Conversely, if β is a root of such an Eisenstein equation over Qp, then Qp(β) is totally ramified over Qp of degree e.

Proof: The ai are symmetric polynomials in the conjugates of α, and since ordp α =

1/e (1/e), all of the conjugates have | |p = 1/p , which implies that |ai|p ≤ 1. But for

e 2 a0 we have |a0|p = |α|p = 1/p so a0 6≡ 0 (mod p ).

Conversely, since an Eisenstein equation is irreducible over Qp, adjoining a root

β gives us an extension of degree e. Since we have a0 = 1, it implies:

1 ord β = − log |β| = − log |a | = (1/e) ord a = (1/e), p p p e p 0 p p 0 so Qp(β) is totally ramified over Qp. 

36 Bibliography

[1] Neal Koblitz p-adic Numbers, p-adic Analysis, and Zeta-Functions, Second Edi- tion Springer-Verlag 1984.

[2] Evan Turner The p-adic Numbers and Fi- nite Field Extensions of Qp March 2014, url: http://www.math.uchicago.edu/ may/VIGRE/VIGRE2011/REUPapers/Turner.pdf

[3] Theodor Christian Herwig The p-adic Comple- tion of Q and Hensel’s Lemma April 2014, url: http://www.math.uchicago.edu/ may/VIGRE/VIGRE2011/REUPapers/Herwig.pdf

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