An Introduction to P-Adic Numbers

An introduction to p-adic numbers

Tudor Ciurca

June 22, 2020 Introduction : let p be a positive integer prime

There are 3 main ways to think about the p-adic numbers. I will go through them starting with the most elementary method. Note that this is a 5-adic integer.

1 : Example

A p-adic number can be thought of as a base-p number where we allow digits to go inﬁnitely far to the left. We can represent 1/3 as a 5-adic integer as follows 1 ... 1313132.0 = ∈ 3 Z5 1 : Example

A p-adic number can be thought of as a base-p number where we allow digits to go inﬁnitely far to the left. We can represent 1/3 as a 5-adic integer as follows 1 ... 1313132.0 = ∈ 3 Z5 Note that this is a 5-adic integer. A p-adic number can be written as

∞ X k akp k=n

where n is an integer and each “digit” ak is one of 0, 1 . . . p − 1. If n is a non-negative integer then we call the p-adic number above a p-adic integer.

1 : p-adic numbers as power series in p

One way to think about the p-adic numbers is as power series in p, just as the real numbers can be thought of as power series in 10−1 [Decimal expansions]. If n is a non-negative integer then we call the p-adic number above a p-adic integer.

1 : p-adic numbers as power series in p

One way to think about the p-adic numbers is as power series in p, just as the real numbers can be thought of as power series in 10−1 [Decimal expansions]. A p-adic number can be written as

∞ X k akp k=n

where n is an integer and each “digit” ak is one of 0, 1 . . . p − 1. 1 : p-adic numbers as power series in p

One way to think about the p-adic numbers is as power series in p, just as the real numbers can be thought of as power series in 10−1 [Decimal expansions]. A p-adic number can be written as

∞ X k akp k=n

where n is an integer and each “digit” ak is one of 0, 1 . . . p − 1. If n is a non-negative integer then we call the p-adic number above a p-adic integer. Let Zp be the p-adic integers.

1 : p-adic numbers as power series in p

For a prime p let Qp be the p-adic numbers. 1 : p-adic numbers as power series in p

For a prime p let Qp be the p-adic numbers. Let Zp be the p-adic integers. To get the ﬁeld of p-adic numbers Qp, we just add the fractions s pn where s is a p-adic integer and n a positive integer.

2 : p-adic integers as an algebraic limit

Another way to construct a p-adic integer is by giving a sequence of elements

2 3 a1 ∈ Z/pZ, a2 ∈ Z/p Z, a3 ∈ Z/p Z ... m which are compatible, so that ak ≡ am (mod p ) if k ≥ m. 2 : p-adic integers as an algebraic limit

Another way to construct a p-adic integer is by giving a sequence of elements

2 3 a1 ∈ Z/pZ, a2 ∈ Z/p Z, a3 ∈ Z/p Z ... m which are compatible, so that ak ≡ am (mod p ) if k ≥ m.

To get the ﬁeld of p-adic numbers Qp, we just add the fractions s pn where s is a p-adic integer and n a positive integer. so the bk which are the partial sums of a, are also the components of a modulo pm.

Relation with the ﬁrst deﬁnition (from 1 to 2)

k k Let us think of Z/p Z as the set of representatives 0, 1 . . . p − 1 P∞ k in Z, and suppose we are given a p-adic integer a = k=0 akp . Then we have

m−1 X k m bm := akp ≡ a (mod p ) k=0 Relation with the ﬁrst deﬁnition (from 1 to 2)

k k Let us think of Z/p Z as the set of representatives 0, 1 . . . p − 1 P∞ k in Z, and suppose we are given a p-adic integer a = k=0 akp . Then we have

m−1 X k m bm := akp ≡ a (mod p ) k=0

so the bk which are the partial sums of a, are also the components of a modulo pm. th ak−ak−1 To get the k coeﬃcient bk we just use pk−1 = bk when k k ≥ 1, and we set b0 = a0. Here we again identify Z/p Z with the representatives 0, 1 . . . pk − 1.

Relation with the ﬁrst deﬁnition (from 2 to 1)

Suppose we are given a compatible sequence k a = {ak ∈ Z/p Z}k≥1. We want to give the power series P∞ k expansion a = k=0 bkp of the corresponding p-adic integer. Relation with the ﬁrst deﬁnition (from 2 to 1)

Suppose we are given a compatible sequence k a = {ak ∈ Z/p Z}k≥1. We want to give the power series P∞ k expansion a = k=0 bkp of the corresponding p-adic integer.

th ak−ak−1 To get the k coeﬃcient bk we just use pk−1 = bk when k k ≥ 1, and we set b0 = a0. Here we again identify Z/p Z with the representatives 0, 1 . . . pk − 1. It can also be written using a power series expansion in 5

1 = 2 + 3 · 5 + 1 · 52 + ... = (... 132.0) 3 5

1 Example: 3 in Z5

Since 3 is coprime to 5 we can ﬁnd its multiplicative inverse modulo 5k for all positive integers k. For example

1 1 1 ≡ 2 (mod 5), ≡ 17 (mod 25), ≡ 42 (mod 125) ... 3 3 3 1 Example: 3 in Z5

Since 3 is coprime to 5 we can ﬁnd its multiplicative inverse modulo 5k for all positive integers k. For example

1 1 1 ≡ 2 (mod 5), ≡ 17 (mod 25), ≡ 42 (mod 125) ... 3 3 3

It can also be written using a power series expansion in 5

1 = 2 + 3 · 5 + 1 · 52 + ... = (... 132.0) 3 5 It can also be written using a power series expansion in 7

2 = 6 + 2 · 7 + 1 · 72 + ... = (... 126.0) 5 7

2 Example: 5 in Z7

Since 5 is coprime to 5 we can ﬁnd its multiplicative inverse modulo 5k for all positive integers k. For example

2 2 2 ≡ 6 (mod 7), ≡ 20 (mod 49), ≡ 69 (mod 343) ... 5 5 5 2 Example: 5 in Z7

Since 5 is coprime to 5 we can ﬁnd its multiplicative inverse modulo 5k for all positive integers k. For example

2 2 2 ≡ 6 (mod 7), ≡ 20 (mod 49), ≡ 69 (mod 343) ... 5 5 5

It can also be written using a power series expansion in 7

2 = 6 + 2 · 7 + 1 · 72 + ... = (... 126.0) 5 7 We select one √ of the representatives for 7:

√ √ √ 7 ≡ 1 (mod 3), 7 ≡ 4 (mod 9), 7 ≡ 13 (mod 27) ...

It can also be written using a power series expansion in 3

√ 2 7 = 1 + 1 · 3 + 1 · 3 + ... = (... 111.0)3

√ Example: 7 in Z3

We will see later that to verify the existence of algebraic numbers in p-adic number systems, it suﬃces to verify their existence modulo p (and some other conditions). It can also be written using a power series expansion in 3

√ 2 7 = 1 + 1 · 3 + 1 · 3 + ... = (... 111.0)3

√ Example: 7 in Z3

We will see later that to verify the existence of algebraic numbers in p-adic number systems, it suﬃces to verify their existence modulo p (and some other conditions). We select one √ of the representatives for 7:

√ √ √ 7 ≡ 1 (mod 3), 7 ≡ 4 (mod 9), 7 ≡ 13 (mod 27) ... √ Example: 7 in Z3

√ √ √ 7 ≡ 1 (mod 3), 7 ≡ 4 (mod 9), 7 ≡ 13 (mod 27) ...

It can also be written using a power series expansion in 3

√ 2 7 = 1 + 1 · 3 + 1 · 3 + ... = (... 111.0)3 Picture of Z3 1 || · || is positive deﬁnite. This means ||q|| ≥ 0 for all q ∈ Q, and equality holds if and only if q = 0.

2 || · || is multiplicative. This means ||q|| · ||r|| = ||q · r|| for all q, r ∈ Q. 3 || · || satisﬁes the triangle inequality. This means ||q + r|| ≤ ||q|| + ||r|| for all q, r ∈ Q.

Absolute values on Q

An absolute value is a nice way of measuring size. It is a function

|| · || : Q → Q satisfying the following conditions: 2 || · || is multiplicative. This means ||q|| · ||r|| = ||q · r|| for all q, r ∈ Q. 3 || · || satisﬁes the triangle inequality. This means ||q + r|| ≤ ||q|| + ||r|| for all q, r ∈ Q.

Absolute values on Q

An absolute value is a nice way of measuring size. It is a function

|| · || : Q → Q satisfying the following conditions: 1 || · || is positive deﬁnite. This means ||q|| ≥ 0 for all q ∈ Q, and equality holds if and only if q = 0. 3 || · || satisﬁes the triangle inequality. This means ||q + r|| ≤ ||q|| + ||r|| for all q, r ∈ Q.

Absolute values on Q

An absolute value is a nice way of measuring size. It is a function

|| · || : Q → Q satisfying the following conditions: 1 || · || is positive deﬁnite. This means ||q|| ≥ 0 for all q ∈ Q, and equality holds if and only if q = 0.

2 || · || is multiplicative. This means ||q|| · ||r|| = ||q · r|| for all q, r ∈ Q. Absolute values on Q

An absolute value is a nice way of measuring size. It is a function

|| · || : Q → Q satisfying the following conditions: 1 || · || is positive deﬁnite. This means ||q|| ≥ 0 for all q ∈ Q, and equality holds if and only if q = 0.

2 || · || is multiplicative. This means ||q|| · ||r|| = ||q · r|| for all q, r ∈ Q. 3 || · || satisﬁes the triangle inequality. This means ||q + r|| ≤ ||q|| + ||r|| for all q, r ∈ Q. The usual absolute value || · ||R, commonly known as | · |.

The trivial absolute value || · ||0 that sends 0 to 0 and everything else to 1. Check that this is an absolute value.

Examples of absolute values on Q

Here are some examples of absolute values on Q. The trivial absolute value || · ||0 that sends 0 to 0 and everything else to 1. Check that this is an absolute value.

Examples of absolute values on Q

Here are some examples of absolute values on Q.

The usual absolute value || · ||R, commonly known as | · |. Examples of absolute values on Q

Here are some examples of absolute values on Q.

The usual absolute value || · ||R, commonly known as | · |.

The trivial absolute value || · ||0 that sends 0 to 0 and everything else to 1. Check that this is an absolute value. It measures how much a is divisible by p. To get the p-adic absolute value, we deﬁne

−vp(a) ||a||p = p

and we set ||0||p = 0, as “0 is divisible by p inﬁnitely often”.

The p-adic absolute values

We start by deﬁning the p-adic valuation vp, which you might be familiar with from Olympiads. For a ∈ Q\{0}, we deﬁne vp(a) as the unique integer so that we can write

q a = pvp(a) · s where q, s are integers coprime to p. The p-adic absolute values

We start by deﬁning the p-adic valuation vp, which you might be familiar with from Olympiads. For a ∈ Q\{0}, we deﬁne vp(a) as the unique integer so that we can write

q a = pvp(a) · s where q, s are integers coprime to p. It measures how much a is divisible by p. To get the p-adic absolute value, we deﬁne

−vp(a) ||a||p = p

and we set ||0||p = 0, as “0 is divisible by p inﬁnitely often”. P∞ k Given a p-adic number a = k=n akp then the partial sums Pm−1 k −m bm = k=n akp satisfy ||a − bm||p ≤ p . This means the partial sums get closer and closer to a as expected. If we deﬁned

vp(a) ||a||p = p then it will not even be an absolute value!

Here we require p to be prime in order for || · ||p to be an absolute value. Due to unique prime factorization, we have vp(ab) = vp(a) + vp(b) for p prime and any a, b ∈ Q.

The p-adic absolute values

We use this deﬁnition because we want numbers to be close together if their diﬀerence is highly divisible by p. Here we require p to be prime in order for || · ||p to be an absolute value. Due to unique prime factorization, we have vp(ab) = vp(a) + vp(b) for p prime and any a, b ∈ Q.

The p-adic absolute values

We use this deﬁnition because we want numbers to be close together if their diﬀerence is highly divisible by p.

P∞ k Given a p-adic number a = k=n akp then the partial sums Pm−1 k −m bm = k=n akp satisfy ||a − bm||p ≤ p . This means the partial sums get closer and closer to a as expected. If we deﬁned

vp(a) ||a||p = p then it will not even be an absolute value! The p-adic absolute values

We use this deﬁnition because we want numbers to be close together if their diﬀerence is highly divisible by p.

vp(a) ||a||p = p then it will not even be an absolute value!

Here we require p to be prime in order for || · ||p to be an absolute value. Due to unique prime factorization, we have vp(ab) = vp(a) + vp(b) for p prime and any a, b ∈ Q. Deﬁnition

A sequence {s0, s1 ... } is Cauchy with respect to || · || if

(∀ ∈ R>0)(∃N ∈ N)(∀i, j ≥ N)(||si − sj|| < )

A Cauchy sequence keeps getting closer to ”something”, but that something may not be there (in Q). We want to add all these missing elements to Q.

3 : p-adic numbers via completions of Q

We can complete Q with respect to any absolute value, generalizing the construction of R using Cauchy sequences. A Cauchy sequence keeps getting closer to ”something”, but that something may not be there (in Q). We want to add all these missing elements to Q.

3 : p-adic numbers via completions of Q

We can complete Q with respect to any absolute value, generalizing the construction of R using Cauchy sequences. Deﬁnition

A sequence {s0, s1 ... } is Cauchy with respect to || · || if

(∀ ∈ R>0)(∃N ∈ N)(∀i, j ≥ N)(||si − sj|| < ) 3 : p-adic numbers via completions of Q

We can complete Q with respect to any absolute value, generalizing the construction of R using Cauchy sequences. Deﬁnition

A sequence {s0, s1 ... } is Cauchy with respect to || · || if

(∀ ∈ R>0)(∃N ∈ N)(∀i, j ≥ N)(||si − sj|| < )

A Cauchy sequence keeps getting closer to ”something”, but that something may not be there (in Q). We want to add all these missing elements to Q. Each Cauchy sequence in Qseq will represent the limit it tends to. However there are too many Cauchy sequences. We write A ∼ B for two Cauchy sequences A, B if their diﬀerence A − B goes to zero. In other words

A ∼ B ⇐⇒ lim (||An − Bn||) = 0 n→∞

3 : p-adic numbers via completions of Q

We start with the ring Qseq of Cauchy sequences under point-wise operations. Check that the sum and product of Cauchy sequences is Cauchy! 3 : p-adic numbers via completions of Q

We start with the ring Qseq of Cauchy sequences under point-wise operations. Check that the sum and product of Cauchy sequences is Cauchy!

Each Cauchy sequence in Qseq will represent the limit it tends to. However there are too many Cauchy sequences. We write A ∼ B for two Cauchy sequences A, B if their diﬀerence A − B goes to zero. In other words

A ∼ B ⇐⇒ lim (||An − Bn||) = 0 n→∞ 3 : p-adic numbers via completions of Q

This means that A and B represent the same limit, so we want them to be the same. ∼ is an equivalence relation. We can ﬁnally deﬁne the completion of Q with respect to || · || as

ˆ Q||·|| = Qseq/ ∼ If Cauchy sequences confuse you, decimal expansions of real numbers are an example of Cauchy sequences. For example {3, 3.1, 3.14, 3.141, 3.1415 ... } approximates π. However Cauchy sequence allow a greater degree of error, as long as eventually we get arbitrarily close to π. Some other completions of Q include

= ˆ R Q||·||R

ˆ Q = Q||·||0

It turns out these are all the completions of Q!

Completions of Q

The p-adic numbers can then be deﬁned as the completions

ˆ Qp = Q||·||p It turns out these are all the completions of Q!

Completions of Q

The p-adic numbers can then be deﬁned as the completions

ˆ Qp = Q||·||p

Some other completions of Q include

= ˆ R Q||·||R

ˆ Q = Q||·||0 Completions of Q

The p-adic numbers can then be deﬁned as the completions

ˆ Qp = Q||·||p

Some other completions of Q include

= ˆ R Q||·||R

ˆ Q = Q||·||0

It turns out these are all the completions of Q! This makes analysis easier in the p-adic world, since convergence is easier to demonstrate. The ultrametric inequality is equivalent to

vp(a + b) ≥ min(vp(a), vp(b))

p-adic analysis is much easier than real analysis

The p-adic absolute value on Qp satisﬁes a stronger form of the triangle inequality, called the ultrametric inequality. This states that for any a, b ∈ Qp we have

||a + b||p ≤ max(||a||p, ||b||p)

and equality can only hold when ||a||p 6= ||b||p. p-adic analysis is much easier than real analysis

The p-adic absolute value on Qp satisﬁes a stronger form of the triangle inequality, called the ultrametric inequality. This states that for any a, b ∈ Qp we have

||a + b||p ≤ max(||a||p, ||b||p)

and equality can only hold when ||a||p 6= ||b||p. This makes analysis easier in the p-adic world, since convergence is easier to demonstrate. The ultrametric inequality is equivalent to

vp(a + b) ≥ min(vp(a), vp(b)) We can see this from our second construction of Qp. From this we can also conclude that

−k k ||x||p ≤ p ⇐⇒ x = up

for some p-adic integer u. We will use these relationships a lot.

Correspondence between p-adic analysis and modular arithmetic

We have the following relationship for p-adic integers x, y

−k k ||x − y||p ≤ p ⇐⇒ x ≡ y (mod p )

where k is a positive integer. for some p-adic integer u. We will use these relationships a lot.

Correspondence between p-adic analysis and modular arithmetic

We have the following relationship for p-adic integers x, y

−k k ||x − y||p ≤ p ⇐⇒ x ≡ y (mod p )

where k is a positive integer. We can see this from our second construction of Qp. From this we can also conclude that

−k k ||x||p ≤ p ⇐⇒ x = up Correspondence between p-adic analysis and modular arithmetic

We have the following relationship for p-adic integers x, y

−k k ||x − y||p ≤ p ⇐⇒ x ≡ y (mod p )

where k is a positive integer. We can see this from our second construction of Qp. From this we can also conclude that

−k k ||x||p ≤ p ⇐⇒ x = up

for some p-adic integer u. We will use these relationships a lot. 0 Then under the condition that ||f (r)||p = 1 we can ﬁnd s ∈ Z so that

−2k −k ||f(s)||p ≤ p and ||s − r||p ≤ p

Thus s is a better approximation, but agrees with r modulo pk. 2k 0 s is also unique modulo p , and satisﬁes ||f (s)||p = 1.

Hensel’s lemma

Let f ∈ Z[x] and r ∈ Z so that

−k ||f(r)||p ≤ p

for a positive integer k ≥ 1. We think of r as an approximate root to f. Thus s is a better approximation, but agrees with r modulo pk. 2k 0 s is also unique modulo p , and satisﬁes ||f (s)||p = 1.

Hensel’s lemma

Let f ∈ Z[x] and r ∈ Z so that

−k ||f(r)||p ≤ p

for a positive integer k ≥ 1. We think of r as an approximate 0 root to f. Then under the condition that ||f (r)||p = 1 we can ﬁnd s ∈ Z so that

−2k −k ||f(s)||p ≤ p and ||s − r||p ≤ p Hensel’s lemma

Let f ∈ Z[x] and r ∈ Z so that

−k ||f(r)||p ≤ p

for a positive integer k ≥ 1. We think of r as an approximate 0 root to f. Then under the condition that ||f (r)||p = 1 we can ﬁnd s ∈ Z so that

−2k −k ||f(s)||p ≤ p and ||s − r||p ≤ p

Thus s is a better approximation, but agrees with r modulo pk. 2k 0 s is also unique modulo p , and satisﬁes ||f (s)||p = 1. Let us do an example.

Let f(x) = x3 − 4 and p = 5. Then r ≡ 4 (mod 5) is a root of f since 43 − 4 = 60 =∼ 0 (mod 5). Now f 0(4) = 3 · 42 = 48 and so 60 we expect 4 − 48 ≡ 4 − 19 · 5 ≡ 9 (mod 25) to be a root of f.

We check that (9)3 − 4 ≡ 729 − 4 ≡ 725 ≡ 0 (mod 25).

Hensel’s lemma

f(r) The explicit formula for s is s = r − f 0(r) [think Newton - Raphson]. We check that (9)3 − 4 ≡ 729 − 4 ≡ 725 ≡ 0 (mod 25).

Hensel’s lemma

f(r) The explicit formula for s is s = r − f 0(r) [think Newton - Raphson]. Let us do an example.

Let f(x) = x3 − 4 and p = 5. Then r ≡ 4 (mod 5) is a root of f since 43 − 4 = 60 =∼ 0 (mod 5). Now f 0(4) = 3 · 42 = 48 and so 60 we expect 4 − 48 ≡ 4 − 19 · 5 ≡ 9 (mod 25) to be a root of f. Hensel’s lemma

f(r) The explicit formula for s is s = r − f 0(r) [think Newton - Raphson]. Let us do an example.

We check that (9)3 − 4 ≡ 729 − 4 ≡ 725 ≡ 0 (mod 25). We f(r) use the substitution s = r − f 0(r) suggested before, and observe

N f(r) f(r) X f(r) f(r − ) = f(r) − f 0(r) + a (− )i = f 0(r) f 0(r) i f 0(r) i=2

N f(r) X f(r) = (− )2 a (− )i−2 f 0(r) i f 0(r) i=2

Proof of Hensel’s lemma

This is analogous to the Newton-Raphson method in R. We write the Taylor expansion to f around r as

N N X i 0 X i f(s) = ai(s − r) = f(r) + (s − r)f (r) + ai(s − r) i=0 i=2 where N is some positive integer, since f is a polynomial. N f(r) X f(r) = (− )2 a (− )i−2 f 0(r) i f 0(r) i=2

Proof of Hensel’s lemma

This is analogous to the Newton-Raphson method in R. We write the Taylor expansion to f around r as

N N X i 0 X i f(s) = ai(s − r) = f(r) + (s − r)f (r) + ai(s − r) i=0 i=2 where N is some positive integer, since f is a polynomial. We f(r) use the substitution s = r − f 0(r) suggested before, and observe

N f(r) f(r) X f(r) f(r − ) = f(r) − f 0(r) + a (− )i = f 0(r) f 0(r) i f 0(r) i=2 Proof of Hensel’s lemma

This is analogous to the Newton-Raphson method in R. We write the Taylor expansion to f around r as

N f(r) f(r) X f(r) f(r − ) = f(r) − f 0(r) + a (− )i = f 0(r) f 0(r) i f 0(r) i=2

N f(r) X f(r) = (− )2 a (− )i−2 f 0(r) i f 0(r) i=2 PN f(r) i−2 Now i=2 ai(− f 0(r) ) is a p-adic integer by assumption, and k f(r) 2 2k since f(r) ≡ 0 (mod p ) we have (− f 0(r) ) ≡ 0 (mod p ). This completes the ﬁrst part of the lemma.

N f(r) X f(r) f 0(r − ) = f 0(r) + ia (− )i−1 f 0(r) i f 0(r) i=2

Proof of Hensel’s lemma

N f(r) f(r) X f(r) f(r − ) = (− )2 a (− )i−2 f 0(r) f 0(r) i f 0(r) i=2 N f(r) X f(r) f 0(r − ) = f 0(r) + ia (− )i−1 f 0(r) i f 0(r) i=2

Proof of Hensel’s lemma

N f(r) f(r) X f(r) f(r − ) = (− )2 a (− )i−2 f 0(r) f 0(r) i f 0(r) i=2

PN f(r) i−2 Now i=2 ai(− f 0(r) ) is a p-adic integer by assumption, and k f(r) 2 2k since f(r) ≡ 0 (mod p ) we have (− f 0(r) ) ≡ 0 (mod p ). This completes the ﬁrst part of the lemma. Proof of Hensel’s lemma

N f(r) f(r) X f(r) f(r − ) = (− )2 a (− )i−2 f 0(r) f 0(r) i f 0(r) i=2

PN f(r) i−2 Now i=2 ai(− f 0(r) ) is a p-adic integer by assumption, and k f(r) 2 2k since f(r) ≡ 0 (mod p ) we have (− f 0(r) ) ≡ 0 (mod p ). This completes the ﬁrst part of the lemma.

N f(r) X f(r) f 0(r − ) = f 0(r) + ia (− )i−1 f 0(r) i f 0(r) i=2 Proof of Hensel’s lemma

N f(r) X f(r) f 0(r − ) = f 0(r) + ia (− )i−1 f 0(r) i f 0(r) i=2 0 Now ||f (r)||p = 1 by assumption and PN f(r) i−1 −1 || i=2 iai(− f 0(r) ) ||p ≤ p and so by the ultrametric 0 f(r) inequality ||f (r − f 0(r) )||p = 1. We are done. Thus we have

Theorem (Hensel’s lifting lemma)

k Let f ∈ Z[x] and r ∈ Zp such that f(r) ≡ 0 (mod p ) and 0 f (r) 6≡ 0 (mod p). Then there is a unique s ∈ Zp such that f(s) = 0 and s ≡ r (mod pk)

Hensel’s lifting lemma

Repeated usage of Hensel’s lemma will give better approximations to p-adic roots of integer polynomials. The 0 condition ||f (s)||p = 1 ensures that we can apply Hensel’s lemma indeﬁnitely. Hensel’s lifting lemma

Theorem (Hensel’s lifting lemma)

k Let f ∈ Z[x] and r ∈ Zp such that f(r) ≡ 0 (mod p ) and 0 f (r) 6≡ 0 (mod p). Then there is a unique s ∈ Zp such that f(s) = 0 and s ≡ r (mod pk) The formula for the better approximation comes out as

−f(3) s = 3 + = 3 + 1 · 7 = 10 f 0(3)

Now we can check that

2 −2 ||10 − 2||7 = ||98||7 = 7

√ Example: 2 exists in Z7

2 This is equivalent to solving f(x) = x − 2 = 0 in Z7. We carry out one iteration of Hensel’s lemma. We start oﬀ with

2 −1 0 ||f(3)||7 = ||3 − 2||7 = 7 and ||f (3)||7 = ||2 · 3||7 = 1 Now we can check that

2 −2 ||10 − 2||7 = ||98||7 = 7

√ Example: 2 exists in Z7

2 This is equivalent to solving f(x) = x − 2 = 0 in Z7. We carry out one iteration of Hensel’s lemma. We start oﬀ with

2 −1 0 ||f(3)||7 = ||3 − 2||7 = 7 and ||f (3)||7 = ||2 · 3||7 = 1

The formula for the better approximation comes out as

−f(3) s = 3 + = 3 + 1 · 7 = 10 f 0(3) √ Example: 2 exists in Z7

2 This is equivalent to solving f(x) = x − 2 = 0 in Z7. We carry out one iteration of Hensel’s lemma. We start oﬀ with

2 −1 0 ||f(3)||7 = ||3 − 2||7 = 7 and ||f (3)||7 = ||2 · 3||7 = 1

The formula for the better approximation comes out as

−f(3) s = 3 + = 3 + 1 · 7 = 10 f 0(3)

Now we can check that

2 −2 ||10 − 2||7 = ||98||7 = 7 The derivative of this polynomial is p−2 (p − 1)x which is non-zero for non-zero x ∈ Z/pZ.

By Hensel’s lifting lemma, there exist p − 1 distinct roots of p−1 th x − 1 in Zp. These are the (p − 1) roots of unity in Zp.

th Example: Zp contains the (p − 1) roots of unity

Every non-zero element of Z/pZ is a root of the polynomial xp−1 − 1, by little Fermat. By Hensel’s lifting lemma, there exist p − 1 distinct roots of p−1 th x − 1 in Zp. These are the (p − 1) roots of unity in Zp.

th Example: Zp contains the (p − 1) roots of unity

Every non-zero element of Z/pZ is a root of the polynomial xp−1 − 1, by little Fermat. The derivative of this polynomial is p−2 (p − 1)x which is non-zero for non-zero x ∈ Z/pZ. th Example: Zp contains the (p − 1) roots of unity

Every non-zero element of Z/pZ is a root of the polynomial xp−1 − 1, by little Fermat. The derivative of this polynomial is p−2 (p − 1)x which is non-zero for non-zero x ∈ Z/pZ.

By Hensel’s lifting lemma, there exist p − 1 distinct roots of p−1 th x − 1 in Zp. These are the (p − 1) roots of unity in Zp. The above is equivalent to ﬁnding an integer x so that

−ki ||x − xi||pi ≤ p ∀i = 1 . . . n It also says that the integers Z are dense in any ﬁnite product n Y Zpi i=1

The Chinese Remainder Theorem restated

Theorem (The Chinese Remainder Theorem)

k1 kn Let {p1 . . . pn } be a ﬁnite set of positive integral powers of distinct primes. Let {x1 . . . xn} be a ﬁnite set of integers. Then there exists an integer x so that

ki x ≡ xi (mod pi ) ∀i = 1 . . . n It also says that the integers Z are dense in any ﬁnite product n Y Zpi i=1

The Chinese Remainder Theorem restated

Theorem (The Chinese Remainder Theorem)

k1 kn Let {p1 . . . pn } be a ﬁnite set of positive integral powers of distinct primes. Let {x1 . . . xn} be a ﬁnite set of integers. Then there exists an integer x so that

ki x ≡ xi (mod pi ) ∀i = 1 . . . n

The above is equivalent to ﬁnding an integer x so that

−ki ||x − xi||pi ≤ p ∀i = 1 . . . n The Chinese Remainder Theorem restated

Theorem (The Chinese Remainder Theorem)

k1 kn Let {p1 . . . pn } be a ﬁnite set of positive integral powers of distinct primes. Let {x1 . . . xn} be a ﬁnite set of integers. Then there exists an integer x so that

ki x ≡ xi (mod pi ) ∀i = 1 . . . n

The above is equivalent to ﬁnding an integer x so that

−ki ||x − xi||pi ≤ p ∀i = 1 . . . n It also says that the integers Z are dense in any ﬁnite product n Y Zpi i=1 Theorem (Weak approximation theorem)

Consider a ﬁnite set {1 . . . n} of positive real numbers, a ﬁnite

set {|| · ||1,... || · ||n} of distinct absolute values and a ﬁnite set ˆ {x1, . . . xn} where xi ∈ Q||·||i . Then there is a rational x so that

||x − xi||i < i

for all i = 1 . . . n.

The weak approximation theorem

The Chinese Remainder Theorem can be restated in a stronger form using p-adic analysis. Here it is The weak approximation theorem

The Chinese Remainder Theorem can be restated in a stronger form using p-adic analysis. Here it is Theorem (Weak approximation theorem)

Consider a ﬁnite set {1 . . . n} of positive real numbers, a ﬁnite

set {|| · ||1,... || · ||n} of distinct absolute values and a ﬁnite set ˆ {x1, . . . xn} where xi ∈ Q||·||i . Then there is a rational x so that

||x − xi||i < i

for all i = 1 . . . n. 2 We strengthen this to show that for a ﬁnite set

{|| · ||1 ... || · ||n} of distinct absolute values we may ﬁnd a

rational y so that ||y||1 > 1 and ||y||i < 1 for all i 6= 1.

3 For a ﬁnite set of absolute values as above, we write yk for

the rational number so that ||yk||k > 1 and ||yk||i < 1 for all i 6= k. Then we have  n yk 1 if i = k lim (|| ||i) = n→∞ n 1 − yk 0 if i 6= k

Proof sketch

1 We first prove that we can ”discern” any two completions of Q. That is, for two distinct absolute values || · ||1, || · ||2 we may find a rational x so that ||x||1 < 1 and ||x||2 > 1. 3 For a finite set of absolute values as above, we write yk for

the rational number so that ||yk||k > 1 and ||yk||i < 1 for all i 6= k. Then we have  n yk 1 if i = k lim (|| ||i) = n→∞ n 1 − yk 0 if i 6= k

Proof sketch

1 We ﬁrst prove that we can ”discern” any two completions of Q. That is, for two distinct absolute values || · ||1, || · ||2 we may ﬁnd a rational x so that ||x||1 < 1 and ||x||2 > 1.

2 We strengthen this to show that for a ﬁnite set

{|| · ||1 ... || · ||n} of distinct absolute values we may ﬁnd a

rational y so that ||y||1 > 1 and ||y||i < 1 for all i 6= 1. Proof sketch

1 We ﬁrst prove that we can ”discern” any two completions of Q. That is, for two distinct absolute values || · ||1, || · ||2 we may ﬁnd a rational x so that ||x||1 < 1 and ||x||2 > 1.

2 We strengthen this to show that for a ﬁnite set

{|| · ||1 ... || · ||n} of distinct absolute values we may ﬁnd a

rational y so that ||y||1 > 1 and ||y||i < 1 for all i 6= 1.

3 For a ﬁnite set of absolute values as above, we write yk for

the rational number so that ||yk||k > 1 and ||yk||i < 1 for all i 6= k. Then we have  n yk 1 if i = k lim (|| ||i) = n→∞ n 1 − yk 0 if i 6= k 5 The rationals are dense in each of the completions and so

we can approximate the elements xi in the theorem arbitrarily well by rational numbers. Then we construct

the sequence of rationals zn above, which for large enough n will satisfy the conditions of the theorem.

Proof sketch

4 It is now easy to show that for any ﬁnite set

{|| · ||1 ... || · ||n} of absolute values and any ﬁnite set

{x1 . . . xn} of rational numbers, we can set

k k X yi zk := xi =⇒ (∀i ∈ {1 . . . n})( lim (||zk−xi||i) = 0) 1 − yk k→∞ i=1 i Proof sketch

4 It is now easy to show that for any ﬁnite set

{|| · ||1 ... || · ||n} of absolute values and any ﬁnite set

{x1 . . . xn} of rational numbers, we can set

k k X yi zk := xi =⇒ (∀i ∈ {1 . . . n})( lim (||zk−xi||i) = 0) 1 − yk k→∞ i=1 i

5 The rationals are dense in each of the completions and so

we can approximate the elements xi in the theorem arbitrarily well by rational numbers. Then we construct

the sequence of rationals zn above, which for large enough n will satisfy the conditions of the theorem. Theorem (Hasse-Minkowski theorem)

The Hasse principle holds for quadratic forms.

So a rational solution to a degree two equation exists if it exists in every completion. Solving equations in complete ﬁelds is easy due to Hensel’s lemma and the Newton-Raphson method.

The Hasse principle

Let f be a multivariate polynomial with integer coefficients. We say f satisfies the Hasse principle when [f has a rational root if and only if it has a root in every completion of Q] holds. So a rational solution to a degree two equation exists if it exists in every completion. Solving equations in complete fields is easy due to Hensel’s lemma and the Newton-Raphson method.

The Hasse principle

Let f be a multivariate polynomial with integer coeﬃcients. We say f satisﬁes the Hasse principle when [f has a rational root if and only if it has a root in every completion of Q] holds.

Theorem (Hasse-Minkowski theorem)

The Hasse principle holds for quadratic forms. The Hasse principle

Let f be a multivariate polynomial with integer coeﬃcients. We say f satisﬁes the Hasse principle when [f has a rational root if and only if it has a root in every completion of Q] holds.

Theorem (Hasse-Minkowski theorem)

The Hasse principle holds for quadratic forms.

So a rational solution to a degree two equation exists if it exists in every completion. Solving equations in complete ﬁelds is easy due to Hensel’s lemma and the Newton-Raphson method. The Hasse principle does not give explicit solutions (there are nice explicit ways to solve degree 2 equations) however it is important in modern number theory. It is part of the philosophy that you can solve problems in number theory and geometry by solving the p-adic versions of the problem.

Counter-example to the Hasse principle

The Hasse principle does not extend to cubic forms. The cubic form 3x3 + 4y3 + 5z3 has a non-zero real root and non-zero roots in Qp for every prime p, but no non-zero rational root. Counter-example to the Hasse principle

The Hasse principle does not extend to cubic forms. The cubic form 3x3 + 4y3 + 5z3 has a non-zero real root and non-zero roots in Qp for every prime p, but no non-zero rational root.

The Hasse principle does not give explicit solutions (there are nice explicit ways to solve degree 2 equations) however it is important in modern number theory. It is part of the philosophy that you can solve problems in number theory and geometry by solving the p-adic versions of the problem. The end

Thank you for listening!