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Adventures in p-adic analysis

Tudor Ciurca

March 4, 2019

Introduction : let p be a positive integer prime

The real numbers R are one way to extend the rationals Q, by

“filling in the gaps” relative to the absolute value | · |, which is

one way to measure size in Q. The p-adic numbers are

constructed by considering another measure of size, that

measures “how much a rational number is divisible by p”.

There are 3 main ways to think about the p-adic numbers. I will

go through them starting with the most elementary method.

Introduction : let p be a positive integer prime

The real numbers R are one way to extend the rationals Q, by

“filling in the gaps” relative to the absolute value | · |, which is

one way to measure size in Q. The p-adic numbers are

constructed by considering another measure of size, that

measures “how much a rational number is divisible by p”.

There are 3 main ways to think about the p-adic numbers. I will

go through them starting with the most elementary method.

1 : p-adic numbers as power series in p

A naıve 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.

A p-adic number can be written as

∞∑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.

series in 10−1. A p-adic number can be written as

∞∑k=n

series in 10−1. A p-adic number can be written as

∞∑k=n

2 : p-adic integers as an algebraic limit

Another way to construct the p-adic numbers is by first

constructing the p-adic integers Zp as the object at the end of

the sequence

Z/pZ← Z/p2Z← Z/p3Z← . . .←

For those familiar with category theory, we are just taking the

limit of this diagram. Giving a p-adic integer is the same as

giving an element bk ∈ Z/pkZ for each positive integer k,

compatible with the quotient maps above.

the sequence

Z/pZ← Z/p2Z← Z/p3Z← . . .← Zp

the sequence

Z/pZ← Z/p2Z← Z/p3Z← . . .← Zp

limit of this diagram.

Giving a p-adic integer is the same as

the sequence

Z/pZ← Z/p2Z← Z/p3Z← . . .← Zp

Picture of Z3

Example: 13 in Z5

Since 3 is coprime to 5 we can find its multiplicative inverse

modulo 5k for all positive integers k. For example

3≡ 2 (mod 5),

3≡ 17 (mod 25),

3≡ 42 (mod 125) . . .

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

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

Example: 13 in Z5

Since 3 is coprime to 5 we can find its multiplicative inverse

modulo 5k for all positive integers k. For example

3≡ 2 (mod 5),

3≡ 17 (mod 25),

3≡ 42 (mod 125) . . .

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

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

Relation with the first definition

Let us think of Z/pkZ as the set of representatives 0, 1 . . . pk − 1

in Z. Then we have

k−1∑i=0

so the bk are just partial sums of the power series of a p-adic

integer. To get the field of p-adic numbers Qp, we just add the

fractions spn where s is a p-adic integer and n a positive integer.

Relation with the first definition

Let us think of Z/pkZ as the set of representatives 0, 1 . . . pk − 1

in Z. Then we have

k−1∑i=0

so the bk are just partial sums of the power series of a p-adic

integer. To get the field of p-adic numbers Qp, we just add the

fractions spn where s is a p-adic integer and n a positive integer.

Absolute values on Q

Recall the definition of an absolute value (on Q). It is a function

|| · || : Q→ R

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.

|| · || : Q→ R

q, r ∈ Q.

||q + r|| ≤ ||q||+ ||r|| for all q, r ∈ Q.

|| · || : Q→ R

q, r ∈ Q.

||q + r|| ≤ ||q||+ ||r|| for all q, r ∈ Q.

|| · || : Q→ R

q, r ∈ Q.

||q + r|| ≤ ||q||+ ||r|| for all q, r ∈ Q.

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.

The p-adic absolute value

We start by defining the p-adic valuation. For a ∈ Q×, we

define vp(a) as the unique integer so that we can write

a = pvp(a) · qs

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

||a||p = p−vp(a)

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

The p-adic absolute value

We start by defining the p-adic valuation. For a ∈ Q×, we

define vp(a) as the unique integer so that we can write

a = pvp(a) · qs

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

||a||p = p−vp(a)

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

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

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

We start with the ring Qseq of Cauchy sequences under

point-wise operations. Check that the sum and product of

Cauchy sequences is Cauchy!

Definition

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

Definition

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

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−Bgoes to zero. In other words

A ∼ B ⇐⇒ limn→∞

(||An −Bn||) = 0

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

Q||·|| = Qseq/ ∼

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−Bgoes to zero. In other words

A ∼ B ⇐⇒ limn→∞

(||An −Bn||) = 0

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

Q||·|| = Qseq/ ∼

Completions of Q

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

Qp = Q||·||p

Some other completions of Q include

R = Q||·||R

Q = Q||·||0

It turns out these are all the completions of Q!

Completions of Q

Qp = Q||·||p

R = Q||·||R

Q = Q||·||0

Completions of Q

Qp = Q||·||p

R = Q||·||R

Q = Q||·||0

Ostrowski’s theorem

If F is a non-trivial completion of Q constructed as in the

previous slides, then F must be one of the following:

Qp for some prime p

The absolute value that was used to construct these fields

extends to their completion. These fields become complete with

respect to their absolute value.

Qp for some prime p

Embedding Q into Qp and Z into Zp

Any rational number a ∈ Q can be sent to the constant sequence

{a, a, a . . . }

This gives a natural embedding of Q into Qp.

Since any p-adic

number is a Cauchy sequence of rational numbers, the rationals

are dense in Qp. If a ∈ Z then we can send it to the p-adic

integer represented by

a (mod p), a (mod p2), a (mod p3), . . .

This gives a natural embedding of Z into Zp. The integers are

dense in Zp because a p-adic integer can be approximated by

the integers corresponding to its value modulo high powers of p.

{a, a, a . . . }

This gives a natural embedding of Q into Qp. Since any p-adic

are dense in Qp.

If a ∈ Z then we can send it to the p-adic

{a, a, a . . . }

This gives a natural embedding of Z into Zp.

The integers are

{a, a, a . . . }

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 = ||b||p.

This makes

analysis easier in the p-adic world. A sequence {s0, s1 . . . } is

Cauchy if and only if limk→∞(||sk+1 − sk||p) = 0.

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 = ||b||p. This makes

analysis easier in the p-adic world. A sequence {s0, s1 . . . } is

Cauchy if and only if limk→∞(||sk+1 − sk||p) = 0.

To see this, a sequence is p-adically Cauchy if and only if

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

So it is necessary that limk→∞(||sk+1 − sk||p) = 0. Also

||si − sj ||p ≤ maxk=i...j−1

(||sk+1 − sk||p)

Since∑j−1

k=i(sk+1 − sk) = sj − si. A series∑∞

k=0 ak therefore

converges if and only if limk→∞(||ak||p) = 0.

Example: p-adic exponential function

We can define the exponential function in Qp as

∞∑k=0

This series converges if and only if

limk→∞

k!||p) = lim

k→∞(pvp(k!)−kvp(x)) = 0

This occurs on the open ball

B(0; p− 1

p−1 ) := {q ∈ Qp : ||q||p < p− 1

p−1 }

∞∑k=0

limk→∞

k!||p) = lim

k→∞(pvp(k!)−kvp(x)) = 0

B(0; p− 1

p−1 ) := {q ∈ Qp : ||q||p < p− 1

p−1 }

∞∑k=0

limk→∞

k!||p) = lim

k→∞(pvp(k!)−kvp(x)) = 0

B(0; p− 1

p−1 ) := {q ∈ Qp : ||q||p < p− 1

p−1 }

Correspondence between p-adic analysis and modular arithmetic

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

||x− y||p ≤ p−k ⇐⇒ x ≡ y (mod pk)

where k is a positive integer.

We can see this from our second

construction of Qp. From this we can also conclude that

||x||p ≤ p−k =⇒ x = upk

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

||x− y||p ≤ p−k ⇐⇒ x ≡ y (mod pk)

where k is a positive integer. We can see this from our second

||x||p ≤ p−k =⇒ x = upk

||x− y||p ≤ p−k ⇐⇒ x ≡ y (mod pk)

where k is a positive integer. We can see this from our second

||x||p ≤ p−k =⇒ x = upk

Hensel’s lemma

Let f ∈ Z[x] and r ∈ Zp so that

||f(r)||p ≤ p−k

for a positive integer k ≥ 1. We think of r as an approximate

root to f .

Then under the condition that ||f ′(r)||p = 1 we can

find s ∈ Zp so that

||f(s)||p ≤ p−2k and ||s− r||p ≤ p−k

Thus s is a better approximation, but agrees with r modulo pk.

s is also unique modulo p2k, and satisfies ||f ′(s)||p = 1.

Hensel’s lemma

||f(r)||p ≤ p−k

root to f . Then under the condition that ||f ′(r)||p = 1 we can

||f(s)||p ≤ p−2k and ||s− r||p ≤ p−k

Hensel’s lemma

||f(r)||p ≤ p−k

root to f . Then under the condition that ||f ′(r)||p = 1 we can

||f(s)||p ≤ p−2k and ||s− r||p ≤ p−k

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

f(s) =

N∑i=0

ai(s− r)i = f(r) + (s− r)f ′(r) +

N∑i=2

ai(s− r)i

where N is a positive integer, since f is a polynomial.

We use

the substitution s = r + upk, because we want ||s− r||p ≤ pk.

f(r + upk) = f(r) + upkf ′(r) +N∑i=2

ai(upk)i

= f(r) + upkf ′(r) + p2kN∑i=2

aiuipk(i−2)

f(s) =

N∑i=0

ai(s− r)i = f(r) + (s− r)f ′(r) +

N∑i=2

ai(s− r)i

where N is a positive integer, since f is a polynomial. We use

f(r + upk) = f(r) + upkf ′(r) +N∑i=2

ai(upk)i

= f(r) + upkf ′(r) + p2kN∑i=2

aiuipk(i−2)

f(s) =

N∑i=0

ai(s− r)i = f(r) + (s− r)f ′(r) +

N∑i=2

ai(s− r)i

f(r + upk) = f(r) + upkf ′(r) +

N∑i=2

ai(upk)i

= f(r) + upkf ′(r) + p2kN∑i=2

aiuipk(i−2)

f(s) =

N∑i=0

ai(s− r)i = f(r) + (s− r)f ′(r) +

N∑i=2

ai(s− r)i

f(r + upk) = f(r) + upkf ′(r) +

N∑i=2

ai(upk)i

= f(r) + upkf ′(r) + p2kN∑i=2

aiuipk(i−2)

f(s) = f(r) + upkf ′(r) + p2kg(u)

for some polynomial g with integer coefficients.

f(s) = zpk + upkf ′(r) + p2kg(u) = pk(z + uf ′(r)) + p2kg(u)

for some p-adic integer z, since ||f(r)||p ≤ p−k by assumption.

For ||f(s)||p ≤ p−2k it is clear that we require

||z + uf ′(r)||p ≤ p−k

f(s) = f(r) + upkf ′(r) + p2kg(u)

||z + uf ′(r)||p ≤ p−k

f(s) = f(r) + upkf ′(r) + p2kg(u)

||z + uf ′(r)||p ≤ p−k

Here it may be easier to use modular arithmetic, although a

purely analytic approach is still possible.

||z + uf ′(r)||p ≤ p−k ⇐⇒ z + uf ′(r) ≡ 0 (mod pk)

The condition ||f ′(r)||p = 1 means that f ′(r) is uniquely

invertible modulo pk and so

u ≡ −zf ′(r)

(mod pk)

This produces u uniquely modulo pk, and so s = r + upk is

unique modulo p2k. Check that ||f ′(s)||p = 1 !

u ≡ −zf ′(r)

(mod pk)

u ≡ −zf ′(r)

(mod pk)

u ≡ −zf ′(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

condition ||f ′(s)||p = 1 ensures that we can apply Hensel’s

lemma indefinitely.

Thus we have

Theorem (Hensel’s lifting lemma)

Let f ∈ Z[x] and r ∈ Zp such that f(r) ≡ 0 (mod pk) and

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

condition ||f ′(s)||p = 1 ensures that we can apply Hensel’s

lemma indefinitely. Thus we have

Theorem (Hensel’s lifting lemma)

Let f ∈ Z[x] and r ∈ Zp such that f(r) ≡ 0 (mod pk) and

f ′(r) 6≡ 0 (mod p). Then there is a unique s ∈ Zp such that

f(s) = 0 and s ≡ r (mod pk)

Example:√

2 exists in Z7

This is equivalent to solving f(x) = x2 − 2 = 0 in Z7. We carry

out one iteration of Hensel’s lemma. We start off with

||f(3)||7 = ||32 − 2||7 = 7−1 and ||f ′(3)||7 = ||2 · 3||7 = 1

The formula for the better approximation comes out as

s = 3 + (−f(3)

7f ′(3)) · 7 = 3 + 1 · 7 = 10

Now we can check that

||102 − 2||7 = ||98||7 = 7−2

Example:√

2 exists in Z7

||f(3)||7 = ||32 − 2||7 = 7−1 and ||f ′(3)||7 = ||2 · 3||7 = 1

s = 3 + (−f(3)

7f ′(3)) · 7 = 3 + 1 · 7 = 10

||102 − 2||7 = ||98||7 = 7−2

Example:√

2 exists in Z7

||f(3)||7 = ||32 − 2||7 = 7−1 and ||f ′(3)||7 = ||2 · 3||7 = 1

s = 3 + (−f(3)

7f ′(3)) · 7 = 3 + 1 · 7 = 10

||102 − 2||7 = ||98||7 = 7−2

Example: Zp contains the (p− 1)th 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− 1)xp−2 which is non-zero for non-zero x ∈ Z/pZ.

By Hensel’s lifting lemma, there exist p− 1 distinct roots of

xp−1 − 1 in Zp. These are the (p− 1)th roots of unity in Zp.

xp−1 − 1, by little Fermat. The derivative of this polynomial is

The Chinese Remainder Theorem

Theorem (The Chinese Remainder Theorem)

Let {pk11 . . . pknn } 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

x ≡ xi (mod pkii ) ∀i = 1 . . . n

The above is equivalent to finding an integer x so that

||x− xi||pi ≤ p−ki ∀i = 1 . . . n

It also says that the integers Z are dense in any finite productn∏

x ≡ xi (mod pkii ) ∀i = 1 . . . n

||x− xi||pi ≤ p−ki ∀i = 1 . . . n

x ≡ xi (mod pkii ) ∀i = 1 . . . n

||x− xi||pi ≤ p−ki ∀i = 1 . . . n

The ring of adeles AQ

Why do analysis on each completion separately when we can do

it on all of them at once!

We define the ring of adeles of Q as

the following restricted topological product (of metric spaces)

AQ = R×∏

p prime

The restriction is made as follows. An element of AQ is given by

a real number r ∈ R and elements qp ∈ Qp for each prime p, so

that all but finitely many of the qp are p-adic integers. This is

done to make AQ a locally compact Hausdorff topological ring.

it on all of them at once! We define the ring of adeles of Q as

AQ = R×∏

p prime

AQ = R×∏

p prime

that all but finitely many of the qp are p-adic integers.

This is

AQ = R×∏

p prime

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 {ε1 . . . εn} of positive real numbers, a finite

set {|| · ||1, . . . || · ||n} of distinct absolute values and a finite set

{x1, . . . xn} where xi ∈ Q||·||i. Then there is a rational x so that

||x− xi||i < εi

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

Theorem (Weak approximation theorem)

Consider a finite set {ε1 . . . εn} of positive real numbers, a finite

set {|| · ||1, . . . || · ||n} of distinct absolute values and a finite set

{x1, . . . xn} where xi ∈ Q||·||i. Then there is a rational x so that

||x− xi||i < εi

for all i = 1 . . . n.

Proof sketch

1 We first prove that we can ”discern” any two completions

of Q. That is, for two distinct absolute values || · ||1, || · ||2we 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

limn→∞

(||ynk

1− ynk||i) =

1 if i = k

0 if i 6= k

Proof sketch

limn→∞

(||ynk

1− ynk||i) =

1 if i = k

0 if i 6= k

Proof sketch

limn→∞

(||ynk

1− ynk||i) =

1 if i = k

0 if i 6= k

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∑i=1

1− yki=⇒ (∀i ∈ {1 . . . n})( lim

k→∞(||zk−xi||i) = 0)

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∑i=1

1− yki=⇒ (∀i ∈ {1 . . . n})( lim

k→∞(||zk−xi||i) = 0)

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.

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.

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.

Theorem (Hasse-Minkowski theorem)

The Hasse principle holds for quadratic forms.

Example: poor man’s Fermat two square theorem

Consider the quadratic form x2 + y2 − q where q is a positive

integer. This will always have real roots so we need only focus

on the p-adic cases. We begin by solving x2 ≡ q − y2 (mod p).

One can check that both x2 and q − y2 take p+12 distinct values

as x and y run through Z/pZ respectively. By the pigeonhole

principle two of these values must agree and so a solution exists.

If q 6≡ 0 (mod p) then (0, 0) cannot be a root so there is a root

(x, y) with either x or y non-zero.

We apply Hensel’s lifting lemma to the variable in which the

root is non-zero, to get a p-adic root. We can do this as long as

p 6= 2. Now we deal with the case q ≡ 0 (mod p).

From now on we assume that q is a product of primes congruent

to 1 modulo 4. In the case that p is a positive prime congruent

to 1 modulo 4, we know that −1 ≡ z2 (mod p) has a root.

Hence x2 + y2 ≡ 0 (mod p) has a non-trivial root, which can be

lifted to Zp by Hensel’s lifting lemma.

We apply Hensel’s lifting lemma to the variable in which the

root is non-zero, to get a p-adic root. We can do this as long as

p 6= 2. Now we deal with the case q ≡ 0 (mod p).

From now on we assume that q is a product of primes congruent

to 1 modulo 4. In the case that p is a positive prime congruent

to 1 modulo 4, we know that −1 ≡ z2 (mod p) has a root.

Hence x2 + y2 ≡ 0 (mod p) has a non-trivial root, which can be

lifted to Zp by Hensel’s lifting lemma.

The case p = 2 requires a generalized version of Hensel’s lifting

lemma, as the partial derivatives of x2 + y2− q vanish modulo 2.

Then by the Hasse-Minkowski theorem, there are rational

numbers x, y so that x2 + y2 = q. This produces a non-trivial

integer solution to the quadratic form x2 + y2 = qz2.

The case p = 2 requires a generalized version of Hensel’s lifting

lemma, as the partial derivatives of x2 + y2− q vanish modulo 2.

Then by the Hasse-Minkowski theorem, there are rational

numbers x, y so that x2 + y2 = q. This produces a non-trivial

integer solution to the quadratic form x2 + y2 = qz2.

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 end

Thank you for listening!

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