Search Results for “chinese remainder theorem”

Source: Chinese remainder theorem

In mathematics, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1).
For example, if we know that the remainder of n divided by 3 is 2, the remainder of n divided by 5 is 3, and the remainder of n divided by 7 is 2, then with no other information, we can determine that the remainder of n divided by 105 (the product of 3, 5, and 7) is 23. Importantly, this tells us that if n is a natural number less than 105, then 23 is the only possible value of n.
It is also known as Sunzi's theorem, as the earliest known statement is by the Chinese mathematician Sunzi in the Sunzi Suanjing in the 3rd to 5th century CE.
The Chinese remainder theorem is widely used for computing with large integers, as it allows replacing a computation for which one knows a bound on the size of the result by several similar computations on small integers.
The Chinese remainder theorem (expressed in terms of congruences) is true over every principal ideal domain. It has been generalized to any ring, with a formulation involving two-sided ideals.

History

Chinese

Statement

Proof

= Uniqueness

remainder

theorem

= Existence (first proof)

= Existence (constructive proof)

= Existence (direct construction)

Computation

and these methods are applied on the example

x

≡
0

(
mod

3
)

x

≡
3

(
mod

4
)

x

≡
4

(
mod

5
)

.

{\displaystyle {\begin{aligned}x&\equiv 0{\pmod {3}}\\x&\equiv 3{\pmod {4}}\\x&\equiv 4{\pmod {5}}.\end{aligned}}}

Several methods of computation are presented. The two first ones are useful for small examples, but become very inefficient when the product

n

1

⋯

n

k

{\displaystyle n_{1}\cdots n_{k}}

is large. The third one uses the existence proof given in § Existence (constructive proof). It is the most convenient when the product

n

1

⋯

n

k

{\displaystyle n_{1}\cdots n_{k}}

is large, or for computer computation.

= Systematic search

=
It is easy to check whether a value of x is a solution: it suffices to compute the remainder of the Euclidean division of x by each ni. Thus, to find the solution, it suffices to check successively the integers from 0 to N until finding the solution.
Although very simple, this method is very inefficient. For the simple example considered here, 40 integers (including 0) have to be checked for finding the solution, which is 39. This is an exponential time algorithm, as the size of the input is, up to a constant factor, the number of digits of N, and the average number of operations is of the order of N.
Therefore, this method is rarely used, neither for hand-written computation nor on computers.

= Search by sieving

=

The search of the solution may be made dramatically faster by sieving. For this method, we suppose, without loss of generality, that

0
≤

a

i

<

n

i

{\displaystyle 0\leq a_{i}
(if it were not the case, it would suffice to replace each

a

i

{\displaystyle a_{i}}

by the remainder of its division by

n

i

{\displaystyle n_{i}}

). This implies that the solution belongs to the arithmetic progression

a

1

,

a

1

+

n

1

,

a

1

+
2

n

1

,
…

{\displaystyle a_{1},a_{1}+n_{1},a_{1}+2n_{1},\ldots }

By testing the values of these numbers modulo

n

2

,

{\displaystyle n_{2},}

one eventually finds a solution

x

2

{\displaystyle x_{2}}

of the two first congruences. Then the solution belongs to the arithmetic progression

x

2

,

x

2

+

n

1

n

2

,

x

2

+
2

n

1

n

2

,
…

{\displaystyle x_{2},x_{2}+n_{1}n_{2},x_{2}+2n_{1}n_{2},\ldots }

Testing the values of these numbers modulo

n

3

,

{\displaystyle n_{3},}

and continuing until every modulus has been tested eventually yields the solution.
This method is faster if the moduli have been ordered by decreasing value, that is if

n

1

>

n

2

>
⋯
>

n

k

.

{\displaystyle n_{1}>n_{2}>\cdots >n_{k}.}

For the example, this gives the following computation. We consider first the numbers that are congruent to 4 modulo 5 (the largest modulus), which are 4, 9 = 4 + 5, 14 = 9 + 5, ... For each of them, compute the remainder by 4 (the second largest modulus) until getting a number congruent to 3 modulo 4. Then one can proceed by adding 20 = 5 × 4 at each step, and computing only the remainders by 3. This gives

4 mod 4 → 0. Continue
4 + 5 = 9 mod 4 →1. Continue
9 + 5 = 14 mod 4 → 2. Continue
14 + 5 = 19 mod 4 → 3. OK, continue by considering remainders modulo 3 and adding 5 × 4 = 20 each time
19 mod 3 → 1. Continue
19 + 20 = 39 mod 3 → 0. OK, this is the result.
This method works well for hand-written computation with a product of moduli that is not too big. However, it is much slower than other methods, for very large products of moduli. Although dramatically faster than the systematic search, this method also has an exponential time complexity and is therefore not used on computers.

= Using the existence construction

=
The constructive existence proof shows that, in the case of two moduli, the solution may be obtained by the computation of the Bézout coefficients of the moduli, followed by a few multiplications, additions and reductions modulo

n

1

n

2

{\displaystyle n_{1}n_{2}}

(for getting a result in the interval

(
0
,

n

1

n

2

−
1
)

{\displaystyle (0,n_{1}n_{2}-1)}

). As the Bézout's coefficients may be computed with the extended Euclidean algorithm, the whole computation, at most, has a quadratic time complexity of

O
(
(

s

1

+

s

2

)

2

)
,

{\displaystyle O((s_{1}+s_{2})^{2}),}

where

s

i

{\displaystyle s_{i}}

denotes the number of digits of

n

i

.

{\displaystyle n_{i}.}

For more than two moduli, the method for two moduli allows the replacement of any two congruences by a single congruence modulo the product of the moduli. Iterating this process provides eventually the solution with a complexity, which is quadratic in the number of digits of the product of all moduli. This quadratic time complexity does not depend on the order in which the moduli are regrouped. One may regroup the two first moduli, then regroup the resulting modulus with the next one, and so on. This strategy is the easiest to implement, but it also requires more computation involving large numbers.
Another strategy consists in partitioning the moduli in pairs whose product have comparable sizes (as much as possible), applying, in parallel, the method of two moduli to each pair, and iterating with a number of moduli approximatively divided by two. This method allows an easy parallelization of the algorithm. Also, if fast algorithms (that is, algorithms working in quasilinear time) are used for the basic operations, this method provides an algorithm for the whole computation that works in quasilinear time.
On the current example (which has only three moduli), both strategies are identical and work as follows.
Bézout's identity for 3 and 4 is

1
×
4
+
(
−
1
)
×
3
=
1.

{\displaystyle 1\times 4+(-1)\times 3=1.}

Putting this in the formula given for proving the existence gives

0
×
1
×
4
+
3
×
(
−
1
)
×
3
=
−
9

{\displaystyle 0\times 1\times 4+3\times (-1)\times 3=-9}

for a solution of the two first congruences, the other solutions being obtained by adding to −9 any multiple of 3 × 4 = 12. One may continue with any of these solutions, but the solution 3 = −9 +12 is smaller (in absolute value) and thus leads probably to an easier computation
Bézout identity for 5 and 3 × 4 = 12 is

5
×
5
+
(
−
2
)
×
12
=
1.

{\displaystyle 5\times 5+(-2)\times 12=1.}

Applying the same formula again, we get a solution of the problem:

5
×
5
×
3
+
12
×
(
−
2
)
×
4
=
−
21.

{\displaystyle 5\times 5\times 3+12\times (-2)\times 4=-21.}

The other solutions are obtained by adding any multiple of 3 × 4 × 5 = 60, and the smallest positive solution is −21 + 60 = 39.

= As a linear Diophantine system

=
The system of congruences solved by the Chinese remainder theorem may be rewritten as a system of linear Diophantine equations:

x

=

a

1

+

x

1

n

1

⋮

x

=

a

k

+

x

k

n

k

,

{\displaystyle {\begin{aligned}x&=a_{1}+x_{1}n_{1}\\&\vdots \\x&=a_{k}+x_{k}n_{k},\end{aligned}}}

where the unknown integers are

x

{\displaystyle x}

and the

x

i

.

{\displaystyle x_{i}.}

Therefore, every general method for solving such systems may be used for finding the solution of Chinese remainder theorem, such as the reduction of the matrix of the system to Smith normal form or Hermite normal form. However, as usual when using a general algorithm for a more specific problem, this approach is less efficient than the method of the preceding section, based on a direct use of Bézout's identity.

Over principal ideal domains

In § Statement, the Chinese remainder theorem has been stated in three different ways: in terms of remainders, of congruences, and of a ring isomorphism. The statement in terms of remainders does not apply, in general, to principal ideal domains, as remainders are not defined in such rings. However, the two other versions make sense over a principal ideal domain R: it suffices to replace "integer" by "element of the domain" and

Z

{\displaystyle \mathbb {Z} }

by R. These two versions of the theorem are true in this context, because the proofs (except for the first existence proof), are based on Euclid's lemma and Bézout's identity, which are true over every principal domain.
However, in general, the theorem is only an existence theorem and does not provide any way for computing the solution, unless one has an algorithm for computing the coefficients of Bézout's identity.

Over univariate polynomial rings and Euclidean domains

The statement in terms of remainders given in § Theorem statement cannot be generalized to any principal ideal domain, but its generalization to Euclidean domains is straightforward. The univariate polynomials over a field is the typical example of a Euclidean domain which is not the integers. Therefore, we state the theorem for the case of the ring

R
=
K
[
X
]

{\displaystyle R=K[X]}

for a field

K
.

{\displaystyle K.}

For getting the theorem for a general Euclidean domain, it suffices to replace the degree by the Euclidean function of the Euclidean domain.
The Chinese remainder theorem for polynomials is thus: Let

P

i

(
X
)

{\displaystyle P_{i}(X)}

(the moduli) be, for

i
=
1
,
…
,
k

{\displaystyle i=1,\dots ,k}

, pairwise coprime polynomials in

R
=
K
[
X
]

{\displaystyle R=K[X]}

. Let

d

i

=
deg
⁡

P

i

{\displaystyle d_{i}=\deg P_{i}}

be the degree of

P

i

(
X
)

{\displaystyle P_{i}(X)}

, and

D

{\displaystyle D}

be the sum of the

d

i

.

{\displaystyle d_{i}.}

If

A

i

(
X
)
,
…
,

A

k

(
X
)

{\displaystyle A_{i}(X),\ldots ,A_{k}(X)}

are polynomials such that

A

i

(
X
)
=
0

{\displaystyle A_{i}(X)=0}

or

deg
⁡

A

i

<

d

i

{\displaystyle \deg A_{i}
for every i, then, there is one and only one polynomial

P
(
X
)

{\displaystyle P(X)}

, such that

deg
⁡
P
<
D

{\displaystyle \deg P
and the remainder of the Euclidean division of

P
(
X
)

{\displaystyle P(X)}

by

P

i

(
X
)

{\displaystyle P_{i}(X)}

is

A

i

(
X
)

{\displaystyle A_{i}(X)}

for every i.
The construction of the solution may be done as in § Existence (constructive proof) or § Existence (direct proof). However, the latter construction may be simplified by using, as follows, partial fraction decomposition instead of the extended Euclidean algorithm.
Thus, we want to find a polynomial

P
(
X
)

{\displaystyle P(X)}

, which satisfies the congruences

P
(
X
)
≡

A

i

(
X
)

(
mod

P

i

(
X
)
)

,

{\displaystyle P(X)\equiv A_{i}(X){\pmod {P_{i}(X)}},}

for

i
=
1
,
…
,
k
.

{\displaystyle i=1,\ldots ,k.}

Consider the polynomials

Q
(
X
)

=

∏

i
=
1

k

P

i

(
X
)

Q

i

(
X
)

=

Q
(
X
)

P

i

(
X
)

.

{\displaystyle {\begin{aligned}Q(X)&=\prod _{i=1}^{k}P_{i}(X)\\Q_{i}(X)&={\frac {Q(X)}{P_{i}(X)}}.\end{aligned}}}

The partial fraction decomposition of

1

/

Q
(
X
)

{\displaystyle 1/Q(X)}

gives k polynomials

S

i

(
X
)

{\displaystyle S_{i}(X)}

with degrees

deg
⁡

S

i

(
X
)
<

d

i

,

{\displaystyle \deg S_{i}(X)
such that

1

Q
(
X
)

=

∑

i
=
1

k

S

i

(
X
)

P

i

(
X
)

,

{\displaystyle {\frac {1}{Q(X)}}=\sum _{i=1}^{k}{\frac {S_{i}(X)}{P_{i}(X)}},}

and thus

1
=

∑

i
=
1

k

S

i

(
X
)

Q

i

(
X
)
.

{\displaystyle 1=\sum _{i=1}^{k}S_{i}(X)Q_{i}(X).}

Then a solution of the simultaneous congruence system is given by the polynomial

∑

i
=
1

k

A

i

(
X
)

S

i

(
X
)

Q

i

(
X
)
.

{\displaystyle \sum _{i=1}^{k}A_{i}(X)S_{i}(X)Q_{i}(X).}

In fact, we have

∑

i
=
1

k

A

i

(
X
)

S

i

(
X
)

Q

i

(
X
)
=

A

i

(
X
)
+

∑

j
=
1

k

(

A

j

(
X
)
−

A

i

(
X
)
)

S

j

(
X
)

Q

j

(
X
)
≡

A

i

(
X
)

(
mod

P

i

(
X
)
)

,

{\displaystyle \sum _{i=1}^{k}A_{i}(X)S_{i}(X)Q_{i}(X)=A_{i}(X)+\sum _{j=1}^{k}(A_{j}(X)-A_{i}(X))S_{j}(X)Q_{j}(X)\equiv A_{i}(X){\pmod {P_{i}(X)}},}

for

1
≤
i
≤
k
.

{\displaystyle 1\leq i\leq k.}

This solution may have a degree larger than

D
=

∑

i
=
1

k

d

i

.

{\displaystyle D=\sum _{i=1}^{k}d_{i}.}

The unique solution of degree less than

D

{\displaystyle D}

may be deduced by considering the remainder

B

i

(
X
)

{\displaystyle B_{i}(X)}

of the Euclidean division of

A

i

(
X
)

S

i

(
X
)

{\displaystyle A_{i}(X)S_{i}(X)}

by

P

i

(
X
)
.

{\displaystyle P_{i}(X).}

This solution is

P
(
X
)
=

∑

i
=
1

k

B

i

(
X
)

Q

i

(
X
)
.

{\displaystyle P(X)=\sum _{i=1}^{k}B_{i}(X)Q_{i}(X).}

= Lagrange interpolation

=
A special case of Chinese remainder theorem for polynomials is Lagrange interpolation. For this, consider k monic polynomials of degree one:

P

i

(
X
)
=
X
−

x

i

.

{\displaystyle P_{i}(X)=X-x_{i}.}

They are pairwise coprime if the

x

i

{\displaystyle x_{i}}

are all different. The remainder of the division by

P

i

(
X
)

{\displaystyle P_{i}(X)}

of a polynomial

P
(
X
)

{\displaystyle P(X)}

is

P
(

x

i

)

{\displaystyle P(x_{i})}

, by the polynomial remainder theorem.
Now, let

A

1

,
…
,

A

k

{\displaystyle A_{1},\ldots ,A_{k}}

be constants (polynomials of degree 0) in

K
.

{\displaystyle K.}

Both Lagrange interpolation and Chinese remainder theorem assert the existence of a unique polynomial

P
(
X
)
,

{\displaystyle P(X),}

of degree less than

k

{\displaystyle k}

such that

P
(

x

i

)
=

A

i

,

{\displaystyle P(x_{i})=A_{i},}

for every

i
.

{\displaystyle i.}

Lagrange interpolation formula is exactly the result, in this case, of the above construction of the solution. More precisely, let

Q
(
X
)

=

∏

i
=
1

k

(
X
−

x

i

)

Q

i

(
X
)

=

Q
(
X
)

X
−

x

i

.

{\displaystyle {\begin{aligned}Q(X)&=\prod _{i=1}^{k}(X-x_{i})\\[6pt]Q_{i}(X)&={\frac {Q(X)}{X-x_{i}}}.\end{aligned}}}

The partial fraction decomposition of

1

Q
(
X
)

{\displaystyle {\frac {1}{Q(X)}}}

is

1

Q
(
X
)

=

∑

i
=
1

k

1

Q

i

(

x

i

)
(
X
−

x

i

)

.

{\displaystyle {\frac {1}{Q(X)}}=\sum _{i=1}^{k}{\frac {1}{Q_{i}(x_{i})(X-x_{i})}}.}

In fact, reducing the right-hand side to a common denominator one gets

∑

i
=
1

k

1

Q

i

(

x

i

)
(
X
−

x

i

)

=

1

Q
(
X
)

∑

i
=
1

k

Q

i

(
X
)

Q

i

(

x

i

)

,

{\displaystyle \sum _{i=1}^{k}{\frac {1}{Q_{i}(x_{i})(X-x_{i})}}={\frac {1}{Q(X)}}\sum _{i=1}^{k}{\frac {Q_{i}(X)}{Q_{i}(x_{i})}},}

and the numerator is equal to one, as being a polynomial of degree less than

k
,

{\displaystyle k,}

which takes the value one for

k

{\displaystyle k}

different values of

X
.

{\displaystyle X.}

Using the above general formula, we get the Lagrange interpolation formula:

P
(
X
)
=

∑

i
=
1

k

A

i

Q

i

(
X
)

Q

i

(

x

i

)

.

{\displaystyle P(X)=\sum _{i=1}^{k}A_{i}{\frac {Q_{i}(X)}{Q_{i}(x_{i})}}.}

= Hermite interpolation

=
Hermite interpolation is an application of the Chinese remainder theorem for univariate polynomials, which may involve moduli of arbitrary degrees (Lagrange interpolation involves only moduli of degree one).
The problem consists of finding a polynomial of the least possible degree, such that the polynomial and its first derivatives take given values at some fixed points.
More precisely, let

x

1

,
…
,

x

k

{\displaystyle x_{1},\ldots ,x_{k}}

be

k

{\displaystyle k}

elements of the ground field

K
,

{\displaystyle K,}

and, for

i
=
1
,
…
,
k
,

{\displaystyle i=1,\ldots ,k,}

let

a

i
,
0

,

a

i
,
1

,
…
,

a

i
,

r

i

−
1

{\displaystyle a_{i,0},a_{i,1},\ldots ,a_{i,r_{i}-1}}

be the values of the first

r

i

{\displaystyle r_{i}}

derivatives of the sought polynomial at

x

i

{\displaystyle x_{i}}

(including the 0th derivative, which is the value of the polynomial itself). The problem is to find a polynomial

P
(
X
)

{\displaystyle P(X)}

such that its j th derivative takes the value

a

i
,
j

{\displaystyle a_{i,j}}

at

x

i

,

{\displaystyle x_{i},}

for

i
=
1
,
…
,
k

{\displaystyle i=1,\ldots ,k}

and

j
=
0
,
…
,

r

j

.

{\displaystyle j=0,\ldots ,r_{j}.}

Consider the polynomial

P

i

(
X
)
=

∑

j
=
0

r

i

−
1

a

i
,
j

j
!

(
X
−

x

i

)

j

.

{\displaystyle P_{i}(X)=\sum _{j=0}^{r_{i}-1}{\frac {a_{i,j}}{j!}}(X-x_{i})^{j}.}

This is the Taylor polynomial of order

r

i

−
1

{\displaystyle r_{i}-1}

at

x

i

{\displaystyle x_{i}}

, of the unknown polynomial

P
(
X
)
.

{\displaystyle P(X).}

Therefore, we must have

P
(
X
)
≡

P

i

(
X
)

(
mod

(
X
−

x

i

)

r

i

)

.

{\displaystyle P(X)\equiv P_{i}(X){\pmod {(X-x_{i})^{r_{i}}}}.}

Conversely, any polynomial

P
(
X
)

{\displaystyle P(X)}

that satisfies these

k

{\displaystyle k}

congruences, in particular verifies, for any

i
=
1
,
…
,
k

{\displaystyle i=1,\ldots ,k}

P
(
X
)
=

P

i

(
X
)
+
o
(
X
−

x

i

)

r

i

−
1

{\displaystyle P(X)=P_{i}(X)+o(X-x_{i})^{r_{i}-1}}

therefore

P

i

(
X
)

{\displaystyle P_{i}(X)}

is its Taylor polynomial of order

r

i

−
1

{\displaystyle r_{i}-1}

at

x

i

{\displaystyle x_{i}}

, that is,

P
(
X
)

{\displaystyle P(X)}

solves the initial Hermite interpolation problem.
The Chinese remainder theorem asserts that there exists exactly one polynomial of degree less than the sum of the

r

i

,

{\displaystyle r_{i},}

which satisfies these

k

{\displaystyle k}

congruences.
There are several ways for computing the solution

P
(
X
)
.

{\displaystyle P(X).}

One may use the method described at the beginning of § Over univariate polynomial rings and Euclidean domains. One may also use the constructions given in § Existence (constructive proof) or § Existence (direct proof).

Generalization to non-coprime moduli

The Chinese remainder theorem can be generalized to non-coprime moduli. Let

m
,
n
,
a
,
b

{\displaystyle m,n,a,b}

be any integers, let

g
=
gcd
(
m
,
n
)

{\displaystyle g=\gcd(m,n)}

;

M
=
lcm
⁡
(
m
,
n
)

{\displaystyle M=\operatorname {lcm} (m,n)}

, and consider the system of congruences:

x

≡
a

(
mod

m
)

x

≡
b

(
mod

n
)

,

{\displaystyle {\begin{aligned}x&\equiv a{\pmod {m}}\\x&\equiv b{\pmod {n}},\end{aligned}}}

If

a
≡
b

(
mod

g
)

{\displaystyle a\equiv b{\pmod {g}}}

, then this system has a unique solution modulo

M
=
m
n

/

g

{\displaystyle M=mn/g}

. Otherwise, it has no solutions.
If one uses Bézout's identity to write

g
=
u
m
+
v
n

{\displaystyle g=um+vn}

, then the solution is given by

x
=

a
v
n
+
b
u
m

g

.

{\displaystyle x={\frac {avn+bum}{g}}.}

This defines an integer, as g divides both m and n. Otherwise, the proof is very similar to that for coprime moduli.

Generalization to arbitrary rings

The Chinese remainder theorem can be generalized to any ring, by using coprime ideals (also called comaximal ideals). Two ideals I and J are coprime if there are elements

i
∈
I

{\displaystyle i\in I}

and

j
∈
J

{\displaystyle j\in J}

such that

i
+
j
=
1.

{\displaystyle i+j=1.}

This relation plays the role of Bézout's identity in the proofs related to this generalization, which otherwise are very similar. The generalization may be stated as follows.
Let I1, ..., Ik be two-sided ideals of a ring

R

{\displaystyle R}

and let I be their intersection. If the ideals are pairwise coprime, we have the isomorphism:

R

/

I

→
(
R

/

I

1

)
×
⋯
×
(
R

/

I

k

)

x

mod

I

↦
(
x

mod

I

1

,

…
,

x

mod

I

k

)
,

{\displaystyle {\begin{aligned}R/I&\to (R/I_{1})\times \cdots \times (R/I_{k})\\x{\bmod {I}}&\mapsto (x{\bmod {I}}_{1},\,\ldots ,\,x{\bmod {I}}_{k}),\end{aligned}}}

between the quotient ring

R

/

I

{\displaystyle R/I}

and the direct product of the

R

/

I

i

,

{\displaystyle R/I_{i},}

where "

x

mod

I

{\displaystyle x{\bmod {I}}}

" denotes the image of the element

x

{\displaystyle x}

in the quotient ring defined by the ideal

I
.

{\displaystyle I.}

Moreover, if

R

{\displaystyle R}

is commutative, then the ideal intersection of pairwise coprime ideals is equal to their product; that is

I
=

I

1

∩

I

2

∩
⋯
∩

I

k

=

I

1

I

2

⋯

I

k

,

{\displaystyle I=I_{1}\cap I_{2}\cap \cdots \cap I_{k}=I_{1}I_{2}\cdots I_{k},}

if Ii and Ij are coprime for all i ≠ j.

= Interpretation in terms of idempotents

=
Let

I

1

,

I

2

,
…
,

I

k

{\displaystyle I_{1},I_{2},\dots ,I_{k}}

be pairwise coprime two-sided ideals with

⋂

i
=
1

k

I

i

=
0
,

{\displaystyle \bigcap _{i=1}^{k}I_{i}=0,}

and

φ
:
R
→
(
R

/

I

1

)
×
⋯
×
(
R

/

I

k

)

{\displaystyle \varphi :R\to (R/I_{1})\times \cdots \times (R/I_{k})}

be the isomorphism defined above. Let

f

i

=
(
0
,
…
,
1
,
…
,
0
)

{\displaystyle f_{i}=(0,\ldots ,1,\ldots ,0)}

be the element of

(
R

/

I

1

)
×
⋯
×
(
R

/

I

k

)

{\displaystyle (R/I_{1})\times \cdots \times (R/I_{k})}

whose components are all 0 except the i th which is 1, and

e

i

=

φ

−
1

(

f

i

)
.

{\displaystyle e_{i}=\varphi ^{-1}(f_{i}).}

The

e

i

{\displaystyle e_{i}}

are central idempotents that are pairwise orthogonal; this means, in particular, that

e

i

2

=

e

i

{\displaystyle e_{i}^{2}=e_{i}}

and

e

i

e

j

=

e

j

e

i

=
0

{\displaystyle e_{i}e_{j}=e_{j}e_{i}=0}

for every i and j. Moreover, one has

e

1

+
⋯
+

e

n

=
1
,

{\textstyle e_{1}+\cdots +e_{n}=1,}

and

I

i

=
R
(
1
−

e

i

)
.

{\displaystyle I_{i}=R(1-e_{i}).}

In summary, this generalized Chinese remainder theorem is the equivalence between giving pairwise coprime two-sided ideals with a zero intersection, and giving central and pairwise orthogonal idempotents that sum to 1.

Applications

= Sequence numbering

=
The Chinese remainder theorem has been used to construct a Gödel numbering for sequences, which is involved in the proof of Gödel's incompleteness theorems.

= Fast Fourier transform

=
The prime-factor FFT algorithm (also called Good-Thomas algorithm) uses the Chinese remainder theorem for reducing the computation of a fast Fourier transform of size

n

1

n

2

{\displaystyle n_{1}n_{2}}

to the computation of two fast Fourier transforms of smaller sizes

n

1

{\displaystyle n_{1}}

and

n

2

{\displaystyle n_{2}}

(providing that

n

1

{\displaystyle n_{1}}

and

n

2

{\displaystyle n_{2}}

are coprime).

= Encryption

=
Most implementations of RSA use the Chinese remainder theorem during signing of HTTPS certificates and during decryption.
The Chinese remainder theorem can also be used in secret sharing, which consists of distributing a set of shares among a group of people who, all together (but no one alone), can recover a certain secret from the given set of shares. Each of the shares is represented in a congruence, and the solution of the system of congruences using the Chinese remainder theorem is the secret to be recovered. Secret sharing using the Chinese remainder theorem uses, along with the Chinese remainder theorem, special sequences of integers that guarantee the impossibility of recovering the secret from a set of shares with less than a certain cardinality.

= Range ambiguity resolution

=

The range ambiguity resolution techniques used with medium pulse repetition frequency radar can be seen as a special case of the Chinese remainder theorem.

= Decomposition of surjections of finite abelian groups

=
Given a surjection

Z

/

n
→

Z

/

m

{\displaystyle \mathbb {Z} /n\to \mathbb {Z} /m}

of finite abelian groups, we can use the Chinese remainder theorem to give a complete description of any such map. First of all, the theorem gives isomorphisms

Z

/

n

≅

Z

/

p

n

1

a

1

×
⋯
×

Z

/

p

n

i

a

i

Z

/

m

≅

Z

/

p

m

1

b

1

×
⋯
×

Z

/

p

m

j

b

j

{\displaystyle {\begin{aligned}\mathbb {Z} /n&\cong \mathbb {Z} /p_{n_{1}}^{a_{1}}\times \cdots \times \mathbb {Z} /p_{n_{i}}^{a_{i}}\\\mathbb {Z} /m&\cong \mathbb {Z} /p_{m_{1}}^{b_{1}}\times \cdots \times \mathbb {Z} /p_{m_{j}}^{b_{j}}\end{aligned}}}

where

{

p

m

1

,
…
,

p

m

j

}
⊆
{

p

n

1

,
…
,

p

n

i

}

{\displaystyle \{p_{m_{1}},\ldots ,p_{m_{j}}\}\subseteq \{p_{n_{1}},\ldots ,p_{n_{i}}\}}

. In addition, for any induced map

Z

/

p

n

k

a

k

→

Z

/

p

m

l

b

l

{\displaystyle \mathbb {Z} /p_{n_{k}}^{a_{k}}\to \mathbb {Z} /p_{m_{l}}^{b_{l}}}

from the original surjection, we have

a

k

≥

b

l

{\displaystyle a_{k}\geq b_{l}}

and

p

n

k

=

p

m

l

,

{\displaystyle p_{n_{k}}=p_{m_{l}},}

since for a pair of primes

p
,
q

{\displaystyle p,q}

, the only non-zero surjections

Z

/

p

a

→

Z

/

q

b

{\displaystyle \mathbb {Z} /p^{a}\to \mathbb {Z} /q^{b}}

can be defined if

p
=
q

{\displaystyle p=q}

and

a
≥
b

{\displaystyle a\geq b}

.
These observations are pivotal for constructing the ring of profinite integers, which is given as an inverse limit of all such maps.

= Dedekind's theorem

=
Dedekind's theorem on the linear independence of characters. Let M be a monoid and k an integral domain, viewed as a monoid by considering the multiplication on k. Then any finite family ( fi )i∈I of distinct monoid homomorphisms fi : M → k is linearly independent. In other words, every family (αi)i∈I of elements αi ∈ k satisfying

∑

i
∈
I

α

i

f

i

=
0

{\displaystyle \sum _{i\in I}\alpha _{i}f_{i}=0}

must be equal to the family (0)i∈I.
Proof. First assume that k is a field, otherwise, replace the integral domain k by its quotient field, and nothing will change. We can linearly extend the monoid homomorphisms fi : M → k to k-algebra homomorphisms Fi : k[M] → k, where k[M] is the monoid ring of M over k. Then, by linearity, the condition

∑

i
∈
I

α

i

f

i

=
0
,

{\displaystyle \sum _{i\in I}\alpha _{i}f_{i}=0,}

yields

∑

i
∈
I

α

i

F

i

=
0.

{\displaystyle \sum _{i\in I}\alpha _{i}F_{i}=0.}

Next, for i, j ∈ I; i ≠ j the two k-linear maps Fi : k[M] → k and Fj : k[M] → k are not proportional to each other. Otherwise fi and fj would also be proportional, and thus equal since as monoid homomorphisms they satisfy: fi (1) = 1 = fj (1), which contradicts the assumption that they are distinct.
Therefore, the kernels Ker Fi and Ker Fj are distinct. Since k[M]/Ker Fi ≅ Fi (k[M]) = k is a field, Ker Fi is a maximal ideal of k[M] for every i in I. Because they are distinct and maximal the ideals Ker Fi and Ker Fj are coprime whenever i ≠ j. The Chinese Remainder Theorem (for general rings) yields an isomorphism:

ϕ
:
k
[
M
]

/

K

→

∏

i
∈
I

k
[
M
]

/

K
e
r

F

i

ϕ
(
x
+
K
)

=

(

x
+

K
e
r

F

i

)

i
∈
I

{\displaystyle {\begin{aligned}\phi :k[M]/K&\to \prod _{i\in I}k[M]/\mathrm {Ker} F_{i}\\\phi (x+K)&=\left(x+\mathrm {Ker} F_{i}\right)_{i\in I}\end{aligned}}}

where

K
=

∏

i
∈
I

K
e
r

F

i

=

⋂

i
∈
I

K
e
r

F

i

.

{\displaystyle K=\prod _{i\in I}\mathrm {Ker} F_{i}=\bigcap _{i\in I}\mathrm {Ker} F_{i}.}

Consequently, the map

Φ
:
k
[
M
]

→

∏

i
∈
I

k
[
M
]

/

K
e
r

F

i

Φ
(
x
)

=

(

x
+

K
e
r

F

i

)

i
∈
I

{\displaystyle {\begin{aligned}\Phi :k[M]&\to \prod _{i\in I}k[M]/\mathrm {Ker} F_{i}\\\Phi (x)&=\left(x+\mathrm {Ker} F_{i}\right)_{i\in I}\end{aligned}}}

is surjective. Under the isomorphisms k[M]/Ker Fi → Fi (k[M]) = k, the map Φ corresponds to:

ψ
:
k
[
M
]

→

∏

i
∈
I

k

ψ
(
x
)

=

[

F

i

(
x
)

]

i
∈
I

{\displaystyle {\begin{aligned}\psi :k[M]&\to \prod _{i\in I}k\\\psi (x)&=\left[F_{i}(x)\right]_{i\in I}\end{aligned}}}

Now,

∑

i
∈
I

α

i

F

i

=
0

{\displaystyle \sum _{i\in I}\alpha _{i}F_{i}=0}

yields

∑

i
∈
I

α

i

u

i

=
0

{\displaystyle \sum _{i\in I}\alpha _{i}u_{i}=0}

for every vector (ui)i∈I in the image of the map ψ. Since ψ is surjective, this means that

∑

i
∈
I

α

i

u

i

=
0

{\displaystyle \sum _{i\in I}\alpha _{i}u_{i}=0}

for every vector

(

u

i

)

i
∈
I

∈

∏

i
∈
I

k
.

{\displaystyle \left(u_{i}\right)_{i\in I}\in \prod _{i\in I}k.}

Consequently, (αi)i∈I = (0)i∈I. QED.

Notes

References

Dauben, Joseph W. (2007), "Chapter 3: Chinese Mathematics", in Katz, Victor J. (ed.), The Mathematics of Egypt, Mesopotamia, China, India and Islam : A Sourcebook, Princeton University Press, pp. 187–384, ISBN 978-0-691-11485-9
Dence, Joseph B.; Dence, Thomas P. (1999), Elements of the Theory of Numbers, Academic Press, ISBN 9780122091308
Duchet, Pierre (1995), "Hypergraphs", in Graham, R. L.; Grötschel, M.; Lovász, L. (eds.), Handbook of combinatorics, Vol. 1, 2, Amsterdam: Elsevier, pp. 381–432, MR 1373663. See in particular Section 2.5, "Helly Property", pp. 393–394.
Gauss, Carl Friedrich (1986), Disquisitiones Arithemeticae, translated by Clarke, Arthur A. (Second, corrected ed.), New York: Springer, ISBN 978-0-387-96254-2
Ireland, Kenneth; Rosen, Michael (1990), A Classical Introduction to Modern Number Theory (2nd ed.), Springer-Verlag, ISBN 0-387-97329-X
Kak, Subhash (1986), "Computational aspects of the Aryabhata algorithm" (PDF), Indian Journal of History of Science, 21 (1): 62–71
Katz, Victor J. (1998), A History of Mathematics / An Introduction (2nd ed.), Addison Wesley Longman, ISBN 978-0-321-01618-8
Libbrecht, Ulrich (1973), Chinese Mathematics in the Thirteenth Century: the "Shu-shu Chiu-chang" of Ch'in Chiu-shao, Dover Publications Inc, ISBN 978-0-486-44619-6
Ore, Øystein (1952), "The general Chinese remainder theorem", The American Mathematical Monthly, 59 (6): 365–370, doi:10.2307/2306804, JSTOR 2306804, MR 0048481
Ore, Oystein (1988) [1948], Number Theory and Its History, Dover, ISBN 978-0-486-65620-5
Pisano, Leonardo (2002), Fibonacci's Liber Abaci, translated by Sigler, Laurence E., Springer-Verlag, pp. 402–403, ISBN 0-387-95419-8
Rosen, Kenneth H. (1993), Elementary Number Theory and its Applications (3rd ed.), Addison-Wesley, ISBN 978-0201-57889-8
Sengupta, Ambar N. (2012), Representing Finite Groups, A Semisimple Introduction, Springer, ISBN 978-1-4614-1232-8
Bourbaki, N. (1989), Algebra I, Springer, ISBN 3-540-64243-9

External links

"Chinese remainder theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Weisstein, Eric W., "Chinese Remainder Theorem", MathWorld
Chinese Remainder Theorem at PlanetMath.
Full text of the Sun-tzu Suan-ching (Chinese) – Chinese Text Project
Chinese remainder theorem at ProofWiki

History

Statement

Proof

= Uniqueness

= Existence (first proof)

= Existence (constructive proof)

= Existence (direct construction)

Computation

= Systematic search

= Search by sieving

= Using the existence construction

= As a linear Diophantine system

Over principal ideal domains

Over univariate polynomial rings and Euclidean domains

= Lagrange interpolation

= Hermite interpolation

Generalization to non-coprime moduli

Generalization to arbitrary rings

= Interpretation in terms of idempotents

Applications

= Sequence numbering

= Fast Fourier transform

= Encryption

= Range ambiguity resolution

= Decomposition of surjections of finite abelian groups

= Dedekind's theorem

See also

Notes

References

Further reading

External links

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