Chinese remainder theorem#Statement for principal ideal domains

The earliest known statement of the problem appears in the 5th-century book Sunzi Suanjing by the Chinese mathematician Sunzi:{{harvnb|Katz|1998|page=197}}

{{blockquote|There are certain things whose number is unknown. If we count them by threes, we have two left over; by fives, we have three left over; and by sevens, two are left over. How many things are there?{{harvnb|Dence|Dence|1999|page=156}}}}

Sunzi's work would not be considered a theorem by modern standards; it only gives one particular problem, without showing how to solve it, much less any proof about the general case or a general algorithm for solving it.{{harvnb|Dauben|2007|page=302}} An algorithm for solving this problem was described by Aryabhata (6th century).{{harvnb|Kak|1986}} Special cases of the Chinese remainder theorem were also known to Brahmagupta (7th century) and appear in Fibonacci's Liber Abaci (1202).{{harvnb|Pisano|2002|pages=402–403}} The result was later generalized with a complete solution called Da-yan-shu ({{lang|zh|大衍術}}) in Qin Jiushao's 1247 Mathematical Treatise in Nine Sections {{harvnb|Dauben|2007|page=310}} which was translated into English in early 19th century by British missionary Alexander Wylie.{{harvnb|Libbrecht|1973}}

File:Disqvisitiones-800.jpg's 1801 book Disquisitiones Arithmeticae.{{Harvnb|Gauss|1986|loc=Art. 32–36}}]]

The notion of congruences was first introduced and used by Carl Friedrich Gauss in his Disquisitiones Arithmeticae of 1801.{{harvnb|Ireland|Rosen|1990|page=36}} Gauss illustrates the Chinese remainder theorem on a problem involving calendars, namely, "to find the years that have a certain period number with respect to the solar and lunar cycle and the Roman indiction."{{harvnb|Ore|1988|page=247}} Gauss introduces a procedure for solving the problem that had already been used by Leonhard Euler but was in fact an ancient method that had appeared several times.{{harvnb|Ore|1988|page=245}}

Let n₁, ..., n_k be integers greater than 1, which are often called moduli or divisors. Let us denote by N the product of the n_i.

The Chinese remainder theorem asserts that if the n_i are pairwise coprime, and if a₁, ..., a_k are integers such that 0 ≤ a_i < n_i for every i, then there is one and only one integer x, such that 0 ≤ x < N and the remainder of the Euclidean division of x by n_i is a_i for every i.

This may be restated as follows in terms of congruences:

If the $n\_i$ are pairwise coprime, and if a₁, ..., a_k are any integers, then the system

:$\backslash begin\{align\}$

x &\equiv a_1 \pmod{n_1} \\

&\,\,\,\vdots \\

x &\equiv a_k \pmod{n_k},

\end{align}

has a solution, and any two solutions, say x₁ and x₂, are congruent modulo N, that is, {{math|x₁ ≡ x₂ (mod N{{hairsp}})}}.{{harvnb|Ireland|Rosen|1990|page=34}}

In abstract algebra, the theorem is often restated as: if the n_i are pairwise coprime, the map

:$x\; \backslash bmod\; N\; \backslash ;\backslash mapsto\backslash ;(x\; \backslash bmod\; n\_1,\backslash ,\; \backslash ldots,\backslash ,\; x\; \backslash bmod\; n\_k)$

defines a ring isomorphism{{harvnb|Ireland|Rosen|1990|page=35}}

:$\backslash mathbb\{Z\}/N\backslash mathbb\{Z\}\; \backslash cong\; \backslash mathbb\{Z\}/n\_1\backslash mathbb\{Z\}\; \backslash times\; \backslash cdots\; \backslash times\; \backslash mathbb\{Z\}/n\_k\backslash mathbb\{Z\}$

between the ring of integers modulo N and the direct product of the rings of integers modulo the n_i. This means that for doing a sequence of arithmetic operations in $\backslash mathbb\{Z\}/N\backslash mathbb\{Z\},$ one may do the same computation independently in each $\backslash mathbb\{Z\}/n\_i\backslash mathbb\{Z\}$ and then get the result by applying the isomorphism (from the right to the left). This may be much faster than the direct computation if N and the number of operations are large. This is widely used, under the name multi-modular computation, for linear algebra over the integers or the rational numbers.

The theorem can also be restated in the language of combinatorics as the fact that the infinite arithmetic progressions of integers form a Helly family.{{harvnb|Duchet|1995}}

The existence and the uniqueness of the solution may be proven independently. However, the first proof of existence, given below, uses this uniqueness.

Suppose that {{mvar|x}} and {{mvar|y}} are both solutions to all the congruences. As {{mvar|x}} and {{mvar|y}} give the same remainder, when divided by {{math|n_i}}, their difference {{math|x − y}} is a multiple of each {{math|n_i}}. As the {{math|n_i}} are pairwise coprime, their product {{math|N}} also divides {{math|x − y}}, and thus {{mvar|x}} and {{mvar|y}} are congruent modulo {{math|N}}. If {{mvar|x}} and {{mvar|y}} are supposed to be non-negative and less than {{math|N}} (as in the first statement of the theorem), then their difference may be a multiple of {{math|N}} only if {{math|1=x = y}}.

The map

:$x\; \backslash bmod\; N\; \backslash mapsto\; (x\; \backslash bmod\; n\_1,\; \backslash ldots,\; x\backslash bmod\; n\_k)$

maps congruence classes modulo {{math|N}} to sequences of congruence classes modulo {{math|n_i}}. The proof of uniqueness shows that this map is injective. As the domain and the codomain of this map have the same number of elements, the map is also surjective, which proves the existence of the solution.

This proof is very simple but does not provide any direct way for computing a solution. Moreover, it cannot be generalized to other situations where the following proof can.

Existence may be established by an explicit construction of {{mvar|x}}.{{harvnb|Rosen|1993|page=136}} This construction may be split into two steps, first solving the problem in the case of two moduli, and then extending this solution to the general case by induction on the number of moduli.

We want to solve the system:

:$$

\begin{align}

x &\equiv a_1 \pmod {n_1}\\

x &\equiv a_2 \pmod {n_2},

\end{align}

where $n\_1$ and $n\_2$ are coprime.

Bézout's identity asserts the existence of two integers $m\_1$ and $m\_2$ such that

:$m\_1n\_1+m\_2n\_2=1.$

The integers $m\_1$ and $m\_2$ may be computed by the extended Euclidean algorithm.

A solution is given by

:$x\; =\; a\_1m\_2n\_2+a\_2m\_1n\_1.$

Indeed,

:$\backslash begin\{align\}$

x&=a_1m_2n_2+a_2m_1n_1\\

&=a_1(1 - m_1n_1) + a_2m_1n_1 \\

&=a_1 + (a_2 - a_1)m_1n_1,

\end{align}

implying that $x\; \backslash equiv\; a\_1\; \backslash pmod\; \{n\_1\}.$ The second congruence is proved similarly, by exchanging the subscripts 1 and 2.

Consider a sequence of congruence equations:

:$$

\begin{align}

x &\equiv a_1 \pmod{n_1} \\

&\vdots \\

x &\equiv a_k \pmod{n_k},

\end{align}

where the $n\_i$ are pairwise coprime. The two first equations have a solution $a\_\{1,2\}$ provided by the method of the previous section. The set of the solutions of these two first equations is the set of all solutions of the equation

:$x\; \backslash equiv\; a\_\{1,2\}\; \backslash pmod\{n\_1n\_2\}.$

As the other $n\_i$ are coprime with $n\_1n\_2,$ this reduces solving the initial problem of {{mvar|k}} equations to a similar problem with $k-1$ equations. Iterating the process, one gets eventually the solutions of the initial problem.

For constructing a solution, it is not necessary to make an induction on the number of moduli. However, such a direct construction involves more computation with large numbers, which makes it less efficient and less used. Nevertheless, Lagrange interpolation is a special case of this construction, applied to polynomials instead of integers.

Let $N\_i\; =\; N/n\_i$ be the product of all moduli but one. As the $n\_i$ are pairwise coprime, $N\_i$ and $n\_i$ are coprime. Thus Bézout's identity applies, and there exist integers $M\_i$ and $m\_i$ such that

:$M\_iN\_i\; +\; m\_in\_i=1.$

A solution of the system of congruences is

:$x=\backslash sum\_\{i=1\}^k\; a\_iM\_iN\_i.$

In fact, as $N\_j$ is a multiple of $n\_i$ for $i\backslash neq\; j,$

we have

:$x\; \backslash equiv\; a\_iM\_iN\_i\; \backslash equiv\; a\_i(1-m\_in\_i)\; \backslash equiv\; a\_i\; \backslash pmod\{n\_i\},$

for every $i.$

Consider a system of congruences:

:$\backslash begin\{align\}$

x &\equiv a_1 \pmod{n_1} \\

&\vdots \\

x &\equiv a_k \pmod{n_k}, \\

\end{align}

where the $n\_i$ are pairwise coprime, and let $N=n\_1\; n\_2\backslash cdots\; n\_k.$ In this section several methods are described for computing the unique solution for $x$, such that

:$$

\begin{align}

x &\equiv 0 \pmod 3 \\

x &\equiv 3 \pmod 4 \\

x &\equiv 4 \pmod 5.

\end{align}

Several methods of computation are presented. The two first ones are useful for small examples, but become very inefficient when the product $n\_1\backslash cdots\; n\_k$ is large. The third one uses the existence proof given in {{slink||Existence (constructive proof)}}. It is the most convenient when the product $n\_1\backslash cdots\; n\_k$ is large, or for computer computation.

It is easy to check whether a value of {{mvar|x}} is a solution: it suffices to compute the remainder of the Euclidean division of {{mvar|x}} by each {{math|n_i}}. Thus, to find the solution, it suffices to check successively the integers from {{math|0}} to {{mvar|N}} until finding the solution.

Although very simple, this method is very inefficient. For the simple example considered here, {{math|40}} integers (including {{math|0}}) have to be checked for finding the solution, which is {{math|39}}. This is an exponential time algorithm, as the size of the input is, up to a constant factor, the number of digits of {{mvar|N}}, and the average number of operations is of the order of {{mvar|N}}.

Therefore, this method is rarely used, neither for hand-written computation nor on computers.

File:Chinese_remainder_theorem_sieve.svg

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

:$a\_1,\; a\_1\; +\; n\_1,\; a\_1+2n\_1,\; \backslash ldots$

By testing the values of these numbers modulo $n\_2,$ one eventually finds a solution $x\_2$ of the two first congruences. Then the solution belongs to the arithmetic progression

:$x\_2,\; x\_2\; +\; n\_1n\_2,\; x\_2+2n\_1n\_2,\; \backslash ldots$

Testing the values of these numbers modulo $n\_3,$ 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 $n\_1>n\_2>\; \backslash cdots\; >\; n\_k.$ 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, {{nowrap|1=9 = 4 + 5}}, {{nowrap|1=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 {{nowrap|1=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.

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 $n\_1n\_2$ (for getting a result in the interval $(0,\; n\_1n\_2-1)$). As the Bézout's coefficients may be computed with the extended Euclidean algorithm, the whole computation, at most, has a quadratic time complexity of $O((s\_1+s\_2)^2),$ where $s\_i$ denotes the number of digits of $n\_i.$

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

:$1\backslash times\; 4\; +\; (-1)\backslash times\; 3\; =\; 1.$

Putting this in the formula given for proving the existence gives

:$0\backslash times\; 1\backslash times\; 4\; +\; 3\backslash times\; (-1)\backslash times\; 3\; =-9$

for a solution of the two first congruences, the other solutions being obtained by adding to −9 any multiple of {{nowrap|1=3 × 4 = 12}}. One may continue with any of these solutions, but the solution {{nowrap|1=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

:$5\backslash times\; 5\; +(-2)\backslash times\; 12\; =1.$

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

:$5\backslash times\; 5\; \backslash times\; 3\; +\; 12\backslash times\; (-2)\backslash times\; 4\; =\; -21.$

The other solutions are obtained by adding any multiple of {{nowrap|1=3 × 4 × 5 = 60}}, and the smallest positive solution is {{nowrap|1=−21 + 60 = 39}}.

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

:$\backslash begin\{align\}$

x &= a_1 +x_1n_1\\

&\vdots \\

x &=a_k+x_kn_k,

\end{align}

where the unknown integers are $x$ and the $x\_i.$ 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.

In {{section link||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 {{math|R}}: it suffices to replace "integer" by "element of the domain" and $\backslash mathbb\; Z$ by {{math|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.

The statement in terms of remainders given in {{slink||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 $R=K[X]$ for a field $K.$ 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 $P\_i(X)$ (the moduli) be, for $i\; =\; 1,\; \backslash dots,\; k$, pairwise coprime polynomials in $R=K[X]$. Let $d\_i\; =\backslash deg\; P\_i$ be the degree of $P\_i(X)$, and $D$ be the sum of the $d\_i.$

If $A\_i(X),\; \backslash ldots,A\_k(X)$ are polynomials such that $A\_i(X)=0$ or i}}, then, there is one and only one polynomial $P(X)$, such that i}}.

The construction of the solution may be done as in {{slink||Existence (constructive proof)}} or {{slink||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 $P(X)$, which satisfies the congruences

:$P(X)\backslash equiv\; A\_i(X)\; \backslash pmod\; \{P\_i(X)\},$

for $i=1,\backslash ldots,k.$

Consider the polynomials

:$\backslash begin\{align\}$

Q(X) &= \prod_{i=1}^{k}P_i(X) \\

Q_i(X) &= \frac{Q(X)}{P_i(X)}.

\end{align}

The partial fraction decomposition of $1/Q(X)$ gives {{mvar|k}} polynomials $S\_i(X)$ with degrees such that

:$\backslash frac\{1\}\{Q(X)\}\; =\; \backslash sum\_\{i=1\}^k\; \backslash frac\{S\_i(X)\}\{P\_i(X)\},$

and thus

:$1\; =\; \backslash sum\_\{i=1\}^\{k\}S\_i(X)\; Q\_i(X).$

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

:$\backslash sum\_\{i=1\}^k\; A\_i(X)\; S\_i(X)\; Q\_i(X).$

In fact, we have

: $\backslash sum\_\{i=1\}^k\; A\_i(X)\; S\_i(X)\; Q\_i(X)=\; A\_i(X)+\; \backslash sum\_\{j=1\}^\{k\}(A\_j(X)\; -\; A\_i(X))\; S\_j(X)\; Q\_j(X)\; \backslash equiv\; A\_i(X)\backslash pmod\{P\_i(X)\},$

for $1\; \backslash leq\; i\; \backslash leq\; k.$

This solution may have a degree larger than $D=\backslash sum\_\{i=1\}^k\; d\_i.$ The unique solution of degree less than $D$ may be deduced by considering the remainder $B\_i(X)$ of the Euclidean division of $A\_i(X)S\_i(X)$ by $P\_i(X).$ This solution is

:$P(X)=\backslash sum\_\{i=1\}^k\; B\_i(X)\; Q\_i(X).$

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

:$P\_i(X)=X-x\_i.$

They are pairwise coprime if the $x\_i$ are all different. The remainder of the division by $P\_i(X)$ of a polynomial $P(X)$ is $P(x\_i)$, by the polynomial remainder theorem.

Now, let $A\_1,\; \backslash ldots,\; A\_k$ be constants (polynomials of degree 0) in $K.$ Both Lagrange interpolation and Chinese remainder theorem assert the existence of a unique polynomial $P(X),$ of degree less than $k$ such that

:$P(x\_i)=A\_i,$

for every $i.$

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

:$\backslash begin\{align\}$

Q(X) &= \prod_{i=1}^{k}(X-x_i) \\[6pt]

Q_i(X) &= \frac{Q(X)}{X-x_i}.

\end{align}

The partial fraction decomposition of $\backslash frac\{1\}\{Q(X)\}$ is

:$\backslash frac\{1\}\{Q(X)\}\; =\; \backslash sum\_\{i=1\}^k\; \backslash frac\{1\}\{Q\_i(x\_i)(X-x\_i)\}.$

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

:$\backslash sum\_\{i=1\}^k\; \backslash frac\{1\}\{Q\_i(x\_i)(X-x\_i)\}=\; \backslash frac\{1\}\{Q(X)\}\; \backslash sum\_\{i=1\}^k\; \backslash frac\{Q\_i(X)\}\{Q\_i(x\_i)\},$

and the numerator is equal to one, as being a polynomial of degree less than $k,$ which takes the value one for $k$ different values of $X.$

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

:$P(X)=\backslash sum\_\{i=1\}^k\; A\_i\backslash frac\{Q\_i(X)\}\{Q\_i(x\_i)\}.$

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 $x\_1,\; \backslash ldots,\; x\_k$ be $k$ elements of the ground field $K,$ and, for $i=1,\backslash ldots,\; k,$ let $a\_\{i,0\},\; a\_\{i,1\},\; \backslash ldots,\; a\_\{i,r\_i-1\}$ be the values of the first $r\_i$ derivatives of the sought polynomial at $x\_i$ (including the 0th derivative, which is the value of the polynomial itself). The problem is to find a polynomial $P(X)$ such that its j{{hairsp}}th derivative takes the value $a\_\{i,j\}$ at $x\_i,$ for $i=1,\backslash ldots,k$ and $j=0,\backslash ldots,r\_j.$

Consider the polynomial

:$P\_i(X)\; =\; \backslash sum\_\{j=0\}^\{r\_i\; -\; 1\}\backslash frac\{a\_\{i,j\}\}\{j!\}(X\; -\; x\_i)^j.$

This is the Taylor polynomial of order $r\_i-1$ at $x\_i$, of the unknown polynomial $P(X).$ Therefore, we must have

:$P(X)\backslash equiv\; P\_i(X)\; \backslash pmod\; \{(X-x\_i)^\{r\_i\}\}.$

Conversely, any polynomial $P(X)$ that satisfies these $k$ congruences, in particular verifies, for any $i=1,\; \backslash ldots,\; k$

:$P(X)=\; P\_i(X)\; +o(X-x\_i)^\{r\_i-1\}$

therefore $P\_i(X)$ is its Taylor polynomial of order $r\_i\; -\; 1$ at $x\_i$, that is, $P(X)$ 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 $r\_i,$ which satisfies these $k$ congruences.

There are several ways for computing the solution $P(X).$ One may use the method described at the beginning of {{slink||Over univariate polynomial rings and Euclidean domains}}. One may also use the constructions given in {{slink||Existence (constructive proof)}} or {{slink||Existence (direct proof)}}.

The Chinese remainder theorem can be generalized to non-coprime moduli. Let $m,\; n,\; a,\; b$ be any integers, let $g\; =\; \backslash gcd(m,n)$; $M\; =\; \backslash operatorname\{lcm\}(m,n)$, and consider the system of congruences:

:$$

\begin{align}

x &\equiv a \pmod m \\

x &\equiv b \pmod n,

\end{align}

If $a\; \backslash equiv\; b\; \backslash pmod\; g$, then this system has a unique solution modulo $M\; =\; mn/g$. Otherwise, it has no solutions.

If one uses Bézout's identity to write $g\; =\; um\; +\; vn$, then the solution is given by

:$x\; =\; \backslash frac\{avn+bum\}\{g\}.$

This defines an integer, as {{mvar|g}} divides both {{mvar|m}} and {{mvar|n}}. Otherwise, the proof is very similar to that for coprime moduli.{{sfn|Ore|1952}}

The Chinese remainder theorem can be generalized to any ring, by using coprime ideals (also called comaximal ideals). Two ideals {{mvar|I}} and {{mvar|J}} are coprime if there are elements $i\backslash in\; I$ and $j\backslash in\; J$ such that $i+j=1.$ 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.{{harvnb|Ireland|Rosen|1990|page=181}}{{harvnb|Sengupta|2012|page=313}}

Let {{math|I₁, ..., I_k}} be two-sided ideals of a ring $R$ and let {{math|I}} be their intersection. If the ideals are pairwise coprime, we have the isomorphism:

:$\backslash begin\{align\}$

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{align}

between the quotient ring $R/I$ and the direct product of the $R/I\_i,$

where "$x\; \backslash bmod\; I$" denotes the image of the element $x$ in the quotient ring defined by the ideal $I.$

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

:$$

I= I_1\cap I_2 \cap\cdots\cap I_k= I_1I_2\cdots I_k,

if {{mvar|I{{sub|i}}}} and {{mvar|I{{sub|j}}}} are coprime for all {{math|i ≠ j}}.

Let $I\_1,\; I\_2,\; \backslash dots,\; I\_k$ be pairwise coprime two-sided ideals with $\backslash bigcap\_\{i\; =\; 1\}^k\; I\_i\; =\; 0,$ and

:$\backslash varphi:R\backslash to\; (R/I\_1)\; \backslash times\; \backslash cdots\; \backslash times\; (R/I\_k)$

be the isomorphism defined above. Let $f\_i=(0,\backslash ldots,1,\backslash ldots,\; 0)$ be the element of $(R/I\_1)\; \backslash times\; \backslash cdots\; \backslash times\; (R/I\_k)$ whose components are all {{math|0}} except the {{mvar|i}}{{hairsp}}th which is {{math|1}}, and $e\_i=\backslash varphi^\{-1\}(f\_i).$

The $e\_i$ are central idempotents that are pairwise orthogonal; this means, in particular, that $e\_i^2=e\_i$ and $e\_ie\_j=e\_je\_i=0$ for every {{mvar|i}} and {{mvar|j}}. Moreover, one has $e\_1+\backslash cdots+e\_n=1,$ and $I\_i=R(1-e\_i).$

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 {{math|1}}.{{harvnb|Bourbaki, N.|1989|page=110}}

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.

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 $n\_1n\_2$ to the computation of two fast Fourier transforms of smaller sizes $n\_1$ and $n\_2$ (providing that $n\_1$ and $n\_2$ are coprime).

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.

{{main article | 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.

Given a surjection $\backslash mathbb\{Z\}/n\; \backslash to\; \backslash mathbb\{Z\}/m$ 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

:$\backslash begin\{align\}$

\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{align}

where $\backslash \{p\_\{m\_1\},\backslash ldots,p\_\{m\_j\}\; \backslash \}\; \backslash subseteq\; \backslash \{\; p\_\{n\_1\},\backslash ldots,\; p\_\{n\_i\}\; \backslash \}$. In addition, for any induced map

:$\backslash mathbb\{Z\}/p\_\{n\_k\}^\{a\_k\}\; \backslash to\; \backslash mathbb\{Z\}/p\_\{m\_l\}^\{b\_l\}$

from the original surjection, we have $a\_k\; \backslash geq\; b\_l$ and $p\_\{n\_k\}\; =\; p\_\{m\_l\},$ since for a pair of primes $p,q$, the only non-zero surjections

:$\backslash mathbb\{Z\}/p^a\; \backslash to\; \backslash mathbb\{Z\}/q^b$

can be defined if $p\; =\; q$ and $a\; \backslash geq\; b$.

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 on the linear independence of characters. Let {{mvar|M}} be a monoid and {{mvar|k}} an integral domain, viewed as a monoid by considering the multiplication on {{mvar|k}}. Then any finite family {{math|( f_i )_i∈I}} of distinct monoid homomorphisms {{math| f_i : M → k}} is linearly independent. In other words, every family {{math|(α_i)_i∈I}} of elements {{math|α_i ∈ k}} satisfying

:$\backslash sum\_\{i\; \backslash in\; I\}\backslash alpha\_i\; f\_i\; =\; 0$

must be equal to the family {{math|(0)_i∈I}}.

Proof. First assume that {{mvar|k}} is a field, otherwise, replace the integral domain {{mvar|k}} by its quotient field, and nothing will change. We can linearly extend the monoid homomorphisms {{math| f_i : M → k}} to {{mvar|k}}-algebra homomorphisms {{math|F_i : k[M] → k}}, where {{math|k[M]}} is the monoid ring of {{mvar|M}} over {{mvar|k}}. Then, by linearity, the condition

:$\backslash sum\_\{i\backslash in\; I\}\backslash alpha\_i\; f\_i\; =\; 0,$

yields

:$\backslash sum\_\{i\; \backslash in\; I\}\backslash alpha\_i\; F\_i\; =\; 0.$

Next, for {{math|i, j ∈ I; i ≠ j}} the two {{mvar|k}}-linear maps {{math|F_i : k[M] → k}} and {{math|F_j : k[M] → k}} are not proportional to each other. Otherwise {{math| f_i }} and {{math| f_j }} would also be proportional, and thus equal since as monoid homomorphisms they satisfy: {{math| f_i{{hairsp}}(1) {{=}} 1 {{=}}  f_j{{hairsp}}(1)}}, which contradicts the assumption that they are distinct.

Therefore, the kernels {{math|Ker F_i}} and {{math|Ker F_j}} are distinct. Since {{math|k[M]/Ker F_i ≅ F_i{{hairsp}}(k[M]) {{=}} k}} is a field, {{math|Ker F_i}} is a maximal ideal of {{math|k[M]}} for every {{mvar|i}} in {{mvar|I}}. Because they are distinct and maximal the ideals {{math|Ker F_i}} and {{math|Ker F_j}} are coprime whenever {{math|i ≠ j}}. The Chinese Remainder Theorem (for general rings) yields an isomorphism:

:$\backslash begin\{align\}$

\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{align}

where

:$K\; =\; \backslash prod\_\{i\; \backslash in\; I\}\backslash mathrm\{Ker\}\; F\_i\; =\; \backslash bigcap\_\{i\; \backslash in\; I\}\backslash mathrm\{Ker\}\; F\_i.$

Consequently, the map

:$\backslash begin\{align\}$

\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{align}

is surjective. Under the isomorphisms {{math|k[M]/Ker F_i → F_i{{hairsp}}(k[M]) {{=}} k,}} the map {{math|Φ}} corresponds to:

:$\backslash begin\{align\}$

\psi: k[M] &\to \prod_{i \in I}k \\

\psi(x) &= \left[F_i(x)\right]_{i \in I}

\end{align}

Now,

:$\backslash sum\_\{i\; \backslash in\; I\}\backslash alpha\_i\; F\_i\; =\; 0$

yields

:$\backslash sum\_\{i\; \backslash in\; I\}\backslash alpha\_i\; u\_i\; =\; 0$

for every vector {{math|(u_i)_i∈I}} in the image of the map {{mvar|ψ}}. Since {{mvar|ψ}} is surjective, this means that

:$\backslash sum\_\{i\; \backslash in\; I\}\backslash alpha\_i\; u\_i\; =\; 0$

for every vector

:$\backslash left(u\_i\backslash right)\_\{i\; \backslash in\; I\}\; \backslash in\; \backslash prod\_\{i\; \backslash in\; I\}k.$

Consequently, {{math|(α_i)_i∈I {{=}} (0)_i∈I}}. QED.

• Covering system
• Hasse principle
• Residue number system

{{Reflist}}

• {{citation|first=Joseph W.|last=Dauben|editor-last=Katz|editor-first=Victor J.|contribution=Chapter 3: Chinese Mathematics|pages=187–384|publisher=Princeton University Press|year=2007|title=The Mathematics of Egypt, Mesopotamia, China, India and Islam : A Sourcebook|isbn=978-0-691-11485-9}}
• {{citation |title=Elements of the Theory of Numbers |first1=Joseph B. |last1=Dence |first2=Thomas P. |last2=Dence |publisher=Academic Press |date=1999 |url=https://books.google.com/books?id=YiYHw7evhjkC&pg=PA156 |isbn=9780122091308}}
• {{citation

| last = Duchet

| first = Pierre

| editor1-last = Graham | editor1-first = R. L.

| editor2-last = Grötschel | editor2-first = M. | editor2-link = Martin Grötschel

| editor3-last = Lovász | editor3-first = L.

| contribution = Hypergraphs

| location = Amsterdam

| mr = 1373663

| pages = 381–432

| publisher = Elsevier

| title = Handbook of combinatorics, Vol. 1, 2

| year = 1995

}}. See in particular Section 2.5, "Helly Property", [https://books.google.com/books?id=5Y9NCwlx63IC&pg=PA393 pp. 393–394].

• {{citation

| last1 = Gauss | first1 = Carl Friedrich

| translator-last = Clarke | translator-first = Arthur A.

| title = Disquisitiones Arithemeticae

| edition = Second, corrected

| publisher = Springer

| location = New York

| year = 1986

| isbn = 978-0-387-96254-2

| url = https://www.springer.com/mathematics/algebra/book/978-0-387-96254-2

}}

• {{citation|first1=Kenneth|last1=Ireland|first2=Michael|last2=Rosen|title=A Classical Introduction to Modern Number Theory|edition=2nd|publisher=Springer-Verlag|year=1990|isbn=0-387-97329-X}}
• {{citation

| last = Kak

| first = Subhash

| author-link = Subhash Kak

| issue = 1

| journal = Indian Journal of History of Science

| pages = 62–71

| title = Computational aspects of the Aryabhata algorithm

| url = http://www.ece.lsu.edu/kak/AryabhataAlgorithm.pdf

| volume = 21

| year = 1986

}}

• {{citation|first=Victor J.|last=Katz|title=A History of Mathematics / An Introduction|edition=2nd|year=1998|publisher=Addison Wesley Longman|isbn=978-0-321-01618-8|url-access=registration|url=https://archive.org/details/historyofmathema00katz}}
• {{citation |first=Ulrich |last=Libbrecht |title=Chinese Mathematics in the Thirteenth Century: the "Shu-shu Chiu-chang" of Ch'in Chiu-shao |publisher=Dover Publications Inc |isbn=978-0-486-44619-6 |date=1973}}
• {{citation

| last = Ore | first = Øystein | author-link = Øystein Ore

| doi = 10.2307/2306804

| journal = The American Mathematical Monthly

| mr = 48481

| pages = 365–370

| title = The general Chinese remainder theorem

| volume = 59

| year = 1952| issue = 6 | jstor = 2306804 }}

• {{citation|first=Oystein|last=Ore|title=Number Theory and Its History|year=1988|orig-year=1948|publisher=Dover|isbn=978-0-486-65620-5|url-access=registration|url=https://archive.org/details/numbertheoryitsh0000orey}}
• {{citation |title=Fibonacci's Liber Abaci |last1=Pisano |first1=Leonardo |author-link1 = Leonardo Pisano |translator-last=Sigler |translator-first=Laurence E. |publisher=Springer-Verlag |year=2002 |isbn=0-387-95419-8 |url=https://www.springer.com/mathematics/book/978-0-387-40737-1 |pages=402–403}}
• {{citation|first=Kenneth H.|last=Rosen|title=Elementary Number Theory and its Applications|edition=3rd|year=1993|publisher=Addison-Wesley|isbn=978-0201-57889-8}}
• {{Citation|title= Representing Finite Groups, A Semisimple Introduction |last=Sengupta |first=Ambar N.|year= 2012 |publisher=Springer |isbn=978-1-4614-1232-8}}
• {{citation|author=Bourbaki, N.|title=Algebra I|publisher=Springer|year=1989|isbn=3-540-64243-9|url=https://books.google.com/books?id=STS9aZ6F204C&q=%22associative+algebra%22}}

• {{citation

| last1 = Cormen

| first1 = Thomas H.

| author-link1 = Thomas H. Cormen

| last2 = Leiserson

| first2 = Charles E.

| author-link2 = Charles E. Leiserson

| last3 = Rivest

| first3 = Ronald L.

| author-link3 = Ronald L. Rivest

| last4 = Stein

| first4 = Clifford

| author-link4 = Clifford Stein

| year = 2001

| title = Introduction to Algorithms

| edition = Second

| publisher = MIT Press and McGraw-Hill

| isbn = 0-262-03293-7

| title-link = Introduction to Algorithms

}}. See Section 31.5: The Chinese remainder theorem, pp. 873–876.

• {{citation

| title = Chinese Remainder Theorem: Applications in Computing, Coding, Cryptography

| last1 = Ding

| first1 = Cunsheng

| last2 = Pei

| first2 = Dingyi

| last3 = Salomaa

| first3 = Arto

| publisher = World Scientific Publishing

| year = 1996

| isbn = 981-02-2827-9

| url = https://archive.org/details/chineseremainder0000ding/page/1

| pages = [https://archive.org/details/chineseremainder0000ding/page/1 1–213]

}}

• {{citation

| title = Algebra

| last = Hungerford

| first = Thomas W.

| author-link = Thomas W. Hungerford

| publisher = Springer-Verlag

| series = Graduate Texts in Mathematics, Vol. 73

| year = 1974

| isbn = 978-1-4612-6101-8

| url = https://www.springer.com/mathematics/algebra/book/978-0-387-90518-1

| pages = 131–132

}}

• {{citation

| last = Knuth

| first = Donald

| author-link = Donald Knuth

| year = 1997

| title = The Art of Computer Programming

| volume = 2: Seminumerical Algorithms

| edition = Third

| publisher = Addison-Wesley

| isbn = 0-201-89684-2

| title-link = The Art of Computer Programming

}}. See Section 4.3.2 (pp. 286–291), exercise 4.6.2–3 (page 456).

• {{springer |title = Chinese remainder theorem |id = p/c022120}}
• {{MathWorld |urlname = ChineseRemainderTheorem |title = Chinese Remainder Theorem|mode=cs2}}
• {{planetmath |urlname = ChineseRemainderTheorem |title = Chinese Remainder Theorem}}
• [http://ctext.org/sunzi-suan-jing Full text of the Sun-tzu Suan-ching] (Chinese){{snd}} Chinese Text Project

{{Number theory}}

{{authority control}}

Category:Articles containing proofs

Sun, Master

Category:Commutative algebra

Category:Modular arithmetic

Category:Theorems in number theory