Recall that the Chinese Remainder Theorem from elementary number
theory asserts that if
are integers that are coprime
in pairs, and
are integers, then there exists an
integer such that
for each
.
In terms of rings, the Chinese Remainder Theorem asserts that the
natural map
is an isomorphism. This result generalizes to rings of integers of
number fields.

*Proof*.
The ideal

is the largest ideal of

that is divisible
by (contained in) both

and

. Since

and

are coprime,

is divisible by

, i.e.,

. By
definition of ideal

, which
completes the proof.

*Remark 9.1.2*
This lemma is true for any ring

and ideals

such that

. For the general proof, choose

and

such that

. If

then

so

, and the other inclusion is obvious by definition.

*Proof*.
First assume that we know the theorem in the case when the

are
powers of prime ideals. Then we can deduce the general case by noting
that each

is isomorphic to a product

, where

, and

is isomorphic to the product of the

,
where the

and

run through the same prime powers as appear
on the right hand side.

It thus suffices to prove that if
are distinct
prime ideals of and
are positive integers,
then

is an isomorphism. Let

be the natural map induced by reduction mod

. Then kernel of

is

,
which by Lemma

9.1.1 is equal to

, so

is injective. Note that the projection

of

onto each factor is obviously
surjective, so it suffices to show that the element

is in the image of

(and the similar elements for the other
factors). Since

is not divisible
by

, hence not contained in

, there is
an element

with

. Since

is maximal,

is a field, so there exists

such that

, for some

. Then

is congruent to

0 mod

for each

since it is in

, and it
is congruent to

modulo

.

*Remark 9.1.4*
In fact, the surjectivity part of the above proof is easy to prove
for any commutative ring; indeed, the above proof illustrates how
trying to prove something in a special case can result in a more
complicated proof!! Suppose

is a ring and

are ideals in

such that

. Choose

and

such that

. Then

maps to

in

and

maps to

in

. Thus the map

is surjective. Also, as mentioned above,

.

*Example 9.1.5*
The

command

`ChineseRemainderTheorem` implements the
algorithm suggested by the above theorem. In the following example,
we compute a prime over

and a prime over

of the ring of
integers of

, and find an element of

that is
congruent to

modulo one prime and

modulo the other.

> R<x> := PolynomialRing(RationalField());
> K<a> := NumberField(x^3-2);
> OK := MaximalOrder(K);
> I := Factorization(3*OK)[1][1];
> J := Factorization(5*OK)[1][1];
> I;
Prime Ideal of OK
Two element generators:
[3, 0, 0]
[4, 1, 0]
> J;
Prime Ideal of OK
Two element generators:
[5, 0, 0]
[7, 1, 0]
> b := ChineseRemainderTheorem(I, J, OK!a, OK!1);
> b - a in I;
true
> b - 1 in J;
true
> K!b;
-4

The element found by the Chinese Remainder Theorem algorithm in
this case is

.

The following lemma is a nice application of the Chinese Remainder
Theorem. We will use it to prove that every ideal of can be
generated by two elements. Suppose is a nonzero integral ideals of
. If , then
, so divides and
the quotient is an integral ideal. The following lemma
asserts that can be chosen so the quotient is coprime to
any given ideal.

**Lemma 9.1.6**
*If are nonzero integral ideals in , then there exists
an such that is coprime to .*
*Proof*.
Let

be the prime divisors of

.
For each

, let

be the largest power of

that divides

. Choose an element

that is not in

(there is such an element
since

, by unique factorization).
By Theorem

9.1.3, there exists

such that

for all

and
also

(We are applying the theorem with the coprime integral ideals

, for

and the integral
ideal

.)

To complete the proof we must show that is not
divisible by any
, or equivalently, that the
exactly divides . Because
, there is
such that
. Since
, it follows that
,
so
divides . If
,
then
, a contradiction, so
does not divide , which completes
the proof.

Suppose is a nonzero ideal of . As an abelian group
is free of rank equal to the degree
of , and is of
finite index in , so can be generated as an abelian group,
hence as an ideal, by
generators. The following proposition
asserts something much better, namely that can be generated *as an ideal* in by at most two elements.

**Proposition 9.1.7**
*
Suppose is a fractional ideal in the ring of integers of a
number field. Then there exist such that .*
*Proof*.
If

, then

is generated by

element and we are done. If

is not an integral ideal, then there is

such that

is
an integral ideal, and the number of generators of

is the same as
the number of generators of

, so we may assume that

is an
integral ideal.

Let be any nonzero element of the integral ideal . We will
show that there is some such that . Let .
By Lemma 9.1.6, there exists such that is
coprime to . The ideal
is the greatest common
divisor of and , so divides , since divides
both and . Suppose
is a prime power that divides
, so
divides both and . Because
and are coprime and
divides , we see that
does not divide , so
must divide . Thus
divides , so as claimed.

We can also use Theorem 9.1.3 to determine the
-module structure of the successive quotients
.

**Proposition 9.1.8**
*
Let
be a nonzero prime ideal of , and let be an
integer. Then
as -modules.*
*Proof*.
(Compare page 13 of Swinnerton-Dyer.)
Since

(by unique factorization), we
can fix an element

such that

. Let

be the

-module morphism defined by

. The kernel of

is

since clearly

and if

then

, so

, so

, since

does not
divide

. Thus

induces an injective

-module
homomorphism

.

It remains to show that is surjective, and this is where we
will use Theorem 9.1.3. Suppose
.
By Theorem 9.1.3 there exists
such that

and

We have

since

and

by the second displayed condition, so

, hence

. Finally

so

is surjective.

William Stein
2004-05-06