The following result was known to

Chinese mathematics
in the first century AD. By way of notation, if

*a,b*
are integers and

*n* is a positive

integer we write

*a* cgt

*b* (mod

*n*) if

*n* divides *a-b*.

**Theorem** Let *n*_{1},....*n*_{m}
be pairwise coprime positive integers, that is,
*(n*_{i},n_{j})=1
for each distinct *i,j*. Then if *a*_{1},...
*a*_{m} are integers the system of equations

*a* cgt *a*_{1} (mod *n*_{1}), ...,
*a* cgt to *a*_{m} (mod *n*_{m})

has an integer solution that is uniquely determined

modulo *n*_{1}...n_{m}.

Here's an example:

*
n*_{1}=3, n_{2}=4, a_{1}=1, a_{2}=2

Then we can take

*a=10* (or

*a=...,-2,22,34,...*).

A generalisation of this result to ring theory nowadays gets
called the Chinese Remainder Theorem. Let's state and prove this
generalisation.

**Theorem** Let *I*_{1},...,*I*_{m}
be ideals in a ring *R* and suppose that *I*_{i}+I_{j}=R
for each distinct *i,j*. Then if *a*_{1},...
*a*_{m} are elements of *R* there exists
*a* in *R* such that the images of *a* and *a*_{i}
in the quotient ring *R/I*_{i} coincide. Further
*a* is uniquely determined up to its coset *a+I*, where
*I* is the intersection of all the *I*_{i}.

**Proof that the first theorem follows from the second:**
Take the ring
*R*=**Z** and consider the ideals
*I*_{i}=n_{i}**Z**. One has
that *(n*_{i},n_{j})=1 iff
*I*_{i}+I_{j}=**Z**. For, if *n*_{i} and
*n*_{j} are coprime then Euclid's algorithm for the
highest common factor supplies *r,s* such that

*
1 = n*_{i}r + n_{j}s

On the other hand, if

*I*_{i}+I_{j}=

**Z**
then we have an equation like the displayed one. Clearly any common
divisor of

*n*_{i} and

*n*_{j} is a

unit.
Finally, note that the quotient ring

**Z**/n

**Z** is the ring of

integers modulo n and that

*n*_{1}...n_{m}**Z**
is the intersection of the

*I*_{i}.

**Proof of the second theorem:**
Consider the inductive statement

**S**(*k*) *R=I*_{1} + I_{2} n....n I_{k}

for

*k=2,....,m*. Here

*n* denotes intersection.
By assumption,

**S**(

*2*) is true. I will
show that

**S**(

*k*) is true for all these values of

*k*, by
induction. So suppose that

**S**(

*k-1*) holds. Thus

*R=I*_{1} + I_{2} n....n I_{k-1}

Since

*R.R=R* (it contains 1) and

*I*_{1}+I_{k}=R
we deduce that

*R=(I*_{1}+I_{k})(I_{1} + I_{2} n....n I_{k-1})

Multiplying out the brackets we see that the LHS is contained in

*I*_{1} + I_{k}(I_{2} n....n I_{k-1})

But the second of thse terms is contained in

*I*_{2} n....n I_{k}, and so we are done.
Thus

**S**(

*m*) is true.

The same argument (with different
labelling of the ideals) shows that *I*_{i} + the intersection
of all *I*_{j}, with *j* different from *i*, is
equal to *R*. Thus, for each i, we can choose *b*_{i}
in *I*_{i} and *c*_{i} in the intersection of
all *I*_{j}, for *j* different from *i*, such that
*a*_{i}=b_{i} + c_{i}.
Now let *a=c*_{1}+...+c_{n}. Think about
the image of *a* in *R/I*_{i}. In this ring each
*c*_{j}, for *j* different from *i*, has
image zero. Also in this ring *c*_{i} has the the same image
as *a*_{i}. Thus *a* has the required property that
it has the same image as *a*_{i} in *R/I*_{i}.
If *a'* also has the same property then *a-a'* lies in each
*I*_{i}, hence the claimed uniqueness.

Finally a corollary that restates the result in terms of direct products
of rings.

**Corollary** With the hypotheses and notation of the theorem
there is an isomorphism of rings between *R/I* and the direct product
*R/I*_{1} x....x R/I_{m}.

**Proof:** Define a ring homomorphism
*f:R-->R/I*_{1} x....x R/I_{m}
by *f(a)=(a+I*_{1},...,a+I_{m}). The theorem shows
that this map is surjective and has kernel *I*, hence we are
done by the first isomorphism theorem.