Theorem 6.10 (Chinese Remainder Theorem)
Let
k >
0
, and let
a
1
, . . . , a
k
∈
D
, and let
n
1
, . . . , n
k
be nonzero polynomials such that
gcd(
n
i
, n
j
) = 1
for all
1
≤
i < j
≤
k
.
Then there
exists a polynomial
x
such that
x
≡
a
i
(mod
n
i
)
(
i
= 1
, . . . , k
)
.
Moreover, any other polynomial
x
0
is also a solution of these congruences if and only if
x
≡
x
0
(mod
n
)
, where
n
:=
Q
k
i
=1
n
i
.
If we set
R
=
D/nD
and
R
i
=
D/n
i
D
for 1
≤
i
≤
k
, then in ringtheoretic language, the
Chinese Remainder Theorem says the homomorphism from the ring
D
to the ring
R
1
× · · · ×
R
1
that sends
x
∈
D
to ([
x
mod
n
1
]
, . . . ,
[
x
mod
n
k
]) is a surjective homomorphism with kernel
nD
,
and hence
R
∼
=
R
1
× · · · ×
R
k
.
Let us recall the formula for the solution
x
(see proof of Theorem 2.6). We have
x
:=
k
X
i
=1
z
i
a
i
,
where
z
i
:=
n
0
i
m
i
,
n
0
i
:=
n/n
i
,
m
i
n
0
i
≡
1 (mod
n
i
)
(
i
= 1
, . . . , k
)
.
Now, let us consider the special case of the Chinese Remainder Theorem where
a
i
∈
K
and
and
n
i
= (
T

b
i
) with
b
i
∈
K
, for 1
≤
i
≤
k
. The condition that gcd(
n
i
, n
j
) = 1 for all
i
6
=
j
is
equivalent to the condition that
b
i
6
=
b
j
for all
i
6
=
j
. Then a polynomial
x
satisfies the system of
congruences if and only if
x
(
b
i
) =
a
i
for 1
≤
i
≤
k
. Moreover, we have
n
0
i
=
Q
j
6
=
i
(
T

b
j
), and
m
i
:= 1
/
Q
j
6
=
i
(
b
i

b
j
) is a multiplicative inverse of
n
0
i
modulo
n
i
. So we get
x
=
k
X
i
=1
a
i
Q
j
6
=
i
(
T

b
j
)
Q
j
6
=
i
(
b
i

b
j
)
.
The reader will recognize this as the LaGrange Interpolation Formula. Thus, the Chinese Remainder
Theorem for polynomials includes LaGrange Interpolation as a special case.