A Chinese Remainder Theorem (CRT) is explained intuitively, which helps reader to understand its connection with Abstract Algebra concepts like Groups and Rings. Here, Author has tried to simplify the concept by explaining sub topic of Groups and Rings which is required by understanding CRT in Abstract Algebra.

1. Chinese Remainder Theorem: The Abstract Algebra Way

2. Normal Subgroup
A subgroup N of a group G is called normal subgroup of G if, g n g-1 𝜖 N ; ꓯ g 𝜖 G and n 𝜖 N , Let’s
understand this more intuitively…
Group: Integers Z under +
Subgroups: Z, 2Z, 3Z, 4Z, 5Z …
5Z = { …-15,-10,-5,0,5,10,15…}  subgroup note: Cosets are not subgroup
1+5Z= {…-14,-9,-4,1,6,11,16…}  coset as they don’t have
2+5Z= {…-13,-8,-3,2,7,12,17…}  coset Identity element, Inverses,
3+5Z= {…-12,-7,-2,3,8,13,18…}  coset and not closed under addition.
4+5Z = {…-11,-6,-1,4,9,14,19…}  coset Cosets play very important role in
error correction and channel coding.
Now original group Z+ is completely covered by a subgroup 5Z and its four cosets.
We can also say, 5Z = 0+ 5Z as a coset.
We use the subgroup 5Z to partition the group Z+ in cosets, because the, cosets form a group we
may call 5Z a normal subgroup.

3. Quotient Group
Group of these cosets is called a Quotient Group and can be written as Z/5Z.
Name Quotient Group is very descriptive, as when you divide one thing by another you get
quotient.
Now we can treat this cosets as elements in a new group, say coset group. So if we add any
element of one group to any element in other group we get element of a another coset.
(1+5Z)+(2+5Z) = 3+5Z
For ex. (-4) + (2) = -2
Remember: Cosets do not form a Group unless it met below condition..
Suppose N ≤ G and its cosets form a Group iff y N y-1 = N , for any y 𝝐 G, this coset group also
called Factor Group.

4. Ring Ideals

5. Chinese Remainder Theorem for Rings
• Let a1,…,an be the ideals of a commutative ring R which satisfy the
condition
(1) ai +aj = R, ꓯ i ≠ j
Then ꓯx1,…,xn 𝝐 R there exist x 𝝐 R such that
(2) x ≡ xi (mod ai) for 1 ≤ i ≤ n
Moreover, x is unique modulo a := a1∩ ⋯ ∩ an
For Example, if we have two Ideals I and J of ring R such that I + J = R , then
any r,s 𝝐 R , there is an x 𝝐 R such that
x ≡ r (mod I) and x ≡ s (mod J)

6. Chinese Remainder Theorem for Rings ..continued
Let us prove: if we have two Ideals I and J of ring R such that I + J = R ,
then any r,s 𝝐 R , there is an x 𝝐 R such that
x ≡ r (mod I) and x ≡ s (mod J)
Let r= ri + rj and s = si+ sj for some si, ri 𝝐 I , and sj, rj 𝝐 J ,
Let x= rj + si then,
x-r = ri – si 𝝐 I and x-s =rj – sj 𝝐 J Thus,
x ≡ r (mod I) and x ≡ s (mod J)

7. Example of first page…
• A bowl contains certain number of Almonds, when we take Almonds in
group of 5 then 3 Almonds left in the bowl,but if we take Almonds in
group of 7 then 4 Almonds left behind. How many Almonds are there in
bowl?
• x ≡ r (mod I) and x ≡ s (mod J)
• Here x ≡ 3 (mod 5) and x ≡ 4 (mod 7)
• Let us write elements of 35 (5 x 7= 35)
• U(35) = {x | 1≤ x ≤35, gcd(x,35)=1}
{1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34}

8. Solving x ≡ 3 (mod 5) and x ≡ 4 (mod 7)
• U(35)={1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34} i.e., x 𝝐 R always!
• we know that 5Z and 7Z are the Ideals of a Ring R.
• lets write the required cosets of 5Z 7Z, i.e., 3+5Z and 4+7Z and find an x which is the unique and
present in both cosets..
• 3+5Z = {…-12,-7,-2,3,8,13,18,…}
• 4+7Z = {…-10,-3,4,11,18,24,32,…}
we can see that x=18.
If x ≡ 2 (mod 5) and x ≡ 𝟑 (mod 7) then we can easily see that x=17
2+5Z = {...-13,-8,-3,2,7,12,17...}
3+7Z = {...,-11,-4,3,10,17,24,...}