A proof problem about congruence relation

Question

Given a ring R and a two-sided ideal I in R, we may define an equivalence relation ~ on R as follows: a ~ b if and only if a − b is in I. Using the ideal properties,

it is not difficult to check that ~ is a congruence relation.

                                                               -------wikipedia

But it's not easy for me.

congruence relations is an equivalence relation such that

Given any elements a, a' , b, and b' of G, if a ~ a' and b ~ b' , then a * b ~ a' * b'

It's obvious that ~ is a equivalence relation,there is some problem with prove:Given any elements a, a' , b, and b' of G, if a ~ a' and b ~ b' , then a * b ~ a' * b'.

I want to know how to prove it

Answer 1

I'm going to assume that you have the equivalence relation part sorted.

If $a-a' \in I$ and $b-b'\in I$, then $ab-a'b' = \underbrace{(a-a')}_{\in \ I}b + a'\underbrace{(b-b')}_{\in \ I}$

Given that $I$ is an ideal:

• For every $x\in I, r\in R$, we have $rx, xr\in I$ (using two-sidedness); and
• For every $x,y \in I$, we have $x+y \in I$

So that the RHS of the above is in $I$, which is precisely what was required for the congruence relation.