Proving Subgroups and Cosets in RingsExploring Subgroups and Cosets in Rings

[DanielThrice]

Original Query: I'm beginning to look at rings for the first time and was given this to start with:
Let R = < R, +, *> (* means multiplication) be a ring, and let I < R be an additive subgroup of < R, + >.
Consider the set of cosets

R/I = {a + I: a is an element of R}

equipped with its own operations + and * defined by

(a + I) + (b + I) = (a + b) + I
(a + I) (b + I) = ab + I

How do we prove that the operations + and
are well-defined <---> the
additive subgroup I satis es the following conditions:

ab is an element of I for all a in R and b in I

ba is an element of I for all a in R and b in I

?

What I understand:
So the question is basically asking, `prove that R/I makes sense if and only if I is an ideal'.

I think I have to prove the following:
a + I = a' + I, b + I = b' + I ---> (a + I) + (b + I) = (a' + I) + (b' + I), (a + I) + (b + I) = (a' + I) (b' + I)
Are we just proving that something is well defined?

And to prove the other part of the `if and only if', this is what I think we should do:
Assume a + I = a' + I and b + I = b' + I but (a + I) (b + I) does not equal (a' + I) (b' + I)
That is, ab + I does not equal a'b' + I, which would imply that ab - a'b' is not an element of I.

I think I want to prove that there exists some r in R and x in I such that xr is not an element of I or rx is not an element of R . T

This needs some patching up, I'm a little new to rings

 

[HallsofIvy]

Yes, proving that something is "well defined" essentially means proving that the definition makes sense!

Here, "a+ I" represents a set of elements, the coset a is in, and a is a "representative" member of that set.

"(a+ I)+ (b+ I)= (a+ b)+ I" really says "select a member of each coset and add them. The sum of the two cosets is the coset a+ b is in". But what if we happened to use different representatives? That is, suppose c was in coset a+ I (which we could also call c+ I) and d was in coset b+ I (so that we could also have called that set d+I). We would not expect c+ d to be the same as a+ b but we must have that c+ d and a+ b are in the same coset. Is that true? If it is then this addition is "well defined".

 

[DanielThrice]

Thanks Ivy, helped a lot I figured it out. I like how you reworded it.