Thread: Well Defined Subgroups

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

?

Originally Posted by DanielThrice

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


You have to probe that

.

I don't understand what you meant with the following part below

Tonio

<---> the
additive subgroup I satises 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

?

.

Originally Posted by tonio

.

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

To the OP, Tonio has told you how to prove something is well defined. So do this. To prove the other part of the `if and only if', you need to first assume that and but .

That is, . This implies that .

You want to prove that there exists some and such that or . To do this, use your assumption that , as well as that and . There is a "trick" involved here, which I will let you figure out for yourself...

So, 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".
What if we used 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. If this is, then this addition is well defined right? Is it?

Yest, it is!

And, here, saying that a and c are in the same coset means that a= c+ x for some x in I. Saying that b and d are in the same coset means that b= d+ y for some y in I. Finally, then ab= (c+ x)(d+ y)= cd+ cx+ dy+ xy. Then ab and cd are in the same coset (and so addition is "well defined" if and onlyif cx+ dy+ xy is in I.