# Well Defined Subgroups

• Feb 16th 2011, 04:11 PM
DanielThrice
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

?
• Feb 16th 2011, 07:00 PM
tonio
Quote:

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 $a+I=a'+I\,,\,b+I=b'+I\Longrightarrow$

$\Longrightarrow (a+I)+(b+I)=(a'+I)+(b'+I)\,,\,\,(a+I)(b+I)=(a'+I)( b'+I)$.

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

?

.
• Feb 17th 2011, 12:59 AM
Swlabr
Quote:

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 $a+I=a^{\prime}+I$ and $b+I=b^{\prime}+I$ but $(a+I)(b+I) \neq (a^{\prime}+I)(b^{\prime} + I)$.

That is, $ab+I \neq a^{\prime}b^{\prime} + I$. This implies that $ab-a^{\prime}b^{\prime} \not\in I$.

You want to prove that there exists some $r \in R$ and $x \in I$ such that $xr \not\in I$ or $rx \not\in R$. To do this, use your assumption that $ab-a^{\prime}b^{\prime} \not\in I$, as well as that $a-a^{\prime} \in I$ and $b-b^{\prime} \in I$. There is a "trick" involved here, which I will let you figure out for yourself...
• Feb 19th 2011, 11:30 AM
DanielThrice
Progress
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?
• Feb 19th 2011, 11:37 AM
HallsofIvy
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.