Quotient Rings .... Remarks by Adkins and Weintraub ....

[Math Amateur]

I am reading "Algebra: An Approach via Module Theory" by William A. Adkins and Steven H. Weintraub ...

I am currently focused on Chapter 2: Rings ...

I need help with fully understanding some remarks by Adkins and Weintraub on quotient rings on page 59 in Chapter 2 ...

The remarks by Adkins and Weintraub on quotient rings read as follows:
View attachment 7948
In the above text from A&W we read the following:

" ... ... All that needs to be checked is that this definition is independent of the choice of coset representatives. To see this suppose $$r + I = r' + I$$ and $$s + I = s' + I$$. Then $$r' = r + a$$ and $$s' = s + b$$ where $$a,b \in I$$. ... ... ... "Can someone please (fully) explain how/why it is that $$r + I = r' + I$$ and $$s + I = s' + I$$ imply that $$r' = r + a$$ and $$s' = s + b$$ where $$a,b \in I$$ ... ... ?
Help will be appreciated ...

Peter

 

[GJA]

Hi, Peter.

Can someone please (fully) explain how/why it is that $$r + I = r' + I$$ and $$s + I = s' + I$$ imply that $$r' = r + a$$ and $$s' = s + b$$ where $$a,b \in I$$ ... ... ?

Here is one possible argument:

By definition, $r + I = \{r+a: a\in I\}$ and $r' + I = \{r'+a: a\in I\}.$ Now, $r'\in r' + I$ because $0\in I$. Since $r'\in r'+I=r+I=\{r+a: a\in I\},$ $r'=r+a$ for some $a\in I$.

 

[Math Amateur]

Hi, Peter.
Here is one possible argument:

By definition, $r + I = \{r+a: a\in I\}$ and $r' + I = \{r'+a: a\in I\}.$ Now, $r'\in r' + I$ because $0\in I$. Since $r'\in r'+I=r+I=\{r+a: a\in I\},$ $r'=r+a$ for some $a\in I$.

Well! Thanks! Really clear ...

Appreciate your help GJA ...

Peter