I 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:

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

 

[fresh_42]

Hi Peter,

you can handle them as usual additions: $r+I=r'+I \Longleftrightarrow r-r'+I = I \Longleftrightarrow a:= r-r' \in I$. If you like you can write those cosets as
$$
r+I = r+ \{i\,\vert \,i \in I\} = \{r+i\,\vert \,i\in I\} \text{ and so } r'+I = r+I \Longleftrightarrow \{r'+i\,\vert \,i\in I\} = \{r+i\,\vert \,i\in I\}
$$
which means a certain $r'+i$ must be of the form $r+j$ and with $r'+i=r+j$ we have $r'-r =j-i \in I$. The set notation is a bit inconvenient, so as soon as we know all those rules mentioned in the text, it's easier to treat them like any addition. The only thing, what really has to be considered is well-definition, because the $r$ in $\{r+i\,\vert \,i\in I\} = [r] = r+I$ isn't uniquely defined as it can differ by elements of $I$.

We can consider $[g] = g\cdot U \in G/U$ for any subgroup $U$ of a (here multiplicative) group $G$, but if and only if $U$ is also a normal subgroup, we can define a group structure on $G/U$, for otherwise it won't be well-defined. As $(R,+)$ is an Abelian additive group, all additive subgroups as $I$ are automatically normal.

 

[Math Amateur]

Thanks for the help, fresh_42 ...

Reflecting on what you have written ...

Thanks again ...

Peter

 

[WWGD]

You can also do a straight subtraction: $r'+I=r+I$ then $r-r' \in I$ so that $r \~ r'$ , so if we have different reps, then the elements are equivalent. Try something similar for the product to show that $r's' \~ rs$.

 

[WWGD]

Hi Peter,

you can handle them as usual additions: $r+I=r'+I \Longleftrightarrow r-r'+I = I \Longleftrightarrow a:= r-r' \in I$. If you like you can write those cosets as
$$
r+I = r+ \{i\,\vert \,i \in I\} = \{r+i\,\vert \,i\in I\} \text{ and so } r'+I = r+I \Longleftrightarrow \{r'+i\,\vert \,i\in I\} = \{r+i\,\vert \,i\in I\}
$$
which means a certain $r'+i$ must be of the form $r+j$ and with $r'+i=r+j$ we have $r'-r =j-i \in I$. The set notation is a bit inconvenient, so as soon as we know all those rules mentioned in the text, it's easier to treat them like any addition. The only thing, what really has to be considered is well-definition, because the $r$ in $\{r+i\,\vert \,i\in I\} = [r] = r+I$ isn't uniquely defined as it can differ by elements of $I$.

We can consider $[g] = g\cdot U \in G/U$ for any subgroup $U$ of a (here multiplicative) group $G$, but if and only if $U$ is also a normal subgroup, we can define a group structure on $G/U$, for otherwise it won't be well-defined. As $(R,+)$ is an Abelian additive group, all additive subgroups as $I$ are automatically normal.

But don't you also have to show that multiplication is well-defined?

 

[fresh_42]

In case of the ring, yes, but this is very similar to the method for addition, especially as I suppose the ring to be commutative. Otherwise left and right ideals, and left and right cosets would have be to distinguished.

In case of a group, there is only one operation, which I took to be the multiplication according to the usual notation. My major point was, that for $U \le G$ one has a disjoint union of equal sized cosets $G/U$ but no well defined group structure, i.e. multiplication on it. On the other hand, this can be done exactly if $U \trianglelefteq G$ is normal. I mentioned it, because it carries the insight why normality is needed, plus the information, that $G/U$ itself doesn't require it - as long as we don't want to make it a group. It was meant to fight the wrong reflex: Consider $G/U \ldots$ - But $U$ isn't a normal subgroup.

 

[WWGD]

In case of the ring, yes, but this is very similar to the method for addition, especially as I suppose the ring to be commutative. Otherwise left and right ideals, and left and right cosets would have be to distinguished.

In case of a group, there is only one operation, which I took to be the multiplication according to the usual notation. My major point was, that for $U \le G$ one has a disjoint union of equal sized cosets $G/U$ but no well defined group structure, i.e. multiplication on it. On the other hand, this can be done exactly if $U \trianglelefteq G$ is normal. I mentioned it, because it carries the insight why normality is needed, plus the information, that $G/U$ itself doesn't require it - as long as we don't want to make it a group. It was meant to fight the wrong reflex: Consider $G/U \ldots$ - But $U$ isn't a normal subgroup.

Yes, story of my life, it is what I am usually told " You isn't normal".

 

[fresh_42]

Yes, story of my life, it is what I am usually told " You isn't normal".

Neither am I, but I can still top that: I've also the strange property, that I seemingly attract especially people, who are definitely even less normal than me: I've looked into so many human abysses, preferably female ones, I should write a book about it.