Definition of a Quotient Ring (need clarification).

Question

My understanding is that we can view the definition of a Quotient ring $R/I$ as a set of cosets.

For example, the ring $Z/(6)$ which I believe is $Z6$, can be viewed like this:

$(6) + 0 = \{...,-12, -6, 0, 6, 12, ...\}$

$(6) + 1 = \{..,-17, -11, -5, 1, 7, 13, 19, ...\}$

$(6) + 2$

$(6) + 3$

$(6) + 4$

$(6) + 5$

My first question is, I realize that we stop at that, because $(6) + 6 = (6) + 0$, but what property do we use to do that exactly?

secondly, how is the set of all the cosets defined above, simply equal to ${0, 1, 2, 3, 4, 5}$ which is $z6$?

Like physically they just don't look like the same set. I'm having some difficulty wrapping my head around it, so any clarification would be appreciated.

Answer 1

First question: do you know (and have some kind of good understanding) of how quotient groups work? If not, you might want to revise them first.

The general idea is that any element of the quotient ring can be written as $a+I$ for some $a$ in your ring. If $a+I=b+I$ then $a-b\in I$ so you need to keep this in mind to check that operations are well defined.

Addition: $(a+I)+(b+I)=(a+b)+I$ so if $a+I=a'+I$ and $b+I=b'+I$ then $a+b-(a'+b')\in I$ this is because (and why) we require that every ideal is a group under addition.

Multiplication: $(a+I)\cdot(b+I)=(ab)+I$ so if $a+I=a'+I$ and $b+I=b'+I$ then $a-a'\in I$ and $b-b'\in I.$ the other property we require of ideals is that for any $i\in I$ and $r\in R$ we have $ri\in I$ so we have $ab-a'b, a'b-a'b\in I$ so adding these we get $ab-a'b'\in I$ so multiplication is well defined.

The way I personally like to think about quotient rings (or anything else algebraic) is that you take your ring and then set everything in the ideal to be 0 and then see what you get. Cosets also work but you have to keep in mind that your choice of representative when you write down an element is arbitrary. When you say $\mathbb Z/(6) = \mathbb Z_6$ then you are either defining $\mathbb Z_6$ or you mean that the two sets are isomorphic. The isomorphism is simply choosing representatives for the different cosets.

Answer 2

To actually answer your direct question as to why we stop at $5$, recall that $$(6)=(6)+0=\{6z|z\in\mathbb Z\}$$ and $$(6)+6=\{6z+6|z\in\mathbb Z\}$$ Put this way it should be clear that these are the same set. Which coset a number is in is determined by the remainder when you divide by $6$.