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I start with our definitions. We defined $R/I :=\{x+I | x \in R\} $ and $x+I := \{y \in R |y-x \in R\} = \{x+a|a\in R\}$.

Now let's say $R=\mathbb{Z}$ and $I= \mathbb{(nZ)}$ and more specific n=3. This would then be $ 0 + \mathbb{3Z} = \{...,-3,0,3,...\} $
$ 1 + \mathbb{3Z} = \{...,-2,1,4,...\} $
$ 2 + \mathbb{3Z} = \{...,-1,2,5,...\} $
$ 3 + \mathbb{3Z} = \{...,-3,0,3,...\} $
and so on...

I understand here that only the equivalence classes of 0,1,2 are of relevance since every further set is the same set as one of these we already have $(0 = 3 \pmod 3)$. As far as I understand $\mathbb{Z}/n\mathbb{Z}$ is a partition on $\mathbb{Z}$ and with that it would make sense that $\mathbb{Z}/n\mathbb{Z}$ is the whole set of natural numbers. But from what I've seen $\mathbb{Z}/n\mathbb{Z} = \{0,1,2\}$.

Do I misunderstand some notation or definition here or do I generally understand something wrong? I've been reading a lot on this and am confused more and more. I still have problems what a modulo I means.

2 Answers 2

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The elements of $\Bbb Z / 3\Bbb Z$ are sets (equivalence classes, in particular). There are $3$ sets in $\Bbb Z/ 3 \Bbb Z$: one element is $[0] = \{\dots,-3,0,3,\dots\}$, and the other elements are $[1]$ and $[2]$. Thus, $\Bbb Z/3\Bbb Z = \{[0],[1],[2]\}$.

It is indeed true that if we consider $[0],[1],[2]$ as sets, then $[0] \cup [1] \cup [2] = \Bbb Z$. However, $\{[0],[1],[2]\}$ and $[0] \cup [1] \cup [2]$ are distinct objects.

It is common to write $\Bbb Z/3 \Bbb Z = \{0,1,2\}$ with the understanding that the elements $0,1,$ and $2$ are merely representatives of their equivalence class (that is, $0$ is a stand-in for $[0]$).

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    so, if $N \subseteq R$ is the set of all nilpotent elements, the statement, that R/N contains no nilpotent element besides 0 should be wrong, right? – 2017-01-18
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    @Smoenybfan For a commutative ring $R$? No: what you are saying is not correct. The **quotient** does not contain any nilpotent elements other than zero. The **quotient's elements** will contain other nilpotent elements, but that does not make the element of the quotient nilpotent. I'm getting the feeling from this question and the OP that you are coping with the existence of membership more than one level deep! – 2017-01-18
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    Note, for example, that $\{\{1\}\}$ does **not** contain $1$ (by the usual terminology) since $1 \notin \{\{1\}\}$. It does contain $\{1\}$, which in turn contains $1$. – 2017-01-18
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You have the right answer embedded in your question:

$\mathbb Z/n\mathbb Z$ is a partition on $\mathbb Z$.

For instance, $$\mathbb Z/3\mathbb Z = \{0+3\mathbb Z,1+3\mathbb Z,2+3\mathbb Z\}.$$ That is, it is a set of sets. Note that this is the same as writing $$\mathbb Z/3\mathbb Z = \{9+3\mathbb Z,4+3\mathbb Z,-1+3\mathbb Z\}$$ or choosing any set of representatives for the equivalence classes.

However, when you are working "inside" one of these quotients, it is fairly typical to write $0$ to indicate the equivalence class $0+3\mathbb Z$ or $1$ to indicate the equivalence class $1+3\mathbb Z$. Often, more explicit notations like $\bar 0$ or $\langle 0\rangle$ are used to highlight that the elements are not literally integers, but classes of integers. This is just because it tends to be easier to work with such notations.

Also, sometimes, one sees that, as a group $(\mathbb Z/n\mathbb Z,+)$ is isomorphic to the group $(\{0,1,\ldots,n-1\},\oplus)$ where $a\oplus b$ is taken as the remainder of $a+b$ when divided by $n$, in which case the members really are integers and the addition is close enough to normal that sometimes one just writes $+$ with the understanding that remainders will be taken. This works similarly if you want to have multiplicative structure on the quotient.