List of all the ideals of $\mathbb{Z}/4 \mathbb{Z}$, deduce that all its ideals are principal but that it's not an integral domain. So I am looking at the solution sheet of the following question: Make a list of all the ideals of $\mathbb{Z}/4 \mathbb{Z}$ and deduce that all ideals of $\mathbb{Z}/4 \mathbb{Z}$ are principal but that $\mathbb{Z}/4 \mathbb{Z}$ is not an integral domain.
The solution goes as follow (I believe it lacks a lot of details and therefore I can't understand it):
$\mathbb{Z}/4 \mathbb{Z} = \{[0], [1], [2], [3] \}$


*

*$(0) = {0}$

*$(1) = \mathbb{Z}/4 \mathbb{Z} = (3)$

*$(2) = \{ [0], [2] \}$ but has 2 dividers of zero.
Now here are my questions: 


*

*What is the following notation: $[i]$?

*What is the following notation: $(i)$?

*And finally, if someone could somehow clarify this solution, it'd be perfect.
 A: $[i]$ is the equivalence class of $i\in \Bbb Z$ under the relation induced by $4\Bbb Z$. Many just ignore the brackets and write $i$ instead.
$(i)$ is the ideal generated by $i$. Because they insist on square brackets on the elements of $\Bbb Z/4\Bbb Z$, they really should've written $([i])$.
The proof is not really a full proof as it stands. They have listed three principal ideals and noted that the ring is not an integral domain, but they have failed to prove that they have all the ideals. It is not difficult to do (heck, it might even suffice to say "and these are all the ideals"), but it has to be done.
A: To add to Arthur's answer, this question has a rather pointed purpose. It seeks to show a quality of rings by example.

Not every ring with only principal ideals is a Principal Ideal Domain.

That is to say, having only principal ideals does not imply a ring is a domain. As for the proof of your example, two facts were used:


*

*Principal ideals are, by definition, the smallest ideals containing a single element. 


To find all of the ideals, first consider the principal ideals, since any other ideals must contain these ones. Since $(1) =(3) = \mathbb{Z} /4\mathbb{Z}$, if an ideal contains these two elements, then it's the whole ring. This implies that any proper ideals will not contain these elements. Then, we have $(0)$ and $(2)$ with the relation that $(0)\subseteq(2)$. Since any proper ideal could only contain $[0]$ and $[2]$, and $(2)$ does contain these elements, no other ideals exist.


*Integral Domains, by definition, have no zero divisors. 


The ring $\mathbb{Z}/4\mathbb{Z}$ does have a zero divisor -- namely $[2]$. This seems to be the missing detail in the given proof.
A: In this solution the possible ideals are enumerated:
First the trivial ideals, generated by $[0]$ (contains only $[0]$) and by $[1]$ (contains every multiple of $[1]$, hence every element of the ring). The latter is equal to the ideal generated by $[3]$ since $[3]=[-1]$.
Therefore a non-trivial ideal can contain only the multiples of $[2]$: $\{[0],[2]\}$.
