What some integer ideals of $\mathbb{Z}$ look like Maybe I'm having a rough day, but I can't seem to wrap my head around what integer ideals of $\mathbb{Z}$  like $2\mathbb{Z}/12\mathbb{Z}$ or $3\mathbb{Z}/12\mathbb{Z}$ look like. I understand that $\mathbb{Z}/2\mathbb{Z}=\left\{\bar{0}, \bar{1} \right\}$, where $\bar{0}=\left\{\ldots , -2,0,2,\ldots \right\}$ and $\bar{1}=\left\{\ldots , -1,1,3,\ldots \right\}$. Could someone show me? Thank you!
 A: It's really hard to tell what, exactly, you're asking.
But from what I can glean, I believe what you're looking here in $m\mathbb Z/n\mathbb Z$ are the elements of $m\mathbb Z$, modulo $n$:
$$2\mathbb Z/12\mathbb Z = \{\overline 0, \overline 2,\overline 4, \overline 6, \overline 8, \overline {10}\}$$
$$3\mathbb Z/12\mathbb Z = \{\overline 0, \overline 3,\overline 6, \overline 9\}$$
A: ideal is a well-established, but not a very descriptive term. you need to hang on to three chief properties that make a subset $I$ of a ring $R$ qualify as an ideal of $R$.


*

*$I$ is an abelian subgroup of the additive group $R$

*$I$ has the property of absorbing elements of $R$ by multiplication: I is a  left  ideal if $RI \subseteq I$. in terms of elements: for any element $r \in R$ and any $x \in I$, we have $rx \in I$ . 
in $\mathbb{Z}$ the set of all integer multiples of any given number, e.g. $6\mathbb{Z} =\{...,-12,-6,0,6,12,...\}$, can easily be shown to have both these properties - it is closed under addition (adding or subtracting two multiples of 6 gives another multiple of 6), inverses are obviously included, and $6\mathbb{Z}$ is absorbing, because if you multiply a multiple of 6 by any integer you get another multiple of 6.
there are also right ideals. if an ideal is both a left  and a right ideal, it is a $two-sided$ ideal. in commutative rings the left and right concepts coincide. 
the importance of two-sided ideals in any ring R  is that they are the kernels of ring homomorphisms $R \rightarrow R'$, and are thus a key element of the ring structure.
