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I am working through Abstract Algebra by Dummit and Foote and want to make sure I understand the notation used in the text. The text is discussing the Isomorphism Theorems for Rings and uses the following example.

"Let $R= \mathbb{Z}$ and let $I$ be the ideal $12\mathbb{Z}$. The quotient ring $\bar{R} = R/I= \mathbb{Z}/12\mathbb{Z}$ has ideals $\bar{R}, 2\mathbb{Z}/12\mathbb{Z}, 3\mathbb{Z}/12\mathbb{Z}, 4\mathbb{Z}/12\mathbb{Z}, 6\mathbb{Z}/12\mathbb{Z}$ and $\bar{0} = 12\mathbb{Z}/12\mathbb{Z}$ corresponding to the ideals in $R = \mathbb{Z}, 2\mathbb{Z}, 3\mathbb{Z}, 4\mathbb{Z}, 6\mathbb{Z}$ and $12\mathbb{Z} = I$ of $R$ containing $I$, respectively. "

What is the meaning of $2\mathbb{Z}/12\mathbb{Z}$? More generally what does $d\mathbb{Z}/n\mathbb{Z}$ mean when $d$ is a divisor of $n$? I know that $\mathbb{Z}/12\mathbb{Z}$ is the set of integers modulo 12. Would $2\mathbb{Z}/12\mathbb{Z}$ then be the multiples of 2 in the set of modulo 2? In other words would $2\mathbb{Z}/12\mathbb{Z} = \{\bar{2}, \bar{4}, \bar{6}, \bar{8}, \bar{10}\}$.

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While @mathslearner98 gives a very nice and very concrete answer, I thought I could add a slightly more high-level answer, to give more context…

Consider a ring $R$ and an ideal $I$ in $R$. This gives rise to a quotient map $\pi:R \rightarrow R/I$.

Given any subset $S\subseteq R$ the notation $S/I$ is a very convenient notation for the image $\pi(S)\subseteq S$ of $S$ under $\pi$, since it reminds us that such an image is of the form $$\pi(S)=S/I=\{s+ I \mid s\in S\}.$$

Now since $\pi$ is surjective we find that every ideal $J\subseteq R$ maps to an ideal $\pi(J)=J/I\subseteq S$. This is precisely what the $d\Bbb Z/n\Bbb Z$ is expressing.

The reason for why we assume that $d$ divides $n$ is the following. Ideals in $R/I$ are in bijection to those ideals in $R$, which contain $I$. In other words the maps $$\begin{array}{rcl} \{\text{ideals }J \subseteq R \mid I \subseteq J\}&\longrightarrow& \{\text{ideals } \widetilde{J}\subseteq R/I\}\\ J & \mapsto & \pi(J)=J/I\\ \pi^{-1}(\widetilde{J}) & \leftarrow & \widetilde{J} \end{array}$$ are mutually inverse. So in our case we not only know that every ideal in $\Bbb Z$ gives an ideal in $\Bbb Z/n\Bbb Z$, but that every ideal in $\Bbb Z/n\Bbb Z$ is of the form $d\Bbb Z/ n\Bbb Z$ for some ideal $d\Bbb Z$ in $\Bbb Z$ containing $n\Bbb Z$, hence for some $d$ dividing $n$!

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$2\mathbb{Z}/12\mathbb{Z}=\{\overline{0},\overline{2},\overline{4},\overline{6},\overline{8},\overline{10}\}$ as a set. To get an idea to understand this sort of a notation in general, note that $2\mathbb{Z}=\{2n:n\in\mathbb{Z}\}$ and $12\mathbb{Z}=\{12n:n\in\mathbb{Z}\}$. So $2\mathbb{Z}/12\mathbb{Z}=\{x+12\mathbb{Z}:x\in2\mathbb{Z}\}$, and since, for instance, $2+12\mathbb{Z}=14+12\mathbb{Z}$, we didn't have to write $\overline{14}$ when writing $2\mathbb{Z}/12\mathbb{Z}$ as a set because $\overline{2}$ is already there.

So in short, a general strategy to understand $d\mathbb{Z}/n\mathbb{Z}$ is first to write down $d\mathbb{Z}$ and $n\mathbb{Z}$ as sets, then think about how would $\{x+n\mathbb{Z}:x\in d\mathbb{Z}\}$ look like.

Here, $x+n\mathbb{Z}$ is the set $\{x+y:y\in n\mathbb{Z}\}$.

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  • $\begingroup$ Another interpretation is $2\mathbb{Z}/12\mathbb{Z}=2(\mathbb{Z}/12\mathbb{Z}) = \{ 2x : x \in 12\mathbb{Z} \}$. $\endgroup$
    – lhf
    Nov 5, 2021 at 10:40

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