# All element of the quotient ring $\mathbb{Z}_m/I$ for some $m \in \mathbb{N}$.

Let $$I = \lbrace \overline{0}, \overline{8}, \overline{16} \rbrace$$ be an ideal in $$\mathbb{Z}_{24}$$. Find all elements of quotient ring $$\mathbb{Z}_{24}/I$$.

The answer is $$\mathbb{Z}_{24}/I = \lbrace I, \overline{1} + I, \overline{2}+I, \dots, \overline{7} + I \rbrace$$. But, I still can't understand how to obtain it. Anyone can explain, please? Thanks in advance.

It results directly from the Third isomorphism theorem: $$I$$ is simply the quotient $$\:8\mathbf Z/24\mathbf Z$$, so $$(\mathbf Z/24\mathbf Z)\big/I=(\mathbf Z/24\mathbf Z)\big/(8\mathbf Z/24\mathbf Z)\simeq \mathbf Z/8\mathbf Z.$$
• Why it is $Z/24Z$ ? Aug 5 '20 at 22:36
• and how can $I = 8Z/24Z$ ? Aug 5 '20 at 22:37
• what if like this: $Z_{24} = \lbrace \overline{0}, \overline{1}, \dots, \overline{23} \rbrace$. Then, $Z_{24}/I = \lbrace \overline{0}+I, \overline{1}+I, \dots, \overline{7}+I \rbrace$ since for $\overline{8}$, it would be $\overline{8}+I = I+I = I$, for $\overline{9}: \overline{9} + I = \overline{1+8} + I = \overline{1} + I$, and so on. Aug 5 '20 at 23:15
• Sorry, I had not seen the above commentsYes, I think you can add these details. $I=8\mathbf Z/24\mathbf Z$ because $I$ is made up of the multiples of $8\bmod 24$ (since $4\cdot\bar 8=\bar 0$). Aug 5 '20 at 23:47