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Let $m$ be a natural number and consider the factor ring

$$\mathbb Z/\langle m\rangle=\mathbb Z_m=\{\bar 0, \bar 1, ... , \overline{m-1}\}.$$

Let $\bar a$ be an element of $\{\bar 0, \bar 1, ... , \overline{m-1}\}$. It is true that we have the isomorphism?

$$\mathbb Z_m/\langle\bar a\rangle\cong\mathbb Z/(\langle a\rangle+\langle m\rangle)$$

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Consider the composition of mappings

$$\mathbb{Z} \to \mathbb{Z}_m \to \mathbb{Z}_m/<a> \,.$$

$x \to (x+<m>) \to (x+<m>)+<a> \,.$

Prove that the kernel is exactly $<a>+<m>$.

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The ideals of $R/I$ correspond to the ideals of $R$ that contain $I$. So the ideals of $\mathbb{Z}/m \mathbb{Z}$ correspond to the ideals $a\mathbb{Z}\supseteq m\mathbb{Z}$, i.e., with $a\mid m$. In particular, $a\mathbb{Z}+m\mathbb{Z}=a\mathbb{Z}$. If $I \le J$ are ideals of $R$, then $(R/I)/(J/I)\simeq R/J$. Hence $$(\mathbb{Z}/m\mathbb{Z})/(a\mathbb{Z}/m\mathbb{Z})\simeq \mathbb{Z}/a\mathbb{Z}.$$

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