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?

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


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 $<a>+<m>$.


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 $$(\mathbb{Z}/m\mathbb{Z})/(a\mathbb{Z}/m\mathbb{Z})\simeq \mathbb{Z}/a\mathbb{Z}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.