# Homomorphic images of $\mathbb{Z}_{n}$

I have just shown that $\mathbb{Z}_{n}$ is a principal ideal ring, by using (1) a previously proven result that $\mathbb{Z}$ is a principal ideal domain (and hence a principal ideal ring), and (2) the result that the image of a principal ideal ring under a ring endomorphism (in this case, the canonical epimorphism $f: \mathbb{Z} \to \mathbb{Z}_{n} \simeq \mathbb{Z}/n\mathbb{Z}$ defined by $a \mapsto a + n\mathbb{Z} \simeq m\mathbb{Z}, \, m \in \mathbb{Z}_{n}$) is also a principal ideal ring.

By this result (or rather as a consequence of the proof of this result), I determined that ideals of $\mathbb{Z}_{n}$ are of the form $f(n\mathbb{Z})$, so does that mean that all the ideals are of the form $m\mathbb{Z}$ as described above in my mention of the canonical epimorphism?

Now, by the First Isomorphism Theorem for Rings, Extended Version, the possible homomorphic images of $\mathbb{Z}_{n}$ are isomorphic to $\mathbb{Z}_{n}/I$, where $I$ is an ideal of $\mathbb{Z}_{n}$. So, once I know how to correctly write down the ideals of $\mathbb{Z}_{n}$, I should be able to describe completely all homomorphic images of $\mathbb{Z}_{n}$, which is what I am looking to do here.

So I suppose my question here is twofold: (1) Have I correctly identified/described the ideals of $\mathbb{Z}_{n}$ here? If not, what should I change? And (2) Once I have correctly done that, is it enough to just say that all homomorphic images are $\mathbb{Z}_{n}/I$ where the $I$ are those ideals, or is there a better way to express them?

Thank you in advance.

## 1 Answer

$\newcommand{\Z}{\mathbb Z}$ By the Third Isomorphism Theorem, ideals $I$ of $\Z/n\Z$ are in bijection of ideals $J$ of $\Z$ that contain $(n)$, and furthermore, $(\Z/n\Z)/I \cong \Z/J$ under this correspondence. So now you just need to figure out what ideals of $\Z$ contain $(n)$.

• how would I figure that out? $(n)$ consists of all finite sums of $mn$, right? Where $m \in \mathbb{Z}$ and $n \in \mathbb{Z}_{n}$? – ALannister Mar 19 '17 at 23:19
• You know that $\mathbb{Z}$ is a principal ideal ring, which means all of its ideals are of the form $(n)$ for some $n$ in $\mathbb{Z}$. The elements of $(n)$ are exactly the multiples of $n$. That is, $\{0,\pm n, \pm 2n, \ldots \}$. When will this be contained in another ideal $(d)$? – user263190 Mar 19 '17 at 23:20
• When $n \vert d$? – ALannister Mar 19 '17 at 23:21
• If $n \vert d$, then $d = mn$ for some $m \in \mathbb{Z}$, which means that $d \in (n)$. But you want to know when $(n)$ is contained in $(d)$, which is equivalent to saying that $n \in (d)$. – user263190 Mar 19 '17 at 23:22
• The homomorphic images of $\mathbb{Z}/n\mathbb{Z}$ are things of the form $(\mathbb{Z}/n\mathbb{Z}) / I$ where $I$ is an ideal of $\mathbb{Z}/n\mathbb{Z}$. But this is isomorphic to $\mathbb{Z}/J$ where $J$ is some ideal of $\mathbb{Z}$ that contains $(n)$. These ideals are of the form $(d)$ where $d \vert n$. – user263190 Mar 19 '17 at 23:26