# Ring map-integers proof

Prove that there is a ring map from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$ iff $m|n$.

I am not able to prove this with the appropriate rigour, could someone show the steps for a proof?

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Clearly, you are assuming that ring homomorphisms between rings with unity must ap the multiplicative identity to the multiplicative identity. Otherwise, the zero map would always work. But see this question – Arturo Magidin Jul 6 '12 at 18:18

$\def\Z{\mathbb Z}$ First suppose that there is a ring morphism $\phi\colon\Z/n\Z \to \Z/m\Z$. Then $0= \phi(0) = \phi(n \cdot 1) = n \cdot \phi(1) = n$, hence $n + m\Z = 0$, which gives $m \mid n$.

If on the other hand $m \mid n$, define $\phi\colon \Z/n\Z \to \Z/m\Z$ by $\phi(a + n\Z) = a + m\Z$, this is well defined: If $a + n\Z = b+n\Z$, then $n \mid a-b$, hence $m \mid a-b$, so $a + m\Z = b+m\Z$. Obviously $\phi$ is a homomorphism.

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I'm assuming you require your maps to send the multiplicative identity $1$ to $1$; otherwise, the statement is false: you can always map all elements of $\mathbb{Z}/n\mathbb{Z}$ to $0+m\mathbb{Z}$ to get an additive and multiplicative morphism. (See this recent question for a discussion of whether the zero map is or is not a ring homomorphism).

Here's a useful "universal property" of the kernel:

Theorem. Let $R$ be a ring, and let $I$ be an ideal. Let $\pi\colon R\to R/I$ be the canonical projection, $r\mapsto r+I$. If $S$ is any ring, and $f\colon R\to S$ is a ring homomorphism, then there exists a unique $\mathfrak{f}\colon (R/I)\to S$ such that $\mathfrak{f}\circ\pi = f$ if and only if $I\subseteq\mathrm{ker}(f)$.

Proof. Suppose that there is a map $\mathfrak{f}\colon (R/I) \to I$ such that $\mathfrak{f}\circ\pi = f$. If $a\in I$, then $f(a) = \mathfrak{f}(\pi(a))$. But $\pi(a) = a+I = 0+I$, so $\mathfrak{f}(0+I) = 0$, so $f(a)=0$, and hence $a\in\mathrm{ker}(f)$. This proves that if a map exists, then $I\subseteq\mathrm{ker}(f)$.

Conversely, suppose that $\mathrm{ker}(f)$ contains $I$. Define $\mathfrak{f}\colon (R/I) \to S$ by $\mathfrak{f}(r+I) = f(r)$ for all $r\in R$. I claim this is well defined: indeed, if $r+I = r'+I$, then $r-r'\in I\subseteq \mathrm{ker}(f)$, so $0=f(r-r')=f(r)-f(r')$. Thus $f(r)=f(r')$, so the value of $\mathfrak{f}$ does not depend on the representative of the coset. Thus, $\mathfrak{f}$ is well-defined.

It is now easy to verify that it is a ring homomorphism: $\mathfrak{f}((r+I) + (r'+I)) = \mathfrak{f}(r+r'+I) = f(r+r') = f(r)+f(r') =\mathfrak{f}(r+I) + \mathfrak{f}(r'+I)$; and $\mathfrak{f}((r+I)(r'+I)) = \mathfrak{f}(rr'+I) = f(rr') = f(r)f(r') = f(r+I)f(r'+I)$.

And if your rings have $1$ and your maps are unital, we also have $\mathfrak{f}(1+I) = f(1) = 1_S$.

Finally, $\mathfrak{f}\circ\pi(r) = \mathfrak{f}(r+I) = f(r)$, so $\mathfrak{f}\circ\pi = f$, as desired.

To finish, we prove that if a map exists, then it is unique. If $\mathfrak{f},\mathfrak{g}\colon R/I \to S$ satisfy $\mathfrak{f}\circ\pi = f = \mathfrak{g}\circ\pi$, then since $\pi$ is onto it can be cancelled on the left to deduce $\mathfrak{f}=\mathfrak{g}$; alternatively, for all $r+I\in R/I$ we have $\mathfrak{f}(r+I) = \mathfrak{f}\circ\pi(r) = \mathfrak{g}\circ\pi(r) = \mathfrak{g}(r+I)$, so $\mathfrak{f}=\mathfrak{g}$. $\Box$

Now to your specific problem: Because homomorphisms between $\mathbb{Z}$ and any ring, or quotients of $\mathbb{Z}$ and any ring, are completely determined by the value at the multiplicative identity, a map $\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}$ will correspond (assuming it must map $1$ to $1$) to a map $\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}$ that "factors through" $\mathbb{Z}/n\mathbb{Z}$. According to the proposition, the map factors through if and only $n\mathbb{Z}$ is contained in the kernel of the map $\mathbb{Z}\to \mathbb{Z}/m\mathbb{Z}$; this is $m\mathbb{Z}$. So thee map factors if and only if $n\mathbb{Z}\subseteq m\mathbb{Z}$, which happens if and only if $n\in m\mathbb{Z}$, which happens if and only if $m|n$.

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Here is a high - level interpretation of proving one direction: Suppose that $m|n$. Then we have that $(n) \subseteq (m)$. Now the kernel of the canonical projection $\pi : \Bbb{Z} \longrightarrow \Bbb{Z}/(m)$ is precisely $(m)$. Because $(n)$ is contained in the $\ker \pi$, a weak form of the first isomorphism theorem tells you that there is a well-defined map $f : \Bbb{Z}/(n) \rightarrow \Bbb{Z}/(m)$ such that

$$f([a]) = \pi(a).$$

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