## If n divides m, then the natural ring homomorphism from ZZ/(m) to ZZ/(n) is surjective on the units

Let $m$ and $n$ be positive integers with $n$ dividing $m$. Prove that the natural surjective ring homomorphism $\varphi : \mathbb{Z}/(m) \rightarrow \mathbb{Z}/(n)$ is also surjective on the units.

[With help from Ravi Vakil’s notes.]

Given a unital ring homomorphism $\varphi : R \rightarrow S$, it is certainly the case that $\varphi[R^\times] \subseteq S^\times$. If we have equality, we say that $\varphi$ is surjective on the units of $S$. If $\varphi$ and $\psi$ are unital ring homomorphisms which are surjective on the units and the composition makes sense, then so is $\varphi \circ \psi$. Clearly isomorphisms are surjective on the units.

We begin with some lemmas.

Lemma 1: Let $\{R_i\}_I$ be a nonempty set of rings with $1 \neq 0$. Then $(a_i) \in \prod R_i$ is a unit if and only if $a_i \in R_i$ is a unit for all $i \in I$. Proof: $(\Rightarrow)$ Let $(a_i)$ be a unit. Then there exists $(b_i) \in \prod R_i$ such that $(a_i)(b_i) = (a_ib_i) = 1$, and thus $a_ib_i = 1$ for all $i$. Similarly, $b_ia_i = 1$. Thus $a_i$ is a unit for each $i$. $(\Leftarrow)$ If $a_i \in R_i$ is a unit, then there exists $b_i \in R_i$ such that $a_ib_i = b_ia_i = 1$. Then $(a_i)(b_i) = (b_i)(a_i) = 1$. $\square$

Lemma 2: Let $p$ be a prime and let $0 \leq s \leq t$. Then the natural surjective ring homomorphism $\pi : \mathbb{Z}/(p^t) \rightarrow \mathbb{Z}/(p^s)$ given by $\pi([a]_{p^t}) = [a]_{p^s}$ is surjective on the units. Proof: Now let $[a]_{p^s}$ be a unit in $\mathbb{Z}/(p^s)$. Then $\mathsf{gcd}(a,p^s) = 1$, so that $\mathsf{gcd}(a,p) = 1$, and thus $\mathsf{gcd}(a,p^t) = 1$. So $[a]_{p^t}$ is a unit in $\mathbb{Z}/(p^t)$. Since $\pi([a]_{p^t}) = [a]_{p^s}$, $\pi$ is surjective on the units. $\square$

Lemma 3: Let $\{R_i\}_I$ and $\{S_i\}_I$ be rings with $1 \neq 0$, and let $\varphi_i : R_i \rightarrow S_i$ be a surjective ring homomorphism which is surjective on the units; that is, $\mathsf{im}\ \varphi|_{R^\times} = S^\times$. Then $\prod \varphi_i$ is surjective on the units. Proof: Follows from Lemma 1. $\square$

Now let $m = \prod p_i^{t_i}$. Since $n|m$, we have $n = \prod p_i^{s_i}$ where $0 \leq s_i \leq t_i$ for all $i$. Now the $(p_i^{t_i})$ are pairwise comaximal, so that by the Chinese remainder Theorem we have isomorphisms $\varphi : \mathbb{Z}/(m) \rightarrow \prod \mathbb{Z}/(p_i^{t_i})$ and $\psi : \mathbb{Z}/(n) \rightarrow \prod \mathbb{Z}/(p_i^{s_i})$ given by $\varphi([a]_m) = ([a]_{p_i^{t_i}})$ and $\psi([a]_n) = ([a]_{p_i^{s_i}})$. Note also that $\pi : \mathbb{Z}/(m) \rightarrow \mathbb{Z}/(n)$ given by $\pi([a]_m) = [a]_n$ is a well-defined ring homomorphism. Likewise we have well-defined surjective ring homomorphisms $\pi_i : \mathbb{Z}/(p_i^{t_i}) \rightarrow \mathbb{Z}/(p_i^{s_i})$ for each $i$ given by $\pi_i([a]_{p_i^{t_i}}) = [a]_{p_i^{s_i}}$. It is clear that $\psi \circ \pi = (\prod \pi_i) \circ \varphi$.

By Lemmas 2 and 3, $\prod \pi_i$ is surjective on the units. Thus $\pi = \psi^{-1} \circ (\prod \pi_i) \circ \varphi$ is surjective on the units.