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.

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