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Not sure if this is the right proof (i found it online):

Since $n\mid m$, if we factor $m = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, then $n = p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k} $ with $\beta_i \leq \alpha_i$. On the other hand, by the Chinese Remainder Theorem: $$(\mathbb Z_m)^\times \cong (\mathbb Z_{p_1^{\alpha_1}})^\times \times (\mathbb Z_{p_2^{\alpha_2}})^\times \times\dots \times (\mathbb Z_{p_k^{\alpha_k}})^ \times $$ $$(\mathbb Z_n)^\times \cong (\mathbb Z_{p_1^{\beta_1}})^\times \times (\mathbb Z_{p_2^{\beta_2}})^\times \times \dots \times(\mathbb Z_{p_k^{\beta_k}})^\times$$

Now, if we define: $$\pi: \mathbb Z_m \longrightarrow \mathbb Z_n,\,\,\, a+(m) \mapsto a+(n)$$

then: $$\pi: \mathbb Z_{p_1^{\alpha_1}} \times \mathbb Z_{p_2^{\alpha_2}} \times \dots \times\mathbb Z_{p_k^{\alpha_k}} \longrightarrow \mathbb Z_{p_1^{\beta_1}} \times \mathbb Z_{p_2^{\beta_2}} \times \dots \times\mathbb Z_{p_k^{\beta_k}}$$ has map: $(a+p_1^{\alpha_1},a+p_2^{\alpha_2}, \dots, a+p_k^{\alpha_k}) \mapsto (a+p_1^{\beta_1},a+p_2^{\beta_2}, \dots, a+p_r^{\beta_k})$

It suffices to show that the statement holds for $n =p^{\beta}$ and $m = p^{\alpha}$ with $\beta \leq \alpha.$ First, we notice that $(a,p^{\alpha})=1 \Leftrightarrow (a, p^{\beta}) =1$, both means that $p \nmid a$. Now, the projection: $$\pi: \mathbb Z/p^{\alpha}\mathbb Z \longrightarrow \mathbb Z/p^{\beta}\mathbb Z$$ maps $(\mathbb Z/p^{\alpha}\mathbb Z)^ \times$ to $(\mathbb Z/p^{\beta}\mathbb Z)^\times$, but $a+(p^{\alpha}) \in (\mathbb Z/p^{\alpha}\mathbb Z)^ \times$ iff $(a,p^{\alpha})=1 \Rightarrow (a, p^{\beta}) =1 $, that is $a+(p^{\beta}) \in (\mathbb Z/p^{\beta}\mathbb Z)^ \times$.

Now it is onto since if $\pi(a+(p^{\alpha})) = a+(p^{\beta}) \in (\mathbb Z/p^{\beta}\mathbb Z)^ \times$, then $(a, p^{\beta}) =1$ and this implies that $(a,p^{\alpha})=1$, so $a+(p^{\alpha}) \in (\mathbb Z/p^{\alpha}\mathbb Z)^ \times$

Then the natural surjective ring projection is also surjective on the units.

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1 Answer 1

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This has been asked on mathoverflow and received a couple of answers:

MO/32875: Lifting units from modulus n to modulus mn.

MO/31495: When does a ring surjection imply a surjection of the group of units?

Your proof has already the right idea: Using the chinese remainder theorem, one easily reduces to the case of powers of a prime $p$. Now if $n \leq m$, then $\mathbb{Z}/p^m \twoheadrightarrow \mathbb{Z}/p^n$ is clearly surjective on unit groups since $z \bmod p^n$ is a unit iff $p \nmid z$.

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  • $\begingroup$ BrandenburgI edited with a better approach,but not sure if this is the right proof $\endgroup$
    – okie
    Nov 28, 2014 at 2:37
  • $\begingroup$ It is identical to the proof I have sketched. $\endgroup$ Nov 28, 2014 at 10:27

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