# Why $f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times$ is surjective? [duplicate]

If $m|n$. Why the map $f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times$ given by $a \mod{n}\mapsto a \mod m$ is a surjective homomorphism of groups?

Attempt: I proved it is well a well defined homomorphism because the canonical projection $\overline{f}\colon \mathbb{Z}_n \to \mathbb{Z}_m$ is a surjective ring homomorphism, it extends $f$ and maps units to units.

But I can't prove surjectiveness. Given $a\in\mathbb{Z}$ such that $(a,m)=1$ I'm looking for a $k\in\mathbb{Z}$ such that $(a+k\cdot m,n)=1$. I don't know why one of the $n/m$ candidates for such $k$ makes the number $a+k\cdot m$ an unit modulo $n$.

• To use a sledgehammer, there is a prime number $p$ of the form $a+km$ that is greater than $n$. Then of course, $(p,n)=1$.
– lhf
Aug 2, 2014 at 3:47
• mathoverflow.net/questions/32875 Aug 2, 2014 at 7:48

When $$n$$ is a power of a prime, it's easy because $$\gcd(a,p^k)>1 \implies \gcd(a,p)>1$$

In the general case, use the Chinese Remainder Theorem.

Let $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ and $m=p_1^{\beta_1}\cdots p_k^{\beta_k}$, with $\alpha_i\geq \beta _i$. Then $$\Bbb Z_n^\times =\prod^{k}_{i=1} \Bbb Z_{p_i^{\alpha_i}}^\times\text{ and }\Bbb Z_m^\times = \prod^{k}_{i=1} \Bbb Z_{p_i^{\beta_i}}^\times$$

The surjectivity of $\Bbb Z_m^\times \to \Bbb Z_n^\times$ follows from the surjectivity of $\Bbb Z^\times_{p_i^{\alpha_i}}\to\Bbb Z^\times_{p_i^{\beta_i}}$.

I don't know an elementary answer, but it is true that any ring surjection $f : R \to S$ where $R$ is a finite ring, induces a surjective group homomorphism $f^\times : R^\times \to S^\times$. For a proof, see my answer on MO (which has a very short proof of a more general statement, using only the Chinese Remainder Theorem).

• what is $\operatorname{mSpec}(R)$ and $V(I)$ ? Aug 2, 2014 at 3:09
• @GastónBurrull: $\text{mSpec}(R)$ is the set of maximal ideals of $R$, and $V(I)$ is the set of prime ideals of $R$ containing $I$. $\text{mSpec}(R) \setminus V(I)$ is just the set of maximal ideals not containing $I$
– zcn
Aug 2, 2014 at 3:18
• I think your answer becomes elementary when we replace $R$ for $\mathbb{Z}$ and the general chinese remainder theorem for its $\mathbb{Z}$ version. Aug 2, 2014 at 5:33
• @GastónBurrull: Yes, when I said elementary, I meant without even the Chinese Remainder Theorem. lhf's answer is more or less my answer in the specific case $R = \mathbb{Z}/n\mathbb{Z}$
– zcn
Aug 2, 2014 at 5:36
• Thanks to provide a more general result, yesterday I tried to prove your claim without the premise "$\operatorname{mSpec}R\setminus V(I)$ finite" and I realized it was false. Nice proof. Aug 2, 2014 at 5:53