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Let $n$ be an integer and $d$ be a divisor of $n$. Then there is a natural group homomorphism: \begin{align}\varphi:(\mathbb{Z}/n\mathbb{Z})^{\times} &\to (\mathbb{Z}/d\mathbb{Z})^{\times}\\ [x] &\mapsto [x] \end{align} I want to know whether $\varphi$ is surjective. This may be a silly question, but I don't know where to look for the answer. Any help would be appreciated.

withoutfeather
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2 Answers2

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$\varphi$ is surjective. By the Chinese remainder theorem it suffices to show this when $n$ and $d$ are powers of the same prime, say $p^a$ and $p^b$ where $b \le a$. In this case it's clear because every lift of a unit $\bmod p^b$ is a unit $\bmod p^a$ (since it's not divisible by $p$).

Qiaochu Yuan
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  • In general we need to be careful because it's not true that if $\phi : R \to S$ is a surjective ring homomorphism then the induced map $R^{\times} \to S^{\times}$ is surjective. However, it is true if $R$ is a local ring and $S$ is the quotient of $R$ by a power of its maximal ideal, which suffices for this case. – Qiaochu Yuan Oct 29 '20 at 21:27
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    I recall that there general case was discussed on MO in 2010, and also somewhere here iirc. – Bill Dubuque Oct 30 '20 at 00:49
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By $\,\rm\color{#c00}{co}$prime Dirichlet: $\,(a,d,n)\overset{d\,\mid\, n}=(a,d)=1\,\Rightarrow\, \exists\, x\in\Bbb Z\!:\ (a+d\:\!x,\,n)=1.\ \ \rm\small QED$

Remark $ $ This theorem on $\rm\color{#c00}{co}$primes in arithmetic progressions has a very short high-school level proof using a simple variation (by Stieltjes) on Euclid's proof that there are infinitely many primes.

Bill Dubuque
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