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I'm looking to construct an isomorphism to show:

$$\mathbb{Z}_{rs}^{\times}\cong \mathbb{Z}_r^{\times} \times \mathbb{Z}_s^{\times}$$

where $\mathbb{Z}_{rs}^{\times}$ represents the invertible elements of $\mathbb{Z}_{rs}$ with multiplication for $r,s$ coprime.

I define $\phi: \mathbb{Z}_{rs}^{\times}\rightarrow \mathbb{Z}_r^{\times} \times \mathbb{Z}_s^{\times}$ as $\phi ([x]_{rs})=([x]_r,[x]_s)$. It's easy to show that such a map is injective. But, surjectivity is giving me trouble. Is this map even surjective? If so, how might I show that?

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  • $\begingroup$ Have you tried googling Chinese Reminder Theorem? :) $\endgroup$ – GSF Feb 18 '17 at 1:29
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This should work, for instance https://www.maths.tcd.ie/pub/Maths/Courseware/NumberTheory/ch05.pdf

Thanks to the CRT, if you have $gcd(r,s)=1$ and $m$ mod$(r)$, $n$ mod$(s)$, there exists an $N$ such that $$N\equiv m \mod(r)$$ $$N\equiv n \mod(s)$$ and it is unique mod$(rs)$. From that follows the bijectivity of the map you gave.

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  • $\begingroup$ Link only answers are discouraged here. Yours, though useful, may be downvoted. $\endgroup$ – Ethan Bolker Feb 18 '17 at 1:32
  • $\begingroup$ Sorry, I did not know that. Let me fix it. $\endgroup$ – GSF Feb 18 '17 at 1:35
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The isomorphism is deduced from the ring isomorphism: \begin{align} \mathbf Z/rs\mathbf Z&\longrightarrow \mathbf Z/r\mathbf Z\times\mathbf Z/s\mathbf Z\\ x\bmod rs&\longmapsto(x\bmod r,\,x\bmod s) \end{align} (An isomorphism between two rings induces by restriction an isomorphism of their groups of units.)

The inverse isomorphism uses a Bézout's relation: $\;ur+vs=1$, and is defined by \begin{align} \mathbf Z/r\mathbf Z\times\mathbf Z/s\mathbf Z&\longrightarrow\mathbf Z/rs\mathbf Z\\ (x\bmod r,\,y\bmod s)&\longmapsto yur+xvs\bmod rs \end{align}

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  • $\begingroup$ It would be useful to say how this restricts to an isomorphism on the groups of units, and how the group of units of a product of rings is the product of the groups of units of the factors. $\endgroup$ – Pedro Tamaroff Feb 18 '17 at 1:56
  • $\begingroup$ Mmm… Isn't all this obvious? Of course, I can add a word about it, but I prefer not to give a fully detailed solution. $\endgroup$ – Bernard Feb 18 '17 at 2:07
  • $\begingroup$ There are many things that are obvious to some and not to others. One should be wise and learn when to distinguish such cases. $\endgroup$ – Pedro Tamaroff Feb 18 '17 at 2:09

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