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The problem is to prove that if $n\mid m$, then the projection $(\mathbb{Z}/m\mathbb{Z})^\times\rightarrow (\mathbb{Z}/n\mathbb{Z})^\times$ is surjective. (Note: I am only looking for hints; see bottom). Here is my proof:

Let $b\in\mathbb{Z}$ with $gcd(b,n)=1$. We are done if we can show that $\exists$ $a\in\mathbb{Z}$ such that $gcd(a,m)=1$ and $a\equiv b$ (mod n). Equivalently, if $\exists$ $x,y,t\in\mathbb{Z}$ such that $(tn+b)x+my=1$. Let $m=nq, q\in\mathbb{Z}$. Then $(tn+b)x+(nq)y=n(tx+yq)+bx=1$. Let $nu+bv=1,u,v\in\mathbb{Z}$. Setting $x=v+ns$, gives $n[t(v+ns)+yq]+b(v+ns)=1$, which has a solution if $t(v+ns)+yq=u-bs$, for some $t,y,s\in\mathbb{Z}$.

Let $d=gcd(v,q)$. Let $s\in\mathbb{Z}$ such that $gcd(d,s)=1$. Then the congruence $dz\equiv u$ (mod s) has a solution $z\in\mathbb{Z}$. Let $dz=sp+u, p\in\mathbb{Z}$. Let $t,y\in\mathbb{Z}$ such that $tv+yq=dz$. Let $p=-b-tn$. Then $tv+yq=s(-b-tn)+u=-bs-tns+u$. Thus $tv+tns+yq=(-bs-tns+u)+tns= u-bs$, which completes the proof.

This is problem #7 in section 7.6 of Dummit and Foote, which is titled The Chinese Remainder Theorem. So I am wondering if there is an easier way to solve this problem using the Chinese Remainder Theorem. I am only looking for a hint as to how use the Chinese Remainder Theorem for this problem. Currently, I do not see a way to set up a system of congruences with relatively prime moduli.

SihOASHoihd
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1 Answers1

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Consider some prime $p$ dividing $n$. Then $p^r$ exactly divides $n$ and $p^s$ exactly divides $m$ where $r\ge s\ge0$ (possibly $s=0$). Then we can choose $a_p$ such that $a_p\equiv b\pmod{p^s}$ and $p\nmid a_p$. When $s>0$ we can take $a_p=b$ and when $s=0$ we can take $a_p=1$. By CRT there is $a$ such that $a\equiv a_p\pmod{p^r}$ for all $p$, etc.

Angina Seng
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