MHB Properties of gcd's and relatively prime integers

[Kiwi1]

I am studying this in the context of group/ring theory, ideals etc. So I post it here and not in the number theory section.

6. Suppose gcd(a,b)=1 and c|ab. Prove That there exist integers r and s such that c=rs, r|a, s|b and gcd (r,s)=1.

One of my attempts:
From gcd(a,b)=1 there exist s',t' such that ar'+bs'=1 and from this:

gcd(a,b)=1
gcd(a,s')=1
gcd(r',b)=1
gcd(a,s')=1

I have the result: if gcd(a,c)=1 and gcd(b,c)=1 then gcd(ab,c)=1. I can use this to show:

gcd(ar',s')=1
gcd(bs',r')=1
gcd(bs',a)=1
gcd(ar',b)=1

This does not seem to be getting me far. I was hoping to show that c=r's'k for some k. and that r'k|a, s'k|b.

 

[Amer]

what is the ring ? is it $\mathbb{Z}$ integers ?

 

[Kiwi1]

Yes the ring is the integers.

For most of the problems so far I haven't needed to use anything more complicated than my Theorem 3 which says that the gcd of a and b can be expressed as a linear combination of a and b.

Some more thoughts. Actually I think I have solved it?

Let r = gcd(a,c) and s = gcd(b,c) then r|a, s|b, r|ab, s|ab and rs|ab

and there exist u,v,w,x: (ua+vc)*(wb+xc)=rs=uwab+uxac+vwbc+vxcc=u'ab+v'c for u'=uw, v'=uxa+vwb+vxc
or
rs=u'ab+v'c. Now since c|ab and c|c we conclude c|rs.

Let gcd(r,s)=k. r|a, s|b so their exist m,n,o,p: r=mk, s=nk, a=ro, b=ps. Therefore: a=mko and b=pnk so k divides both a and b.
Since gcd(a,b)=1 K|1 and it must be that k=1. i.e. gcd(r,s)=1.

Now r|c, s|c, gcd(r,s)=1 so using a previous result (question C3) rs|c.

Since rs|c and c|rs we get c=rs.