Timeline for Is $gcd(zx,zy)=zgcd(x,y)$ (i.e. does the left hand side of this equality 'exist' if the right hand side does).
Current License: CC BY-SA 3.0
4 events
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| Apr 20, 2012 at 18:38 | comment | added | Ralph | If you define $gcd(x,y)=Rx+Ry=(x,y)$ then it's always defined (though no element in R but an ideal) and $(x,y)=1=(x,z)$ implies $x,yz)=1$. | |
| Apr 20, 2012 at 11:43 | comment | added | Martin Brandenburg | I agree with Mark. The gcd is not well behaved in arbitrary integral domains. | |
| Apr 20, 2012 at 11:20 | comment | added | Mark Grant | I think the answers are yes and yes. By the way: en.wikipedia.org/wiki/GCD_domain | |
| Apr 20, 2012 at 10:17 | history | asked | Mikhail Bondarko | CC BY-SA 3.0 |