Anyone have an idea on the explanation in the back of the book? I dont get where if gcd(k,n)=1 that means its a unit...and why does k(n/d)=0? I may just be not seeing something...any suggestions? -K. Brumbaugh
Here is a suggestion: The book says that the set of units of Zn is U(n). That is why I thought that they could write the gcd(k,n)=1 part. As far as the k(n/d)=0 part, they have in the back: k(n/d)=sd(n/d)=sn=0. What I understood from that was that sn is equal to zero because Zn is a set under (addition/multiplication?) modulo n, so k is a zero divisor by definition of zero divisor. That is what I tried to piece together about it. -Josie
Consider an element $ \scriptstyle a $ of $ \scriptstyle Z_n $. If $ \scriptstyle gcd(a,n)\ =\ 1 $, then $ \scriptstyle a\ \in\ U(n)\ \subseteq\ Z_n $. We know that under multiplication modulo $ \scriptstyle n $, $ \scriptstyle U(n) $ is a group, and therefore $ \scriptstyle a $ has an inverse $ \scriptstyle a^{-1} $ such that $ \scriptstyle a\cdot a^{-1}\ =\ 1 $. Since $ \scriptstyle a $ has a multiplicative inverse, it must be a unit of the ring $ \scriptstyle Z_n $. (As a side note, is this perhaps where the name "unit group" of $ \scriptstyle U(n) $ came from??)
If instead $ \scriptstyle gcd(a,n)\ =\ d\ \neq\ 1 $, then we know that $ \scriptstyle a\ =\ bd $ and $ \scriptstyle n\ =\ md,\ b,m\in Z_n $. Then, $ \scriptstyle am\ =\ bdm\ =\ bmd $ (because $ \scriptstyle Z_n $ is commutative) $ \scriptstyle=\ bn\ \equiv\ 0 $. Thus in this case, $ \scriptstyle a $ must be a zero-divisor of the ring $ \scriptstyle Z_n $.
- --Nick Rupley 04:21, 12 March 2009 (UTC)
