(New page: 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 sugge...) |
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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 | 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 | ||
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| + | 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 | ||
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| + | Consider an element <math>\scriptstyle a</math> of <math>\scriptstyle Z_n</math>. If <math>\scriptstyle gcd(a,n)\ =\ 1</math>, then <math>\scriptstyle a\ \in\ U(n)\ \subseteq\ Z_n</math>. We know that under multiplication modulo <math>\scriptstyle n</math>, <math>\scriptstyle U(n)</math> is a group, and therefore <math>\scriptstyle a</math> has an inverse <math>\scriptstyle a^{-1}</math> such that <math>\scriptstyle a\cdot a^{-1}\ =\ 1</math>. Since <math>\scriptstyle a</math> has a multiplicative inverse, it must be a unit of the ring <math>\scriptstyle Z_n</math>. (As a side note, is this perhaps where the name "unit group" of <math>\scriptstyle U(n)</math> came from??) | ||
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| + | If instead <math>\scriptstyle gcd(a,n)\ =\ d\ \neq\ 1</math>, then we know that <math>\scriptstyle a\ =\ bd</math> and <math>\scriptstyle n\ =\ md,\ b,m\in Z_n</math>. Then, <math>\scriptstyle am\ =\ bdm\ =\ bmd</math> (because <math>\scriptstyle Z_n</math> is commutative) <math>\scriptstyle=\ bn\ \equiv\ 0</math>. Thus in this case, <math>\scriptstyle a</math> must be a zero-divisor of the ring <math>\scriptstyle Z_n</math>. | ||
| + | :--[[User:Narupley|Nick Rupley]] 04:21, 12 March 2009 (UTC) | ||
Latest revision as of 23:21, 11 March 2009
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)
