Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this excercise, I need your help on the third point:

i) Determine two integers $\alpha$ and $\beta$ such that $12\alpha + 7\beta = 1$

Answer: $\alpha = 3$ and $\beta = -5$

ii) Determine all the solutions of $$7x\equiv 1 (mod. 12)$$

Answer: $[-5]_{12} = \{-5+12k, k\in\mathbb{Z}\}$

iii) determine invertible elements (for product) for $(\mathbb{Z}_{12}, +, \cdot)$, and zero divisors

Answer: Here I need your help! How can I determine all invertible elements and all zero divisors?

iv) determine, if exists, a class $[a]\in\mathbb{Z}_{12}$ such that $[a][6]=[2].$

Answer: No, doesn't exist. $gcd(6,12)\neq 1$

share|improve this question
    
You might want to do something about that 20% accept rate. –  Gerry Myerson Mar 6 '12 at 22:52
    
I have to pass the algebra exam. Successively i will try to improve that number. :) –  Mariano Mar 7 '12 at 12:13
add comment

3 Answers

There is the brute force method: try them all!

2 (2x6=12=0), 3 (3x4=12=0), 4 (4x3=12=0), 6 (6x2=12=0), 8 (8x6=48=0), 9 (9x4=36=0), 10 (10x6=60=0) are zero divisors

1, 5 (5 x 5 = 25 = 1), 7 (7 x 7 = 49 = 1), 11 (11x11=121=1) are units.

Notice that the units have gcd 1 with 12...

share|improve this answer
    
I've alredy thinked of that. Is there something much elegant? –  Mariano Mar 6 '12 at 23:00
1  
This is quite elegant as it is. The units are relatively prime to 12, and the zero-divisors are not. –  Brett Frankel Mar 6 '12 at 23:03
    
and what if the same question is for $\mathbb{Z}_{200}$ or something bigger? –  Mariano Mar 6 '12 at 23:37
1  
If the ring is $\mathbb{Z}_{200}$ and the question is "list all invertible elements and zero divisors", then there are going to be 199 answers no matter what you do. But it should be easy enough to list all the zero divisors of $\mathbb{Z}_{200}$: its prime factors are 2 and 5, so list all the multiples of 2 and 5. –  Tanner Swett Mar 7 '12 at 0:18
add comment

On iv), your answer is right, but the justification is not. For example, it is easy to see that there are $[a]$ such that $[a][6]=[6]$, even though $\gcd(6,12)\ne 1$. For $a=1$ will work, so will $a=3$, as will $a=9$.

The reason that we do not have an $a$ such that $[a][6]=[2]$ is that $\gcd(6,12)$ does not divide $2$. For if $[a][6]=[2]$, then $6a\equiv 2 \pmod {12}$, or equivalently $12$ divides $6a-2$. If this happens, then $6$ divides $6a-2$, which implies that $6$ divides $2$. That is clearly not the case.

Or else we could say that there is no $a$ such that $[a][6]=[2]$ because we tried everything from $a=0$ to $a=11$, and nothing worked. What's fine for a small modulus like $12$, but one would not care to do that with $1200$.

share|improve this answer
add comment

Hint $\rm\ n\:$ invertible in $\rm\mathbb Z_{12}\iff \exists\: a\!:\: an\equiv 1\iff \exists\: a,b\!:\: an+12b = 1\iff gcd(n,12)=1$

Conversely, $\rm\: gcd(n,12) = c > 1\:\Rightarrow\: n(12/c) \equiv 12(n/c) \equiv0\:\Rightarrow\: n\:$ is a zero-divisor, if $\rm n\not\equiv 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.