MHB Find all the zero divisors in a ring

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Ring Zero
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! :D
I am given the following exercise:Find all the zero divisors in the ring $\mathbb{Z}_{20}$.
For each zero divisor $[a]$,find an element $ \neq [0]$ such that $[a]=[0]$.
That's what I did..Could you tell me if it is right??
Zero divisors at the ring $\mathbb{Z}_{20}$: $\{ [2], [4], [5], [6], [8], [10], [12], [14], [15], [16], [18]\}$
The couples are:
$([2],[10]),([10],[2])$
$([4],[5]),([5],[4])$
$([4],[10]),([10],[4])$
$([4],[15]),([15],[4])$
$([5],[8]),([8],[5])$
$([5],[12]),([12],[5])$
$([5],[16]),([16],[5])$
$([6],[10]),([10],[6])$
$([8],[10]),([10],[8])$
$([8],[15]),([15],[8])$
$([10],[12]),([12],[10])$
$([10],[14]),([14],[10])$
$([10],[16]),([16],[10])$
$([10],[18]),([18],[10])$
$([12],[15]),([15],[12])$
$([14],[15]),([15],[14])$
$([15],[16]),([16],[15])$
 
Physics news on Phys.org
evinda said:
Hello! :D
I am given the following exercise:Find all the zero divisors in the ring $\mathbb{Z}_{20}$.
For each zero divisor $[a]$,find an element $ \neq [0]$ such that $[a]=[0]$.
That's what I did..Could you tell me if it is right??
Zero divisors at the ring $\mathbb{Z}_{20}$: $\{ [2], [4], [5], [6], [8], [10], [12], [14], [15], [16], [18]\}$
The couples are:
$([2],[10]),([10],[2])$
$([4],[5]),([5],[4])$
$([4],[10]),([10],[4])$
$([4],[15]),([15],[4])$
$([5],[8]),([8],[5])$
$([5],[12]),([12],[5])$
$([5],[16]),([16],[5])$
$([6],[10]),([10],[6])$
$([8],[10]),([10],[8])$
$([8],[15]),([15],[8])$
$([10],[12]),([12],[10])$
$([10],[14]),([14],[10])$
$([10],[16]),([16],[10])$
$([10],[18]),([18],[10])$
$([12],[15]),([15],[12])$
$([14],[15]),([15],[14])$
$([15],[16]),([16],[15])$


Yep. It is right! (Cool)
 
I like Serena said:
Yep. It is right! (Cool)

Great!Thanks a lot! (Giggle)
 
In fact, it is not hard to see that in $\Bbb Z_{20}$ we have:

$[k]$ is a zero divisor if and only if $\text{gcd}(k,20) > 1$.

We also have the following fact (which is not true for rings in general, but IS true for cyclic rings):

$[k]$ is a zero divisor, or $[k]$ is a unit, and never both.

This turns out to be very useful when examining properties of integers in general (often, we "reduce mod $n$" and then "lift" what we have learned in $\Bbb Z_n$ to $\Bbb Z$).
 
The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...

Similar threads

Back
Top