Find all the zero divisors in a ring

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SUMMARY

The zero divisors in the ring $\mathbb{Z}_{20}$ are identified as $\{ [2], [4], [5], [6], [8], [10], [12], [14], [15], [16], [18]\}$. For each zero divisor $[a]$, there exists a corresponding non-zero element $[b]$ such that $[a][b]=[0]$. The relationship is established through the condition that $[k]$ is a zero divisor if and only if $\text{gcd}(k,20) > 1$. Additionally, in cyclic rings, an element $[k]$ is either a zero divisor or a unit, but not both.

PREREQUISITES
  • Understanding of ring theory, specifically zero divisors
  • Knowledge of the structure of $\mathbb{Z}_{n}$ rings
  • Familiarity with the concept of greatest common divisor (gcd)
  • Basic principles of modular arithmetic
NEXT STEPS
  • Study the properties of zero divisors in other rings, such as $\mathbb{Z}_{n}$ for different values of n
  • Explore the implications of the gcd condition in various algebraic structures
  • Learn about units in rings and their relationship with zero divisors
  • Investigate the concept of cyclic rings and their characteristics
USEFUL FOR

Mathematicians, algebra students, and educators focusing on abstract algebra, particularly those interested in ring theory and modular arithmetic.

evinda
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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:
$([2],[10]),([10],[2])$
$([4],[5]),([5],[4])$
$([4],[10]),([10],[4])$
$([4],[15]),([15],[4])$
$([5],[8]),([8],[5])$
$([5],[12]),([12],[5])$
$([5],[16]),([16],[5])$
$([6],[10]),([10],[6])$
$([8],[10]),([10],[8])$
$([8],[15]),([15],[8])$
$([10],[12]),([12],[10])$
$([10],[14]),([14],[10])$
$([10],[16]),([16],[10])$
$([10],[18]),([18],[10])$
$([12],[15]),([15],[12])$
$([14],[15]),([15],[14])$
$([15],[16]),([16],[15])$
 
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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 $\Bbb Z$).
 

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