Find all the zero divisors in a ring

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Discussion Overview

The discussion revolves around identifying all zero divisors in the ring $\mathbb{Z}_{20}$. Participants share their findings and reasoning regarding the zero divisors and their corresponding pairs that yield zero when multiplied.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant lists the zero divisors in $\mathbb{Z}_{20}$ as $\{ [2], [4], [5], [6], [8], [10], [12], [14], [15], [16], [18]\}$ and provides pairs of elements that multiply to zero.
  • Another participant confirms the correctness of the first participant's findings without providing additional reasoning.
  • A later reply introduces a condition stating that $[k]$ is a zero divisor if and only if $\text{gcd}(k,20) > 1$, and notes that in cyclic rings, an element is either a zero divisor or a unit, but not both.

Areas of Agreement / Disagreement

There is a general agreement on the identification of zero divisors in $\mathbb{Z}_{20}$, but the later introduction of conditions regarding zero divisors and units suggests that there may be additional nuances to explore.

Contextual Notes

The discussion includes assumptions about the properties of zero divisors and units in cyclic rings, which may not apply universally to all rings.

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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