Is Every Non-Unit Element in Zn a Zero Divisor? Proof and Explanation

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Homework Help Overview

The discussion revolves around the proof that a non-zero element A in Zn (integers mod n) that is not a unit must be a zero divisor. Participants are exploring the definitions and implications of units and zero divisors in modular arithmetic.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants question the rigor of the original proof attempt, particularly regarding the transition between Z/nZ and Z, and the implications of A being a multiple of n.
  • There is a discussion about the definition of zero divisors and units, with references to coprimality and the use of Euclid's algorithm.
  • One participant suggests a proof structure that involves showing the relationship between coprimality and being a unit, while another seeks clarification on proving the initial assumption regarding units.

Discussion Status

The discussion is active, with participants providing hints and suggestions for proving key assumptions. There is a focus on ensuring the rigor of the arguments presented, and multiple interpretations of the definitions are being explored. The original poster is encouraged to refine their proof based on feedback.

Contextual Notes

Participants note the importance of definitions and theorems related to coprimality and zero divisors, as well as the need for clarity in the proof structure. There is an acknowledgment of the potential confusion arising from the relationship between different mathematical structures.

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Homework Statement


Prove that if A, a non zero element in Zn (integers mod n) is not a unit then A is a zero divisor in Zn.



The Attempt at a Solution


[AB] does not equal 1 mod n for some A and all B in Zn
=> the element A is a multiple of n so n divides A because otherwise there should be a remainder of 1.
=> AB=0 for all B in Zn

So A is a zero divisor in Zn.

I feel the first implication is a bit unrigorous.
 
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It is very unrigorous. You are flitting between Z/nZ, and Z at will and making no comment about what you're doing.

2 is a zero divisor mod 4. Are you claiming 2 is a multiple of 4 (in the integers)? There is also no reason why AB=0 for all B. Indeed this is clearly false if you let B=1 (since A is not zero). Notice that your other thread on zero divisors now contradicts the definition you need here. 0 is obviously not a unit, and by your previous thread, not a zero divisor either. In fact what you've just argued is that the only zero divisor is 0 (you just asserted in your first step that n divides A, i.e. A=0 mod n).

Something in Z is a unit when reduced mod n if and only if it is coprime to n. That is a starting hint.
 
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Note that A must be nonzero.

Your hint is useful. If we assume it and use it's negatation then

A in Z is not a unit when reduced mod n if and only if it is not coprime to n.

=> There exists b such that b|a and b|n. for b>1.
=> A=bi, n=bj, 1<j<n so j is an element in Z mod n
=> n|Aj
=> Aj mod n = 0
=> A is a 0 divisor in Z mod n

Correct?

If so then I like to prove the assumption we used which was also your hint. I seem to have some trouble doing this. Could yoy maybe give a hint of how to prove that?
 
Assumption? That the units mod n are the residues coprime with n? This is the first thing you shuold have proved, and is a simple consequence of Euclid's algorithm. It involves coprimality. There isn't a great deal else you can do than invoke Euclid/highest common factor stuff.
 
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I will do the proof from the beginning

First prove
A in Zn is coprime to n
=> gcd(A,n)=1
=> Use theorem involving Eulid's algorithm that there is a B in Zn and k such that AB+kn=1
=>Ab is congruent to 1 mod n
=>A is a unit

So from the above lemma we can turn it around so that
A is not a unit
=> A is not coprime to n
=> gcd(A,n)=d>1
=> Suppose A=ad and n=bd
=> Ab-na=adb-bda=0 since multiplication is commutative
=> Ab=0 mod n for some non zero b in Zn
=> A is a zero divisor in Zn

QED

Rigorous enough?
 
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