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Proof of zero divisor

  1. Mar 11, 2007 #1
    1. The problem statement, all variables and given/known data
    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.

    3. 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.
  2. jcsd
  3. Mar 11, 2007 #2

    matt grime

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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.
    Last edited: Mar 11, 2007
  4. Mar 11, 2007 #3
    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


    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?
  5. Mar 12, 2007 #4

    matt grime

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    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.
    Last edited: Mar 12, 2007
  6. Mar 12, 2007 #5
    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


    Rigorous enough?
    Last edited: Mar 12, 2007
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