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Math Help - Zero divisors and cancelation as technique for solving equations ...

  1. #1
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    Zero divisors and cancelation as technique for solving equations ...

    I have the feeling that this question will be another manifestation of grand-scale stupidity, but anyways ...

    I'm trying to put it all together. In a ring R which has zero divisors, what and when can we cancel, when trying to solve an equation?

    Known facts:
    1. An element is either zero or zero divisor.
    2. If an element is unit, we can always cancel it.

    Unknowns:
    1. If an element is the only zero divisor in the ring, can we cancel it from equations like ab=ac ?
    2. How about when we are sure ab\ne0

    Easy examples:
    1. In Z/14Z 2 and 7 are the only zero divisors. And we cannot cancel because 2 * 0 \equiv 2 * 7 but  0 \not\equiv 7 .
    This example is trivial to rework into general proof.
    2. In Z/14Z,  2 * 8 \equiv 2 * 1 but 8 \not \equiv 1, so we cannot cancel the zero divisor even when we know that the multiple of it and the unknown is not zero. Same for 2*9 and 2*2, 2*10 and 2*3, 2*11 and 2*4, 2*12 and 2*5, 2*13, and 2*6.

    How to rework this example into proof. Alternatively this is like asking the question on which exact conditions we can cancel zero divisor.

    1. Never ?
    2. On some conditions ?
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    Re: Zero divisors and cancelation as technique for solving equations ...

    The above post has multiple errors:

    In known facts 1 the corrected statement should be:
    An non-zero element is either unit or zero divisor.

    In unknowns 1 and example 1:
    The example is for commutative ring and should be mentioned that in such ring the number of zero divisors is always even. And even if
    we have non-commultative ring with a as only zero divisor, we still cannot cancel, because ab=ac is always true for c = 0 \ne b

    Other than this, I think I solved it:

    If a is zero divisor in R, then in R there is b\ne 0, s.t. ab=0. Then if c is another element of R, ac=ac-0=ac-ab=a(c-b), so for any c, ac=a(c-b). We can cancel the zero divisor only if c=c-b, or -b=0, or b=0. b is never zero by definition, so for any c, c\ne b-c. We can never cancel zero divisor.
    Last edited by mrproper; July 28th 2014 at 12:50 AM.
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  3. #3
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    Re: Zero divisors and cancelation as technique for solving equations ...

    It is not necessarily the case that a non-zero, non zero-divisor is a unit.

    To see this, consider the ring $\Bbb Z_6[x]$, and the polynomial $x$. This is not a unit, nor is it a zero-divisor.

    If $a$ is a zero divisor is a ring, you cannot, in general, cancel it. There is typically more than one element of $R$ equal to $ab$, the map $b \mapsto ab$ is not injective.

    If $a$ is NOT a zero divisor, we CAN cancel it. For suppose $ab = ac$ and thus $a(b - c) = 0$. Since $a$ is not a zero-divisor, it must be that $b - c = 0 \implies b = c$. Here, the map $b \mapsto ab$ is injective.

    While for the rings $\Bbb Z/n\Bbb Z$, it is true the units and the zero-divisors (plus 0) are complementary sets, in general rings the most we can say is that they are disjoint.
    Thanks from mrproper
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    Re: Zero divisors and cancelation as technique for solving equations ...

    I knew that dammit .... One thing is to know facts about something, other is to use this knowledge to reason about something DD.

    Thanks, Deveno!

    ...and to finish. If element is nor unit, nor zero divisor, we can cancel it. And the proof of it is fundamental for the definition of integral domain. It's there in D&F but one should "decode" it.

    If (non-zero) a is not zero divisor, then there is no b\ne0, s.t. ab=0. So if ab=ac we have ab-ac=0, a(b-c)=0. So b-c must be zero, therefore b=c
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    Re: Zero divisors and cancelation as technique for solving equations ...

    Given any binary operation on a set $S$ we can re-write $\ast:S\times S \to S$ as:

    $a \ast b = L_a(b)$ where $L_a: S \to S$.

    We usually call this "multiplying $b$ (on the left) by $a$".

    "Cancelling", is the process of "undoing $L_a$", that is, from $c = a\ast b$ we seek to recover $b$. This means we seek a left-inverse to the function $L_a$, a function:

    $g: L_a(S) \to S$ such that $g \circ L_a = 1_S$. This is possible if and only if the function $L_a$ is injective, which happens if and only if $a$ is not a zero-divisor.

    Of course, if $a$ is a unit, the function $L_{a^{-1}}$ works as such a left-inverse.

    Now why $\Bbb Z/n\Bbb Z$ is special, is because it is finite, and for FINITE sets, we have: injective functions $S \to S$ are bijective. In a finite commutative ring $R$, this means for a non-zero divisor $a$ there exists $b$ such that $L_a(b) = 1_R$, since $L_a$ is surjective, and thus $a$ is a unit.
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