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Fields and Nilpotent Elements

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data

    Through my exploration/experimentation, I came up with this little conjecture: Let ##F## be a field and ##x## some element in the field. If there exists a natural number ##m## such that ##x^m = 0##, then ##x=0##. In other words, a field contains no nontrivial nilpotent elements.

    2. Relevant equations


    3. The attempt at a solution

    Suppose that such a natural number ##m## exists for which ##x^m = 0##, or ##x \cdot x^{m-1}##. Since fields do not contain zero divisors, either ##x=0## or ##x^{m-1}##. If the latter holds, we are finished, so assume ##x^{m-1} = 0## or ##x \cdot x^{m-1} =0##. This implies ##x^{m-2} = 0##, etc.

    It seems that I could repeat this process until I obtain ##x=0##, suggesting that the conjecture is true (unless I made some elementary mistake, which is not unlikely!). However, is there anyway of making this process/argument more rigorous?
     
  2. jcsd
  3. Dec 1, 2016 #2

    PeroK

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    It's a bit simpler to consider the inverse of any non-zero element.
     
  4. Dec 1, 2016 #3
    So, are you saying I do the following: Suppose there exists a nonzero element ##x \in F## such that ##x^m = 0## for some ##m \in \mathbb{N}##. Since ##F## has multiplicative inverses, ##x^{-m}## exists and therefore ##x^{-m} x^m = x^{-m} 0## or ##1 = 0##, which is a contradiction. Hence, ##x = 0##.

    Does that sound right?
     
  5. Dec 1, 2016 #4

    fresh_42

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    If ##x^m=0## then you can't invert it. But you can invert each ##x## ##m-##times under the assumption it isn't zero.
     
  6. Dec 1, 2016 #5
    Do you mean something like ##x^m = 0## is ##\underbrace{xx...x}_{|m|-times} = 0## and multiplying the equation by ##x^{-1}## ##|m|##-times gives us ##(\underbrace{x^{-1}x^{-1}...x^{-1}}_{|m|-times} (\underbrace{xx...x}_{|m|-times}) = 0## or ##1 = 0##, the desired contradiction.
     
  7. Dec 1, 2016 #6

    fresh_42

    Staff: Mentor

    Yes, because the assumption on ##x## (+ associativity) allows you to do that while ##x^m=0## doesn't.
    However, your original proof by induction on ##m## has been valid, too. Except you had some wording issues in the middle:
    It should have been:
    and now you could have used your induction hypothesis, that ##x^{m-1}=0## already implies ##x=0##. (The induction base ##x^1=0## is trivial.)

    Edit: Induction is the mathematical formalism for "and so on".
     
  8. Dec 1, 2016 #7

    PeroK

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    Alternatively, if ##x^m = 0## then we can take ##m## to be the lowest natural number for which this holds. You can then derive a contradiction from applying ##x^{-1}## once.
     
  9. Dec 1, 2016 #8
    Oh, yes! The well-ordering property, right? Very clever! So, I just need to prove that if ##x \neq 0##, then ##x^{-1} \neq 0##. Here is my proof. Suppose that ##x \neq 0## yet ##x^{-1} = 0##. Then ##xx^{-1} = 1## would imply ##x \cdot 0 = 1## or ##0 = 1##, a contradiction. Does this seem right?
     
  10. Dec 1, 2016 #9

    PeroK

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    If ##x^{-1} = 0## then ##xx^{-1} = 0## which is absurd. Also, it would mean that ##0## is invertible, as ##0^{-1} = x##.

    PS there's no reason to consider ##x^{-1} = 0## in any case.
     
    Last edited: Dec 1, 2016
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