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Is {0} a field?

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Is {0} a field?


    2. Relevant equations
    field is ring where every non-zero a has a multiplicative inverse


    3. The attempt at a solution
    I thought {0} was a field, since it seems trivially true that every non-zero element has a multiplicative inverse. Since there are no non-zero elements in {0}. But today I came across the proposition that if R is a field, it has exactly two ideals. Which suggests to me that {0} is not a field, because then R would have only one ideal.
     
  2. jcsd
  3. Aug 18, 2010 #2

    radou

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    Every field has at least two distinct elements (i.e. the additive and the multiplicative identity), so {0} can't be a field.
     
  4. Aug 18, 2010 #3

    Hurkyl

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    It depends on how you define field. It's rather typical to include [itex]0 \neq 1[/itex] as a field axiom.
     
    Last edited: Aug 18, 2010
  5. Aug 18, 2010 #4

    Fredrik

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    When 0≠1 is not included in the axioms, the characteristic of the field is defined as the smallest positive integer p such that [itex]\sum_{i=1}^p 1=0[/itex], or as 0 if no such integer exists. The field of real numbers has characteristic 0. The only field with characteristic 1 is the trivial field {0}. (Because 1=0 implies x=1x=0x=0). The characteristic is always 0,1 or a prime number.
     
  6. Aug 18, 2010 #5
  7. Aug 18, 2010 #6
    radou: I think we must show that {0} is not a field first, then use that to prove that every field must have at least two elements.

    I guess the reason why {0} is not a field is because (according to Artin) part of the definition of a field is that if F is a field then F-{0} is a group. But if {0} were a group I would have an empty group, which does not contain an identity element.

    Yes, I think some books state 1 is not 0 in their definition of field, but in my notes it just said a ring where "every non-zero element has an inverse." I don't think that definition is correct. It should have said a non-zero ring where every non-zero element has an inverse, or something like that.

    Thanks, guys.
     
  8. Aug 18, 2010 #7

    Fredrik

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    The first of those two references includes the requirement 1≠0. The other one doesn't. As I (and Hurkyl) said above, both definitions make sense, but are not equivalent. {0} is a field according to one of the definitions, but not the other. I don't see a problem with letting {0} be considered a field. It's already considered a group, a vector space, etc.
     
  9. Aug 18, 2010 #8
    an Empty Set has no elements that could satisfy any of the additive and multiplicative rules !
    thus it is not a field :)

    *** BUT If by {0} you meant not an empty set but a set with one element 0 (zero), it could be a field :)
    I think it obeys all the rules..

    P.S. the additive and multiplicative identities don't have to be 1 and 0 (could be any other element), they just have to exist in the set (field).
     
    Last edited: Aug 18, 2010
  10. Aug 19, 2010 #9

    Fredrik

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    Yes, {0} is a set with one element. The empty set is written either as [tex]\emptyset[/tex] or {}.
     
  11. Aug 20, 2010 #10
    A field is a commutative ring with unity such that all nonzero elements are units.

    {} is a commutative ring with no nonzero elements, so it does not violate this definition.

    Why doesn't this suffice? Or does the empty set, by the same reasoning as above, not have a 1?
     
  12. Aug 20, 2010 #11

    Hurkyl

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    {} has neither a 0 nor a 1, so it's not a ring. :tongue:

    (The zero ring does have a 1. It's just equal to 0)
     
  13. Aug 21, 2010 #12

    HallsofIvy

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    What "empty" group are you talking about? The set of objects, {0}, with addition is a group as is the set of invertible members with multiplication. (Note that 0*0= 0 and if 0 is the only member then 0 is also the multiplicative identity so 0 has a multiplicative inverse.)

     
  14. Aug 21, 2010 #13

    I misunderstood {0} as being an empty set {}.
     
    Last edited: Aug 21, 2010
  15. Aug 21, 2010 #14

    Office_Shredder

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    The empty group isn't a group! It needs an identity element
     
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