Is {0} a Field? Exploring the Definition and Properties of Fields in Mathematics

[murmillo]

Homework Statement

Is {0} a field?

Homework Equations

field is ring where every non-zero a has a multiplicative inverse

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.

 

[radou]

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

 

[Hurkyl]

It depends on how you define field. It's rather typical to include $0 \neq 1$ as a field axiom.

 

[Fredrik]

When 0≠1 is not included in the axioms, the characteristic of the field is defined as the smallest positive integer p such that $\sum_{i=1}^p 1=0$, 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.

 

[gomunkul51]

{0} is not a proper algebraic field.
every field must have at least 2 elements.

think about it, there is no sense in a field that is an empty group :)

reference:
http://mathworld.wolfram.com/Field.html
http://mathworld.wolfram.com/FieldAxioms.html

 

[murmillo]

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.

 

[Fredrik]

{0} is not a proper algebraic field.
every field must have at least 2 elements.

think about it, there is no sense in a field that is an empty group :)

reference:
http://mathworld.wolfram.com/Field.html
http://mathworld.wolfram.com/FieldAxioms.html

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.

 

[gomunkul51]

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.

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).

 

[Fredrik]

Yes, {0} is a set with one element. The empty set is written either as $$\emptyset$$ or {}.

 

[dmatador]

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?

 

[Hurkyl]

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.

{} 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)

 

[HallsofIvy]

{0} is not a proper algebraic field.
every field must have at least 2 elements.

think about it, there is no sense in a field that is an empty group :)

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.)

reference:
http://mathworld.wolfram.com/Field.html
http://mathworld.wolfram.com/FieldAxioms.html

 

[gomunkul51]

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.)

I misunderstood {0} as being an empty set {}.

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).

 

[Office_Shredder]

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